1
EIE209 Basic Electronics
Transistor Devices
Prof. C.K. Tse: Transistor devices
Contents• BJT and FET• Characteristics• Operations
2Prof. C.K. Tse: Transistor devices
What is a transistor?Three-terminal device
whose voltage-currentrelationship is controlled bya third voltage or current
We may regard a transistor as acontrolled voltage or currentsource.
i
+v–+
vc–
ic
3Prof. C.K. Tse: Transistor devices
Types of transistorsAccording to the physics of the device, we canclassify transistors into two main classes:
1. Bipolar junction transistors (BJT)2. Field effect transistors (FET)
Diode-based devicewhich is usually blockedunless the controlterminals are forward-biased. So, the control isa current, and BJT is acurrent amplifier bynature.
Conduction is controlledby electric field which isproduced by voltageapplied to the controlterminals. So, the controldraws no current andFET is a voltage-controlled device.
4Prof. C.K. Tse: Transistor devices
Bipolar junction transistor (BJT)a bit of physics…
collector
emitter
collector
emitter
base
C
B
E
C
B
E
npn transistor pnp transistor
B
C
E
B
C
E
IC
IBI
I
npn pnp
C
B
Basic model
2 types of BJT devicesConsider the npn BJT. The collector-basejunction is reverse-biased. So, no current canflow down.
But if the base-emitter junction is forward-biased (≈0.6V), then the diode “contactpotential barrier” can be overcome. Electronscan go to base — called base injection.
These electrons are minority carriers, whichare strongly attracted/captured by thecollector. Hence, current flows down fromcollector to emitter.
THUS, we use a small base current to induce alarge collector current.
This large collector current is proportional tothe base injection. IC = b IB
5Prof. C.K. Tse: Transistor devices
Simple BJT model
Base
Collector
Emitter
Consider npn transistor.
Collector is more positive than Emitter.
B-E and B-C junctions are pn junctions, like diodes.In normal operation, B-E is forward biased and B-C isreverse biased.
Collector
Emitter
Basenpn
Main relation:
†
IC = bIB
IC
IB
b ≈ 100 typically
IC
IB
6Prof. C.K. Tse: Transistor devices
Some properties
Base
Collector
Emitter
IC
IB
VBE ≈ 0.6 V when the transistor turns on.
Never try to stick a large voltage across VBEbecause it may produce enormous currentor may just kill the device!
b is a “bad” parameter. Don’t trust thedatabook. Its value can vary to ±50% or more.
IC = bIB holds only under some carefully setconditions. We’ll look at it later.
ICIB
IE = IB + IC
7Prof. C.K. Tse: Transistor devices
Typical operations
1. Cut-off
2. Active operation
3. Saturation+10 V
IC
VBE
+
–
Determining factors:• How large is IB or VBE• How large is RL
RL
IB
8Prof. C.K. Tse: Transistor devices
Cut-off
When the B-E junction is not forward-biased, thetransistor is basically not doing anything.
This is called CUT-OFF. +10 V
0A
VBE = 0
9Prof. C.K. Tse: Transistor devices
Active operation
+10 V
IC
VBE
+
–
RL
IB
†
IC = bIB
When the following holds:
the BJT is said to be in activeoperation.
This is the case of currentamplification.
But we need ICRL < 10V
10Prof. C.K. Tse: Transistor devices
Condition for active operation: ICRL<VCC
Let b = 100.
+10 V
IC=1mA1kΩ
IB=10µA
VCE = 9V
+
–
+10 V
IC=1mA5kΩ
IB=10µA
VCE = 5V
+
–
+10 V
IC=1mA10kΩ
IB=10µA
VCE = 0V
+
–
How about 11kΩ?
11Prof. C.K. Tse: Transistor devices
Saturation
When VCE is reduced to 0, the BJT is saturated.
+10 V
IC=0.6667mA15kΩ
IB=10µA
VCE = 0V
+
–
IC cannot be 1mA!!In fact, it must drop in orderto make up for the totalvoltage.
In this case,
IC = 10V/15kΩ = 0.6667mA
†
IC = bIB
12Prof. C.K. Tse: Transistor devices
What makes it saturate?
Large RL Large IB
+10 V
IC=10mA1kΩ
IB=100µA
VCE = 0V
+
–just saturate!
+10 V
IC=1mA10kΩ
IB=10µA
VCE = 0V
+
–just saturate!
