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– Assignments are 20% of the final grade– Exam questions very similar (30%)

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• Come and talk to me if you need help.• mid-term: Thursday, October 27th

Adam Riess, Brian Schmidt and Saul Perlmutter

Lecture 10 Overview

• Transistors (continued)

• The common-emitter amplifier

• Amplifier parameters

• Black box amplifier models

Summary of useful equations

• Basic DC operating conditions:

CCCCC

VVSC

CE

BC

CBE

RIVV

eII

II

II

III

TBE

/

1

• Add a small signal:

T

Cm

bembeT

Cc

V

Ig

vgvV

Ii

mme

bee

mb

be

ggi

vr

gi

vr

1

Using small signal models

1) Determine the DC operating conditions (in particular, the collector current, IC)

2) Calculate small signal model parameters: gm, rπ, re

3) Eliminate DC sources: replace voltage sources with shorts and current sources with open circuits

4) Replace BJT with equivalent small-signal model. Choose most convenient depending on surrounding circuitry

5) Analyze

e.g. Tmodel

Voltage gain with small signal model

Find the gain using a small signal model:

vbe vbe

+

-

+

-

ic

re

RC

Cmbe

c

mme

e

C

ee

Ce

be

c

eebe

CeCe

Ccc

Rgv

v

ggr

r

R

ri

Ri

v

v

riv

RiRi

Riv

gain voltageso

1

gain Voltage

ic

vc

eliminate DC sources and apply T-model

How to build a Real Common emitter amplifier

• Why bother with 2 voltage supplies? • Use a voltage divider R2/R1 to provide base-emitter voltage to correctly bias the transistor.

DC condition: the voltage divider

• The voltage divider should provide sufficient voltage to place the transistor in active mode (base-emitter forward biased):

Cbb I

RR

V10

21

V6.0

V6.0

EEB

EBBE

RIV

VVV

• Current through resistors should be >10 times base current for stability

BB

EE

BBB

V

RI

RR

R

RR

RVV

V6.0 so

now

21

1

21

1

Amplifier specifications

• What other parameters of an amplifier do we care about?– Voltage gain– Dynamic range– Frequency response (bandwidth)– Input impedance– output impedance

Voltage Gain• Voltage gain• Use small signal model (short Voltage sources and capacitors)

1 since

)(

CeCcc

Eeeb

RiRiv

Rriv

Ee

C

b

c

in

out

Rr

R

v

v

v

v

E

C

b

c

R

R

v

v

voltage gain

usually re<<RE

• Voltage gain is only defined by resistors RC and RE

ground

ground

αie

Frequency response (Bandwidth)

• Normally interested in providing a small, AC signal to the base

• Use capacitors to remove ("block") any low frequency (DC) component ("capacitively couple the signal to the base") which could affect the bias condition

• C1 forms a high-pass filter with R1in parallel with R2 (Assuming the AC impedance into the base is large).

• Cut off frequency ω0=1/RC, so to remove frequencies <fmin:

21

21min

1

2

1

RRRR

f

C

Frequency response (Bandwidth)

• Also worthwhile to place a capacitor on the output

• C2 forms a high pass filter with RL.

• Cut off frequency ω0=1/RC, so to remove frequencies <fmin:

LRfC

min2 2

1

Dynamic Range

• Maximum voltage output = Vbb

• Minimum = 0• Beyond this the signal becomes 'clipped' or distorted• To get the maximum possible voltage swing, both positive and negative, set VC=0.5 VBB • Maximum 'dynamic range'

VC

rb

rOUT

Input impedance• Consider the circuit without the voltage divider resistors. What's the small signal (AC) input impedance at the base, rb?

• Including voltage divider resistors in parallel

• Input signal sees a total input impedance rIN= R1 // R2 // rb

RB

rb

ROUT

Output impedance

Co

Co

o

oOUT R

i

Ri

i

vr

• If RL=10kΩ and we want a low frequency cutoff of 20Hz, What is C2?• If VBB=15V and IC=2mA what is the output impedance?

Co

Co

o

oOUT R

i

Ri

i

vr

BBC VV 5.0

LRfC

min2 2

1

21

21min

1

2

1

RRRR

f

C

E

C

b

c

R

R

v

v

BB

EE

V

RI

RR

R V6.0

21

1

Cbb I

RR

V10

21

DC condition

Frequency responseImpedance Gain/Dynamic range

• Why do we care about the input and output impedance?

• Simplest "black box" amplifier model:

Impedances

RIN

ROUT

VIN AVINVOUT

• The amplifier measures voltage across RIN, then generates a voltage which is larger by a factor A

• This voltage generator, in series with the output resistance ROUT, is connected to the output port.

• A should be a constant (i.e. gain is linear)

• Attach an input - a source voltage VS plus source impedance RS

Impedances

RIN

ROUT

VINAVIN

VOUT

• Note the voltage divider RS + RIN.

• VIN=VS(RIN/(RIN+RS)

• We want VIN = VS regardless of source impedance

• So want RIN to be large.

• The ideal amplifier has an infinite input impedance

VS

RS

• Attach a load - an output circuit with a resistance RL

Impedances

• Note the voltage divider ROUT + RL.

• VOUT=AVIN(RL/(RL+ROUT)

• Want VOUT=AVIN regardless of load

• We want ROUT to be small.

• The ideal amplifier has zero output impedance

RIN

ROUT

VINAVIN VOUTVS

RS

RL

Operational Amplifier

• Integrated circuit containing ~20 transistors

Operational Amplifier

• An op amp is a high voltage gain amplifier with high input impedance, low output impedance, and differential inputs.• Positive input at the non-inverting input produces positive output, positive input at the inverting input produces negative output. • Can model any amplifier as a "black-box" with a parallel input impedance Rin, and a voltage source with gain Av in series with an output impedance Rout.

Ideal op-amp• Place a source and a load on the model

• Infinite internal resistance Rin (so vin=vs).• Zero output resistance Rout (so vout=Avvin).• "A" very large• No saturation• iin=0; no current flow into op-amp

-

+

voutRL

RS

So the equivalent circuit of an ideal op-amp looks like this:

Schematics An amplifier will not work without a power supply. And a more complete diagram looks like the figure below, which also indicates the standard pin configuration.

-

+

out

V+

V-

2

3

4

6

7 1

4 5

8

top view

PinFunction

2 Inverting input

3 Non-inverting input

4 V- supply

6 Output

7 V+ supply

Measuring Impedances

• Assuming you can only vary RL and RS, how would you measure the input and output impedances of the amplifier?

RIN

ROUT

VINAVIN VOUTVS

RS

RL

Measuring Impedances

• With the black box model, it is simple to measure the input and output impedances of an amplifier

• To measure the input impedance, vary RS until the output voltage has dropped to half ; then RS=RIN= input impedance

•To measure the output impedance, vary RL until the output voltage has dropped to half ; then RL=ROUT= output impedance

RIN

ROUT

VINAVIN VOUTVS

RS

RL

Cascaded Amplifiers

• Easiest way to increase amplification is to link amplifiers together

RIN1

ROUT1

VIN1A1VIN1 VOUT1 RIN2

ROUT2

VIN2A2 VIN2 VOUT2

• Ideal amplifiers; VOUT2=A1A2VIN1

• In reality, take account of voltage divider action due to input and output impedances