Transistor Devices: Explaining BJT and MOSFET

7/24/2019 Transistor Devices: Explaining BJT and MOSFET

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EIE209 Basic Electronics

Transistor Devices

Prof. C.K. Tse: Transistor devices

Contents BJT and FET Characteristics

Operations

http://cktse.eie.polyu.edu.hk/eie209/


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2Prof. C.K. Tse: Transistor devices

What is a transistor?Three-terminal device

whose voltage-current

relationship is controlled by

a third voltage or current

We may regard a transistor as acontrolled voltage or currentsource.

i

+v

+

vc

ic


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Types of transistorsAccording to the physics of the device, we can

classify 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 a

current amplifier bynature.

Conduction is controlledby electric field which isproduced by voltageapplied to the control

terminals. So, the controldraws no current andFET is a voltage-controlled device.


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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 diodecontactpotential barriercan be overcome. Electrons

can 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= bIB


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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

Base

np

n

Main relation:

IC =bIB

IC

IB

b100 typically

IC

IB


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Some properties

Base

Collector

Emitter

IC

IB

VBE

0.6 V when the transistor turns on.

Never try to stick a large voltage across VBE

because it may produce enormous current

or may just kill the device!

bis a bad parameter. Dont trust the

databook. Its value can vary to 50% or more.

IC= bI

Bholds only under some carefully set

conditions. Well look at it later.

IC

IB

IE= IB + IC


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Typical operations

1. Cut-off

2. Active operation

3. Saturation+10 V

IC

VBE

+

Determining factors: How large is I

Bor V

BE

How large is RL

RL

IB


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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


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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 current

amplification.

But we need ICRL< 10V


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Condition for active operation: ICRL


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SaturationWhen V

CEis reduced to 0, the BJT is saturated.

+10 V

IC=0.6667mA15k

IB=10A

VCE = 0V

+

ICcannot be 1mA!!In fact, it must drop in order

to make up for the totalvoltage.

In this case,

IC= 10V/15k= 0.6667mA

IC =bIB


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What makes it saturate?

Large RL Large IB

+10 V

IC=10mA1k

IB=100A

VCE = 0V

+

just saturate!

+10 V

IC=1mA10k

IB=10A

VCE = 0V

+

just saturate!


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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, IC0.1A.

Light bulb turns on.

100

IB= IC= 0. Light bulb turns off.


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Detailed BJT characteristics

Obviously, VBE

andIBare related by diode characteristic.

IB = Iss eVBE/VT -1

VT=kT

q

Boltzmans constant

absolute temperature

electronic charge

thermal voltage

25 mV @room temp

VBE

IB

0.6

Input characteristic (IBversus VBE)


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Also, ICis just I

Bmultiplied by b.

IC =bIss eVBE/VT -1

= Is eVBE/VT -1

VT=kT

q

Boltzmans constant

absolute temperature

electronic charge

thermal voltage

25 mV @room temp

VBE

IC

= bIB

0.6

Transfer characteristic (ICversus VBE)

same shape as IB


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ICis nearly flat unless near saturation.

Output characteristic (ICversus V

CE)

IC

VCE

IC

VCE

+

Not Ohms law!!

for one particularchoice of I

Bor V

BE


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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 =IC10.025

W1

m

0.65 0.68 V

BE

BE

BE

Different IB(or V

BE) has

different output characteristic.

A range of VBE

corresponds toa range of I

C.

Transconductance:

gm =DIC

DVBE

= slopeon the transfer char.


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What is gm?

gm =DIC

DVBE A simple differentiation gives

gm = dIC

dVBE= d

dVBEbIss(e

VBE/VT -1)

=bIss(eVBE/VT )

1

VT

IC

VT or

IC

25mV

at room tem


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A bit more preciseAt saturation, V

CEis not really 0, it is about 0.2 V.

IC

VCE

for one particularchoice of I

Bor V

BE

0.2V

+10 V

IC=0.98mA not 1mA!!

10k

IB=10A

VCE = 0.2V

+


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A bit more preciseIn active region, ICis not really flat. It goes up gently! This iscalled Early Effect!

IC

VCE

for one particularchoice of IBor VBE

0.2V

slope IC/ VA

Early voltage

typically 100V

IC


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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


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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


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n-channel MOSFETG

SiO2insulator

p

nn

body or substrate

When gate is +ve,electrons are attracted toit and this becomes n-type

conducting channel.

This action is calledchannel enhancement.

S D

The channel is notconducting initially whengate is zero volt.


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n-channel MOSFET characteristicDrain

Source

Gate

Characteristics:

Gate current = 0 (always)The channel conduction is determined by VGS

ID

VDS

VGS=2V

VGS=1.9V

VGS=1.8VVGS=1.7VThreshold voltage

Vth= 1.7 V, for example.

saturation

(like active in BJT)

triode

VGS

+

VDS

+

ID


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Saturation regionDrain

Source

Gate

So, it looks like the npn BJT!! But if we lookcloser, we find that the saturation current isproportional to (VGSVth)

2.

ID

VDS

VGS=2V

VGS=1.9V

VGS=1.8VVGS=1.7VThreshold voltage

Vth= 1.7 V, for example.

saturation


ID=K (V

GSV

th)2 for saturation region

VGS

+

VDS

+

ID


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Saturation regionDrain

Source

GateID

ID=K (VGSVth)2 for saturation region

VGS

+

VDS

+

ID

VGS

If we plot the saturation IDversus VGS,we have aquadratic (parabolic) curve.

