Lecture 1
DC Circuits: Ohm’s Law
Series and parallel circuit
1
SSCP 2313 - Basic electronics2012/13 -1
Voltage
Review of V, I, and R
the amount of energy per charge available to
move electrons from one point to another in a
circuit and is measured in volts (V).
Current the rate of charge flow and is measured in
amperes (A)
Resistancethe opposition to current and is
measured in ohms ()
SSCP 2313- Basic electronics 2012/13-1 2
3
Ohm’s Law
Defines the relationship between voltage, current,
and resistance in an electric circuit
Ohm’s Law:
Current in a resistor is directly
proportional to the voltage across it and is
inversely proportional to the resistance.
Stated mathematically:
R
VI
V
I R
+ -
Where: I is the current (A)
V is the potential difference (V)
R is the resistance ()
V
I R
SSCP 2313- Basic electronics 2012/13-1
What is the voltage across a 680 resistor if the
current is 26.5 mA?
Example 1:
4SSCP 2313- Basic electronics 2012/13-1
5
Series Circuits
There is only a single path for current to flow.The amount of current is the same at all points.
R1
VSR2
R3
+
Total resistances in series:
Rtotal = R1 + R2 + R3...
V1 = IR1, V2 = IR2 and V3= IR3.
The voltage across each resistor:
Vs = V1 + V2 + V3
Total voltage:
= I (R1+ R2 + R3)
I
V1
SSCP 2313- Basic electronics 2012/13-1
SSCP 2313- Basic electronics 2012/13-1 6
Parallel Circuits
The voltage across every parallel component is equal.
The total current in the circuit is the sum of the currents in all the
branches.
I = I1 + I2 + ……. + In
nTotal RRRR
1......
111
21
RNR2R1Vs
IT
I3I2I1
Total resistances in parallel:V1 = V2 = Vs
Voltage drop in each resistor :
Special case for resistance of two
parallel resistors
R1 R2T
1 2
1
1 1R
R R
1 2
T
1 2
R RR
R R
or
7SSCP 2313- Basic electronics 2012/13-1
PT = P1 + P2 + P3 +…..+ Pn
Or PT = Vs IT
thus
T
sT
R
VP
2
Power in series - Parallel circuits
8SSCP 2313- Basic electronics 2012/13-1
House circuits contain parallel circuits
The parallel circuit will continue to operate even though one component may be open
Only the open or defective component will no longer continue to operate.
Advantages of parallel circuit
9SSCP 2313- Basic electronics 2012/13-1
Example 2:
5 25
30 15
5 V 15 15
(Ans: Rs = 25 , Is = 0.2A)
Calculate: a) the total resistance.
b) curent flowing through the circuit.
10SSCP 2313- Basic electronics 2012/13-1
Example 3: If Vs = 5 V, calculate: a) the total resistance.
b) curent flowing through the circuit.
11SSCP 2313- Basic electronics 2012/13-1
Lecture 2
DC Circuits: Voltage Divider Rule
Current Divider Rule
1
SSP 2313- Basic electronics2012/13-1
The voltage drop across any given resistor in a series
circuit is equal to the ratio of that resistor to the total
resistance, multiplied by source voltage.
Voltage Divider Rule (VDR)
A potentiometer can act
as a variable voltage
divider, to control a
voltage.
I = Vin/ (R1 + R2)
V1 = I R1
21
1
RR
RVin
2SSCP 2313- Basic electronics 2012/13-1
Example 1
The output voltage from the voltage divider is
a. 2.0 V
b. 4.0 V
c. 12 V
d. 20 V VS
VOUT
R2
R1
+
2.0 kW
10 kW
24 V
3SSCP 2313- Basic electronics 2012/13-1
R1Vs
IT
I2I1
+
V1
-
+
V2
-
TR
R
21
21
RI T
R
R
21
12
RI
Current Divider Rule (CDR)
R2
2211 RIRIV
21 IIIT
21
21
RR
RRRT
21
21
RR
RRIRIV TTT
To determine how current entering a node is split between the various parallel resistors connected to the node.
4SSCP 2313- Basic electronics 2012/13-1
Quiz
The current in R1 is
a. 6.7 mA
b. 13.3 mA
c. 20 mA
d. 26.7 mA
R1 R2
200 W100 WI = 20 mA
Example 2
5SSCP 2313- Basic electronics 2012/13-1
Series voltage sources
Parallel current sources
Review
6SSCP 2313- Basic electronics 2012/13-1
Voltage Sources in Parallel
Voltage sources with different potentials should never be connected in parallel
When two equal sources are connected in parallel
Each source supplies half the required current
If two unequal sources are connected
Large currents can occur and cause damage
7SSCP 2313- Basic electronics 2012/13-1
Find: a) Voltage across R2
b) Current flows through R2.
Ans: VR2 = 3.3 V, IR2 = 0.335 mA
Example 3
8SSCP 2313- Basic electronics 2012/13-1
If Vs = 5 V, calculate: a) voltage across AB
b) current flows through R6
Example 4
9SSCP 2313- Basic electronics 2012/13-1
Ans: RT = 148.4 W IT = 0.034 A VAB= 1.63 V I6 = 0.02 A
Have a nice day….
Lecture 3
DC Circuits:
Kirchoff’s Law
1
SSP 2313- Basic electronics
2012/13-1
Kirchoff’s Law
Kirchhoff’s Voltage Law
(KVL)
Kirchhoff’s Current Law
(KCL)
2SSCP 2313- Basic electronics 2012/13-1
Gustav Robert Kirchhoff
(12 March 1824 – 17 Oct
1887)
Kirchhoff’s Voltage Law (KVL)
VS = VR1 + VR2 + VR3
R1
I
VS
R3
VR1
VR3+-
VR2
+ - +-
R2
In any closed loop network, the total voltage source
around the loop is equal to the sum of all the voltage
drops within the same loop.
3SSCP 2313- Basic electronics 2012/13-1
IRVs
The sum of the currents entering a junction is
equal to the sum of the currents leaving the
junction.
Kirchhoff’s Current Law (KCL)
I1
I2 I3
I1 = I2 + I3
4SSCP 2313- Basic electronics 2012/13-1
4Ω
6Ω 12Ω
R1
R3
R2I1 I2
I3
V1 V2
36 V
72 V
Loop 1 Loop 2
Two loop and two-supply circuit
Referring to the circuit above, calculate:a) Current I1 , I2 , I3
b) Total power dissipated in the circuit.
5SSCP 2313- Basic electronics 2012/13-1
KCL:
Current entering node A =
Current leaving node A
I1 + I2 = I3 (1)
KVL:
Loop 1: V1 = VR1 + VR3
36 = 6I1+ 4I3 (2)
Loop 2: V2 = VR2 + VR3
72 = 12I2 + 4I3 (3)
Substitute (1) into (2):
36 = 6 (I3 – I2) + 4I3
36 = 10 I3 - 6I2 (4)
(4) x -2 :
-72 = -20I3 + 12I2 (5)
(3)-(5):
144 = 24 I3
I3 = 6 ASubstitute I3 into (2) to get I1 and (2) to get I2.
4Ω
6Ω 12Ω
R1
R3
R2I1 I2
I3
V1 V2
36 V
72 V
A
Loop1 Loop 2
4Ω
6Ω 12Ω
R1
R3
R2I1 I2
I3
V1 V2
36 V
72 V
A
Loop 1 Loop 2
I1 = 2A and I2 = 4 A 66SSCP 2313- Basic electronics 2012/13-1
Power dissipated by each resistor
PR1 = I12 R1 = 24 W
PR2 = I22 R2 = 192 W
PR3 = I32R3 = 144 W
Total power dissipated
PT = PR1 + PR2 + PR3
= 360 W
7SSCP 2313- Basic electronics 2012/13-1
Find the current flowing in the 40 Ω resistor, R3
8SSCP 2313- Basic electronics 2012/13-1
I1= -1/7 A , I2 = 3/7 A, I3= 2/7 A
KCL:
)1(321 III
KVL:
IRVs
Loop 1:
)3(41
10)2(401010
31
31
II
II
Loop 2:
)5(31
20)4(6020
40)(20
402020
31
31
313
32
II
II
III
II
Solve eq (3) and (5)9SSCP 2313- Basic electronics 2012/13-1
Thank you…
Lecture 4
DC Circuits:
Thevenin’s Theorem
1
SSP 2313- Basic electronics
2012/13-1
M. Leon Thévenin (1857-1926),
published his famous theorem in 1883.
Thevenin’s Theorem
It is possible to simplify any linear circuit (network), no
matter how complex, to an equivalent circuit with just a
single voltage source (VTH) and a series resistance
(RTH) connected to a load.
Network •
•
A
B
VTH
RTH
A
B
+_
VTH: Thevenin’s voltage and
RTH : Thevenin’s resistance
Thevenin’s equivalent
circuit
3SSCP 2313- Basic electronics 2012/13-1
Determination
of the
Thevenin’s
Voltage
VTH = Open circuit voltage
with load (RL) removed.
Determination
of the
Thevenin’s
Resistance
RTH = Total resistance in
open network with
sources were set to zero.
* If a voltage source - short circuit.
* If a current source - open circuit.
4SSCP 2313- Basic electronics 2012/13-1
Find the voltage across
load resistor RL.
i) Disconnect RL to find VTH = VAB
5SSCP 2313- Basic electronics 2012/13-1
V
RR
R
VV RAB
24
3621
2
2
ii) Short-circuit V to find RTH =RAB
iii) Draw the Thevenin’s equivalent circuit
2// 21 RRRTH
6SSCP 2313- Basic electronics 2012/13-1
iv) Put back RL. Calculate the IL and VL.
7SSCP 2313- Basic electronics 2012/13-1
ARR
VI
LTH
THL 6
Find the Thevenin’s equivalent circuit for:
i) VTH = Vab = 32 V
ii) RTH = Rab = 8
8SSCP 2313- Basic electronics 2012/13-1
Lecture 5
DC Circuits:
Norton’s Theorem
1
SSCP 2313- Basic electronics2012/13-1
Edward Lawry Norton
1898-1983
Any two-terminal linear network can be replaced by
an equivalent circuit consisting of a current source (IN)
and a parallel resistor (RN).
Norton’s Theorem
IN: the output current when the output terminals or RL are
shorted.
RN: the total resistance between the two output terminals
when all sources have set to zero.
RNIN
Network •
•
A
B
Norton’s equivalent circuitRTH=RN and IN= VTH/RTH
3SSCP 2313- Basic electronics 2012/13-1
3Ω
2Ω
10V
10Ω
RL
2Ω
Find Norton’s equivalent circuit and find the current that
passes through RL when RL = 1Ω
i) Short circuit RL to find IN = IAB
3Ω
2Ω
10V
10Ω
2Ω
IscIN
IN = 0.45 A
4SSCP 2313- Basic electronics 2012/13-1
3Ω
2Ω 10Ω
2Ω
ii) Short-circuit V to find RN =RAB
RN = 13.2
RL13.20.45
iii) Draw the Norton’s equivalent circuit
5SSCP 2313- Basic electronics 2012/13-1
iv) Put back RL. Calculate the IL and VL.
IL = 0.42 A
6SSCP 2313- Basic electronics 2012/13-1
RL13.20.45 1Ω
Find the current in RL, if RL =2 k.
7SSCP 2313- Basic electronics 2012/13-1
10 mA1 k
A
3 k
B
RL
5 k
IL= 0.833 mA
Maximum Power Transfer
The maximum amount of power will be dissipated by a
load resistance when that load resistance is equal to
the Thevenin’s/Norton’s resistance of the network
supplying the power.
8SSCP 2313- Basic electronics 2012/13-1
RTH
VTH 12 V
10
10
WR
VP 6.3
10
622
The power delivered to
the matching load is
Lecture 6
DC Circuits:
Superposition’s Theorem
1
SSCP 2313- Basic electronics2012/13-1
A circuit can be analyzed with only one source of
power at a time, the corresponding component
voltages and currents algebraically added to find
out what they'll do with all power sources in
effect.
Superposition Theorem
ii) Remove the rest:
Voltage – replace with short circuit
Current – replace with open circuit
Step to follow:
i) Pick one source at one time
iii) Sum the responses
2SSCP 2313- Basic electronics 2012/13-1
R3V1 V2
100 W 20 W
10 W
15 V 13 V
R1 R2
Find the current flowing through R3 using
superposition theorem.
R1 R2
R3V1
100 W 20 W
10 W
15 VV2 shorted
REQ = 106.7 W, IT = 0.141 A and I’R3= 0.094 A
3SSCP 2313- Basic electronics 2012/13-1
REQ = 29.09 W, IT = 0.447 A and I’’R3= 0.406 A
R1 R2
R3V2
100 W 20 W
10 W
13 V
V1 shorted
4SSCP 2313- Basic electronics 2012/13-1
R1 R2
V1 V2
100 W 20 W15 V 13 V
Adding the currents gives IR3= 0.5 A
REQ ’ = 106.7 W, IT ’ = 0.141 A and I’R3
= 0.094 A
REQ’’ = 29.09 W, IT ’’ = 0.447 A and I’’
R3= 0.406 A
With V2 shorted
With V1 shorted
0.094 A 0.406 A
Therefore;
5SSCP 2313- Basic electronics 2012/13-1
Using the superposition theorem, calculate the current in the R2.
6SSCP 2313- Basic electronics 2012/13-1
I2’= 1 A , I2” = 1.2 A thus; I2= 0.2 A
With current source opened
With voltage source shorted
7SSCP 2313- Basic electronics 2012/13-1
Thank you…
Lecture 7
•AC Current
•Phasor Diagram
•Capacitive Reactance
•Inductive Reactance
1
SSCP 2313- Basic electronics2012/13-1
AC CURRENT
An alternating current has no direction in the sense that
direct current has. An “AC” circuit is one in which the
voltage and the current are sinusoidal in time.
Vp
Ip
time, t
V = Vp sin
=Vp sin (wt)
i = Ip sin (wt)
AC-voltage and current
Where : is angle in rad or degrees, w = 2π f (rad/s) and f =1/T (Hz).
