4. lecture
• Diodes - applications
• Rectifiers
Jiří Jakovenko – Electronics and Microelectronics - Department of Microelectronics – CTU
Electronics and Microelectronics AE4B34EM
Junction breakdown or reverse breakdown
An applied reverse bias (voltage) will result in a small current to flow through the device.
At a particular high voltage value, which is called as breakdown voltage VB, large currents start to flow.
If there is no current limiting resistor which is connected in series to the diode, the diode will be destroyed.
There are two physical effects which cause this breakdown:
1) Zener breakdown is observed in highly doped p-n junctions and occurs for voltages of about 5 V or less.
2) Avalanche breakdown is observed
in less doped p-n junctions.
Jiří Jakovenko – Electronics and Microelectronics - Department of Microelectronics – CTU
Junction breakdown or reverse breakdown
Avalanche breakdown mechanism occurs when electrons and holes moving through the DR and acquire sufficient energy from the electric field to break a bond i.e. create electron-hole pairs by colliding with atomic electrons within the depletion region.
The newly created electrons and holes move in opposite directions due to the electric field and thereby add to the existing reverse bias current. This is the most important breakdown mechanism in p-n junction.
Zener breakdown occurs at highly doped p-n junctions with a tunneling mechanism. In a highly doped p-n junction the conduction and valance bands on
opposite side of the junction become so close during the reverse-bias that the electrons on the p-side can tunnel from directly VB into the CB on the n-side.
Jiří Jakovenko – Electronics and Microelectronics - Department of Microelectronics – CTU
Diode Circuit Models
The Ideal Diode Model The diode is designed to allow current to flow in only one direction. The perfect 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.
Example: Assume the diode in the circuit below is ideal. Determine the value of ID if a) VA = 5 volts (forward bias) and b) VA = -5 volts (reverse bias)
+
_ VA
ID
RS = 50
a) With VA > 0 the diode is in forward bias and is acting like a perfect conductor so:
ID = VA/RS = 5 V / 50 = 100 mA
b) With VA < 0 the diode is in reverse bias and is acting like a perfect insulator, therefore no current can flow and ID = 0.
Jiří Jakovenko – Electronics and Microelectronics - Department of Microelectronics – CTU
The Ideal Diode with Barrier Potential This model is more accurate than the simple ideal diode
model because it includes the approximate barrier potential voltage. Remember the barrier potential voltage is the voltage at which appreciable current starts to flow.
Example: To be more accurate than just using the ideal diode model include the barrier potential. Assume V = 0.3 volts (typical for a germanium diode) Determine the value of ID if VA = 5 volts (forward bias).
+
_ VA
ID
RS = 50
With VA > 0 the diode is in forward bias and
is acting like a perfect conductor so write a
KVL equation to find ID:
0 = VA – IDRS - V
ID = VA - V = 4.7 V = 94 mA
RS 50
V
+
V
+
Diode Circuit Models
Jiří Jakovenko – Electronics and Microelectronics - Department of Microelectronics – CTU
The Ideal Diode with Barrier Potential and
Linear Forward Resistance
This model is the most accurate of the three. It includes a linear forward resistance that is calculated from the slope of the linear portion of the transconductance curve. However, this is usually not necessary since the RF (forward resistance) value is pretty constant. For low-power germanium and silicon diodes the RF value is usually in the 2 to 5 ohms range, while higher power diodes have a RF value closer to 1 ohm.
Linear Portion of transconductance curve
VD
ID
VD
ID
RF = VD
ID
+ V
RF
Diode Circuit Models
Jiří Jakovenko – Electronics and Microelectronics - Department of Microelectronics – CTU
Rs D
D+Rs
The Ideal Diode with Barrier Potential and Linear Forward Resistance
Example: Assume the diode is a low-power diode with a forward resistance value of 5 ohms. The barrier potential voltage is still: V = 0.3 volts (typical for a germanium diode) Determine the value of ID if VA = 5 volts.
+
_ VA
ID
RS = 50
V
+
RF
Once again, write a KVL equation for the circuit: 0 = VA – IDRS - V - IDRF
ID = VA - V = 5 – 0.3 = 85.5 mA
RS + RF 50 + 5
Diode Circuit Models
Jiří Jakovenko – Electronics and Microelectronics - Department of Microelectronics – CTU
Values of ID for the Three Different Diode Circuit Models
Ideal Diode
Model
Ideal Diode
Model with
Barrier
Potential
Voltage
Ideal Diode
Model with
Barrier
Potential and
Linear Forward
Resistance
ID 100 mA 94 mA 85.5 mA
These are the values found in the examples on previous slides where the applied voltage was 5 volts, the barrier potential was 0.3 volts and the linear forward resistance value was assumed to be 5 ohms.
Diode Circuit Models
Jiří Jakovenko – Electronics and Microelectronics - Department of Microelectronics – CTU
ID (mA)
VD (Volts)
2
4
6
8
10
12
0.2 0.4 0.6 0.8 1.0 1.2 1.4
The transconductance curve below is for a Silicon diode. The Q point in this example is located at 0.7 V and 5.3 mA.
4.6
0.7
5.3
Q Point: The intersection of the load
line and the
transconductance curve.
The Q Point
Jiří Jakovenko – Electronics and Microelectronics - Department of Microelectronics – CTU
+
_ VA
= 6V
ID
RS = 1000
V
+ Rectifiers
Jiří Jakovenko – Electronics and Microelectronics - Department of Microelectronics – CTU
TYPES OF RECTIFIERS
Rectifier
Half-wave Rectifier Full-wave Rectifier
Centre-tape
full-wave rec.
