working of a transistor

7/31/2019 Working of a Transistor

1/43

Slide 1

6.071 Bipolar Transistors 1

emitter

collector

base

npn bipolar transistor

The Transistor as a voltage-controlled resister

control knob

current in

current

out

e b c

2N2222


2/43

Slide 2


The Diode

V (volts)

I(A) 0.6V

As described, but turns on at 0.6V (for a Si diode).


3/43

Slide 3


Normally off (base/emitter

reverse biased), small input

current and voltage relative

to emitter turns it on, switching

and amplifying

emitter

collector

base

Transistor Properties

IC = IB

IE = IC+ IB = (1+)IB

VBE = VB VE = +0.6V

~ 100, but changes withtemperature and with VCE


4/43

Slide 4


10 k

Transistor SwitchVCC

R

on

off

IC IB

collector current depends onthe voltage drop across the

bulb.

IB =VCCVBE

R

since the transistor state

depends on the base current,

leaving the base open circuited

would eventually shut down the

transistor, but this is sloppy.


5/43

Slide 5


Transistor Switch

VCC

IBR

VBE

VCC= IR+VBE

IB =VCCVBE

R

IC=VCCVBE

R

VBE=0.6V


6/43

Slide 6


Emitter Follows as a Current Source #1

Vin

+10V

-10V

1k1k

load

+10V

-10V

Vin

What is Vout?

Vout = Vin - 0.6V

If the base/emitter is forward biased


7/43

Slide 7



Vin

+10V

-10V

1k1k

What is when the transistor is off?Vout

1k 1k

-10V

-5V

+10V

-10V

1k1k

high

impedance


8/43

Slide 8



Vin

+10V

-10V

1k1k

At what base voltage does it turn off?

VBE = 0.6V

Vin = 4.4V


9/43

Slide 9



Vin

+10V

-10V

1k1k

+10V

-10V

Vin

Vin

Vout

Vout=

Vin 0.6V;

5V;

Vin 4.4V

Vin < 4.4V


10/43

Slide 10


Biasing 1

Often signals are AC (or capacitively) coupled into amplifier

stages. Note a single sided voltage supply can not amplify

the negative inputs.

Vin

+VCC

VoutR

Vin

VoutVB


11/43

Slide 11


Biasing 2

Vin

Solve this by adding a DC to the base to shift the signal so that there

is no clipping and AC coupling the output.

RE

R1

R2

VCC

Vout

select

want

VCC = 15V

R1 || R2


12/43

Slide 12


Biasing 3

Vin

RE

R1

R2

VCC

Vout

VCC

=15V, R1|| R

2


13/43

Slide 13


oscillator

Frequency

amplitude

trigger

Input #1 Input #2

oscilloscope

time-base display

Vout1k

dual voltage source

12V

33k

Vin

time

Vout

time

Vin

Demo: BJT Emitter Follower #1


14/43

Slide 14


Demo: BJT Emitter Follower #2

oscillator

Frequency

amplitude

Input #1 Input #2

oscilloscope

x,y display

Vout1k

dual voltage source

12V

33k

Vout

Vin

Vin


15/43

Slide 15


Emitter Follower

VCC

Rload

Rin = Rload

The output voltage is almost the

base voltage, with a 0.6 V cutoff.Notice the change in impedance.

VB

Vout = VB 0.6V


16/43

Slide 16


The Emitter Follows Has Unit Voltage Gain;

Is it Useless?

Vin

+VCCV

in= V

out

VCC

VoutR

Note: IE =VoutR

=VinR

and IB =IE

1 + =

VinR 1+ ( )

Pin =Vin

2

R(1+);Pout =

Vin2

RP= IV

There is a gain in Power of.

The effective base resistance is R.


17/43

Slide 17


Dont design with

Vin

VCC

VoutRE

RE

Here the quiescent point was

selected by bleeding a small

amount of current into the base.

Now the operating point depends

critically on which variestremendously from device to

device and with temperature.


18/43

Slide 18


Problem: Look at last circuit as is varied

from 100 to 200.


19/43

Slide 19


Current SourceVCC

Rload

Iload = IC =VB 0.6V

R

IC IE, for large

The base voltage controls

the current through the loadup to the limit of VCC.

R

VB

VE

VE = VB 0.6V

IE =VER


20/43

Slide 20


Common Emitter #1

The common emitter configuration

provides a (negative) voltage gain. (1)Set the quiescent current such that

VC = VCC/2 Want a voltage drop of

VC over RC.

VCC

Vin

RC

VC

RER2

R1

IC = Iq =VCRC

=VoutRC

;

IB =VinRE

=IC

VoutVin

=RCRE


21/43

Slide 21


Common Emitter

So RC = VCC/(2 Iq).

Gain = -RC/RE

RE is necessary for stability,

otherwise there is a small resistance

rtr~ 0.026V/IE, but this is very

temperature sensitive.

