Transistors I

7/27/2019 Transistors I

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Introduction to Transistors• A transistor is a device with three separate layers of

semiconductor material stacked together – The layers are made of n –type or p –type material in the

order pnp or npn

– The layers change abruptly to form the pn or np junctions

– A terminal is attached to each layer (The Art of Electronics, Horowitzand Hill, 2nd Ed.)

(Introductory Electronics, Simpson, 2nd Ed.)



Introduction to Transistors• Thus when a transistor is

off it behaves like atwo –diode circuit

• A transistor operates (or

turns on) when the base –emitter junction is forward

biased and the base –collector junction is reversedbiased (“biasing”)

(Electronic Devices and

Circuits, Bogart, 1986)

(The Art of Electronics,

Horowitz and Hill, 2nd Ed.)

(Lab 4 –1)



Transistor Biasing (npn Transistor)• Electrons are constantly

supplied to the emitter bythe battery with voltage V EE

• These electrons can:1. Recombine with holes in

the base, giving rise to I B 2. Diffuse across base and be swept (by electric field at

base –emitter junction) into collector, then diffuse around

and eventually recombine with holes injected into

collector, giving rise to I C • Since the base region is designed so thin, process 2

dominates (no time for #1 to occur as often) – In an actual npn transistor, 98 or 99% of the electrons that

diffuse into the base will be swept into the collector



Current Flow Inside a Transistor • Current flow for an npn transistor (reverse for pnp):

– From conservation of current( I E = I B + I C ) we can obtain the following

expressions relating the currents:

where b ≈ 20 – 200

(depends on emitter current)

• b increases as I E increases (for very small I E ) since there is less

chance that recombination will occur in the base

• b decreases slightly (10 –20%) as I E increases beyond several mA

due to increased base conductivity resulting from larger number of

charge carriers in the base

• Thus b is not a constant for a given transistor!

• An average value of 100 is typically used

(Electronic Devices and Circuits,

Bogart, 1986)

BC I I b B E I I 1 b

(and thus I C ≈ I E )

(Lab 4 –5)



Transistor Current Amplification• If the “input” current is I B and the “output” current is

I C , then we have a current amplification or gain – Happens because base –emitter junction is forward-biased

– Forward bias ensures that the base –emitter junction

conducts (transistor is turned on)

– Reverse bias ensures that most of the large increase inthe base –emitter current shows up as collector current

(Student Manual for The Art

of Electronics, Hayes and

Horowitz, 2nd Ed.)

Thus small gains in I B

result in large gains in I E

and hence I C



Basic Transistor Switch Circuit• Transistor “switch” circuit:

– With switch open, transistor is off and lamp is off

– With switch closed, I B = (10 – 0.6) V / 1k = 9.4 mA

– However, I C = b I B 940 mA (assuming b = 100)• When collector current I C = 100 mA, lamp has 10V across it

• To get a higher current, collector would need to be below ground

• Transistor can’t do this, so it goes into saturation

• Collector voltage gets as close to emitter voltage as it can (about0.2 V higher) and I C remains constant ( I C is “maxed out”)

(The Art of Electronics, Horowitz and Hill, 2nd Ed.) 0 V

0.6 V0.2 V

(BC junction forward biased)

V B V C

V E

(Lab 4 –9)



Emitter Follower • Output “follows” the input: only

difference is a 0.6 V diode drop – True for V in > 0.6 V

– If V in < 0.6 V, transistor turns off (no

current – “valve” is closed) and V out = 0

– Data with R E = 3.3k : (The Art of Electronics, Horowitz

and Hill, 2nd Ed.)

BE

C

E

(Lab 4 –2)

V in

V out



Emitter Follower • By returning the emitter

resistor to a negativesupply voltage, you can

obtain negative voltage

swings as well

– Data with R E = 3.3k :(The Art of Electronics,




Emitter Follower Biasing

• You must always provide a DC path for base bias

current, even if it is just through a resistor to ground – HW Problem 2.5

(The Art of Electronics, Horowitz and Hill, 2nd Ed.)



Emitter Follower Biasing• With R B included in the previous circuit:

f = 1 kHz



Emitter Follower Biasing• Without R B included in the previous circuit:

(Here there is no DC base bias current, so transistor is off.)



Emitter Follower Biasing• To obtain symmetric output waveforms without

“clipping,” provide constant DC bias using a voltagedivider

– Capacitors block “outside” DC current, which may affect

quiescent (no input) values (“AC-coupled follower”)


(Lab 4 –4)



Emitter Follower Impedance• The usefulness of the emitter follower can be seen

by determining its input and output impedance: – Input impedance (i.e. the impedance looking into the base

of the transistor):

• Details of proof given in class – Output impedance (i.e. the impedance looking into the

emitter of the transistor):

• Details provided by you in the homework!

• Thus the input impedance is much larger than the

output impedance

loadloadin 1 Z Z Z b b

b b

sourcesourceout

1

Z Z Z



Emitter Follower Impedance• Thus the input and output “sees” what it wants to

see on the other side of the transistor:

• Using an emitter follower, a given signal sourcerequires less power to drive a load than if the sourcewere to drive the load directly – Very good, since in general we want

Z out (stage n) << Z in (stage n + 1) (by at least a factor of 10)

– An emitter follower has current gain, even though it has novoltage gain

– The emitter follower has power gain



Horowitz, 2nd Ed.)



