transistor devices - city ucktse.eie.polyu.edu.hk/eie209/4.transistordevices.pdf · prof. c.k. tse:...

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


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.


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


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


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


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


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


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


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?


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


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!


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.


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)



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



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


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.


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


A bit more precise

At saturation, VCE is not really 0, it is about 0.2 V.

IC

VCE


0.2V

+10 V

IC=0.98mA not 1mA!!

10kΩ

IB=10µA

VCE = 0.2V

+

–


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

IC

VCE


0.2V

slope ≈ IC / VA

Early voltagetypically 100V

IC


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


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


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.


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


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.


ID = K (VGS–Vth)2 for saturation region

VGS

+

–

VDS

+

–

ID


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


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


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:


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



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


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


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


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.


n

Junction FET (JFET)Drain

Source

Gate np p

Drain

Source

Gatep pmore

–ve voltage–ve voltageapplied toreducecurrent Channel

becomesnarrower


Pinch off in JFET

np p

Drain

Source

Gate

more–ve voltage

Channelpinch off;Current stops

Pinch-offvoltage

Vp

ID

VGSVp

surely depletion type


n-channel JFET characteristicsDrain

Source

Gate

ID

VGS

+

–

VDS

+

–

ID

VDS

VGS=2V

VGS=1V

VGS=0V

VGS=–2V


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.


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.



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!



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



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


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


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


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.

transistor devices - city ucktse.eie.polyu.edu.hk/eie209/4.transistordevices.pdf · prof. c.k. tse:...

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