V and i : the instantaneous voltage and current.
Vp and Ip : the peak current and voltage.
T
2SSCP 2313- Basic electronics 2012/13-1
3
• The relationship between period and frequency is very useful to
know when displaying an AC voltage or current waveform on an
oscilloscope screen.
• By measuring the period of the wave on the horizontal axis of the
oscilloscope screen and reciprocating that time value (in
seconds), we can determine the frequency in Hertz.
3SSCP 2313- Basic electronics 2012/13-1
4
• Other “waveforms” of AC produced within
electronic circuitry:
4SSCP 2313- Basic electronics 2012/13-1
Peak-to-peak: is twice the peak value.Vpp = 2 Vp
5
AC voltage alternates in polarity and AC current alternates in
direction
One way to express magnitude (also called the amplitude), of
an AC quantity is to measure its peak height on a waveform
graph. This is known as the peak of an AC waveform.
• Another way is to measure the total height between opposite peaks.
This is known as the peak-to-peak (P-P) value of an AC waveform
Measurement of AC Magnitude
5SSCP 2313- Basic electronics 2012/13-1
6
p
p
rms II
I 707.02
p
p
rms VV
V 707.02
Pave = Irms Vrms
RMS QUANTITIES IN AC CIRCUITS
The “RMS” stands for Root Mean Square, the
algorithm used to obtain the DC equivalent value from
points on a graph .
Sometimes the alternative terms equivalent or DC
equivalent are used instead of “RMS,” but the quantity
and principle are both the same.
6SSCP 2313- Basic electronics 2012/13-1
7
Average Amplitude
The average value of a sine wave is defined over a half cycle rather than over a full cycle.
Vave = 0.637 Vp
Iave = 0.637 Ip
Average = 0 ; Peak = + / -
because all the positive points cancel out all the
negative points over a full cycle
7SSCP 2313- Basic electronics 2012/13-1
8
8SSCP 2313- Basic electronics 2012/13-1
9
PHASE SHIFT
Phase is always a relative measurement between two
waveforms
9SSCP 2313- Basic electronics 2012/13-1
Phasor Diagrams
The phasor rotates counterclockwise about the origin with the angular frequency of the AC quantity.Phasor diagrams are useful in solving complex AC circuits.
The sine wave can be represented as the projection of a
vector rotating at a constant rate. This rotating vector is
called a phasor.
450 900 1350
1800 2700 3600
V
Radius = Vp
v = Vp sin
10
10SSCP 2313- Basic electronics 2012/13-1
11SSCP 2313- Basic electronics 2012/13-1
I-V RELATIONSHIPS IN AC CIRCUITS: RESISTORS
V and I “In-phase”, there are
no phase difference, (f = 0).
V
wtp
I
2p
VR
~R
I
V
I
V
rms
rms
p
p
R is the Resistance, ()
VR12
12SSCP 2313- Basic electronics 2012/13-1
I-V relationships in AC circuits: Inductor
L~Where XL is the Inductive Reactance, ()
V
wtp 2p
I
V and I “out of phase” by 90º.
The voltage leads the current by 90o.
(phase difference, f = +90).
L
rms
rms
p
pX
I
V
I
V
fLXL p2
where L is inductance (H).
13SSCP 2313- Basic electronics 2012/13-1
V
~
C
I-V RELATIONSHIPS IN AC CIRCUITS: CAPACITOR
c
rms
rms
p
pX
I
V
I
V
fCXC
p2
1
V
wtp 2p
I
Where Xc is the Capacitive Reactance, ()
V and I “out of phase” by 90º.
The voltage lags the current by 90o
(phase difference, f = -90).
where C is capacitance (F).
IC
VC 14
14SSCP 2313- Basic electronics 2012/13-1
The voltages across a resistor, an
inductor and a capacitor for the
same sinusoidal current.
15
15SSCP 2313- Basic electronics 2012/13-1
Example:Consider a purely inductive circuit. The voltage across a 0.2 H
inductance is VL = 100 sin (400t + 70o)V. Determine iL and sketch it.
Solution:rad/s400w
Therefore,
8020400 .LXL w
AX
VI
L
mm 251
80
100.
The current lags voltage by 900, therefore
( )AtIL
020400251 sin.
16SSCP 2313- Basic electronics 2012/13-1
17
Lecture 8
SSCP 2313- Basic electronics2012/13-1
1
Complex Number Representation in Rectangular
Form and Trigonometry Form
A complex number W with magnitude M and direction
can be written as:
z = x + jy = r (cos + j sin ) Imaginary
real
r
z
0 x
jy
Complex Number
Rectangular
formTrigonometry
form
1j
j is the imaginary unit
The x-axis - real axis
The y-axis - imaginary axis
2SSCP 2313- Basic electronics 2012/13-1
r
re
jyx)sinj(cosrz
j
The complex conjugate of a complex number, z = x + jy,
denoted by z* , is given by
z* = x – jy.
Polar
form
3SSCP 2313- Basic electronics 2012/13-1
With siny r
x
y1tan
cos ,x r
r is called the absolute value or modulus or
magnitude of z and is denoted by |z|.
22 yxrz
)sin(cos jrz
Pz = x + iy
x
y
O
Im
Re
|z| =
r
θ
4SSCP 2313- Basic electronics 2012/13-1
jyxz 22 yxr
)(tan 1
x
y
rej
x = r cos
z = rej atau r < x +jy
y = r sin
CONVERTING from Rectangular to Polar and Polar to Rectangular
Rectangular to Polar
Polar to Rectangular
5SSCP 2313- Basic electronics 2012/13-1
Suppose we have 2 complex numbers, z1 and z2 given by :
2
1
2222
1111
j
j
erjyxz
erjyxz
2121
221121
yyjxx
jyxjyxzz
Easier with normal
form than polar form
Addition and Subtraction of Complex Numbers
2121
221121
yyjxx
jyxjyxzz
6SSCP 2313- Basic electronics 2012/13-1
Example:
2. In figure below, if i2(t) = 2 cos( wt – 90) and I1 = 3 – j4,
calculate i3(t).
i1 i
2
i3
1. Calculate: (4 6i) + (3 + 7i)
= (4 + (-3) + i(-6 +7)
= 1 + i
Solution
From i2(t) = 2 cos( wt – 90), then
209022 jI
7.336.3
3
2tan23
23
)20()43(
122
213
j
jj
III
)7.33cos(6.3)(3w tti
7SSCP 2313- Basic electronics 2012/13-1
)(
21
2121
21
21
j
jj
err
ererzz
magnitudes multiply! phases add!
Easier with polar form
than normal form
Multiplying
8SSCP 2313- Basic electronics 2012/13-1
Example
45077
9010
.Z
and I
Solution:
45770
904507710
450779010
.
)().()(
).)((V
or
5050
55100
j
)j)(j(V
1. Find V, if given in a circuit
9SSCP 2313- Basic electronics 2012/13-1
For a complex number z2 ≠ 0,
)(j
j
j
er
r
er
er
z
z21
2
1
2
1
2
1
2
1
magnitudes divide!
phases subtract!
Dividing
)( 21
2
1 r
r
10SSCP 2313- Basic electronics 2012/13-1
Example
467130 .V
1535 .Z
)120(261.535
4.67130
Z
VI
5.12026
4.222.13
169
)200360()480150(
43
43
43
12050
j
j
j
j
j
j
Z
VI
Solution:
OR
12050 j
43 j
11SSCP 2313- Basic electronics 2012/13-1
1. Convert i) z = 3 + j4 to polar form.
Exercise:
3)
j
ezii
to rectangular form.
2. If z1 = 3+ j5 dan z2 = 5 + j4 . Calculate:
i) z1 + z2 ii) z1z2 iii) z1z1*
)33
26()
65
43( :Calculate.3
j
j
j
j
12SSCP 2313- Basic electronics 2012/13-1
13SSCP 2313- Basic electronics 2012/13-1
RL AND RC SERIES CIRCUITS impedance,
Low-pass and high-pass filter.
Lecture 9
SSCP 2313- Basic electronics2012/13-1
1
RC Series CircuitR
CVi
Total Impedance,= R –jXC
=
Magnitude impedance,Z =
Phase angle, =
R
XXR c
c
122 tan
22
cXR
R
XC1tan
R
XZ
V
Z
V
Ic
o
i
i
1tan
0
R
X
Z
V ci 1tan
Vi = Vi 0.
If input voltage, Vi has phase angle 0.
R
XCZ
The input voltage lags current by .
Impedance triangle:
VC
VR
Vi
Phasor diagram:
2SSCP 2313- Basic electronics 2012/13-1
cOIZV
R
X
Z
V ci 1tan o
cX 90
R
X
Z
VXcoic 1tan90
Z
VX ic
R
Xtan90 c1o
To determine Vo:
I)If Vo is taken across capacitor, it is called low- pass filteror RC lag network.
The magnitude of the output voltage,Vo =
and the phase lag between the output & input voltage, =
3SSCP 2313- Basic electronics 2012/13-1
Cut-off freq, fc- The freq at which Xc = R .fc=
RC2
1
Vo versus frequency for a low-pass R-C filter
As freq increases, Vout decreases low-pass filtering.
4SSCP 2313- Basic electronics 2012/13-1
IRVO
ii) If Vo is taken across resistor, it is called high-pass filter
or RC lead network.
0tan 1 RR
X
Z
V ci
R
X
Z
RV ci 1tan
orZ
RVV i
o
R
Xtan c1
i
c
VXR
RV
220
where
and the phase lead, =
Vo increases with freq, because as Xc
becomes smaller, more of the Vi is
dropped across the resistor.5SSCP 2313- Basic electronics 2012/13-1
R
Vi
Example 1:Determine:i) the output voltage ii)the amount of phase lag from input to output in the lag network:iii)Draw the input and output waveforms showing the proper relationship.
Cf =1kHz
10 Vrms
680
0.1F
V0o
Z
VXVi ic
0)
Xc = 1/2fC = 1/2(1 kHz)(0.1 F) = 1.6 k
Z = R2 + Xc2 = 1.73 k
Vo = 9.2 Vrms
Solution:
6SSCP 2313- Basic electronics 2012/13-1
iii) The waveforms:
ii) The phase lag,
2.23
tan90 1
R
Xco
7SSCP 2313- Basic electronics 2012/13-1
Impedance,
= R +jXL
Magnitude impedance, Z =
Phase angle, =
R
XtanXR L
L122
2L
2 XR
R
Xtan L1
Vi
R
L
RL Series Circuit
Phasor diagram:
Impedance Triangle:
VR
VLVi
XL
Z
R
The input voltage leads the current by . 0ii VV
R
XtanZ
V
Z
V
IL
oi
i
1
0
R
Xtan
Z
V Li 1
If the input voltage Vi has phase angle 0.
=
8SSCP 2313- Basic electronics 2012/13-1
IRVo
01 RR
Xtan
Z
V Li
R
Xtan
Z
RV Li 1
If Vo is taken across R is called low- pass filter
(phase lag)
R
VL
Vout
L
VoutVin
Vin
Vout
Vin
9SSCP 2313- Basic electronics 2012/13-1
Lo IXV
901
LLi X
R
Xtan
Z
V
R
Xtan
Z
XV LoLi 190
If Vo is taken across L is called high- pass filter.
q
(phase lead)
R
VR
Vout
L VoutVin
Vin
Vout Vin
10SSCP 2313- Basic electronics 2012/13-1
R
0.068 F
10 k
90 V
200 Hz
i) Determine I, Z, VC and VR
ii) Draw the waveforms for Vin, VR and Vc
(Ans: I= 5.84 mA, Z= 15.4 k,
VC = 68.3 V
40 V
50 kHz
R
30 mH
6.8 k
i)Determine I, Z, VL , VR.ii)Draw the waveforms for Vin, VR
and VL
(Ans: I= 3.44 mA, Z= 11.62 k,
VL = 32.4 V , VR = 23.4 V )
Problems
11SSCP 2313- Basic electronics 2012/13-1
12SSCP 2313- Basic electronics 2012/13-1
LECTURE 10
SERIES RCL•Impedance
•Series resonance•Quality factor1
SSCP 2313- Basic electronics2012/13-1
THE SERIES RLC CIRCUIT
• This circuit acts as
both a low pass
and a high pass
filter at the same
time.
• It only allows
signal to pass from
a narrow range of
frequencies.
2 2SSCP 2313- Basic electronics 2012/13-1
R L C
VS
The total impedance for the RLC circuit is given
by
L CR jX jX Z
In polar form, this is written
22 1tan tot
L C
XR X X
R
Z
Impedance of series RLC circuits
X XL C
Z
R
“ Impedance Triangle”
3
3SSCP 2313- Basic electronics 2012/13-1
Variation of XL and XC with frequency
In a series RLC circuit, the circuit can be capacitive or inductive,
depending on the frequency.
At the frequency where
XC=XL, the circuit is at series
resonance.
Below the resonant
frequency, the circuit is
predominantly capacitive.
Above the resonant
frequency, the circuit
is predominantly
inductive.
Rea
ctan
ce
f
XC XL
Series
resonance
XC=XL
XC>XL XL>XC
4SSCP 2313- Basic electronics 2012/13-1
What is the total impedance and phase angle of the series
RLC circuit if R= 1.0 kW, XL = 2.0 kW, and XC = 5.0 kW?
VS
R L C
1.0 kW XC =
5.0 kW
XL =
2.0 kW
The circuit is capacitive, so I leads V by 71.6o.
The impedance is
WW
kjk
XXjRZ cL
)3(1
)(
3.16 kW
The total impedance is
2 2 2 21.0 k +3.0 ktot totZ R X W W
The phase angle is
1 1 3.0 ktan tan
1.0 k
totX
R W
W 71.6o
W 6.7116.3 kZ
5SSCP 2313- Basic electronics 2012/13-1
What is the total impedance for the circuit below?
R L C
VS
330 mH
f = 400 kHz
470 W 2000 pF
786 53.3 W Z
The circuit is inductive.6
6SSCP 2313- Basic electronics 2012/13-1
XL=XC
Notice that there is a frequency at which XC = XL.
This condition is called series resonance.