13Prof. C.K. Tse: Transistor devices
Application: BJT as switch
C
B
E
10V
10V 0.1Alamp
1kW
Situation 1
Situation 2
Saturation
Cut-off
†
IB =(10 - 0.7)V
1kW= 9.3mA
IC = 100x9.3 = 0.93A which istoo large and surely saturatesthe BJT!!! So, IC ≈ 0.1A.
Light bulb turns on.
100Ω
IB = IC = 0. Light bulb turns off.
14Prof. C.K. Tse: Transistor devices
Detailed BJT characteristics
Obviously, VBE and IB are related by diode characteristic.
†
IB = Iss eV BE /VT -1Ê Ë Á
ˆ ¯ ˜
†
VT =kTq
Boltzman’s constant
absolute temperature
electronic charge
thermal voltage
≈25 mV @room temp
VBE
IB
0.6
Input characteristic (IB versus VBE)
15Prof. C.K. Tse: Transistor devices
Detailed BJT characteristics
Also, IC is just IB multiplied by b.
†
IC = bI ss eVBE /VT -1Ê Ë Á
ˆ ¯ ˜ = Is eV BE /VT -1
Ê Ë Á
ˆ ¯ ˜
†
VT =kTq
Boltzman’s constant
absolute temperature
electronic charge
thermal voltage
≈25 mV @room temp
VBE
IC = b IB
0.6
Transfer characteristic (IC versus VBE)
same shape as IB
16Prof. C.K. Tse: Transistor devices
Detailed BJT characteristics
IC is nearly flat unless near saturation.
Output characteristic (IC versus VCE)
IC
VCE
IC
VCE
+
–
Not Ohm’s law!!
for one particularchoice of IB or VBE
17Prof. C.K. Tse: Transistor devices
Important small-signal characteristic
0.6 V
IB
VBE
C IC
VCE
V =0.68 VBE
V =0.65 V
V =0.60 V
V =0 (cut-off)
0.2 V
active
saturation
0.6
VBE
I
C1I
slope = g =IC1
0.025 W–1
m
0.65 0.68 V
BE
BE
BE
Different IB (or VBE) hasdifferent output characteristic.
A range of VBE corresponds toa range of IC.
Transconductance:
†
gm =DIC
DVBE
= slopeon the transfer char.
18Prof. C.K. Tse: Transistor devices
What is gm?
†
gm =DIC
DVBE A simple differentiation gives
†
gm =dIC
dVBE
=d
dVBE
bI ss(eVBE /VT -1)Ê
Ë Á
ˆ ¯ ˜
= bIss (eVBE /VT )1
VT
ªICVT
or IC
25mV at room temp
19Prof. C.K. Tse: Transistor devices
A bit more precise
At saturation, VCE is not really 0, it is about 0.2 V.
IC
VCE
for one particularchoice of IB or VBE
0.2V
+10 V
IC=0.98mA not 1mA!!
10kΩ
IB=10µA
VCE = 0.2V
+
–
20Prof. C.K. Tse: Transistor devices
A bit more preciseIn active region, IC is not really flat. It goes up gently! This iscalled Early Effect!
IC
VCE
for one particularchoice of IB or VBE
0.2V
slope ≈ IC / VA
Early voltagetypically 100V
IC
21Prof. C.K. Tse: Transistor devices
Field Effect Transistors (FET)
Two kinds of channels:
n-channel FETp-channel FET
Two kinds of gate electrodes:
Junction FET (JFET)Metal-oxide-semiconductor FET (MOSFET)
Gate
Drain
Source
Current goes down from D to S,controlled by the gate voltage at G.
channel
22Prof. C.K. Tse: Transistor devices
Terminology confusionBefore we move on, it is important to clarify some possible confusionsdue to terminology difference.
BJT
FET
saturation region
triode region
active region
saturation region
cut-off
cut-off
23Prof. C.K. Tse: Transistor devices
n-channel MOSFET
G
SiO2 insulator
pnn
body or substrate
When gate is +ve,electrons are attracted toit and this becomes n-typeconducting channel.
This action is calledchannel enhancement.
S D
The channel is notconducting initially whengate is zero volt.
24Prof. C.K. Tse: Transistor devices
n-channel MOSFET characteristic
Drain
Source
Gate
Characteristics:
Gate current = 0 (always)The channel conduction is determined by VGS
ID
VDS
VGS=2V
VGS=1.9V
VGS=1.8V
VGS=1.7VThreshold voltageVth = 1.7 V, for example.
saturation(like “active” in BJT)
triode
VGS
+
–
VDS
+
–
ID
25Prof. C.K. Tse: Transistor devices
Saturation region
Drain
Source
Gate
So, it looks like the npn BJT!! But if we lookcloser, we find that the saturation current isproportional to (VGS–Vth)2.