Vth


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Triode region

ID= aVDS(2M VDS)

VDS2M

ID

Triode region like a quadratic (parabolic) function

K (VGSVth)2

So, the equation is:

y = a x(2M x)

Obviously, M= VGS Vth, a = K,

ID= KVDS[2 (VGS Vth) VDS]M= VGSVth for triode region


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n-channel MOSFET characteristic

VDS

ID

K (VGSVth)2

ID= KVDS[2 (VGS Vth) VDS]

VGSVth

triode region (quadratic in VDS)

ID=K (VGSVth)2saturation region (flat)

Complete model summary:


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Example (biasing in saturation)

VDS

ID

53 = 2V

Vth= 3V10V

5V

K = 0.5 mAV2

ID

2k

ID= 0.5x22

= 2mA

load line

slope = 1/2k

10V6V

By using load line


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VDS

ID

53 = 2V

Vth= 3V10V

5V

K = 0.5 mAV2

ID

2k

VDS= 10 2x2 = 6Vwhich is > 2

OKAY!

6VWhat happen if a 4.5kis used?

ID= 0.5x22

= 2mA

By analysis


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Example (biasing in triode)

VDS

ID

53 = 2V

Vth= 3V10V

5V

K = 0.5 mAV2

ID4.5k

So, it is in the triode region.

ID= KVDS[2 (VGS Vth) VDS]

= 0.5 VDS (4 VDS) = 0.5 (104.5 ID)(410+ 4.5 ID)i.e., 10.125 ID

2 35ID+ 30 = 0ID= 1.8845 mA or 1.5722 mA And 1.88mA gives VDS= 104.5x1.88=1.54V.

But 1.57mA gives VDS= 104.5x1.57=2.95V!!

ID= 0.5x22

= 2mA

VDS= 10 4.5x2 = 1VOops!!

?

ID= 1.88mAVDS= 1.54V


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Enhancement and depletion

MOSFETWhat we have just studied is the enhancementMOSFET.

Enhancement the channel is originally notconducting when gate voltage is 0, and we have to

apply 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


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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.


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n

Junction FET (JFET)Drain

Source

Gate np p

Drain

Source

Gatep pmore

ve voltage

ve voltageapplied toreducecurrent Channel

becomes

narrower


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Pinch off in JFET

np p

Drain

Source

Gate

moreve voltage

Channel

pinch off;Current stops

Pinch-offvoltage

Vp

ID

VGSVp

surely depletion type


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n-channel JFETcharacteristicsDrain

Source

Gate

ID

VGS

+

VDS

+

ID

VDS

VGS=2V

VGS=1V

VGS=0V

VGS=2V

saturation


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 Vpinstead of threshold voltage.


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n-channel JFETcharacteristicsDrain

Source

Gate

ID

VGS

+

VDS

+

Pinch-off voltage of thisJFET is Vp= 2 V

Everything is almost the same!!

VDS

ID

K (VGSVp)2

ID= KVDS[2 (VGSVp) VDS]

VGSVp

triode region (quadratic in VDS)

ID=K (VGSVp)2saturation region (flat)

Be careful about sign!VGScan be negative or positive,but Vpis negative.


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ID

VDS

+

Vp= 2 VK = 0.2mA/V2

VDS

ID10V

0(2)=2V

ID = K (VGSVp)2

= 0.2(2)2

= 0.8mA10k

VDS= 10 10x0.8 = 2Vjust okay in saturation!

But if the resistor ismore than 10k, it willbe in triode region!


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ID

VDS

+

Vp= 2 VK = 0.2mA/V2

VDS

ID10V

0(2)=2V

ID = K (VGSVp)2

= 0.2(2)2

= 0.8mA12k

VDS= 10 12x0.8

= 0.4V

< 2VSo, it cant be in saturation!

Recalculate ID:ID= KVDS[2 (VGS Vp) VDS] = 0.2 (1012 ID)[2x2(1012 ID)]i.e., 28.8 I

D

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= 0.7195mA

VDS= 1.366V


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ID

VDS

+

Vp= 2 VK = 0.2mA/V2

VDS

ID10V

0(2)=2V

ID = K (VGSVp)2

= 0.2(2)2

= 0.8mA12k

ID= 0.7195mA

VDS= 1.366V

Of course, youmay also solveit by using

load line.

Load line

10V


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Important small-signal characteristicID

VDS

VGS=2V

VGS=1.9V

VGS=1.8V

VGS=1.7V

Consider only the saturation region.

If we change VGSin a small range, thenIDalso changes in a range. The ratio ofthe change in IDto the change in VGSiscalled transconductance.

Similar to BJT!!!

gm =DID

DVGS

which is the slope of the curveIDversus VGS, or analytically,

gm =dID

dVGS

=d

dVGS

K(VGS-Vth)2

= 2K(VGS -Vth)

= 2 K K(VGS -Vth)

= 2 K ID


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Other FETsSo far, we have only talked about

1. 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


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p-channel FETsOperation 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 lower

and minority carrier lifetime shorter.

Consequence: p-channel devices are usually POORER!higher threshold voltage, higher resistance, and lower

current capability.

Documents

Transistor Devices: Explaining BJT and MOSFET