Series resonance
XL
f
XC
X
Series Resonance
1
2rf
LC
The resonant frequency:
At series resonance, XC and XL
cancel. VC and VL also cancel
because the voltages are equal
and opposite. The circuit is
purely resistive at resonance,
and Z = R..
- the total impedance is a
minimum.
Z = R
7SSCP 2313- Basic electronics 2012/13-1
8
Resonance occurs at the frequency o where the current has its maximum value
8SSCP 2313- Basic electronics 2012/13-1
What is the resonant frequency for the circuit?
R L C
VS
330 mH470 W 2000 pF
fr =196 kHz
9
9SSCP 2313- Basic electronics 2012/13-1
• The sharpness of the resonance curve is usually described
by a dimensionless parameter known as the quality factor,
Q.
• For a series RLC circuit the quality factor is
– is the bandwidth of the curve, measured between the
two values of for which Pav has half its maximum value
• These points are called the half-power points
Quality Factor
10
C
L
R
R
Lf
R
LQ
s
s
1
2
0
10SSCP 2313- Basic electronics 2012/13-1
The bandwidth (BW) of the
resonant circuit is the range of
frequencies for which the output
is 70.7% of the maximum current.
rfBWQ
RIPwherePPHPF
2
maxmaxmax2
1
f1 and f2 are referred to as the
critical frequencies, cutoff
frequencies or half-power
frequencies.
11SSCP 2313- Basic electronics 2012/13-1
12SSCP 2313- Basic electronics 2012/13-1
2
2:
2
1
12
fff
fffwhere
Q
ffffBW
r
r
r
• A high-Q circuit responds
only to a narrow range of
frequencies
– Narrow peak
• A low-Q circuit can detect
a much broader range of
frequencies
• Typical Q values in
electronics range from 10
to 100
13 13SSCP 2313- Basic electronics 2012/13-1
a. For the series resonant circuit, find I, VR, VL, and VC at resonance.
b. What is the Q of the circuit?
c. If the resonant frequency is 5000Hz, find the bandwidth.
d. What is the power dissipated in the circuit at the half-power frequencies?
14SSCP 2313- Basic electronics 2012/13-1
9050
9050
010
05.
VV
VV
VV
AIa
C
L
R
WPd
HzBWc
Qb
HPF
s
25.
1000.
5.
LECTURE 11
PARALLEL RCL•Conductance, Susceptance, Admittance
•Resonant frequency•Series-parallel RLC circuit1
SSCP 2313- Basic electronics2012/13-1
1 1
Impedance of parallel RLC circuits
1 1 1 1
0 90 90L CR X X
Z
R L CVS
)11
(1
1111
LC
LC
XXj
R
jXjXRZ
Admittance, Y =
Conductance (G), Admittance (Y) and Susceptance (B)
1 1
0R
G
RConductance,
Capacitive Susceptance, 1
90C
CX
B
cjX
1
Inductive Susceptance, 1
90L
LX
B
LjX
1
22SSCP 2313- Basic electronics 2012/13-1
R
10
Xc
5 Vi
10 V
XL
10
Determine the equivalent impedance for the circuit :
)XX
(jRZ
YLC
1111
)(j10
1
5
1
10
1
= 0.1 + j 0.1
= 0.141 45
45 0.141Z
1
= 7.1 - 45
IT = V Y = 10 (0.141 45)
= 1.41 A 45 ,
it lags the input voltage by 45.
3 3SSCP 2313- Basic electronics 2012/13-1
Draw the admittance phasor diagram for the circuit.
1 11.0 mS
1.0 kG
R
1
0.629 mS2 10 kHz 25.3 mH
LB
2 22 2 + 1.0 mS + 0.629 mS 1.18 mSLY G B
VSL
25.3 mHf = 10 kHz
R
1.0 k
Y =
1.18 mS
G = 1.0 mS
BL =
0.629 mS
4 4SSCP 2313- Basic electronics 2012/13-1
For parallel RLC circuits, the current phasors can be
obtained directly from Ohm’s law.
SR
VI
R
SC
C
V
IX
SL
L
V
IX
and that IR is plotted along the positive real
axis.
and that IC is plotted along the positive j axis.
and that IL is plotted along the negative j
axis.
+90o
90o
IC
IR
IL
The total current is given by:
22 1tan CL
tot R C L
R
II I I
I
I
5 5SSCP 2313- Basic electronics 2012/13-1
Parallel resonance
Ideally, at parallel resonance, IC and IL cancel because the
currents are equal and opposite. The circuit is purely
resistive at resonance.
The resonant frequency in both parallel and
series circuits is the same
1
2rf
LC (ideal case)
• Capacitive and inductive susceptance
are equal.
• Total impedance is a maximum
(ideally infinite).
• The current is minimum.
6 6SSCP 2313- Basic electronics 2012/13-1
f
Zmax
0.707Zmax
f1 fr f2
BW
Ztot
rfBWQ
Series-parallel RLC circuits
VS =R L
C
200 kHz
700 mH1.0 k
2700 pFCalculate the total
impedance of the circuit.
The impedance of the yellow box,
7 7SSCP 2313- Basic electronics 2012/13-1
k2.0j43.0
49k66.0Z
L
L1
X
1j
R
1
jX
1
R
1
Z
1Y The total impedance , ZT
c1 jXZ
9.24k47.0
8
LECTURE 12
Power in AC Circuit•Real, Reactive and Apparent Power
•Power Factor•Maximum Power Transfer Theorem1
SSCP 2313- Basic electronics2012/13-1
1 1SSCP 2313- Basic electronics 2012/13-1
Power in Resistive Components
• Suppose a voltage v = Vp sin t is applied across a
resistance R. The resultant current i will be
• The power p :
• The average value of (1 - cos 2t) is 1, so
tIR
tV
R
vi P
P
sinsin
)2
2cos1()(sinsinsin 2 t
IVtIVtItVvip PPPPPP
rmsrmsPP
PP IVIV
IVP 222
1Power Average
2 2SSCP 2313- Basic electronics 2012/13-1
Relationship between v, i and p in a resistor
3 3SSCP 2313- Basic electronics 2012/13-1
Power in Capacitors
• In capacitors, the current leads the voltage by 90.
Therefore, if a voltage v = Vp sin t is applied across a
capacitance C, the current will be given by i = Ip cos t.
• Then
• The average power is zero
)2
2sin(
)cos(sin
cossin
tIV
ttIV
tItV
vip
PP
PP
PP
4 4SSCP 2313- Basic electronics 2012/13-1
Relationship between v, i and p in a capacitor
5 5SSCP 2313- Basic electronics 2012/13-1
Power in Inductors
• In inductors, the current lags the voltage by 90.
Therefore, if a voltage v = Vp sin t is applied across an
inductance L, the current will be given by i = -Ip cos t
• Therefore
• Again the average power is zero
)2
2sin(
)cos(sin
cossin
tIV
ttIV
tItV
vip
PP
PP
PP
6 6SSCP 2313- Basic electronics 2012/13-1
Relationship between v, i and p in an inductor
7 7SSCP 2313- Basic electronics 2012/13-1
Circuit with Resistance and Reactance
• When a sinusoidal voltage v = Vp sin t is applied
across a circuit with resistance and reactance, the
current i = Ip sin (t - ).
• Therefore, the instantaneous power, p is
)2cos(2
1cos
2
1
)2cos(cos2
1
)sin(sin
tIVIVp
tIV
tItV
vip
PPPP
PP
PP
8 8SSCP 2313- Basic electronics 2012/13-1
• The expression for p has two components
• The second part oscillates at 2 and has an average
value of zero over a complete cycle
– this is the power that is stored in the reactive
elements and then returned to the circuit within each
cycle
• The first part represents the power dissipated in resistive
components. Average power dissipation is
)2cos(2
1cos
2
1 tIVIVp PPPP
cos)(cos22
)(cos2
1VI
IVIVP PPPP
9 9SSCP 2313- Basic electronics 2012/13-1
• The average power dissipation
is termed the real power in units watts (W).
• The product of the r.m.s. voltage and current VI is termed
the apparent power, S in units volt amperes (VA).
cos)(cos2
1VIIVP PP
cos
cos
S
VIP
• This cosine is referred to as the power factor
cosfactor Power S
P
10 10SSCP 2313- Basic electronics 2012/13-1
Active and Reactive Power
• When a circuit has resistive and reactive parts, the
resultant power has 2 parts:
– The first is dissipated in the resistive element. This is
the real power, P
– The second is stored and returned by the reactive
element. This is the reactive power, Q , which has
units of volt amperes reactive (VAR).
Reactive power (inductive)
QL = VI = I 2 XL (VAR)
Reactive power (capacitive)
QC = VI = I 2 XC (VAR)
11 11SSCP 2313- Basic electronics 2012/13-1
Power triangle
S = VI
S = VI
S = P – jQC
S = P + jQL
12 12SSCP 2313- Basic electronics 2012/13-1
Real Power P = I2R = VI cos watts
Reactive Power Q = VI sin var
Apparent Power S = VI VA
S 2 = P 2 + Q 2
Note: V and I are in rms value
)(2CL XXIQ
Summary
13 13SSCP 2313- Basic electronics 2012/13-1
Maximum power transfer theorem
• When the output of a circuit has a reactive element
maximum power transfer is achieved when the load
impedance is equal to the complex conjugate of the
output impedance.
14 14SSCP 2313- Basic electronics 2012/13-1
Determine the total PT
and QT for the circuit.
Sketch the power triangle.
15 15SSCP 2313- Basic electronics 2012/13-1
16
1
Analysis of Series RCL Circuit :
Kirchhoff’s Laws
Lecture 13
1
SSCP 2313- Basic electronics2012/13-1
2
Kirchhoff’s Laws in Complex FormKirchhoff’s Current Law (KCL):
At any junction in a circuit with impedances, the total complex
current is zero. 0I
Kirchhoff’s Voltage Law (KVL):
For any closed loop with impedances, the total complex
e.m.f. is equal to the total potential differences across the
circuit components in the loop.
IZV
Based on Kirchhoff’s law:
kVVVV ....0 21
2SSCP 2313- Basic electronics 2012/13-1
3
Z1
Z2
Z3
V1
V2
V3
V• From KVL:
eqZI
)ZZZ(I
VVVV
321
321
where Zeq is the equivalent impedance. Thus
321 ZZZZ eq
Analysis of Series RCL Circuit Using Kirchhoff’s Laws
3SSCP 2313- Basic electronics 2012/13-1
Complex Power
*IVS, PowerComplex
jQP
4SSCP 2313- Basic electronics 2012/13-1
5
For the circuit shown below, calculate:
(a) the impedances Z1, Z2 dan Z3
(b) the total impedance of the circuit
(c) the current I
(d) the voltage V across Z1
(e) the power dissipated in Z1
Z1 Z2
Z3
0.1 H 10 5 10-3
F
0.2 H
5
V = 28 V
f = 50/ Hz
Example 1:
5SSCP 2313- Basic electronics 2012/13-1
6
(a) The impedances Z1, Z2 dan Z3 are
4514
1010
)1.0)(100(10
111
j
j
LjRZ
4.6311
105
10100
15
1
3
22
j
j
CjRZ
7621
205
)2.0)(100(5
333
j
j
LjRZ
Solution:
6SSCP 2313- Basic electronics 2012/13-1
(b) The total impedance is
453.28
2020
)205()105()1010(
ZZZZ 321
j
jjj
eq
(c) The current I is
451453.28
028
Z
VI
eq
A
The current I is 1A rms and lagged behind the voltage by 45.
(d) The voltage across Z1 is
0V41)4514)(451(ZIV 111
The voltage across Z1 is 14V in phase with the source voltage.
7SSCP 2313- Basic electronics 2012/13-1
8
(e) The dissipated power on Z1 is
1010
4514
)451014(
11
j
)(
IVS
The apparent power is 14 VA,
The real power is 10 W and
The reactive power is 10 VAR
8SSCP 2313- Basic electronics 2012/13-1
Analysis of Parallel RCL Circuit Using Kirchhoff’s Laws
Z1 Z2 Z3
I1 I2 I3
IT
V
321 IIII • From KCL:
321
111
ZZZV=
eqZ
V=
321
1111
ZZZZeq
where Zeq is the equivalent impedance. Thus
I
200 V5F
400 f= 1000/ Hz
I1 I2
Example 2:
For the circuit shown below, calculate:
(a) the total impedance of the circuit
(b) the current I1, I2 dan I.
(c) the total power dissipated in the circuit
0.15 H
Cj
CjZ
111
Solution:
= -j100 = 100 -90o
LjRZ 2 = 400 + j300 = 500 37o
21
111
ZZZeq
(a)
300400
1
100
1
jj
Zeq = 112 -79.6o
o
o
o
Z
V
I 90290100
0200
11
A,(b)
AZ
V
I o
o
o
374.037500
0200
22
AZ
V
I o
o
o
6.7979.16.79112
0200
*IVS
= 200 0o (1.79 -79.6o )= 358 -79.6o
= 64.6 –j352
(c)
S =358 VA, P = 64.6 W andQ = 352 VAR (capacitive).
13SSCP 2313- Basic electronics 2012/13-1
Lecture 14
Analysis of RCL Circuit :
Mesh Current
SSCP 2313- Basic electronics2012/13-1
2
RCL circuit analysis using: Mesh Current
1Z
2Z
3Z
aI cI
bI
1V 2V
• We can use Kirchhoff Laws to obtain three equations
and the values of Ia, Ib and Ic can be solved.
• This problem can be simplified by using mesh current
analysis method
2SSCP 2313- Basic electronics 2012/13-1
3
Loop 1:
1Z
2Z
3Z
1I 2I1V 2V
22211
221111
)(
)(
ZIZZI
ZIIZIV
Loop 2:
21322
221322
)(
)(
ZIZZI
ZIIZIV
• Only two equations required to solve the values of I1 and I2 , to obtain the values of Ia, Ib and Ic .
