NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 1 Topic 6 Analog Circuits Ideal Operational Amplifiers One...

NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 1

Topic 6 Analog Circuits

Ideal Operational AmplifiersOne Transistor Amplifier

ECE 271

Electronic Circuits I

NJIT ECE 271 Dr. Serhiy Levkov

Example of Analog Electronic System: FM Stereo Receiver

• Linear functions: Radio and audio frequency amplification, frequency selection (tuning), impedance matching, local oscillator

• Nonlinear functions: DC power supply(rectification), frequency conversion (mixing), detection/demodulation

• The characteristics of signals are most often manipulated with linear amplifiers.

• Much information in the world (temperature, pressure, light intensity, sound, etc.) is analog in nature.

• In electric form those signals are transformed by different linear and nonlinear functions.

Topic 6 - 2


Amplification: Introduction

A complex periodic signal can be represented as the sum of many individual sine waves, one component of which has amplitude Vi = 1 mV and frequency ws with 0 phase (signal is used as reference):

After amplification, linear amplifier output is sinusoidal with same frequency but different amplitude VO and phase :

sini i s

v V t

(sin )o o s

v V t

Example: Audio amplifier. The input is a form of complicated periodic signal.

Topic 6 - 3

Ri

ii

vi


Amplification: Introduction (cont.)

Amplifier output power is:

If we need to deliver PO = 100 W and have RL = 8 , W then

This power results in output current:

where

Input current is given by (Vi = 1 mV )

and the phase is zero because circuit is purely resistive.

PO

Vo

2

21

RL

2 2 100 8 40Vo O L

V P R

(sin )o o si I t

Io V

oR

L

40V8 5A

-3810 V 1.82 10 A

5k 50ki

ii in

VI

R R

Topic 6 - 4


Amplification: Gain

The main parameter of an amplifier is the gain. Using phasor representation , we can introduce three types of gain.

• Voltage Gain (complex number):

Magnitude and phase of voltage gain are given by

and

For our example,

• Current Gain:

Magnitude of current gain is given by

0o o o

vi ii

V VA

VV

vv

ov

i

VA

V vA

43

40V 4 1010 V

ov

i

VA

V

0o o o

iii i

I IA

II

ii

8-8

5A 2.75 101.82 10 A

oi

i

IA

I

V v

Topic 6 - 5


Amplification: Gain (cont.)

• Power Gain:

For our example,

The gain is often expressed in decibel scale:

2 2

2 2

o o

o oOv iP

i ii i i

V IP V IA A AP V I V I

AP 40 5

10 3 1.82 10 8 1.10 1013

AvdB

20logAv

AidB 20logA

i

APdB

10logAP

Topic 6 - 6


Two-port Model for Amplifier

• Simplifies amplifier-behavior modeling in complex systems.• Two-port (four terminal) models are linear network models, valid only

under small-signal conditions.• Represented by g-, h-, y- and z-parameters.• (v1, i1) and (v2, i2) represent signal components of voltages and

currents at the network ports.

Type of input-output model, black box that relates outputs to inputs.

input output

x1x2

Amplifier

Topic 6 - 7


Two-port Model Parameters• Theoretically, two-port model could be described fully by a 4x4 matrix or 16

parameters.• In practice, several sets of 4 parameters are used.

Topic 6 - 8


Two-port Model Parameters

Impedance (z)-parameters Admittance (y)-parameters

• Theoretically, two-port model could be described fully by a 4x4 matrix or 16 parameters.

• In practice, several sets of 4 parameters are used.