Full-wave
Bridge rec.
Jiří Jakovenko – Electronics and Microelectronics - Department of Microelectronics – CTU Jiří Jakovenko – Electronics and Microelectronics - Department of Microelectronics – CTU
Half-wave rectifier
DC
230 V RMS 50 Hz
diode
RL Rectifier vL
1:a
Transformer Load
+
-
t
v L
t 0
DC
Average
VMax VRMS 2Vm for sineorcosinewave
VM VRMS 2 a
VL is the DC part of vL
aVL
.2230
Average value of half-wave rectified
π
V
θcos2π
V
dθ0dθθsinV2π
1
dtVT
1V
P
π
0
P
2π
0p
T
0DC
Jiří Jakovenko – Electronics and Microelectronics - Department of Microelectronics – CTU
RMS Value of half-wave rectified
4
V
θsinθπ4
V
dθ2θcos12
1
π2
V dθθ
π2
V
dθ0dθθsinVπ2
1
dtVT
1V
P
π
0
P
P π
0
P
2π
0
2
p
T
0
2
rms
2
2
0
2
2
2
2
2
22
1
sin
Jiří Jakovenko – Electronics and Microelectronics - Department of Microelectronics – CTU
VDC
Vrms
t0
t1
t2
t3
Vp
VO
t
Half-wave rectified
Jiří Jakovenko – Electronics and Microelectronics - Department of Microelectronics – CTU
VDC – Average value Vrms – RMS value
A center-tapped full-wave rectifier
Jiří Jakovenko – Electronics and Microelectronics - Department of Microelectronics – CTU
Jiří Jakovenko – Electronics and Microelectronics - Department of Microelectronics – CTU
diode 1
RL
vL
1:a
+
diode 2
100 RMS 50 Hz
-
D1 D1 D1 D2 D2
vL
t(ms)
1/50 1/100
VL
DC
0
aVL
21002
L
LR
aI
21002
Full-wave rectifier Average Value of center-tapped full-wave rectifier
π
V2
θcosπ
V
dθθsinVπ
1
dtVT
1V
P
π
0
P
π
0p
T
0DC
Jiří Jakovenko – Electronics and Microelectronics - Department of Microelectronics – CTU
RMS Value of center-tapped full-wave rectifier
2
V
θsinθπ2
V
dθ2θcos12
1
π
V
dθθsinVπ
1
dtVT
1V
P
π
0
P
P
π
0
2
p
T
0
2
rms
2
2
0
2
2
2
22
1
Jiří Jakovenko – Electronics and Microelectronics - Department of Microelectronics – CTU
Full-wave Bridge rectifier
Jiří Jakovenko – Electronics and Microelectronics - Department of Microelectronics – CTU
Full-wave Bridge rectifier
Jiří Jakovenko – Electronics and Microelectronics - Department of Microelectronics – CTU Jiří Jakovenko – Electronics and Microelectronics - Department of Microelectronics – CTU
diode-1
diode-2
diode-4
RL
vL
100 RMS
50 Hz
+ -
diode-3
t(s) 1/50 1/100
VL
DC or
Average
D1 D3
D4 D2
0
Full-wave Bridge rectifier
aVL
21002
L
LR
aI
21002
Jiří Jakovenko – Electronics and Microelectronics - Department of Microelectronics – CTU
Low Pass Filter To smooth the bumps (ripple)
Jiří Jakovenko – Electronics and Microelectronics - Department of Microelectronics – CTU
RL
diode
Rectifier vL
1:a
Transformer Load 230 V RMS
50 Hz +
-
C
Filter
iC iRL
Diode; Reverse Biased RL
vL
-
C
+
-
Low Pass Filter To smooth the bumps (ripple)
Capacitor
Charge
Capacitor
discharge
t(s) 0
v L
VL
DC or
Average
∆V
same filter capacitor and load and derived from the same sinusoidal input voltage
Comparison of ripple voltages for half-wave and full-wave rectified
Jiří Jakovenko – Electronics and Microelectronics - Department of Microelectronics – CTU Jiří Jakovenko – Electronics and Microelectronics - Department of Microelectronics – CTU
t(s) 0
v L
VL C1
C2 >C1
∆V1
t1 t2 t3
T
iD
VL C2
C1
Vm
ID DC
∆V2
T
t3 t1 t2
VL Vm(1T
2RLC)
v Vm 1 eT /(RLC )
v VmT
RLC
If T <<< RLC
T = T/2 for Full
wave
Jiří Jakovenko – Electronics and Microelectronics - Department of Microelectronics – CTU
Full wave:
vVmT
RLC
If T <<< RLC
Full Wave: vL
t(ms) 1/50 1/100
DC
Effects of RL
and C
C=1000µF C=470µF
C=100µF
R=1500Ω
R=1000Ω
R=500Ω
(a) RL fixed (b) C fixed
Jiří Jakovenko – Electronics and Microelectronics - Department of Microelectronics – CTU
COMPARISON OF RECTIFIERS
Half-wave Centre-tap Bridge type
No. of diode 1 2 4
Transformer
necessary
No Yes No
Maximum
efficiency
40.6% 81.2% 81.2%
Jiří Jakovenko – Electronics and Microelectronics - Department of Microelectronics – CTU Jiří Jakovenko – Electronics and Microelectronics - Department of Microelectronics – CTU
Rectifier
Half-wave
or
Full-wave
Low pass
Filter
Voltage
regulator
RL
-
+
IL
Power Supply
VBR