3SSCP 2313- Basic electronics 2012/13-1
Cramer’s Rule:
a11 x1 + a12 x2 = k1
a21 x1 + a22 x2 = k2
Can be written in matrix forms:
a11 a12 x1 = k1a21 a22 x2 = k2
x1 , x2 and x3 can be found by :
x1 = k1 a12k2 a22
a
4SSCP 2313- Basic electronics 2012/13-1
x2 = a11 k1a21 k2
a
where a = a11 a12
a21 a22
5SSCP 2313- Basic electronics 2012/13-1
Referring to the figure below, calculate the values of Ia, Iband Ic
2 - j5 j6
4 010 305
Ia
Ib
Ic
6SSCP 2313- Basic electronics 2012/13-1
Example 1
Solution
The circuit can be redrawn as shown below:
621 jZ += 53 jZ -=
42 =Z
o
010
o305
1I 2I
Loop 1:
21
21
211
4)66(
4)642(
4)()62(010
II
II
III
j
j
j
7SSCP 2313- Basic electronics 2012/13-1
Loop 2:
12
212
4)54(
4)()5(305
II
III
j
j
The two simultaneous equations can be solved using
Cramer rules to find I1 and I2.
305
010
544
466
2
1
I
I
j
j
95.38
638
(4)(4)-j5)-j6)(4-(6
544
466
j
j
j
8SSCP 2313- Basic electronics 2012/13-1
9
Thus,
3.60A67.1
95.38
60j7.22
95.38
)305)(4()5j4)(010(
95.38
5j4305
4010
I1
9SSCP 2313- Basic electronics 2012/13-1
10
Therefore:
45A3.1
95.38
7.533.50
95.38
)010)(4()305)(6j6(
95.38
3054
0106j6
I2
1.53A95.2III
45A3.1II
3.60A67.1II
21b
2c
1a
10SSCP 2313- Basic electronics 2012/13-1
11SSCP 2313- Basic electronics 2012/13-1
Thank you…
Lecture 15
RCL circuit analysis using:
Thevenin’s Theorem
SSCP 2313- Basic electronics2012/13-1
Any circuit having two output terminals A and B, can be
replaced by an equivalent Thevenin’s circuit having a
Thevenin’s voltage source VTH in series with a Thevenin’s
impedance, ZTH.
• Thevenin’s output voltage, VTH is the open circuit
voltage measured at terminals A and B.
• Thevenin’s equivalent impedance, ZTH is the impedance
measured at the terminals A and B where the voltage
source is shorted and the current source is opened.
A
B
CIRCUIT
A
B
VTH
zTH
RCL circuit analysis : Thevenin’s Theorem
22SSCP 2313- Basic electronics 2012/13-1
• If a load resistor, RL is connected at the output terminals A
and B, the current in RL :
LT
TL
RZ
VII
A
B
RL
ZT IL
VT
I
33SSCP 2313- Basic electronics 2012/13-1
Example 1:
Determine the Thevenin’s equivalent voltage VTH and the
Thevenin’s equivalent impedance ZTH for the circuit below:
010
16j 10
10
9010
4j
5jA'
B'
A
B
I
44SSCP 2313- Basic electronics 2012/13-1
Solution:
VT = VAB = VA’B’
Loop 1,
ZIE
A 5.0
20j20
10j10
20j20
9010010I
)4j1016j10(I9010010
V 4.632.11
10j5
j10-(10)0.5
9010)10(IVT
55SSCP 2313- Basic electronics 2012/13-1
The value of ZTH can be calculated by shorting all the
voltage sources:
16j 10
10
4j
5jA'
B'
A
B
6.188.7
5.2j4.7
5j1020j10
)20j10(10
5j]10||)20j10[(ZTH
66SSCP 2313- Basic electronics 2012/13-1
If the terminals A and B are connected to an impedance ,
ZL = 10 + j10, what is the current flowing in ZL ?
.
A 7.8659.0
5.7j4.17
4.632.11
)10j10()5.2j4.7(
4.632.11
RZ
VII
LTH
THL
A
B
VTH
zTH
ZL= 10 +j10
Draw the Thevenin’s equivalent circuit:
By using Kirchhoff’s voltage law,
77SSCP 2313- Basic electronics 2012/13-1
-j2
3
5 0o
j4
ZL= 5
Find current in ZL using Thevenin Theorem.
Example 2:
Ans:( IL = 1.07 43.5o )
88SSCP 2313- Basic electronics 2012/13-1
99SSCP 2313- Basic electronics 2012/13-1
Lecture 16
RCL circuit analysis using:
Norton’s Theorem
SSCP 2313- Basic electronics2012/13-1
• Any circuit having two output terminals A and B, can be
replaced by an equivalent Norton’s circuit having a
Norton’s current source IN in parallel with a Norton’s
impedance, ZN, :
A
B
CIRCUIT
A
B
ZN Z
LIN
RCL circuit analysis : Norton’s Theorem
NLN
NL I
ZZ
ZI
2SSCP 2313- Basic electronics 2012/13-1
Example 1 :
For the circuit shown below, calculate IN , ZN and IL
9050
5
3
4j
5jA
B
050
33SSCP 2313- Basic electronics 2012/13-1
• The Norton’s current source, IN is the current which flows
through the short circuit at the terminals A and B.
• The Norton’s impedance, ZN is the equivalent impedance
between terminals A and B when the current source is
opened and the voltage source is shorted.
To calculate ZN, terminals A and B are opened and
all sources are shorted:5 5jA
B
ZN = ((5 + j0) // (0 + j5)
=3.53 45
44SSCP 2313- Basic electronics 2012/13-1
Solution:
Loop 1:
9050
5
050
5jA
B
Loop 1 Loop 2
I1 I2
IN
90A10
5
9050I
)5)(I(9050
1
1
55SSCP 2313- Basic electronics 2012/13-1
To calculate IN, terminals A and B are shorted.
Loop 2 :
90A10
905
050
5j
050I
)5j)(I(050
2
2
90A20
20j
10j10j
90109010
III 21N
66SSCP 2313- Basic electronics 2012/13-1
To calculate IL, draw the Norton equivalent circuit:
A
B
IN
9020
ZN
4553.3 Z
L
Using the current divider rule, the load current IL
2.58A3.8
)4j3(5453.3
5453.39020
ZZ
ZII
LN
NNL
77SSCP 2313- Basic electronics 2012/13-1
-j2
3
5 0o
j4
ZL= 5
Find current in ZL using Norton’s Theorem.
Example 2:
Ans:( IL = 1.07 A 43.5o )
88SSCP 2313- Basic electronics 2012/13-1
Semiconductor
LECTURE 17
SSCP 2313- Basic electronics2012/13-1
• Atom is the smallest particle for an element which
maintain the characteristics of the element.
• Each atom has an orbital type structure consisting of
a nucleus surrounded by orbiting electrons.
• The nucleus consists of protons and neutrons.
• Electrons are negatively charged particles and
protons are positively charged particles
• Neutrons are particles with no charge.
• The number of electrons in an electrically neutral
atom of a material gives the atomic number for that
material
Atomic Structure
22SSCP 2313- Basic electronics 2012/13-1
• The goal of electronic materials is to generate andcontrol the flow of an electrical current.
• Electronic materials include:
1. Conductors: have low resistance which allowselectrical current flow
2. Insulators: have high resistance whichsuppresses electrical current flow
3. Semiconductors: can allow or suppresselectrical current flow
Electronic Materials
33SSCP 2313- Basic electronics 2012/13-1
• Good conductors have low resistance so
electrons flow through them with ease.
• Best conductors include:
– Copper, silver, gold, aluminum, & nickel
• Alloys are also good conductors:
– Brass & steel
• Good conductors can also be liquid:
– Salt water
Conductors
44SSCP 2313- Basic electronics 2012/13-1
• The atomic structure of good
conductors usually includes
only one electron in their
outer shell.
– It is called a valence
electron.
– It is easily striped from
the atom, producing
current flow.
Conductor Atomic Structure
Copper Atom
55SSCP 2313- Basic electronics 2012/13-1
• Insulators have a high resistance so current
does not flow in them.
• Good insulators include:
– Glass, ceramic, plastics, & wood
• Most insulators are compounds of several
elements.
• The atoms are tightly bound to one another so
electrons are difficult to strip away for current
flow.
Insulators
66SSCP 2313- Basic electronics 2012/13-1
• Semiconductors are materials that essentially
can be conditioned to act as good conductors, or
good insulators, or any thing in between.
• Common elements : carbon (C), silicon (Si),
and germanium (Ge)
• Silicon is the best and most widely used
semiconductor.
Semiconductors
77SSCP 2313- Basic electronics 2012/13-1
The name “semiconductor” implies that it conducts
somewhere between the two cases (conductors or
insulators)
Conductivity :
σmetals ~1010 /Ω-cm
σinsulators ~ 10-22 /Ω-cm
The conductivity (σ) of asemiconductor (S/C) lies between these two extreme cases.
S/C
8 8SSCP 2313- Basic electronics 2012/13-1
Semiconductor Valence Orbit
• The main characteristic of a semiconductor element is
that it has four electrons in its outer or valence orbit.
99SSCP 2313- Basic electronics 2012/13-1
• The unique capability of semiconductor atoms is theirability to link together to form a physical structure calleda crystal lattice.
• The atoms link together with one another sharing theirouter electrons.
• These links are called covalent bonds.
Crystal Lattice Structure
1010SSCP 2313- Basic electronics 2012/13-1
Intrinsic Semiconductor
Pure (good insulating) semiconductor material, i.e.
undopped semiconductor
1111SSCP 2313- Basic electronics 2012/13-1
Since the outer valence electrons of each atom aretightly bound together with one another, theelectrons are difficult to dislodge for current flow.
Silicon is a great insulator.
Types of Semiconductor Materials
Doping forms two types of semiconductor:
i. N type :-The Si doped with extra e (donor).
-“N” is for -ve, which is the charge of
an electron.
ii. P type” :- Si doped with material missing e that
produce holes (acceptors).
- “P” is for +ve, which is the charge
of a hole
Extrinsic Semiconductor Dopped semiconductor materials
To make the semiconductor conduct electricity, otheratoms called impurities must be added.This processis called doping.
1212SSCP 2313- Basic electronics 2012/13-1
An impurity, like arsenic or phosphorous, has 5 valenceelectrons.
Adding arsenic (doping) will allow four of the As valenceelectrons to bond with the neighboring Si atoms.
The one electron left over for each As atom becomesavailable to conduct current flow.
The free electrons are the majority carriers.
N type
1313SSCP 2313- Basic electronics 2012/13-1
P type
An impurity, like Boron, has 3 valence electrons.
The 3 e in the outer orbit form covalent bonds with its
neighboring semiconductor atoms. But one e is missing from
the bond.
This place where a fourth electron, is referred to as a hole.
The hole assumes a + charge so it can attract e from some
other source. Thus, holes are the majority carriers to
support current flow.
1414SSCP 2313- Basic electronics 2012/13-1
• If you use lots of arsenic atoms for doping, there
will be lots of extra electrons so the resistance of
the material will be low and current will flow freely.
• If you use only a few boron atoms, there will be
fewer free electrons so the resistance will be high
and less current will flow.
• By controlling the doping amount, any resistance
can be achieved.
Resistance Effects of Doping
1515SSCP 2313- Basic electronics 2012/13-1
Current Flow in N-type
Semiconductors
• The battery has a + terminal
that attracts the free electrons in
the semiconductor and pulls
them away from their atoms
leaving the atoms charged
positively.
• Electrons from the - terminal of
the supply enter the
semiconductor material and are
attracted by the + charge of the
atoms missing one of their
electrons.
• Current (electrons) flows from
the + terminal to the – terminal.
1616SSCP 2313- Basic electronics 2012/13-1
Current Flow in P-type
Semiconductors
• Electrons from the - battery
are attracted to the positive
holes and fill them.
• The + terminal of the
supply pulls the electrons
from the holes leaving the
holes to attract more
electrons.
• Current (electrons) flows
from the - terminal to the +
terminal.
1717SSCP 2313- Basic electronics 2012/13-1
Thank You…
LECTURE 18PN JUNCTION- Diode
SSCP 2313- Basic electronics2012/13-1
• A diode is made by joining p and n-type semiconductor
materials
• Diodes are unidirectional devices that allow current to
flow in one direction.
Structure of PN junction diode
Symbol of PN junction diode
2SSCP 2313- Basic electronics 2012/13-1
The PN Junction Diode
The Open-Circuited p-n JunctionA p-n junction is the junction between an n-type
semiconductor and a p-type semiconductor.
Near the junction, electrons diffuse across to combine
with holes, creating a "depletion region".
3SSCP 2313- Basic electronics 2012/13-1
In the p-type region there are holes from the
acceptor impurities and in the n-type region there
are extra electrons.
When a p-n junction is formed, some of the
electrons from the n-region which have reached
the conduction band are free to diffuse across the
junction and combine with holes.
Filling a hole makes a - ion and leaves behind a +
ion on the n-side. A space charge builds up,
creating a depletion region which inhibits any
further electron transfer unless it is helped by
putting a forward bias on the junction.
Depletion Region
4SSCP 2313- Basic electronics 2012/13-1
The Biased p-n Junction: A. Forward Bias
Because + charge is repelled by a + voltage and - charge is
repelled by a - voltage, both the free electrons in the n-type and the
holes in the p-type are forced toward the junction.
This causes the width of the depletion region to decrease.
The potential across the junction which opposes diffusion is
decreased by the applied bias to the value VB − VS, where VB is the
built-in potential and Vs is the biased voltage.
This caused the diffusion current
increases rapidly as VS is increased.
Because a large current flows, the
junction is said to be forward biased.
current is permitted
5SSCP 2313- Basic electronics 2012/13-1
B. Reverse Bias
Because - charge is attracted by a + voltage and + charge is
attracted by a - voltage, both the free electrons in the n-type
and the holes in the p-type are pulled away from the junction.
This causes the width of the depletion region to increase.
The potential across the junction which opposes diffusion is
increased by the applied bias to the value VB + VS.
This causes a very small
current to flow, thus the
junction is said to be reverse
biased.
so current is prevented
66SSCP 2313- Basic electronics 2012/13-1
Forward Biased diode
The diode behaves like a ‘ON’ switch in this mode. A
conduction diode has approximately a constant voltage
drop across it. It’s called turn-on voltage.