Topic 6 - 9






Hybrid (h)-parameters Inverse hybrid (g)-parameters

Topic 6 - 10






Hybrid (h)-parameters Inverse hybrid (g)-parameters

Transfer (A,B,C,D) -parameters

Topic 6 - 11


g-parameters

Use open-circuit (i = 0) and short-circuit (v = 0) terminal conditions2221212

2121111

ivvivi

gggg

111

12

112

21

0

0

g

g

iv

i

ii v

Open-circuit input conductance

Reverse short-circuit current gain

Forward open-circuit voltage gain

Short-circuit output resistance

221

12

222

21

0

0

g

g

vv i

vi v

Topic 6 - 12


g-parameters

Use open-circuit (i = 0) and short-circuit (v = 0) terminal conditions2221212

2121111

ivvivi

gggg

111

12

112

21

0

0

g

g

iv

i

ii v

Open-circuit input conductance

Reverse short-circuit current gain

Forward open-circuit voltage gain

Short-circuit output resistance

221

12

222

21

0

0

g

g

vv i

vi v

If g12 = 0Norton transformation

Topic 6 - 13


g-parameters: Example

Problem: Find g-parameters.Approach: Apply specified boundary conditions for each g-parameter, use circuit analysis.

For g11 and g21: apply voltage v1 to input port and open circuit output port. For g12 and g22: apply current i2 to output port and short circuit input port.

Topic 6 - 14


g-parameters: Example

Problem: Find g-parameters.Approach: Apply specified boundary conditions for each g-parameter, use circuit analysis.

For g11 and g21: apply voltage v1 to input port and open circuit output port. For g12 and g22: apply current i2 to output port and short circuit input port.

g11i

1v

1 i20

1210451(200k)

g119.7910 8S

g21v

2v

1 i20

g11

(51)(200k)0.998

222

2 v 01

112

2 v 01

v 1 391i 1 51

200k 20ki 391 0.0196i 20k

g

g

8 21 1 2

22 1 2

i 9.79 10 v 1.96 10 i

v 0.998v 3.91 10 i

Topic 6 - 15

Ideal Voltage Amplifier

1

1

Lo

out L

ins

S in

o in Lv

s S in out L

Rv AvR RRv v

R RV R RA AV R R R R

Consider amplifier with g12 = 0 (typical voltage amplifier) that includes source and load resistances and calculate the voltage gain from the source voltage to load voltage.



1

1

Lo

out L

ins

S in

o in Lv

s S in out L

Rv AvR RRv v





1

1

Lo

out L

ins

S in

o in Lv

s S in out L

Rv AvR RRv v

R Rv R RA Av R R R R




1

1

Lo

out L

ins

S in

o in Lv

s S in out L

Rv AvR RRv v




Conclusion: full gain depends on source and load parameters.


1

1

Lo

out L

ins

S in

o in Lv

s S in out L

Rv AvR RRv v





How to make it non- dependent?


1

1

Lo

out L

ins

S in

o in Lv

s S in out L

Rv AvR RRv v


Rin >> Rs and Rout<< RL,

Av A

Ideal voltage amplifier: Rout = 0,

Rin






1

1

Lo

out L

ins

S in

o in Lv

s S in out L

Rv AvR RRv v


Rin >> Rs and Rout<< RL,

Av A

Ideal voltage amplifier: Rout = 0,

Rin

1

o

o L o S in S ini v

s s L L

S in

VV R R R RI RA AV R RVI

R R




For current:


Other Amplifier Types


Voltage amplifier

0,i o

R R 0

(V/V)ovo

i io

vA v

Current amplifier

0

(A/A)ois

i vo

iA i

0,

i oR R

Transconductance amplifier

0

(A/V)om

i vo

iG v

,i o

R R

Transresistance amplifier

0

(V/A)om

i io

vR i

00,i o

R R



Voltage amplifier

0,i o

R R 0

(V/V)ovo

i io

vA v

Current amplifier

0

(A/A)ois

i vo

iA i

0,

i oR R


0

(A/V)om

i vo

iG v

,i o

R R


0

(V/A)om

i io

vR i

00,i o

R R

Operational Amplifier (Op Amp)


• Op amp is a fundamental building block in electronic design. They are cheap mass produced IC and can be used to build many different types of electronic devices like instrumentation amplifiers, active filters, rectifiers, A/D and D/A converters, and others.

• Typical integrated circuit amplifier: LM386N (~ $1.00 each)• The practical op amp is a form of differential amplifier: responds to a difference

of two input signals. Typically: VCC >0, VEE <0 – so the voltages are symmetric 5V, 10V, 18V.