ID
VDS
VGS=2V
VGS=1.9V
VGS=1.8V
VGS=1.7VThreshold voltageVth = 1.7 V, for example.
saturation(like “active” in BJT)
ID = K (VGS–Vth)2 for saturation region
VGS
+
–
VDS
+
–
ID
26Prof. C.K. Tse: Transistor devices
Saturation region
Drain
Source
GateID
ID = K (VGS–Vth)2 for saturation region
VGS
+
–
VDS
+
–
ID
VGS
If we plot the saturation ID versus VGS, we have aquadratic (parabolic) curve.
Vth
27Prof. C.K. Tse: Transistor devices
Triode region
ID = a VDS (2M – VDS)
VDS2M
ID
Triode region — like a quadratic (parabolic) function
K (VGS–Vth)2
So, the equation is:
y = a x (2M – x)
Obviously, M = VGS – Vth , a = K,
ID = K VDS [2 (VGS – Vth) – VDS]M
= VGS–Vth for triode region
28Prof. C.K. Tse: Transistor devices
n-channel MOSFET characteristic
VDS
ID
K (VGS–Vth)2
ID = K VDS [2 (VGS – Vth) – VDS]
VGS–Vth
triode region (quadratic in VDS)
ID = K (VGS–Vth)2 saturation region (flat)
Complete model summary:
29Prof. C.K. Tse: Transistor devices
Example (biasing in saturation)
VDS
ID
5–3 = 2V
Vth = 3V10V
5V
K = 0.5 mAV–2
ID
2kΩ
ID = 0.5x22
= 2mA
load lineslope = –1/2k
10V6V
By using load line
30Prof. C.K. Tse: Transistor devices
Example (biasing in saturation)
VDS
ID
5–3 = 2V
Vth = 3V10V
5V
K = 0.5 mAV–2
ID
2kΩ
VDS = 10 – 2x2 = 6Vwhich is > 2OKAY!
6VWhat happen if a 4.5kΩ is used?
ID = 0.5x22
= 2mA
By analysis
31Prof. C.K. Tse: Transistor devices
Example (biasing in triode)
VDS
ID
5–3 = 2V
Vth = 3V10V
5V
K = 0.5 mAV–2
ID4.5kΩ
So, it is in the triode region.
ID = K VDS [2 (VGS – Vth) – VDS] = 0.5 VDS (4 – VDS) = 0.5 (10–4.5 ID)(4–10+ 4.5 ID)i.e., 10.125 ID
2 – 35ID + 30 = 0ID = 1.8845 mA or 1.5722 mA And 1.88mA gives VDS = 10–4.5x1.88=1.54V.
But 1.57mA gives VDS = 10–4.5x1.57=2.95V!!
ID = 0.5x22
= 2mAVDS = 10 – 4.5x2 = 1VOops!!
?
ID = 1.88mAVDS = 1.54V
32Prof. C.K. Tse: Transistor devices
Enhancement and depletionMOSFET
What we have just studied is the enhancementMOSFET.
Enhancement — the channel is originally notconducting when gate voltage is 0, and we have toapply a positive gate voltage (bigger than a thresholdVth) to make it conduct (enhance it).
Depletion — In fact, we also have another kind ofMOSFET, in which the channel can conduct even whengate voltage is not applied. Then, we need to applyreverse gate voltage to cut it off. This is calleddepletion MOSFET.
NOTE THAT DUE TO A SEMICONDUCTOR DOPING PROPERTY,For n-channel MOSFET, both enhancement and depletiontypes can be made.For p-channel MOSFET, only enhancement type can be made.
ID
VGSVth
Enhancement mode
Vth
Depletion mode
33Prof. C.K. Tse: Transistor devices
Junction FET (JFET)
np p
Drain
Source
Gate
depletion region width depends on themagnitude of the gate reverse bias
Current can flow initially because plentyof electrons are available in the channel.
Gate : Apply negative voltage to increasethe depletion width, so as to reduce thecurrent. When the gate voltage isnegative enough, current will stop.
Hence, this is a depletion device.