Resistance R and diode’s body resistance limits the
current through the diode
Vs has to overcome VB in order for the diode to
conduct.
VV
VV
onD
onD
25.0
7.0
)(
)(
For silicon
For germanium
77SSCP 2313- Basic electronics 2012/13-1
Reverse Biased diode
The diode behaves like a ‘OFF’ switch in this
mode.
If we continue to increase reverse voltage Vs
breakdown voltage of the diode is reached.
Once breakdown voltage is reached, diode
conducts heavily causing its destruction.
88SSCP 2313- Basic electronics 2012/13-1
Breakdown diode
Diode breakdown is caused by thermally generated
electrons in the depletion region.
When the reverse voltage across diode reaches
breakdown voltage, these electrons will get sufficient
energy to collide and dislodge other electrons.
The number of high energy electrons increases
rapidly, leading to an avalanche effect causing heavy
current and ultimately destruction of diode.
99SSCP 2313- Basic electronics 2012/13-1
In the forward bias
region, current increases
rapidly after the barrier
potential (0.7 V for Si) is
reached.
The voltage across the
diode remains equal to the
barrier potential.
The reverse-biased diode
effectively acts as an
insulator until breakdown is
reached.
I-V Characteristics Curve
VR VF
IF
IRReverse voltage
Forward
voltage
0.7 V
Barrier potential
VBR (breakdown voltage)
1010SSCP 2313- Basic electronics 2012/13-1
Leakage current
Avalanche current
Diode models
The ideal model assumes the diode is either an open or closed switch.
VR
VF
IF
IR
Reverse
bias
Forward
bias
The complete model includes the forward resistance of the diode.
The practical model includesthe barrier voltage in theapproximation.
0.7 V
The diode is designed to allow current to flow in only one
direction. The ideal diode would be a perfect conductor in one
direction (forward bias) and a perfect insulator in the other
direction (reverse bias). In many situations, using the ideal
diode approximation is acceptable.
1111SSCP 2313- Basic electronics 2012/13-1
Thank you….
Chapter 19
i) Rectifier Circuits
SSCP 2313- Basic electronics2012/13-1
• Power supply is a group of circuits that convert ac
energy to dc energy.
• Two types:
– linear power supply: provides constant current path
between its input and its load.
- switching power supply: provides intermittent current
path between its input and its output.
Introduction
2SSCP 2313- Basic electronics 2012/13-1
Linear Power Supply
Basic components:1. Rectifier – diode circuit that converts the ac to what is called
a pulsating dc.2. Filter – circuit that reduces the variations in the output of
the rectifier.3. Voltage regulator – maintain a constant power supply output
voltage.
3SSCP 2313- Basic electronics 2012/13-1
2 types:
Half-wave rectifier
Full-wave rectifier
Rectifier
4SSCP 2313- Basic electronics 2012/13-1
Half-wave Rectifier
5SSCP 2313- Basic electronics 2012/13-1
The effect of the barrier potential on the half-wave rectified output voltage
6SSCP 2313- Basic electronics 2012/13-1
Average and RMS value of HWR
π
VV P
DC 2
VV P
rms
VDC
Vrms
t0
t1
t2
t3
Vp
VO
t
7SSCP 2313- Basic electronics 2012/13-1
8SSCP 2313- Basic electronics 2012/13-1
A center-tapped full-wave rectifier.
9SSCP 2313- Basic electronics 2012/13-1
Basic operation of a center-tapped full-wave rectifier.
10SSCP 2313- Basic electronics 2012/13-1
Center-tapped full-wave rectifier with a
transformer turns ratio of 1.
Vp(pri) is the peak value of the primary voltage.
11SSCP 2313- Basic electronics 2012/13-1
Center-tapped full-wave rectifier with a
transformer turns ratio of 2.
12SSCP 2313- Basic electronics 2012/13-1
Full - wave Bridge Rectifier
can buy pre-packaged diode bridges
13SSCP 2313- Basic electronics 2012/13-1
Full-wave Rectifier - operation
π
V2V P
DC
2
VV P
rms
14SSCP 2313- Basic electronics 2012/13-1
15SSCP 2313- Basic electronics 2012/13-1
The End
16SSCP 2313- Basic electronics 2012/13-1
Lecture 20
Capacitive filter in power supply
SSCP 2313- Basic electronics2012/13-1
Transformer Rectifier FilterInput Voltage
240 V 50 Hz
DC Output
Voltage
(Unregulated)
Unregulated Power Supply
A simplest power supply consists of a transformer,
a rectifier and a filter.
In a power supply, a transformer is required to step-up or
step-down mains voltage source of 240 Vrms at 50 Hz to a
suitable voltage.
The function of a rectifier is to convert the two way current
from the transformer to a one way current
2SSCP 2313- Basic electronics 2012/13-1
A
C
B
D
AC source
load
capacitor
An RC circuit has a time constant = RC
The function of a filter is to remove ripples on the rectifier
output and create a power supply with a constant DC
value.
The simplest filter consists of a large capacitor which is
connected in parallel with a load resistor RL .
3SSCP 2313- Basic electronics 2012/13-1
Filter
Vr = ripple voltage
4SSCP 2313- Basic electronics 2012/13-1
How the Filter Works?
When the diode conducts, thecapacitor charges up to thepeak of the sine wave.
When the sine voltage drops,the charge on the capacitorremains. Since the capacitor islarge it forms a long timeconstant with the load resistor.The capacitor slowlydischarges into the loadmaintaining a more constantoutput.
The next positive pulse comesalong, recharging the capacitorand the process continues.
5SSCP 2313- Basic electronics 2012/13-1
Large capacitor capable of storing charge in longer time, and will not
discharge too much during the time between t1 and t2.
If RL is small, the capacitor will discharge faster because the
discharge current through RL is large.
- the more charge will be removed from the capacitor and the larger
will be the ripple
Effect of RL and CF on the ripple
6SSCP 2313- Basic electronics 2012/13-1
Filter Output Voltage
Vpk : peak rectifier output voltage
Vdc : average (or dc) value
Vr : ripple
Example:
Assuming the line frequency is 60 Hz, the time between charging peak for half-wave rectifier,
t = 1/f = 1/60 = 16.7 ms
For full-wave rectifier, the frequency is 2x60 = 120 Hz. Therefore,
t = 1/120 = 8.33 ms
7SSCP 2313- Basic electronics 2012/13-1
Ripple Voltage
)fw(RCf2
V
C
tIV
L
LL)pp(r
where IL = the dc load current
t = the time between charging peaks
C = the capacitance (in Farads)
8SSCP 2313- Basic electronics 2012/13-1
DC Output Voltage with Capacitor Filter
For Full-wave:
Cf4
IV
VVV
Lp
)p(rpDC
For Half-wave:
Cf2
IVV L
pDC
Vr and VDC determine the ripple factor,r
Vp(rect) : unfiltered peak rectified voltage
Vr(pp) : peak to peak ripple voltage
VDC : average value of ripple voltage
rhwr Vr( pp)
VDC
1
2 3 fRLC
rfwr Vr(pp)
VDC
1
4 3 fRLC
10SSCP 2313- Basic electronics 2012/13-1
The full-wave rectifier has approximately half the ripple output
produced by the half-wave rectifier. This is due to the shortened
time period between capacitor charging pulses. Therefore, full-wave
rectifier are typically used in power supplies.11SSCP 2313- Basic electronics 2012/13-1
Regulated Power Supply
Unregulated power supply gives an output voltage which
is not constant at the value that has been specified.
The variation of the output voltage can be caused by
several factors.
1. Variation of Load Current
The current flows in the load varies when the load
resistance is varied.
2. Variation of Input Voltage
The primary voltage for the transformer is normally
taken from mains supply.
-This voltage varies between 220 V to 260 V
depending on the usage of electric power in an
area.
12SSCP 2313- Basic electronics 2012/13-1
Although the variation of the output voltage is small, it can
cause problem to some electrical equipments:
- test instruments and computer circuits.
To overcome this problem a voltage regulator circuit is
added to the output of the unregulated power supply.
- to maintain the output voltage of the power supply at
the required value despite of the variation of the input
voltage or the variation of load current.
Transformer Rectifier Filter
input voltage
240 V 50 Hz
unregulated output
voltage
regulated output
voltage
Voltage
Regulator
13SSCP 2313- Basic electronics 2012/13-1
Parameters for a Regulated Power Supply:
Load Regulation
100% V
VVRegulation Load %
load) (full o
load) (full oload)(without o
Line Regulation ( Source Regulation)
ILILIH
OLOLOH
V/)VV(
V/)VV(
variationageinput volt %
variation voltageload %Regulation Line
14SSCP 2313- Basic electronics 2012/13-1
Determine the peak to peak ripple voltage and the ripple
factor for a bridge rectifier using capacitive filter of 100 F.
The load resistance is 2 k and the voltage across load is
12 V. Assume supply freq to be 50 Hz and ideal diodes.
Answer:
%44.1r,V6.0V )pp(r
15SSCP 2313- Basic electronics 2012/13-1
16SSCP 2313- Basic electronics 2012/13-1
LECTURE 21
Application of diodes:
ii) Voltage Regulator
1
SSCP 2313- Basic electronics2012/13-1
Many of the fixed voltage regulator ICs have 3 leads and look
like power transistors. Ex:
7805 (+5V) , 7905 (-5 V) 7812 (+12 V), 7912 (-12V).
(a) Voltage Regulator IC
2SSCP 2313- Basic electronics 2012/13-1
For low current power supplies, a
simple voltage regulator can be made
with a resistor and a zener diode
connected in reverse.
Zener diodes are rated by their
breakdown voltage Vz and maximum
power Pz (typically 400mW or 1.3W).
The resistor limits the current. The
current through the resistor is constant,
so when there is no output current all
the current flows through the zener
diode and its power rating Pz must be
large enough to withstand this.
(b) Zener diode regulator
3SSCP 2313- Basic electronics 2012/13-1
4
Circuit Symbol :
• When forward-biased, zener diode behave as standard
rectifying diodes: it has a forward voltage drop about 0.7 V.
• In reverse-bias mode, it does not conduct until the applied
voltage reaches the zener voltage- at which point the diode is
able to conduct substantial current.
• As the power dissipated by this reverse current does not
exceed the diode's thermal limits, the diode will not be
harmed.
Zener Diode
4SSCP 2313- Basic electronics 2012/13-1
Choosing a zener diode
1.The zener voltage Vz is the output voltage required.
2.The input voltage Vs must be a few volts greater than Vz
(this is to allow for small fluctuations in Vs due to ripple) .
3.The maximum current Imax is the output current required
plus 10% .
4.The zener power Pz is determined by the maximum
current: Pz > Vz × Imax or P = I2 Z
5SSCP 2313- Basic electronics 2012/13-1
Example:
6
A 1N754A Zener diode has a dc power dissipation rating of 500 mW and a nominal Zener voltage of 6.8 V. What is the value of IZM for the device?
(max) 500mW73.5mA
6.8V
D
ZM
Z
PI
V
Solution:
6SSCP 2313- Basic electronics 2012/13-1
Zener Diodes – Operating Range
A zener diode is much
like a normal diode, the
exception being is that it
is placed in the circuit in
reverse bias and
operates in reverse
breakdown.
Note:
Its forward characteristics
are just like a normal
diode.Operating range
7SSCP 2313- Basic electronics 2012/13-1
Zener Diodes – Breakdown Characteristics
Note very small reverse current (before “knee”).
Breakdown occurs @ knee.Breakdown Characteristics:• VZ remains near constant• VZ provides:
-Reference voltage-Voltage regulation
• IZ escalates rapidly• IZMAX is achieved quickly• Exceeding IZMAX is fatal
8SSCP 2313- Basic electronics 2012/13-1
Zener Diodes – Voltage Regulation
Regulation occurs between:
VZK - knee voltage
to
VZM - Imax
9SSCP 2313- Basic electronics 2012/13-1
Zener Diodes – Equivalent Circuit
• Ideal Zener exhibits a
constant voltage,
• Ideal Zener exhibits no
resistance characteristics.
10SSCP 2313- Basic electronics 2012/13-1
Cathode
Anode
Reverse
biasing
Zener Diodes – Equivalent Circuit
Zener exhibits a near
constant voltage, varied by
current draw through the
series resistance ZZ.
As Iz increases, Vz also
increases.
11SSCP 2313- Basic electronics 2012/13-1
Zener Diodes – Characteristic Curve
Vz results from Iz.
Range at which voltage is maintained while current is
changing
Maximum current the diode can handle
Minimum current required to operate in breakdown
region
12SSCP 2313- Basic electronics 2012/13-1
Zener Diode - Applications
Regulation with varying input voltage
The zener diode will “adjust” its impedance based on
varying input voltages. Zener current will increase or
decrease directly with voltage input changes. The zener
current, Iz, will vary to maintain a constant Vz.
Note: The zener has a finite range of current operation.
Vo remains constant
13SSCP 2313- Basic electronics 2012/13-1
Zener Diode - Applications
Regulation with varying load
The zener diode will “adjust” its impedance based on varying
input voltages and loads (RL) to be able to maintain its
designated Vz. The zener current will increase or decrease
inversely with varying loads. The zener has a finite range of
operation.
VZ
remains constant
14SSCP 2313- Basic electronics 2012/13-1
Example 2:
Vinmin= VRmin + VZ = 2.26V + 10V = 12.26V.
Vin(max) = VRmax + VZ = 24.2V + 10V = 34.2V
VReg is ≈12.3 V to 34.2 V.
1N4740
PDMAX = 1W, VZ = 10V, RL= 1 k
IZmin = 0.25mA to
IZmax = 100mA.
Calculate the range of input voltage,
VReg for which output will remain
constant.
IL= Vz/RL=10 mA
VRmin = IR= (Izmin+ IL) x 220
= 2.26 V
VRmax = I R= (Izmax+ IL) x 220
= 24.2V.
Solution:
15SSCP 2313- Basic electronics 2012/13-1
Thank you….