Topic 6 - 28


Differential Amplifier – Signal Amplification

A - open-circuit voltage gain,

A =10, Adb =20log(10) = 20 db

vID = (v+-v--) - differential input signal

vID = VID + vid = 1 + 0.25sin t, where

VID – dc value, vid – signal component

vO = VO + vo = 5 + 2.5sin t

The typical voltage transfer characteristic for a differential amplifier biased by two symmetric power supplies:

Topic 6 - 29


Distortion in Amplifiers• If the ac input signal exceeds 0.5V, the output signal will be clipped off (see vO2 ) and

get distorted. • Different gains for positive and negative values of input also cause distortion in output.• Total Harmonic Distortion (THD) is a measure of signal distortion that compares

undesired harmonic content of a signal to the desired component.

v(t) VO V

1(sin

ot

1) V

2(sin2

ot

2) V

3(sin3

ot

3) ...

dcdesired output

2nd harmonic distortion

3rd harmonic distortion

THD100% V

i

2

i 2

V1

Topic 6 - 30


Differential Amplifier Model: Basic

A - open-circuit voltage gain

vid = (v+-v-) = differential input signal voltage

Rid - amplifier input resistance

Ro - amplifier output resistance

Signal developed at amplifier output is in phase with the voltage applied at + input (non-inverting) terminal and 180o out of phase with that applied at - input (inverting) terminal.

The basic model of differential amplifier is represented by:

Topic 6 - 31


Differential Amplifier Model: With Source and Load

RL = load resistanceRS = Thevenin equivalent resistance of

signal sourcevs = Thevenin equivalent voltage of

signal sourceL

o L

RA

o id R R

v v

Topic 6 - 32





signal sourceL

o L

RA

o id R R

v v

Rid

id s R Rid s

v vand

Topic 6 - 33





signal sourceL

o L

RA

o id R R

v v

Rid

id s R Rid s

v vand

Topic 6 - 34

vo

vs

R Rid LA A

v R R R Rid S o L


Differential Amplifier Model: With Source and Load (Example)

• Problem: Calculate voltage gain• Given Data: A=100, Rid =100kW, Ro = 100W, RS =10kW, RL =1000W

o

s

100k 100038.3dB

10k 100k 100 1000100 82.6

R Rid LA A

v R R R Rid S o L

v

v

Topic 6 - 35


Differential Amplifier Model: With Source and Load (Example)

• Problem: Calculate voltage gain• Given Data: A=100, Rid =100kW, Ro = 100W, RS =10kW, RL =1000W

• Ideal amplifier’s output depends only on input voltage difference and not on source and load resistances. This can be achieved by using resistance condition (Rid >> RS or infinite Rid and Ro << RL or zero Ro ):

A - open-loop gain (maximum voltage gain available from the device)

o

s

100k 100038.3dB

10k 100k 100 1000100 82.6

R Rid LA A

v R R R Rid S o L

v

v

o

idA A

v

v

v

Topic 6 - 36


Ideal Operational Amplifier• Ideal op amp is a special case of ideal differential amplifier with:

– - infinite gain, – - infinite Rid ,

– - zero Ro .

Topic 6 - 37

A

idR

0oR




– - zero Ro .

• Thus:

1. Input voltage is zero, vid =0 for any finite output voltage (since A is infinite).

2. Input currents are zero, i+ =0 and i-=0 (since Rid is infinite).

,vov v 0limid idA A

Topic 6 - 38

A

idR

0oR




– - zero Ro .

• Thus:

1. Input voltage is zero, vid =0 for any finite output voltage (since A is infinite).

2. Input currents are zero, i+ =0 and i-=0 (since Rid is infinite).

• Additional properties:– infinite common-mode rejection (rejects input signals common to both inputs)

– infinite open-loop bandwidth (frequency magnitude response is flat everywhere with zero phase shift)

– infinite output voltage range ( in practice limited by VS+ and VS- )

– infinite output current capability – infinite slew rate (rate of change of output voltage)

– zero input-offset voltage (when the input terminals are shorted, vout = 0)

• The major benefit of ideal op-amp: the same standard op-amp allows to build various circuits with very specific properties by using external to op-amp circuit elements. The classic op-amps circuits are considered further.