34Prof. C.K. Tse: Transistor devices
n
Junction FET (JFET)Drain
Source
Gate np p
Drain
Source
Gatep pmore
–ve voltage–ve voltageapplied toreducecurrent Channel
becomesnarrower
35Prof. C.K. Tse: Transistor devices
Pinch off in JFET
np p
Drain
Source
Gate
more–ve voltage
Channelpinch off;Current stops
Pinch-offvoltage
Vp
ID
VGSVp
surely depletion type
36Prof. C.K. Tse: Transistor devices
n-channel JFET characteristicsDrain
Source
Gate
ID
VGS
+
–
VDS
+
–
ID
VDS
VGS=2V
VGS=1V
VGS=0V
VGS=–2V
saturation(like “active” in BJT)
triode
Pinch-off voltage of thisJFET is Vp = –2 V
The characteristics are very similar to those of MOSFET. But,now the threshold is a negative value, which is called thepinch-off voltage Vp instead of threshold voltage.
37Prof. C.K. Tse: Transistor devices
n-channel JFET characteristicsDrain
Source
Gate
ID
VGS
+
–
VDS
+
–
Pinch-off voltage of thisJFET is Vp = –2 V
Everything is almost the same!!
VDS
ID
K (VGS–Vp)2
ID = K VDS [2 (VGS – Vp) – VDS]
VGS–Vp
triode region (quadratic in VDS)
ID = K (VGS–Vp)2
saturation region (flat)
Be careful about sign!VGS can be negative or positive,but Vp is negative.
38Prof. C.K. Tse: Transistor devices
Example (biasing in saturation)
ID
VDS
+
–
Vp = –2 VK = 0.2mA/V2
VDS
ID10V
0–(–2)=2V
ID = K (VGS–Vp)2
= 0.2(2)2
= 0.8mA10kΩ
VDS = 10 – 10x0.8 = 2Vjust okay in saturation!
But if the resistor ismore than 10kΩ, it willbe in triode region!
39Prof. C.K. Tse: Transistor devices
Example (biasing in triode)
ID
VDS
+
–
Vp = –2 VK = 0.2mA/V2
VDS
ID10V
0–(–2)=2V
ID = K (VGS–Vp)2
= 0.2(2)2
= 0.8mA12kΩ
VDS = 10 – 12x0.8 = 0.4V < 2VSo, it can’t be in saturation!
Recalculate ID:ID = K VDS [2 (VGS – Vp) – VDS] = 0.2 (10–12 ID)[2x2–(10–12 ID)]i.e., 28.8 ID
2 – 37.4 ID + 12 = 0 ID = 0.7195mA or 0.5791mA
And, ID = 0.7195mA gives VDS = 1.366V ---okayBut, ID = 0.5791mA gives VDS = 3.051V --- reject!
ID = 0.7195mAVDS = 1.366V
40Prof. C.K. Tse: Transistor devices
Example (biasing in triode)
ID
VDS
+
–
Vp = –2 VK = 0.2mA/V2
VDS
ID10V
0–(–2)=2V
ID = K (VGS–Vp)2
= 0.2(2)2
= 0.8mA12kΩ
ID = 0.7195mAVDS = 1.366V
Of course, youmay also solveit by usingload line.
Load line
10V
41Prof. C.K. Tse: Transistor devices
Important small-signal characteristic
ID
VDS
VGS=2V
VGS=1.9V
VGS=1.8V
VGS=1.7V
Consider only the saturation region.
If we change VGS in a small range, thenID also changes in a range. The ratio ofthe change in ID to the change in VGS iscalled transconductance.
Similar to BJT!!!
†
gm =DID
DVGS
which is the slope of the curveID versus VGS , or analytically,
†
gm =dID
dVGS=
ddVGS
K (VGS -Vth )2
= 2K (VGS -Vth )
= 2 K K (VGS -Vth )
= 2 K ID
42Prof. C.K. Tse: Transistor devices
Other FETs
So far, we have only talked about1. n-channel MOSFET (enhancement type)2. n-channel JFET (depletion type)
Other FETs:
MOSFET
JFET
FET
n-channel MOSFET
p-channel MOSFET
n-channel JFET
p-channel JFET
enhancement
depletion
enhancement
depletion
depletion
similar to npn BJT
43Prof. C.K. Tse: Transistor devices
p-channel FETs
Operation is almost the same as n-channel FETs.
Voltage polarity and current direction reversed.
BUT… for p-channel devices,
the carriers are holes (not electrons). So, mobility is lowerand minority carrier lifetime shorter.
Consequence: p-channel devices are usually POORER!higher threshold voltage, higher resistance, and lowercurrent capability.