16SSCP 2313- Basic electronics 2012/13-1
Lecture 22
Diode applications:
iii) Clippers
SSCP 2313- Basic electronics2012/13-1
Clipper Circuit
• Clipper circuits have the ability to „clip‟ off a portion of
the input signal without distorting the remaining part
of the alternating waveform.
2SSCP 2313- Basic electronics 2012/13-1
There are several types of clippers:
Series clippers
Parallel clippers
Biased clippers
Zener clippers
Two stages clippers
(i) Series Clipper
The diode in a series clipper “clips” any voltage that
does not forward bias.
The diode is in series with load.
RL
Vi
Vo
Negative Series Clipper
RL
Vi
Vo
Positive Series Clipper
Forward biased: Vo = Vi – 0.7
Reverse biased: Vo = 0
3SSCP 2313- Basic electronics 2012/13-1
(ii) Parallel Clippers (Shunt)
The diode in a parallel clipper circuit “clips” any voltage that forward bias it.
The diode is in parallel with load. The series resistor RS acts as a
current limiter to avoid any damage to the diode or any component in
the circuit.
RS
Vi
RL
Vo
Positive Shunt Clipper
RS
Vi
RL
Vo
Negative Shunt Clipper
iSL
Lo V
RR
RV
Reverse biased:
Forward biased: Vo = 0 or 0.7 V
4SSCP 2313- Basic electronics 2012/13-1
(iii) Biased Series Clippers
Adding a DC
source in series
with the clipping
diode changes the
effective forward
bias of the diode.
5SSCP 2313- Basic electronics 2012/13-1
Biased Parallel Clipping Circuit
RL
Vi
RS
D1
VB
VB + 0.7 V
i) Positive biased clipperD1
VB
RL
RS
- (VB + 0.7 V)
ii) Negative biased clipper
6SSCP 2313- Basic electronics 2012/13-1
iv) Positive and Negative biased clipper
Two Batteries
7SSCP 2313- Basic electronics 2012/13-1
Batteries arereplaced byZener diodes.
8SSCP 2313- Basic electronics 2012/13-1
v) Zener Clipper
The End
9SSCP 2313- Basic electronics 2012/13-1
Lecture 23
Diode applications:
iv) Clamper
SSCP 2313- Basic electronics2012/13-1
Clamper
A clamping circuit is a circuit to shift a waveform either
above or below a different dc level without distorting
the waveform.
The network has a capacitor, a diode and a resistor.
The magnitude of R and C must be chosen such that thetime constant t = RC is large enough to ensure that thevoltage across the capacitor does not dischargesignificantly during the interval the diode is non-conducting.
2SSCP 2313- Basic electronics 2012/13-1
Two Types of Clamper:
Simple Clamper
Biased Clamper
Negative Clampers
Negative clamper shifts the input waveform so that the
positive peak voltage of the output waveform will about
the same as the DC reference voltage of the clamper.
3SSCP 2313- Basic electronics 2012/13-1
Operation of negative clamper+ ve region
+
R Vo
-
+
-
Vi
C • 0 - T/2: Diode is ON (shorted)• Assume RC time is small and capacitor charge to V volts very quickly• Vo=0 V (ideal diode)
- ve region • T/2 -T: Diode is OFF (opened)• Both for the stored voltage across capacitor and applied signal current through cathode to anode
•KVL: Vo = -2V
+
R Vo
-
+
-
V
C
Vo
-
+V
Vi
t0 T/2 T
V
-V
4SSCP 2313- Basic electronics 2012/13-1
Positive clamper shifts the input waveform so that the
negative peak voltage of the output waveform will about
the same as the DC reference voltage of the clamper.
Positive Clampers
5SSCP 2313- Basic electronics 2012/13-1
Operation of a positive clamper circuit:
RL
D1
ON
C1
+ 2.5 V
- 2.5 V
0 V
(a) square wave input waveform
(b) Capacitor charging path
D1
OFF
RL
I
C1
+ 5.0 V
0 V
(d) Output of clamper circuit
(c) Small discharge through RL
6SSCP 2313- Basic electronics 2012/13-1
KVL: Vo = 2V
• Biased clamper enable us to shift a waveform so
that it occurs above or below a DC reference voltage
other than 0 V.
Biased Clamping Circuits
(a) Positive biased clamper
7SSCP 2313- Basic electronics 2012/13-1
Vs
VB
Vo
Vm
-Vm
VB
2Vm+VB
C
The End
8SSCP 2313- Basic electronics 2012/13-1
LECTURE 24
Diode Applications:
iv) Voltage multiplier
SSCP 2313- Basic electronics2012/13-1
Voltage-Multiplier Circuits
• Voltage Doubler
• Voltage Tripler
• Voltage Quadrupler
Voltage multiplier circuits use a combination of diodes and
capacitors to step up the output voltage of rectifier circuits.
VOLTAGE MULTIPLIERS are used primarily to develop high
voltages where low current is required, such as picture tube
in TV receivers, oscilloscopes, etc.
2SSCP 2313- Basic electronics 2012/13-1
Two types:
Half wave voltage doubler
Full wave voltage doubler
Voltage Doubler
A voltage doubler is an electronic circuit which charges
capacitors from the input voltage and switches these
charges - voltage is twice at the output (in the ideal case)
The switching elements are diodes. The output is taken
across the second capacitor.
3SSCP 2313- Basic electronics 2012/13-1
Half wave voltage doubler
• +ve Half-Cycle
o D1 conducts
o D2 is switched off
o Capacitor C1
charges to Vm
• -ve Half-Cycle
o D1 is switched off
o D2 conducts
o Capacitor C2
charges to 2Vm
Vout = VC2 = 2Vm
4SSCP 2313- Basic electronics 2012/13-1
Full wave voltage doubler
+ve half cycle:
D1 conducts, charging C1
to the peak voltage Vm.
-ve half cycle:
D2 conducts, charging C2
to Vm,
Since both capacitors
C1 and C2 are in
series, the output
voltage is
approximately 2Vm.
5SSCP 2313- Basic electronics 2012/13-1
Voltage Tripler
First +ve half cycle:
D1 conducts, charging the
C1 to Vm.
First -ve half cycle:
D2 conducts charging the
C2 to 2Vm.
second +ve half cycle:
D3 and D1 will conduct,
charging C1 and the voltage
across C2 charges C3 to
the same value 2Vm
Thus the voltage across C1
and C3 is 3Vm.
D1D2
6SSCP 2313- Basic electronics 2012/13-1
Voltage Quadrupler
D1D2
Second -ve half cycle:
D2 and D4 will conduct, leading to C3 charging C4 to the
same peak value 2Vm.
Thus the voltage across C2 and C4 is 4Vm.
7SSCP 2313- Basic electronics 2012/13-1
The End…
8SSCP 2313- Basic electronics 2012/13-1
LECTURE 25
BIPOLAR JUNCTION TRANSISTOR (BJT)
SSCP 2313- Basic electronics2012/13-1
Introduction
• The famous and commonly use of semiconductor device isBJTs (Bipolar Junction Transistors).
• It was invented by Shockley in 1951.
• It can be used as amplifier ( radio and TV signals) and logicswitches.
• BJT consists of three terminal:
collector : C
base : B
emitter : E
• A BJT consists of a three-layer "sandwich" of doped (extrinsic) semiconductor materials:
P-N-P or N-P-N.
2SSCP 2313- Basic electronics 2012/13-1
The Two Types of BJT Transistors:
3SSCP 2313- Basic electronics 2012/13-1
BJTs – Practical Aspects
Heat sink
4SSCP 2313- Basic electronics 2012/13-1
• An NPN transistor can be considered as two diodes with a shared
anode.
• In typical operation, the base-emitter junction JE of the NPN
transistor is forward biased and the base-collector junction JC is
reverse biased.
Operation of NPN BJT
• Due to f-b, e in N-type emitter
flow towards the base and
constitues the IE. These e flow
through P-type base where
they tend to combine with holes
(less than 5%) to constitutes
base current, IB. Remaining e
crossover into the collector
region which constitutes
collector current, Ic.
5SSCP 2313- Basic electronics 2012/13-1
Transistor Currents
• Emitter current (IE) is the sum of the collector
current (IC) and the base current (IB) . • Kirchhoff’s current law;
BCE III
6SSCP 2313- Basic electronics 2012/13-1
Collector Current (IC)
• Collector current (IC) comprises two components;
majority carriers (electrons) from the emitter
minority carriers (holes) from reverse-biased BC junction
→ leakage current, ICBO
• Total collector current (IC);
• Since leakage current ICBO is usually so small, it can be
ignored.
EC IImajority
CBOC II minority
CBOEC III
7SSCP 2313- Basic electronics 2012/13-1
…Collector Current (IC)
• Then;
E
C
I
I
• The value of DC represents the quality of a transistor.
• Usually its value lies between 0.95 to 0.99, which means
95% to 99% of the emitter current reach the collector.
8SSCP 2313- Basic electronics 2012/13-1
Base Current (IB)
• IB is very small compared to IC;
• The ratio of IC to IB is the dc current gain of a transistor,
called beta (β)
B
C
I
I
• Typical value of : 50 – 400, and usually designated
as an equivalent hybrid (h) parameter, hfe on transistor
datasheets.
• Beta can vary with temperature and also varies from
transistor to transistor.
9SSCP 2313- Basic electronics 2012/13-1
Circuit Configuration
Common Base (CB)
Common Emitter (CE)
Common Collector (CC)
10SSCP 2313- Basic electronics 2012/13-1
Common Base (CB)
The base is common to both
the input signal and output
signal.
The collector current is less
than the emitter current –
current gain less than 1, it
attenuates the input signal.
is not very common due to its
high voltage characteristics.
It has high voltage gain but
small current gain.
11SSCP 2313- Basic electronics 2012/13-1
Common Emitter (CE)
is generally used for current
amplifiers - highest voltage
gain and power gain of all
three configurations.
is an inverting amplifier –
output signal being 180 out
of phase with input signal.
12SSCP 2313- Basic electronics 2012/13-1
Common Collector (CC) is known as a voltage followeror emitter follower circuit - itsoutput is taken from the emitterresistor.
is very useful for impedance
matching because of its high
input impedance and low output
impedance.
is a non-inverting amplifier-
input signal is in phase with the
output signal.
It has low voltage gain (less
than 1) but high current gain.13SSCP 2313- Basic electronics 2012/13-1
Thank you….
14SSCP 2313- Basic electronics 2012/13-1
LECTURE 26
Static Characteristics of BJT
SSCP 2313- Basic electronics2012/13-1
• When a transistor is operating as an amplifier, a signal given at the input side and a signal gain is obtained at the output side
• The static characteristics of a BJT is the characteristics of the device when it is given bias voltages at the junctions without any signal at the input side of a circuit containing the transistor.
• There are 3 important static characteristics:
1. Input Characteristics – this characteristic relates the input current to the input voltage
2. Output Characteristics – this characteristic relates the output current to the output voltage
3. Transfer Characteristics – this characteristic relates the output current to the input current.
Introduction
2SSCP 2313- Basic electronics 2012/13-1
• The input characteristic
curve is almost similar
to the forward bias
operation curve of a
PN junction diode.
Input Characteristics
Input characteristic: input current (IB) against input voltage (VBE) for several output voltage (VCE)
From the graph :IB = 0 A VBE < 0.7V (Si)IB = value VBE > 0.7V (Si)
3SSCP 2313- Basic electronics 2012/13-1
• Output characteristic:
Output current (IC) against output voltage (VCE) for
several input current (IB).
• In this circuit, VBB is used to set IB, while VCC is used to varyVCE.
• For each value of IB, the value of VCE is varied and IC isrecorded.
mA
IB
IC
VCC
VBB
VCE
u A
Circuit used to
obtain the output
characteristics of
a BJT.
Output Characteristics
4SSCP 2313- Basic electronics 2012/13-1
For other values of IB, the
current IC saturated to a value
which is independent of VCE and
only controlled by IB .
As IB become larger the slope of
the characteristic curve
increases.
This slope is due to the base
width modulation effect which is
known as Early effect.
In theory, CEOBC III
For IB = 0, CEOC II
That is the reverse saturation
current.
ICEO
IB = 0
IB1
IB2
IB3
IB4
IB5
VCE
IC
IB
increases
0
The output characteristics of a BJT
transistor
5SSCP 2313- Basic electronics 2012/13-1
Operating Regions of BJT Transistor
IB = 0
IB1
IB2
IB3
IB4
IB5
VCE
IC
0
cut-off region
saturation region
power limit region
active region
Four operating
regions:
1. Cutoff region
2. Active region
3. Saturation region
4. Power limit region
6SSCP 2313- Basic electronics 2012/13-1
• The active region of operation is the region where thecollector curves are almost horizontal.
• Active region – in which the transistor can act as alinear amplifier, where the BE junction is forward-biased and BC junction is reverse-biased.
• The collector current, IC varies with base current IB as
IC = IB .
Active Region
7SSCP 2313- Basic electronics 2012/13-1
• Saturation region – in which both junctions areforward-biased and IC increase linearly with VCE
• In this region the relationship IC = IB does not occur.
– When the IB is high enough, the entire Vcc isdropped across Rc, with no voltage between CE.This condition is called saturation.
– The saturation current,
Saturation Region
CR
CCV
CsatI
V 0CEV
8SSCP 2313- Basic electronics 2012/13-1
Cut-Off Region
In this region the transistor is said to be in a non-operatingmode.
Cut-off region – when IB = 0, no IC flows, IC is essentiallyzero with increasing VCE, therefore no voltage drop acrossRc. As a result, the VCE is nearly equal to Vcc.
In this condition the collector-emitter circuit is open circuit.
The saturation mode and cut-off mode are useful in digitalelectronic where the transistor is acting as a switch.
In saturation mode the transistor is operating as a closedswitch, while in cut-off mode it is operating as an openswitch.
9SSCP 2313- Basic electronics 2012/13-1
When the collector-base voltage is too large, the
collector-base junction will breakdown, giving rise to a
large and unwanted collector current to flow.
This large current generates large heat which may
damage the transistor.