,vov v 0limid idA A

Topic 6 - 39

A

idR

0oR


Inverting Amplifier: Configuration

• Non-inverting input is grounded.• Feedback network (resistor R2 ) is connected between

inverting input and output• Input network (resistor R1 ) is connected between

inverting input and signal source.

Topic 6 - 40


Inverting Amplifier:Voltage Gain

1 2 2KVL: - 0

S Oi iv R Rs v

Topic 6 - 41



1 2 2KVL: - 0

S Oi iv R Rs v

Topic 6 - 42

22KCL: 0ii i i ss i



1 2 2KVL: - 0

S Oi iv R Rs v

Topic 6 - 43

22KCL: 0ii i i ss i since 0 by 2i



1 2 2KVL: - 0

S Oi iv R Rs v

Topic 6 - 44


1 1S

v vsiR

vsR



1 2 2KVL: - 0

S Oi iv R Rs v

Topic 6 - 45


1 1S

v vsiR

vsR

since 0 and 0 by 1

idv v



1 2 2KVL: - 0

S Oi iv R Rs v

Topic 6 - 46


1 1S

v vsiR

vsR

since 0 and 0 by 1

idv v

Substituting iS and i2 into KVL:

2

1

2

1

orS O

vv RoA

v v Rs

RvR

1 2 1 1 2 1

1

( ) 0 ( ) 0SS O S S O

v R R v R R R v RR

v v v



1 2 2KVL: - 0

S Oi iv R Rs v

Topic 6 - 47


1 1S

v vsiR

vsR

since 0 and 0 by 1

idv v

Substituting iS and i2 into KVL:

2

1

2

1

orS O

vv RoA

v v Rs

RvR

Conclusions:• The gain of the inverting op-amp circuit is determined by R1 and R2 .

• The gain is a result of imposing the negative feedback.• Gain is greater than 1 if R2 > R1. Gain is less than 1 if R1 > R2

• Negative voltage gain implies 1800 phase shift between input and output signals.• Inverting input of op amp is at ground potential (not connected directly to ground) and is

said to be at virtual ground.

1 2 1 1 2 11

( ) 0 ( ) 0SS O S S O

v R R v R R R v RR

v v v


Inverting Amplifier: Input and Output Resistances

1nin i

vsRis

R R

Topic 6 - 48


Inverting Amplifier: Input and Output Resistances

1nin i

vsRis

R R 2 2 1 10KVL: v i R i Rx

But i1=i2 (i- =0) 1 2 1( )v i R Rx

Since v- = 0, i1=0 and vx = 0 irrespective of the value of ix ,

0x

x

vi

Rout

Topic 6 - 49

Rout is found by applying a test current (or voltage) source to amplifier output, turning off all independent sources and determining the voltage (or current):


Inverting Amplifier: Example

• Problem: Design an inverting amplifier with

Av=40 dB, Rin =20kW,

• Assumptions: Ideal op amp

Topic 6 - 50

• Convert the gain from db:

• Input resistance is controlled by R1 and voltage gain is set by R2 / R1.

Thus

and

40/20[db] 20log 40 10 100v

A A Av v

120k

inR R

22 1

1

100 2RA R R Mv R


Summing Amplifier

• Scale factors for the 2 inputs can be independently adjusted by proper choice of R2 and R1.

• Any number of inputs can be connected to summing junction through extra resistors.

• This is an example of a simple digital-to-analog converter.

11

1

vi

R 2

22

vi

R 3

3

Ov

iR

Since negative amplifier input is at virtual ground (vid = 0):

Since i-=0, i3= i1 + i2,

3 31 2

1 2

R Rv v vo R R

Topic 6 - 51

vid


Non-inverting Amplifier: Configuration

• Input signal is applied to the non-inverting input terminal.• Portion of the output signal is fed back to the negative input terminal.• Analysis is done by relating voltage at v1 to input voltage vs and output

voltage vo .

