The Transistor as a voltage-controlled resister › bitstream › handle › 1721.1 › 35722 ›...

6.071 Bipolar Transistors 1

emitter

collector

base

npn bipolar transistor

The Transistor as a voltage-controlled resister

control knob

current in

currentout

e b c

2N2222


The Diode

V (volts)

I(A) 0.6V

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


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 = βIBIE = IC + IB = (1+ β)IBVBE =VB −VE = +0.6V

β ~ 100, but changes with temperature and with VCE


10 kΩ

Transistor SwitchVCC

R

on

off

IC ≅ βIB

collector current depends on the voltage drop across the bulb.

IB =VCC−VBE

R

since the transistor statedepends on the base current,leaving the base open circuitedwould eventually shut down the transistor, but this is sloppy.


Transistor Switch

VCC

IB

R

VBE

VCC = IR+VBE

IB =VCC −VBER

IC = β VCC−VBER

VBE =0.6V


Emitter Follows as a Current Source #1

Vin

+10V

-10V

1kΩ1kΩ

load

+10V

-10V

Vin

What is Vout?

Vout = Vin - 0.6VIf the base/emitter is forward biased



Vin

+10V

-10V

1kΩ1kΩ

What is when the transistor is off?Vout

⇒1kΩ 1kΩ

-10V

-5V

⇒

+10V

-10V

1kΩ1kΩ

highimpedance



Vin

+10V

-10V

1kΩ1kΩ

At what base voltage does it turn off?

VBE = 0.6V∴ Vin = −4.4V



Vin

+10V

-10V

1kΩ1kΩ

+10V

-10V

Vin

Vin

Vout

Vout = Vin − 0.6V;

−5V;Vin ≥ −4.4VVin < −4.4V


Biasing 1

Often signals are AC (or capacitively) coupled into amplifierstages. Note a single sided voltage supply can not amplifythe negative inputs.

Vin

+VCC

VoutR

Vin

VoutVB


Biasing 2

Vin

Solve this by adding a DC to the base to shift the signal so that thereis no clipping and AC coupling the output.

RE

R1

R2

VCC

Vout

select

want

VCC =15V

R1 || R2 <<βRER1|| R2 is the impedance of thecurrent source used to drive the transistor, βRE is the effective impedance of the base of the transistor.

Rule: make the impedanceof the source smallcompared to the loadit drives.


Biasing 3

Vin

RE

R1

R2

VCC

Vout

VCC=15V, R1|| R2<<βRE• want the output to swing ±7.5V• select RE. RE = 7.5kΩ. ∴ quiescent current into base = 1mA.• VE @ (= 0) = 7.5V (allows Vout

±7.5V) then VB = VE + 0.6V = 8.1VR2

R1 + R2

= 8.1V15V

R1R2

=1

1.17or

• R1|| R2<<βRE<<100 × 7.5kΩ.R1 = 130kΩ, R2 = 150kΩ.


oscillatorFrequencyamplitude

trigger

Input #1 Input #2

oscilloscopetime-base display

Vout1kΩ

dual voltage source

12V

33kΩ

Vin

time

Vout

time

Vin

Demo: BJT Emitter Follower #1


Demo: BJT Emitter Follower #2

oscillatorFrequencyamplitude

Input #1 Input #2

oscilloscopex,y display

Vout1kΩ

dual voltage source

12V

33kΩ

Vout

Vin

Vin


Emitter Follower

VCC

Rload

Rin = βRload

The output voltage is almost the base voltage, with a 0.6 V cutoff. Notice the change in impedance.

VB

Vout =VB − 0.6V


The Emitter Follows Has Unit Voltage Gain;

Is it Useless?

Vin

+VCC ∆Vin = ∆Vout

−VCC

VoutR

Note: ∆IE =∆VoutR

=∆VinR

and ∆IB =∆IE

1 + β=

∆VinR 1 + β( )

∴ Pin =Vin

2

R(1+ β);Pout =

Vin2

RP = IV

There is a gain in Power of β.

The effective base resistance is βR.


Don’t design with β

Vin

VCC

VoutRE

βREHere the quiescent point was selected by bleeding a small amount of current into the base.Now the operating point depends critically on β which varies tremendously from device to device and with temperature.


Problem: Look at last circuit as β is varied from 100 to 200.


Current SourceVCC

Rload

Iload = IC = VB − 0.6VR

IC ≅ IE , for large β

The base voltage controls the current through the load up to the limit of VCC.

R

VB

VE

VE =VB −0.6V

IE = VER


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 ofVC over RC.