Emitter Follower Impedance• When measuring the input and output impedance of

the emitter follower, it is useful to think about theThévenin equivalent circuit as “seen” at the input

and the output:

– Input impedance seen by the source:

– Output impedance seen by the load:

V in

Z source

~

Z out Z load

Z in

V B

V out, load V out, no load

in

insource

inB V

Z Z

Z V

loadnoout,

loadout

loadloadout, V

Z Z

Z V


of Electronics, Hayes andHorowitz, 2nd Ed.)

(Lab 4 –3)



Emitter Follower With Load• Consider the following circuit:

– V out and V in waveforms:

V in (V) V out (V) I E (mA)

+9.4 8.8 27.6

5 4.4 18.8

0 – 0.6 8.8 – 3 – 3.6 2.8

– 4.4 – 5.0 0.0

– 5 – 5.0 0.0

– 10 – 5.0 0.0

(The Art of Electronics,


V in

V out

V in V out I E

(HW 2.2)



Emitter Follower With Load• Thus the npn emitter follower can only “source”

current (supply current to something like a load)

• It cannot “sink” current (draw current from somethinglike a load)

• In this example, the transistor turns off when

V in = –

4.4 V (V out = –

5.0 V) – Then I E = 0 and the base –emitter junction becomes reversebiased

– As V in increases further, a rather large reverse biasdevelops across this junction which could result in

breakdown

• The output could swing more negative than –5 V byreducing the R E = 1k resistor, but this increasespower consumption in both the resistor and transistor



Zener Diodes as Voltage Regulators• Zener diodes “like” to break down at

a particular reverse bias: – When reverse biased, they provide a

constant voltage drop over a wide range

of currents

• Zeners thus provide a means of voltage regulation

– We choose the specifications for the zener based on:

(max)outoutminin, I

R

V V

zener minout,

min

outmaxin,zener V I R

V V P

(The Art of Electronics, Horowitz

and Hill, 2nd Ed.)



Horowitz, 2nd Ed.)



Example Problem 2.3

Solution details given in class.

Design a +10 V regulated supply for loadcurrents from 0 to 100 mA; the input

voltage is +20 to +25 V. Allow at least

10 mA zener current under all (worst-case)

conditions. What power rating must thezener have?



Emitter Followers as Voltage Regulators

• However, the zener current can change significantly

depending on the load, affecting regulation

performance

• A better voltage regulator would incorporate an

emitter follower:

– Here the zener current is more constant, relatively

independent of load current since changes in I E (or I load)

produce only small changes in I B

• Load current determined from (V B – 0.6 V) / Rload

(The Art of

Electronics, Horowitz

and Hill, 2nd Ed.)

(HW 2.4)



Transistors as Current Sources• A transistor can be used as a

current source with the setup at

right:

– Note that I C is independent of V C as long as V C > V E + 0.2 V(i.e., the transistor is not saturated)

• The output voltage (V load or V C ) range over which I load

(= I C ) is (nearly) constant is called the output

compliance

E

B

E

E

C E

R

V

R

V I I

6.0


V C

V E

(Lab 4 –6)



Deficiencies of Current Sources• The load current will still vary somewhat, even when

the transistor is “on” and not in saturation

• There are two kinds of effects that cause this: – V BE varies somewhat with collector-to-emitter voltage for a

given collector current (Early effect), as does b • DV BE ≈ – 0.0001 DV CE

• We assume V BE = constant = 0.6 V in the basic transistor model

– V BE and b depend on temperature• DV BE ≈ – 2.1 mV/0C

• We neglect changes in b by assuming I C = I E

• To minimize DV BE from both effects, choose V E largeenough ( 1V) so that DV BE 10 mV will not result inlarge fractional changes in the voltage across R E – V E too large will result in decreased output compliance,

however (V C range for transistor “on” state decreases)



Common –Emitter Amplifier • Consider a transistor current

source with a resistor RC

as

load, and block unwanted

DC at the base input (V in is

an AC signal):

so where

– Now imagine we apply a base wiggle v B via the input signal

– The emitter follows the wiggle so v E = v B

– Then the wiggle in the emitter current is:

C R f

eq

dB32

1

eq32

1

R f C

dB

E R R R R b 21eq

C

E

B

E

E

E i R

v

R

v

i


(lower-case letters

represent smallchanges, or “wiggles”)

(Note DC quiescent

output voltage of 10 V)

(Lab 4 –7)



Common –Emitter Amplifier – Now V C = V CC – I C RC so vC = – iC RC = – v B( RC / R E )

– Since vin

= v B

and vout

= vC

, we have a voltage amplifier,

with a voltage gain of:

– Minus sign means that a positive wiggle at the input getsturned into a negative wiggle at the output

• Input and output impedance:

– Z in = R1 R2 b R E ≈ 8k (see figure on previous slide)

– Z out = RC (impedance looking into collector) = RC (high Z current source) ≈ RC = 10k (see figure on previous slide)

• Be careful to choose R1 and R2 correctly so that

design is not b dependent ( R1 R2 << b R E )

E

C

R

R

v

vG

in

out

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Transistors I

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