Topic 6 - 52


Non-inverting Amplifier: Analysis

Since i-=0, (voltage divider) 11

1 2

Rv vo R R

Topic 6 - 53

+- vid

i-

i+



Since i-=0,

From KVL

11

1 2

Rv vo R R

10

S idv v v

Topic 6 - 54

+- vid

i-

i+



Since i-=0,

From KVL , since vid =0

and

11

1 2

Rv vo R R

10

S idv v v

1v vs

1 2

1

R Rv vo s R

Topic 6 - 55

+- vid

i-

i+



Since i-=0,


and

.

11

1 2

Rv vo R R

10

S idv v v

1v vs

1 2

1

R Rv vo s R

Topic 6 - 56

21 2

1 1

1v R R RoA

v Rv Rs

+- vid

i-

i+



Since i-=0,


and

.

11

1 2

Rv vo R R

10

S idv v v

1v vs

1 2

1

R Rv vo s R

Topic 6 - 57

21 2

1 1

1v R R RoA

v Rv Rs

+- vid

i-

i+

Since i+=0, . in

vsRi



Since i-=0,


and

.

11

1 2

Rv vo R R

10

S idv v v

1v vs

1 2

1

R Rv vo s R

Since i+=0, .

Rout is found by applying a test current source to amplifier output and setting vs = 0 and is identical to the output resistance of inverting amplifier i.e. Rout =0.

Topic 6 - 58

21 2

1 1

1v R R RoA

v Rv Rs

+- vid

i-

i+

in

vsRi

21 2

1 1

1v R R RoA

v Rv Rs



Since i-=0,


and

.

11

1 2

Rv vo R R

10

S idv v v

1v vs

1 2

1

R Rv vo s R

Since i+=0, .

Rout is found by applying a test current source to amplifier output and setting vs = 0 and is identical to the output resistance of inverting amplifier i.e. Rout =0.

Topic 6 - 59

21 2

1 1

1v R R RoA

v Rv Rs

+- vid

i-

i+

in

vsRi

21 2

1 1

1v R R RoA

v Rv Rs

2

1

,1 RAv R , 0

in outR R


Non-inverting Amplifier: Example

• Problem: Determine the characteristics of given non-inverting amplifier.

• Given Data: R1= 3kW, R2 =43kW, vs=+0.1 V

• Assumptions: Ideal op amp.• Analysis:

Since i-=0,

2

1

15.343k1 13k

(15.3)(0.1 ) 1.53V

RAv R

v A v Vo sv

2 1

1.53 33.343 3

OO

v Vi AR R k k

Topic 6 - 60

+- vid

i-

i+

io


Voltage Follower (Unity-gain Buffer)

Topic 6 - 61


Voltage Follower (Unity-gain Buffer)

• A special case of non-inverting amplifier, also called voltage follower with infinite R1 and zero R2. Hence

• Provides excellent impedance-level transformation while maintaining signal voltage level.

• Ideal voltage buffer does not require any input current and can drive any desired load resistance without loss of signal voltage.

• Unity-gain buffer is used in many sensor and data acquisition systems.

Topic 6 - 62

2 1/1 1

vRA R


Difference Amplifier

Topic 6 - 63

A combination of inverting and non-inverting amplifier.



Topic 6 - 64


2 2 1 2

1 21 - 22 1

11 1

(since 0)- -

( )- -

iv v i R v i RoR Rv v Rv R v v

RR R



2 2 1 2

1 21 - 22 1

11 1

(since 0)- -

( )- -


RR R

Also, (voltage division)

Topic 6 - 65


22

1 2

Rv v

R R



Since v-= v+, 2 ( )1 21

Rv v vo R

For R2= R1, ( )1 2

v v vo

Topic 6 - 66


2 2 1 2

1 21 - 22 1

11 1

(since 0)- -

( )- -


RR R

22

1 2

Rv v

R R




• The circuit amplifies difference between input signals (differential subtractor).

Since v-= v+, 2 ( )1 21

Rv v vo R

For R2= R1, ( )1 2

v v vo

Topic 6 - 67


2 2 1 2

1 21 - 22 1

11 1

(since 0)- -

( )- -


RR R

22

1 2

Rv v

R R




• The circuit amplifies difference between input signals (differential subtractor).