VCC

Vin

RC

VC

RER2

R1


22/43


23/43

Slide 23


Voltage Transfer

VL =RL

RL + RthVth

Rth

RLVth so if VL~Vth

Rth


24/43

Slide 24


Current Transfer

IL =RN

RL + RNIN

RLIN so if IL~IN

RN >> RL

RN

IL

For efficient current transfer keep the load impedance small

compared to the source impedance.


25/43

Slide 25


Power Transfer #1

VL =RL

RL + RthVth

Rth

RLVthI=

Vth

(RL + Rth )

PL = VLI=RL

RL + Rth( )Vth

Vth

RL + Rth( )=

RLVth2

RL + Rth( )2

PL

Rth RL


26/43

Slide 26


Power Transfer #2

IL =RN

RL + RN

IN

RLIN RN

IL

PL = IL2RL =

RN2 IN

2 RL

RL + RN( )2

PL

RN RL


27/43

Slide 27


Unity-gain phase splitter

Vin (+)

goal: from an AC signal generate a copy and its inverse.

Vout = Vin(+)

Vout+ =Vin (+)

4.7k

4.7k

56k

150k

20V

Vout+

Vout

emitter follows

unity gain

common emitter

with RC=RE, gain=-1


28/43

Slide 28


Biasing Unity-gain phase splitter

4.7k

4.7k

56k

150k

20V

15V

5V

5.6V

choose VE = 5V

VB = 5.6V

since IC ? IE, there is a drop

of 5V over both 4.7kresistors.


29/43

Slide 29


Darlington Pair

Useful for large current applications,

and high input impedance. They are

slow, however. Base to Emitter drop

is 1.2 V.

2

11 2


30/43

Slide 30


Transistor AND Gate

Ain

Bin

out

6V

Ain Bin out

low low low

low high low

high low low

high high high

Transistors form thebasis of logic gates, and

of integrated circuits.


31/43

Slide 31


Transistor OR Gate

Ain

Bin

out

6V

Ain Bin out

low low low

low high high

high low high

high high high


32/43

Slide 32


Properties of Bipolar Transistors

(current gain) is not a parameter, it varies with

everything.IC,max - maximum collector current rating.

BVCBO - maximum collector to base voltage.

BVCEO - maximum collector to emitter voltage.

VEBO - emitter to base breakdown voltage.

PD - maximum collector power dissipation.


33/43

Slide 33


Datasheet of 2N2222 (1 of 3)


34/43

Slide 34




35/43

Slide 35




36/43

Slide 36


Voltage Regulator

Vin

Vout = VzenerREA simple voltage regulator.

Poor ripple suppression,

requires a zener with high

power rating, and variations

with load impedance.

Real

Zener

Ideal

Zener

I

V

Zeners are diodes that have

variable resistance. Specifically,

zeners have a constant currentoutput over a range of input

voltages. Thus, by providing a

constant current to a circuit,

zeners can be used as voltage

regulators.


37/43

Slide 37


Voltage Regulator

Vin

Vout

= Vzener

-0.6 V

R

Emitter follower configuration.

Base current is only 1/ of supplycurrent.

RC filter reduces ripple.

VzenerC

Rload


38/43

Slide 38


Switching inductive loads

The voltage kick from

interrupting current flow in

an inductor can lead to

voltage breakdown in the

transistor. A backwards

diode across the inductive

load shorts this out.


39/43

Slide 39


Common Collector AmplifierVCC

Rload

An amplifier with a current gain

(no voltage gain) and offset toavoid clipping negative inputs.

VB

R1

R2

C1C2

RE

R1 and R2provide the

DC offset and C1 acts

as a filter (so inputs do

not disturb quiescent

point).


40/43

Slide 40



1. choose a quiescent current, 1 mA

VB

R1

R2

C1C2

RE

2. VE = Vcc/2(allows the largest symmetric input).

RE =VCC 2

IQ=VCC 2

1mA

VCC

Rload

R1

R2

C1C2

RE


41/43

Slide 41


VCC

Rload

R1

R2

C1C2

RE


3. Set the quiescent current via R1 & R2.

R1R2

=VCC VB

VB=VCCVE 0.6V

VE+ 0.6V

VE = VCC 2recall:

So, forget0.6V and R1 = R2

Note that Rbase = RE,

so R1||R2


42/43

Slide 42



Rload

VB

R1

R2

C1C2

RE

4. Choose coupling capacitors

The effective AC input resistance

Rin = R1 R2 RE Rload( )

C1 and Rin form a high-pass filter

C1 =1

3dBRin

C2 and Rload also form a high-pass filterC2 =

1

3dBRload

VCC

Rload

R1

R2

C1C2

RE


43/43

Slide 43


Common Collector Amplifier

VCC

Rload

VB

R1

R2

C1C2

RE

base voltage

output

frequency

response

working of a transistor

Documents