VCC

Vin

RC

VC

RER2

R1

∴ IC = Iq = VCRC

= VoutRC

;

IB = VinβRE

= ICβ

∴ VoutVin

= −RCRE


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


Rule for robust control

source load

If the source output impedance is much lower than the loadinput impedance than the circuit performance will beindependent of load variation. Therefore in a multi-stage device, if we use a FET asthe building blocks of the load, the input impedance of the load will be high and we will have a robust circuit.


Voltage Transfer

VL =RL

RL + RthVth

Rth

RLVth so if VL~Vth

∴ Rth << RL

For efficient voltage transfer, keep the load impedance largerthan the source impedance.

Two exceptions:• Radio Frequency circuits, Zsource=Zload

(provide maximum power transfer)• Coupling currents rather than voltages.


Current Transfer

IL =RN

RL + RNIN

RLIN so if IL~IN

∴ RN >> RL

RN

IL

For efficient current transfer keep the load impedance smallcompared to the source impedance.


Power Transfer #1

VL =RL

RL + RthVth

Rth

RLVthI =

Vth(RL + Rth )

∴ PL = VLI =RL

RL + Rth( )Vth ⋅Vth

RL + Rth( ) =RLVth

2

RL + Rth( )2

PL

Rth RL


Power Transfer #2

IL =RN

RL + RNIN

RLIN RN

IL

PL = IL2RL =

RN2 IN

2 RLRL + RN( )2

PL

RN RL


Unity-gain phase splitter

Vin (+)

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

Vout− = −Vin(+)

Vout+ =Vin (+)

4.7kΩ

4.7kΩ

56kΩ

150kΩ

20V

Vout+ ≡

Vout− ≡

emitter follows∴ unity gain

common emitter∴ with RC=RE, gain=-1


Biasing Unity-gain phase splitter

4.7kΩ

4.7kΩ

56kΩ

150kΩ

20V

15V5V

5.6V

• choose VE = 5V∴ VB = 5.6V

since IC ? IE, there is a dropof 5V over both 4.7kΩresistors.


Darlington Pair

Useful for large current applications, and high input impedance. They are slow, however. Base to Emitter drop is 1.2 V.

β2

β1β1 β2


Transistor AND Gate

Ain

Bin

out

6V

Ain Bin outlow low lowlow high lowhigh low lowhigh high high

Transistors form the basis of logic gates, and of integrated circuits.


Transistor OR Gate

Ain

Bin

out

6V

Ain Bin outlow low lowlow high highhigh low highhigh high high


Properties of Bipolar Transistors

β (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.


Datasheet of 2N2222 (1 of 3)


Voltage Regulator

Vin

Vout = VzenerREA simple voltage regulator.

Poor ripple suppression, requires a zener with high power rating, and variations with load impedance.

RealZener

IdealZener

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 voltageregulators.


Voltage Regulator

Vin

Vout = Vzener-0.6 V

R

Emitter follower configuration.

Base current is only 1/β of supply current.

RC filter reduces ripple.

VzenerC

Rload


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.


Common Collector AmplifierVCC

Rload

An amplifier with a current gain (no voltage gain) and offset to avoid clipping negative inputs.

VB

R1

R2

C1C2

RE

R1 and R2 provide the DC offset and C1 acts as a filter (so inputs do not disturb quiescent point).



1. choose a quiescent current, 1 mA

VB

R1

R2

C1C2

RE

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

RE = VCC 2IQ

= VCC 21mA

VCC

Rload

R1

R2

C1C2

RE


VCC

Rload

R1

R2

C1C2

RE


3. Set the quiescent current via R1 & R2.

R1R2

= VCC −VBVB

= VCC −VE − 0.6VVE + 0.6V

VE =VCC 2recall:

So, forget 0.6V and R1 = R2

Note that Rbase = β RE,

so R1||R2<< β RE

prevents the quiescent point from shifting with load.



Rload

VB

R1

R2

C1C2

RE

4. Choose coupling capacitorsThe effective AC input resistance

Rin = R1 R2 β RE Rload( )C1 and Rin form a high-pass filter

C1 = 1ω3dB Rin

C2 and Rload also form a high-pass filter C2 = 1ω3dB Rload

VCC

Rload

R1

R2

C1C2

RE


Common Collector Amplifier

VCC

Rload

VB

R1

R2

C1C2

RE

base voltage

output

frequencyresponse

The Transistor as a voltage-controlled resister › bitstream › handle › 1721.1 › 35722 ›...

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