• To understand better, use superposition:

For v2=0, Rin1= R1, the circuit reduces to an inverting amplifier with the gain – R2 /R1 :

vO1= (– R2 /R1 ) v1

For v1=0, v2 is attenuated by the voltage divider and then amplified by non-inverting gain 1+R2 /R1:

vO2= (R2 /(R1 +R1 ))(1+ R2 /R1 ) v2

Then vO = vO1 + vO2,

Since v-= v+, 2 ( )1 21

Rv v vo R

For R2= R1, ( )1 2

v v vo

Topic 6 - 68


2

1

v (v v )1 2

Ro R

2 2 1 2

1 21 - 22 1

11 1

(since 0)- -

( )- -


RR R

22

1 2

Rv v

R R



Difference Amplifier: Example

Topic 6 - 69

Do example on the board


Difference Amplifier: Example

• Problem: Determine Vo, V+, V-, Io, I1, I2, I3 .

• Given Data: R1= 10kW, R2 =100kW, V1= 5V, V2=3V

• Assumptions: Ideal op amp.

Hence, V-= V+ and I-= I+= 0.

• Analysis:

1002 3 2.732- 10 1001 2

R kV V V V VR R k k

5 -2.731 - 227

1 2 101

V V V VI I AR k

5 (227 )(110 ) 20.01 1 1 2 2

V V I R I R V A k Vo

2272

I I Ao

Topic 6 - 70

0 32 273 1101 2

V VI AR R k


Integrator

Topic 6 - 71


Integrator

• Feedback resistor R2 in the inverting amplifier is replaced by capacitor C.

• The circuit uses frequency-dependent feedback.

Rsv

si dt

odvCci

Topic 6 - 72


Integrator



Rsv

si dt

odvCci

Topic 6 - 73

Since ic= is ,1

sv

RC

dvodt


Integrator



Rsv

si dt

odvCci

Since ic= is ,

1( ) ( ) (0)0

(0) (0)

1

tv t v d vo s

dv v dso RC

oRC

v Vo c

Topic 6 - 74

1s

vRC

dvodt

Voltage at the circuit’s output at time t is given by the initial capacitor voltage plus integral of the input signal from start of

integration interval, here, t=0.


Integrator



Rsv

si dt

odvCci

Since ic= is ,

1( ) ( ) (0)0

(0) (0)

1

tv t v d vo s

dv v dso RC

oRC

v Vo c

Topic 6 - 75

1s

vRC

dvodt





Integration of an input step signal results in a ramp at the output.


Differentiator

Topic 6 - 76


Differentiator

Topic 6 - 77

• Input resistor R1 in the inverting amplifier is replaced by capacitor C.


Differentiator


Rov

Ri dt

sdvCsi

Since iR= is

dtsdv

RCov

Topic 6 - 78


Differentiator


Rov

Ri dt

sdvCsi

Since iR= is

dtsdv

RCov

Output is scaled version of derivative of input voltage.

Topic 6 - 79


Differentiator


Rov

Ri dt

sdvCsi

Since iR= is

dtsdv

RCov

Topic 6 - 80

Derivative operation emphasizes high-frequency components of input signal, hence is less often used than the integrator.

Output is scaled version of derivative of input voltage.


Op-Amps – What’s Next

• Frequency dependent feedback (cont. ch 10):– high, low and band pass amplifiers (filters)– integrators and differentiators

Topic 6 - 81




• Analysis of non-ideal amplifiers (ch. 11)• Amplifier frequency response and stability (ch. 11)

Topic 6 - 82




• Analysis of non-ideal amplifiers (ch. 11)• Amplifier frequency response and stability (ch. 11)• Op-amps applications (ch. 12)

– instrumentation amplifiers – active filters– D/A and A/D converters– Oscillators– Nonlinear circuits (precision rectifiers)– Circuits with positive feedback (comparators, triggers)

Topic 6 - 83






• Single transistor amplifiers – linear amplification and small signal model (ch. 13)

Topic 6 - 84






• Single transistor amplifiers – linear amplification and small signal model (ch. 13)• Single transistor amplifiers – detailed design (ch. 14)

Topic 6 - 85






• Single transistor amplifiers – linear amplification and small signal model (ch. 13)• Single transistor amplifiers – detailed design (ch. 14)• Op-amps design (multistage amplifiers) (ch. 15)

What do we study in this course?