This region of operation should be avoided in transistor
circuits.
Power Limit Region
10SSCP 2313- Basic electronics 2012/13-1
Note: VCE is at maximum and IC is at minimum (ICmax=ICEO) in the cutoff
region. IC is at maximum and VCE is at minimum (VCE max = VCEsat = VCEO) in
the saturation region. The transistor operates in the active region between
saturation and cutoff.11SSCP 2313- Basic electronics 2012/13-1
Because a transistor's collector current is
proportionally limited by its base current, it can be
used as a sort of current-controlled switch.
Transistor as Switch
A transistor when used as a switch is simply being biased so that it
is in cutoff (switched off) or saturation (switched on). Remember
that the VCE in cutoff is VCC and 0 V in saturation.
12SSCP 2313- Basic electronics 2012/13-1
• The BJT can be used as a switch by driving it back and
forth between saturation (closed) and cutoff (open).
The BJT as a switch
13SSCP 2313- Basic electronics 2012/13-1
Thank you…
Lecture 27
Transistor Bias Circuit
•The DC Operating Point
•Fixed Base Bias Configuration
SSCP 2313- Basic electronics2012/13-1
The DC Operating point
The analysis or design of any electronic amplifier has two
components: DC and AC.
A transistor must be properly biased with a dc voltage in
order to operate as a linear amplifier.
Bias establishes the dc operating point.
A dc operating point must be set so that signal variations at
a terminal input are amplified and accurately reproduced at
the output terminal.
If not, it can go into saturation or cutoff region.
The dc operating point is often referred to as Q-point
(quiescent point).
2SSCP 2313- Basic electronics 2012/13-1
DC Load Line
The straight line is known as
the DC load line.
The dc load line is a graph that
represents all the possible
combinations of IC and VCE
for a given amplifier. VCE
IC
C
CCC
R
VI (sat)
CCCE VV (off)
• This load line is
obtained from the
output equation when
IC = 0 and VCE = 0 V.
3SSCP 2313- Basic electronics 2012/13-1
Selection of Q-Point
• The intersection of the dc bias
value of IB with the dc load line
determines the Q-point.
• It is desirable to have the Q-point
centered on the load line. Why?
• When a circuit is designed to have
a centered Q-point, the amplifier
is said to be midpoint biased.
• Midpoint biasing allows optimum
ac operation of the amplifier.
• This includes the AC voltage gain
and the maximum AC output
voltage swing allowed by the
amplifier.
4SSCP 2313- Basic electronics 2012/13-1
When the Q-point is selected in the middle of a load line,
the voltage VCEQ will have a value half of VCC and ICQ will
be half of IC(sat) .
5SSCP 2313- Basic electronics 2012/13-1
VCE
V
IC
mA
IB = 0 A
IB = 10 A
IB = 20 A
IB = 30 A
IB = 40 A
Q
0
2
4
6
8
10
2 4 6 8 10
(sat)21
CI
(sat)CI
0
CCV210 (off)CEV
• If the Q-point is at the upper side of the load line center, theinput can cause the transistor to saturate.
• When this occurs, part of the output sinusoidal waveform will beclamped.
VCE V
IB = 0 A
IB = 10 A
IB = 20 A
IB = 30 A
IB = 40 A
Q
0
2
4
6
8
10
2 4 6 8 10
(sat) CI
0
0aCEV Q
Qa
(off) CEV
aCQI
IC mA
6SSCP 2313- Basic electronics 2012/13-1
If the Q-point is at the lower side of the load line center, theinput can cause the transistor to be in cut-off.
When this occurs, part of the output sinusoidal waveform willalso be clamped.
VCE V
IC mA
IB = 0 A
IB = 10 A
IB = 20 A
IB = 30 A
IB = 40 A
Q
0
2
4
6
8
10
2 4 6 8 10
n)(saturatio CI
0
0bCEV Q
Qb
(off) CEV
bCQI
7SSCP 2313- Basic electronics 2012/13-1
The selection of IBQ, ICQ and VCEQ will influence
the performance of the transistor when it is used
to amplify an AC signal.
The external circuit required to be connected to a
transistor to produce the values of IBQ, ICQ and
VCEQ is called the biasing circuit.
8SSCP 2313- Basic electronics 2012/13-1
• The purpose of biasing is to fix the operation of a transistorin a linear condition.
• Although the operating point has been fixed, there aresome transistor parameters which can vary and causingthe selected operating point to shift.
• There are two factors affecting an operating point of atransistor:
- transistor replacement (the change in value)
- temperature.
Stability of Biasing Circuit
9SSCP 2313- Basic electronics 2012/13-1
• The value of for a transistor differs from other
transistors, although they are from the same type.
• For example, transistors 2N2219 have values between
50 and 200.
• The collector current :
COBC III )1(
• So, if the transistor that has been biased in a circuit isdamaged and replaced by other transistor from thesame type, the value of is different from the originaltransistor.
• This causes the operating point of the circuit to change.
1. Transistor Replacement
10SSCP 2313- Basic electronics 2012/13-1
• Parameters vBE, and ICBO of a transistor amplifier circuit are very sensitive to temperature.
• vBE reduces by 2.5 mV/C when temperature increases
• increases linearly with the increase in temperature
• ICBO doubles for every 10C temperature increase for temperature above 25C.
• ICBO is the most sensitive to temperature change.
• The effect of IC against temperature variation is given by the stability factor S.
• Thus, to enable the stability factor of a bias circuit to be evaluated, all equations of IC should include the leakage current (ICBO) term.
2. Temperature
11SSCP 2313- Basic electronics 2012/13-1
• Solve the circuit using
KVL to find Q-point:
• 1st step: Replace
capacitors with an open
circuit
• 2nd step: Locate 2 main
loops which;
BE loop
CE loop
FIXED BASE BIAS CIRCUIT
12SSCP 2313- Basic electronics 2012/13-1
FIXED BASE BIAS CIRCUIT
• 1st step: Replace capacitors with an open circuit
13SSCP 2313- Basic electronics 2012/13-1
• 2nd step: Locate 2 main loops.
12
1
2
BE Loop CE Loop
FIXED BASE BIAS CIRCUIT
14SSCP 2313- Basic electronics 2012/13-1
FIXED BASE BIAS CIRCUIT
• BE Loop Analysis
1 From KVL;
IB
B
BECCBQ
BEBBCC
R
VVI
0VRIV
15SSCP 2313- Basic electronics 2012/13-1
FIXED BASE BIAS CIRCUIT
• CE Loop Analysis
From KVL;
As we known;
BQCQ II
2
ICCCQCCCEQ
CECCCC
RIVV
0VRIV
16SSCP 2313- Basic electronics 2012/13-1
Load Line Analysis
ICsat = VCC / RC
VCEcutoff = VCC
The dc load line are:
17SSCP 2313- Basic electronics 2012/13-1
FIXED BASE BIAS CIRCUIT
• Example 1 Find the Q-point for the
circuit below and draw
the dc load line:
Answers;
ICQ = 2.35 mA
IBQ = 47.08 μA
VCEQ = 6.83V
18SSCP 2313- Basic electronics 2012/13-1
Thank you…..
LECTURE 28
BJT BIASING CIRCUIT
ii) Fixed Bias with Emitter Resistor Circuit
SSCP 2313- Basic electronics2012/13-1
FIXED BASE BIAS CIRCUIT
• DISADVANTAGE
Unstable – because it is too dependent on β and
produce width change of Q-point
For improved bias stability , add emitter resistor to
dc bias.
2SSCP 2313- Basic electronics 2012/13-1
FIXED BIAS WITH EMITTER RESISTOR
• An emitter resistor, RE is
added to improve stability
• Solve the circuit using KVL
• 1st step: Replace capacitors
and with an open circuit
• 2nd step: Locate 2 main loops
which;
BE loop
CE loop
Resistor, RE added
3SSCP 2313- Basic electronics 2012/13-1
FIXED BIAS WITH EMITTER RESISTOR
• 1st step: Replace capacitors with an open circuit
4SSCP 2313- Basic electronics 2012/13-1
FIXED BIAS WITH EMITTER RESISTOR
• 2nd step: Locate 2 main loops.
2
BE Loop CE Loop
1
12
5SSCP 2313- Basic electronics 2012/13-1
• BE Loop Analysis From KVL;
Since
Substitute for IE
0 EEBEBBCC RIVRIV
EB
BECCBQ
EBBEBBCC
R)1(R
VVI
0RI)1(VRIV
BE II )1(
1
6SSCP 2313- Basic electronics 2012/13-1
FIXED BIAS WITH EMITTER RESISTOR
FIXED BIAS WITH EMITTER RESISTOR
• CE Loop Analysis
From KVL;
Assume;
Therefore:
Also:
ECQ II
0 EECECCCC RIVRIV
2)RR(IVV ECCCCCEQ
EBEBBCCB
CCCCECEC
EEE
V V RI– V V
RI - V V V V
RI V
7SSCP 2313- Basic electronics 2012/13-1
Saturation Level
VCEcutoff: ICsat:
The endpoints can be determined from the load line.
mA 0 I
V V
C
CCCE
ERCR
CCV
CI
CE V 0V
8SSCP 2313- Basic electronics 2012/13-1
FIXED BIAS WITH EMITTER RESISTOR
• Example 2 Find the Q-point of
the circuit below and
draw the dc load
line:
Answers;
ICQ = 2.01 mA
IBQ = 40.1 μA
VCEQ = 13.97V
9SSCP 2313- Basic electronics 2012/13-1
Thank you…..
Lecture 29
BJT Biasing Circuit:
iii) Voltage Divider Bias Circuit
SSCP 2313- Basic electronics2012/13-1
• Provides good Q-point stabilitywith a single polarity supply voltageand independent of .
• Most widely used.
• Solve the circuit using KVL:
• 1st step: Replace capacitors with anopen circuit
• 2nd step: Simplified circuit usingThevenin Theorem
• 3rd step: Locate 2 main loopswhich;
BE loop
CE loop
VOLTAGE DIVIDER BIAS CIRCUIT
2SSCP 2313- Basic electronics 2012/13-1
• 1st step: Replace capacitors with an open circuit
VOLTAGE DIVIDER BIAS CIRCUIT
3SSCP 2313- Basic electronics 2012/13-1
Simplified Circuit
Thevenin Theorem;
2nd step: : Simplified circuit using Thevenin Theorem
21
2121 //
RR
RRRRRTH
CCTH VRR
RV
21
2
From Thevenin Theorem;
VOLTAGE DIVIDER BIAS CIRCUIT
4SSCP 2313- Basic electronics 2012/13-1
• 2nd step: Locate 2 main loops.
1
2
BE Loop CE Loop
1
2
VOLTAGE DIVIDER BIAS CIRCUIT
5SSCP 2313- Basic electronics 2012/13-1
• BE Loop Analysis From KVL;
Recall;
Subtitute for IE
0 EEBETHBTH RIVRIV
ERTH
BETHBQ
EBBETHBTH
R)1(R
VVI
0RI)1(VRIV
BE II )1(
1
VOLTAGE DIVIDER BIAS CIRCUIT
6SSCP 2313- Basic electronics 2012/13-1
• CE Loop Analysis
From KVL;
Assume;
Therefore;
ECQ II
0 EECECCCC RIVRIV
)RR(IVV ECCQCCCEQ
2
VOLTAGE DIVIDER BIAS CIRCUIT
7SSCP 2313- Basic electronics 2012/13-1
VOLTAGE DIVIDER BIAS CIRCUIT
Example 3: Find the Q-point of the
circuit below and draw
the dc load line:
Answers;
RTH = 3.55 kΩ
VTH = 2V
IBQ = 6.05 μA
ICQ = 0.85 mA
VCEQ = 12.22V
8SSCP 2313- Basic electronics 2012/13-1
Thank you….
9SSCP 2313- Basic electronics 2012/13-1
Lecture 30
BJT Biasing Circuit:
vi) Collector feedback bias
v) Emitter bias
SSCP 2313- Basic electronics2012/13-1
Another way to
improve the stability
of a bias circuit is to
add a feedback path
from collector to
base.
In this bias circuit
the Q-point is only
slightly dependent
on the transistor
beta, .
iv) Collector to Base Bias Circuit (Collector Feedback)
2SSCP 2313- Basic electronics 2012/13-1
BE Loop
)R(RR
VVI
ECB
BECCBQ
From KVL:
0RI–V–RI–RI– V EEBEBBCCCC
Where IB << IC:
CI
BI
CI
CI'
Knowing IC = IB and IE IC, the
loop equation becomes:
0RIVRIRI– V EBBEBBCBCC
Solving for IB:
3SSCP 2313- Basic electronics 2012/13-1
CE Loop
IE RE+ VCE + I’CRC = VCC
Since IC IC and IC = IB:
IC(RC + RE) + VCE – VCC =0
Solving for VCE:
VCEQ = VCC – ICQ(RC + RE)
From KVL:
4SSCP 2313- Basic electronics 2012/13-1
Example 4:
5
Determine the values of ICQ and VCEQ for the amplifier shown.
RB
RC
1.5 k
+10 V
180 k
hFE = 100
1
10V 0.7V28.05μA
180kΩ 101 1.5kΩ
CC BEB
B FE C
V VI
R h R
( 1)
10V 101 28.05μA 1.5kΩ
5.75V
CEQ CC FE B CV V h I R
Answer:
100 28.05μA
2.805mA
CQ FE BI h I
5SSCP 2313- Basic electronics 2012/13-1
v) Emitter bias.
6
RC
RE
RB
IC
IE
IB
Q1
Input
Output
+VCC
-VEE
0.7V
1
EEB
B FE E
VI
R h R
C FE BI h I
CE CC C C E E EEV V I R I R V
CE CC C C E EEV V I R R V
• Emitter bias circuit uses both +ve and
-ve power supply.
• Provides excellent bias stability.
Loop BE: KVL
EEBEBBEE RIVRIV
Loop CE: KVL
ECQ II:Assume
6SSCP 2313- Basic electronics 2012/13-1
Example 5:
RC
750
RE
1.5k
RB
100
IC
IE
IB
Q1
Input
Output
+12 V
-12 V
hFE
= 200
Determine the values of ICQ and VCEQ for the amplifier shown.