Topic 6 - 86






• Single transistor amplifiers – linear amplification and small signal model (ch. 13)• Single transistor amplifiers – detailed design (ch. 14)• Op-amps design (multistage amplifiers) (ch. 15)

Topic 6 - 87

NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 88NJIT ECE 271 Dr. Serhiy Levkov

Transistors as Amplifiers


Transistors as AmplifiersIn this section we will study • The general techniques for employing individual transistors as amplifiers


Transistors as AmplifiersIn this section we will study • The general techniques for employing individual transistors as amplifiers • Operation of common source MOSFET amplifier


Transistors as AmplifiersIn this section we will study • The general techniques for employing individual transistors as amplifiers • Operation of common source MOSFET amplifier• Operation of common emitter BJT amplifier



What do we know:• MOSFET (FET) can be used as amplifier if operated in saturation region (SR)

In this section we will study • The general techniques for employing individual transistors as amplifiers • Operation of common source MOSFET amplifier• Operation of common emitter BJT amplifier



What do we know:• MOSFET (FET) can be used as amplifier if operated in saturation region (SR) • BJT can be used as an amplifier when biased in the forward-active region (FAR)




What do we know:• MOSFET (FET) can be used as amplifier if operated in saturation region (SR) • BJT can be used as an amplifier when biased in the forward-active region (FAR)• We will refer to the FAR and SR as to the “active region”




What do we know:• MOSFET (FET) can be used as amplifier if operated in saturation region (SR) • BJT can be used as an amplifier when biased in the forward-active region (FAR)• We will refer to the FAR and SR as to the “active region” • In these regions, transistors can provide high voltage, current and power gains




What do we know:• MOSFET (FET) can be used as amplifier if operated in saturation region (SR) • BJT can be used as an amplifier when biased in the forward-active region (FAR)• We will refer to the FAR and SR as to the “active region” • In these regions, transistors can provide high voltage, current and power gains• Bias is provided to stabilize the operating point in a desired operation region




What do we know:• MOSFET (FET) can be used as amplifier if operated in saturation region (SR) • BJT can be used as an amplifier when biased in the forward-active region (FAR)• We will refer to the FAR and SR as to the “active region” • In these regions, transistors can provide high voltage, current and power gains• Bias is provided to stabilize the operating point in a desired operation region• The Q-point also determines

– Small-signal parameters of transistor– Voltage gain, input resistance, output resistance– Maximum input and output signal amplitudes– Power consumption



BJT Amplifier

Consider a BJT , biased in active region by dc voltage source VBE = 0.7V.

1510 , 100, 0.025)(s F T

VI


BJT Amplifier


From the simplified FAR model we get:

1510 , 100, 0.025)(s F T

VI

15μA 1.5mA 10 3.3 10 3.3 1.5 5

VBEVS T

B C F CE CBF

II e I I V K I V


BJT Amplifier



1510 , 100, 0.025)(s F T

VI

15μA 1.5mA 10 3.3 10 3.3 1.5 5

VBEVS T

B C F CE CBF

II e I I V K I V


BJT Amplifier



Thus the Q-point is set at (IC, VCE) = (1.5 mA, 5 V)

1510 , 100, 0.025)(s F T

VI

15μA 1.5mA 10 3.3 10 3.3 1.5 5

VBEVS T

B C F CE CBF

II e I I V K I V


BJT Amplifier

Now let’s inject the sin signal with the amplitude 8 mV by adding the voltage source vbe The total base-emitter voltage becomes:

vBEV

BEv

be


BJT Amplifier


From the FAR model for VBEh = 0.708V and VBEl = 0.692V we will get

vBEV

BEv

be

μA and20 10μABh Bl

I I


BJT Amplifier



vBEV

BEv

be

Thus 8 mV peak change in vBE : 5 mA change in iB and 0.5 mA change in iC.