12V 0.7V
( 1)
11.3V37.47μA
100Ω+201 1.5kΩ
B
B FE E
IR h R
200 37.47μA
7.49mA
CQ FE BI h I
( )
24V 7.49mA 750Ω 1.5kΩ
7.14V
CEQ CC C C E EEV V I R R V
CE CC C C E EEV V I R R V
7SSCP 2313- Basic electronics 2012/13-1
Load Line forEmitter-Bias Circuit
8
(sat)
( )CC EE CC EEC
C E C E
V V V VI
R R R R
( )CE off CC EE CC EEV V V V V
VCE
IC
IC(sat)
VCE(off)
8SSCP 2313- Basic electronics 2012/13-1
The end…
9SSCP 2313- Basic electronics 2012/13-1
Lecture 31
SSCP 2313- Basic electronics2012/13-1
BJT AC Analysis: Circuit 1 and 2
2SSCP 2313- Basic electronics 2012/13-1
3SSCP 2313- Basic electronics 2012/13-1
4SSCP 2313- Basic electronics 2012/13-1
5SSCP 2313- Basic electronics 2012/13-1
6SSCP 2313- Basic electronics 2012/13-1
7SSCP 2313- Basic electronics 2012/13-1
8SSCP 2313- Basic electronics 2012/13-1
9SSCP 2313- Basic electronics 2012/13-1
10SSCP 2313- Basic electronics 2012/13-1
Example 1
11SSCP 2313- Basic electronics 2012/13-1
Answer
12SSCP 2313- Basic electronics 2012/13-1
13SSCP 2313- Basic electronics 2012/13-1
14SSCP 2313- Basic electronics 2012/13-1
Example 2
circuit in Fig. E
Answer
19SSCP 2313- Basic electronics 2012/13-1
Lecture 32
BJT AC Analysis: Circuit 3 and 4
SSCP 2313- Basic electronics2012/13-1
1SSCP 2313- Basic electronics 2012/13-1
Example 2
2SSCP 2313- Basic electronics 2012/13-1
Answer
3SSCP 2313- Basic electronics 2012/13-1
4SSCP 2313- Basic electronics 2012/13-1
5SSCP 2313- Basic electronics 2012/13-1
6SSCP 2313- Basic electronics 2012/13-1
7SSCP 2313- Basic electronics 2012/13-1
Example 3
8SSCP 2313- Basic electronics 2012/13-1
Answer
9SSCP 2313- Basic electronics 2012/13-1
10SSCP 2313- Basic electronics 2012/13-1
11SSCP 2313- Basic electronics 2012/13-1
12SSCP 2313- Basic electronics 2012/13-1
Junction Field Effect Transistor
(JFET)
Lecture 33
SSCP 2313- Basic electronics2012/13-1
Introduction
1. Field effect transistors control current by voltage
applied to the gate.
2. The FET’s major advantage over the BJT is high input
resistance.
3. Overall, the purpose of the FET is the same as that of
the BJT.
2SSCP 2313- Basic electronics 2012/13-1
N-channel Junction field-effect transistor (JFET)
3SSCP 2313- Basic electronics 2012/13-1
BJT vs JFETBipolarJunctionTransistor
• Current-based device
• IBase controls ICollectorEmitter
JunctionFieldEffectTransistor
•voltage controlled devices
•VGate controls IDrainSource
ID
4SSCP 2313- Basic electronics 2012/13-1
The JFET – Primary Characteristics
• The JFET uses voltage to control the current flow.
• You will recall, the transistor uses current flow through the
base-emitter junction to control current.
• JFETs can be used as an amplifier just like the BJT.
• VGG voltage level controls current flow in the VDD, RD circuit.
5SSCP 2313- Basic electronics 2012/13-1
There are 2 types of JFET
n-channel JFET
p-channel JFET
Three Terminal
Drain – D
Gate -G
Source – S
6SSCP 2313- Basic electronics 2012/13-1
– Major structure is n-type material (channel) betweenembedded p-type material to form 2 p-n junction.
– In the normal operation of an n-channel device, the Drain(D) is positive with respect to the Source (S). Currentflows into the Drain (D), through the channel, and out ofthe Source (S)
– Because the resistance of the channel depends on thegate-to-source voltage (VGS), the drain current (ID) iscontrolled by that voltage.
N-channel JFET
7SSCP 2313- Basic electronics 2012/13-1
The JFET - Biasing
• The “source-drain” current is
controlled by a voltage field at
the “gate”.
• That field is developed by the
reverse biased gate-source
junction.
• With more VGG (reverse bias)
the field grows larger.
• This field or resistance limits
the amount of current flow
through RD.
• With low or no VGG current
flow is at maximum.8SSCP 2313- Basic electronics 2012/13-1
JFET Characteristics and Parameters Ohmic
Let’s first take a look at the effects with a VGS of 0V. ID increases proportionally with increases of VDD (VDS increases as VDD is increased). This is called the ohmic region (point A to B).
• The N-channel JFET can be explained by the output
characteristic curve.
• The output characteristics curve shows the relationship
between iD and vDS for different values of vGS .
• In this characteristics curve it can be identified that there are
three regions of operation:
• cut-off region,
• ohmic region and
• saturation region
9SSCP 2313- Basic electronics 2012/13-1
JFET Characteristic Curve
• To start, suppose VGS=0
• Then, when VDS is increased, ID increases. Therefore,ID is proportional to VDS for small values of VDS
• For larger value of VDS, as VDS increases, the depletionlayer become wider, causing the resistance of channelincreases.
• After the pinch-off voltage (Vp) is reached, the IDbecomes nearly constant (called as ID maximum, IDSS-Drain to Source current with Gate Shorted)
10SSCP 2313- Basic electronics 2012/13-1
Pinch-off (VGS = 0 V, VDS = VP).
11SSCP 2313- Basic electronics 2012/13-1
JFET Characteristics and Parameters Pinch-Off
The point when ID stop to increase regardless of VDD increases (constant
current source) is called the pinch-off voltage (point B) (Note: VGS = 0).
This current is called maximum drain current (IDSS).Breakdown (point C) is reached when too much voltage is applied. This is
undesirable, so JFETs operation is always well below this value.
• In saturation condition,
one end (at the drain
end) the JFET will
become pinch-off due
to the depletion region.
12SSCP 2313- Basic electronics 2012/13-1
JFET Characteristics and Parameters Drain CurvesFrom this set of curves you can see increased negative voltage applied to
the gate (-VGS) produces no change in ID. ID is limited and the pinch-off
voltage (VP) is reduced. Note: VGS controls IDSS
p(off) GSGS VVV
If VGS is less than pinch-off voltage, the resistance becomes an
open-circuit ;therefore the device is in cutoff (VGS=VGS(off))
13SSCP 2313- Basic electronics 2012/13-1
Pinch-Off “Vp” vs Cutoff “VGS(off)”
• VGS(off) and VP are always = and opposite is sign
• VP(pinch off) - the value of VDS where ID becomes constant
with VGS = 0.
• VP(pinch off) also occurs for VDS<VP if VGS ≠ 0.
Note: Although VP is constant, VDSmin where ID is constant, varies.
VGS(off)= -5VVp=+5V
14SSCP 2313- Basic electronics 2012/13-1
JFET Characteristics and Parameters
The transfer characteristic curve illustrates the control VGS has on
ID from cutoff (VGS(off) ) to pinch-off (VP). Note the parabolic shape. The
formula below can be used to determine drain current.
Square-law device: Parabolic curve of the JFET Transfer Characteristic
Curve.
VGS(OFF)
Vp
2
(off)
1
GS
GSDSSD
V
VIi
Note:(VGS = 0 to VGS(off)
controls ID)
JFET Characteristics and Parameters
gm2
gm1
The transfer curve is also known as the transconductance
curve.The value of drain current when vGS = 0 is marked as
IDSS .
JFET Transfer Characteristic Curve JFET Characteristic Curve
16SSCP 2313- Basic electronics 2012/13-1
• Defined by Shockley’s equation:
• Relationship between ID and VGS.
• Obtaining transfer characteristic curve axis point from
Shockley:
– When VGS = 0 V, ID = IDSS
– When VGS = VGS(off) or Vp, ID = 0 mA
)(
2
)(
1offGSP
offGS
GSDSSD VV
V
VII
Transfer Characteristics…
17SSCP 2313- Basic electronics 2012/13-1
Exercise 1
Sketch the transfer defined by IDSS = 12 mA dan VGS(off) = Vp= - 6V
Answer
2GS
D DSS
P
V I = I 1 -
V
DGS P
DSS
I V = V 1 -
I
VGS ID
0 IDSS
0.3Vp IDSS/2
0.5Vp IDSS/4
Vp 0 mA
VGS =0.3VP
VGS =0.5VP
IDSS/2
IDSS/4
18SSCP 2313- Basic electronics 2012/13-1
Exercise 2
DGS P
DSS
I V = V 1 -
I
Sketch the transfer defined by IDSS = 4 mA and
VGS(off) = 3V
2GS
D DSS
P
V I = I 1 -
V
VGS =0.5VP
VGS =0.3VP
VP
IDSS/2
IDSS/4
Answer
19SSCP 2313- Basic electronics 2012/13-1
Thank you…
Lecture 34
JFET Biasing Circuit:• Fixed – Bias
• Self-Bias
SSCP 2313- Basic electronics2012/13-1
• The general relationship of all FET amplifiers for dc Biasing analysis:
• For the JFET, the relationship between input and output quantities is nonlinear due to the squared term in Shockley’s equation:
• Nonlinear functions results in curves as obtained for transfer characteristic of a JFET.
• JFETs differ from BJTs:
• Nonlinear relationship between input (VGS) and output (ID)
• JFETs are voltage controlled devices, whereas BJTs are current controlled
Introduction to FET Biasing
2
(off)
1
GS
GSDSSD
V
VIi
SDG IIandA0I
2SSCP 2313- Basic electronics 2012/13-1
Common FET Biasing Circuits:
i) Fixed – Bias or Gate Biasedii) Self-Biasiii) Voltage-Divider Bias
Remember: the purpose of biasing is to set a point of
operation (Q-point).
3SSCP 2313- Basic electronics 2012/13-1
• The simplest biasing arrangement.
• The configuration includes the ac levels Vi and Vo
and the coupling capacitors.
Fixed-Bias Configuration
4SSCP 2313- Basic electronics 2012/13-1
• For the DC analysis,
• Capacitors are open circuits
• The zero-volt drop across RG permits
replacing RG by a short-circuit
Applying KVL:VGS= -VGG
5SSCP 2313- Basic electronics 2012/13-1
2 k
1 M
2 V
IDSS = 10 mA
VP = -8 V
Determine the following of the network and plot the transfer
characteristic and load line :
i) VGSQ
ii) IDQ
iii)VDS
iv)VG
Answer:
V2VV)i GGGSQ
mA625.5
V
V1Ii)ii
2
(off) GS
GSDSSDQ
V75.4
RIVV)iii DDDDDS
V2VV)iv GSG
Example 1
6SSCP 2313- Basic electronics 2012/13-1
• The self-bias configuration eliminates the need for
two dc supplies.
• The controlling VGS is now determined by the
voltage across the resistor RS
Self- Bias Configuration
7SSCP 2313- Basic electronics 2012/13-1
Self-bias is the most common type of biasing method for
JFETs. No voltage is applied to the gate. The voltage to
ground from here will always be 0V.
ID = IS for all JFET circuits.
VGS
Self bias is more stable than Fixed bias.
SD
SGGS
RI
VVV
The drain voltage, VD = VDD - IDRD
Since VS= IDRS, the drain to source voltage,
VDS is
)RR(IV
VVV
SDDDD
SDDS
8SSCP 2313- Basic electronics 2012/13-1
Example:
3.3 k
1 M1 k
Determine the following of the network and plot the transfer characteristic
and load line :
i) VGSQ
ii) IDQ
iii) VDS
iv) VS
v) VD
Answer:
V8V
,mA8IChoosing
V4V
,mA4IChoosing
RIV)i
GS
D
GS
D
SDGS
ii) At the Q- point:IDQ = 2.6 mA
mA82.8
)RR(IVV)iii SDDDDDS
iv) VS = IDRS= 2.6 V
v) VD = 11.42 V
IDSS = 8 mAVP = -6 V
Thus; At the Q- point: VGSQ = -2.6 V
9SSCP 2313- Basic electronics 2012/13-1
10SSCP 2313- Basic electronics 2012/13-1
Lecture 35
JFET Biasing Circuit:
Voltage Divider Bias
SSCP 2313- Basic electronics2012/13-1
• The arrangement is the same as BJT but the DC
analysis is different
• In BJT, IB provide link to input and output circuit, in
FET VGS does the same
Voltage-Divider Bias
2SSCP 2313- Basic electronics 2012/13-1
IG = 0A ,Kirchoff’s current law requires that IR1= IR2 and
the series equivalent circuit appearing to the left of
the figure can be used to find the level of VG.
SDGGS RIVV
DDDDD RIVV
)RR(IVV SDDDDDS
DC equivalent circuit3SSCP 2313- Basic electronics 2012/13-1
Example:
Determine the following of the network and plot the transfer
characteristic and load line :
i) VGSQ and IDQ
ii) VS
iii)VD
iv)VDS
16 V
2.4 k
2.1 M
270 k
1.5 k
IDSS = 8 mA
VP = -6 V
4SSCP 2313- Basic electronics 2012/13-1
Answer:
V82.1
RR
VRV)i
21
DD2G
and
)k5.1(IV82.1
RIVV
D
SDGGS
When ID=0 mA, VGS= +1.82 V
VGS = 0V, ID=1.21 mA
Thus VGSQ= -1.8 V and IDQ = 2.4 mA
ii) VS= 3.6 V
iii) VD= 10.24 V
iv) VDS= VD - VS = 6.64 V
Plot the
transfer
curve
5SSCP 2313- Basic electronics 2012/13-1
6SSCP 2313- Basic electronics 2012/13-1
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