20μA and 10μABh Bl

I I


BJT Amplifier



vBEV

BEv

be

Thus 8 mV peak change in vBE : 5 mA change in iB and 0.5 mA change in iC.

From the load line equation this will produce the 1.65V change in vCE . 330010CE C

v i

20μA and 10μABh Bl

I I


If changes in operating currents and voltages are small enough, then iC and vCE waveforms are undistorted replicas of input signal.

Small voltage change at base causes large voltage change at collector. Voltage gain is given by:

Minus sign indicates 1800 phase shift between input and output signals.

Av

VceV

be

1.65 1800.008 0

206 180 206

BJT Amplifier


MOSFET Amplifier

Consider MOSFET biased in active region by dc voltage source VGS = 3.5 V, which

Will set the Q-point at (ID, VDS) = (1.56 mA, 4.8 V).

Then we introduce the signal vGS and total gate-source voltage becomes

vGS V

GS v

gs

Doing similar analisis as for BJT, we get :

1 V change in vGS 1.25 mA change in iD 4 V change in vDS.


Coupling and Bypass Capacitors

• The constant voltage biasing is not a good method, 4 resistor is better.



• The constant voltage biasing is not a good method, 4 resistor is better.• We need to introduce the input signal without disturbing the bias point.



• The constant voltage biasing is not a good method, 4 resistor is better.• We need to introduce the input signal without disturbing the bias point.• AC coupling through capacitors is used to inject ac input signal and extract output

signal• Capacitors provide negligible impedance at frequencies of interest and provide open

circuits at dc.





circuits at dc.

C1 and C3 are large-valued coupling capacitors or dc blocking capacitors whose reactance at the signal frequency is designed to be negligible.





circuits at dc.

C1 and C3 are large-valued coupling capacitors or dc blocking capacitors whose reactance at the signal frequency is designed to be negligible.

C2 is a bypass capacitor that provides a low impedance path for ac current from emitter to ground thereby removing RE from the circuit when ac signals are considered.


DC and AC Analysis

• DC analysis:– Make dc equivalent circuit by replacing all capacitors by open circuits

and inductors by short circuits.– Find Q-point from dc equivalent circuit by using appropriate large-

signal transistor model.


DC and AC Analysis




• AC analysis:– Make ac equivalent circuit by replacing all capacitors by short circuits,

inductors by open circuits, dc voltage sources by ground connections and dc current sources by open circuits.


DC and AC Analysis






– Replace transistor by small-signal model


DC and AC Analysis






– Replace transistor by small-signal model– Use small-signal ac equivalent to analyze ac characteristics of amplifier.


DC and AC Analysis






– Replace transistor by small-signal model– Use small-signal ac equivalent to analyze ac characteristics of amplifier.– Combine end results of dc and ac analysis to yield total voltages and

currents in the network.


DC Equivalent for MOSFET Amplifier

dc equivalent

ac equivalent Simplified ac equivalent

Full circuit

Make dc equivalent circuit by replacing all


dc equivalent


Full circuit

Make dc equivalent circuit by replacing all - capacitors by open circuits



dc equivalent


Full circuit

Make dc equivalent circuit by replacing all - capacitors by open circuits - inductors by short circuits



AC Equivalent for MOSFET Amplifier

dc equivalent


Full circuit

Make ac equivalent circuit by replacing all - capacitors by short circuits



dc equivalent


Full circuit

Make ac equivalent circuit by replacing all - capacitors by short circuits - inductors by open circuits



dc equivalent


Full circuit

Make ac equivalent circuit by replacing all - capacitors by short circuits - inductors by open circuits - dc voltage sources by ground connections



dc equivalent


Full circuit

Make ac equivalent circuit by replacing all - capacitors by short circuits - inductors by open circuits - dc voltage sources by ground connections- dc current sources by open circuits


dc equivalent


Full circuit


Simplify the ac equivalent circuit


DC and AC Equivalents for MOSFET Amplifier

dc equivalent


Full circuit


DC Equivalent for BJT Amplifier

All capacitors in original amplifier circuits need to be replaced by open circuits, disconnecting vI , RI , and R3 from circuit.



All capacitors in original amplifier circuits need to be replaced by open circuits, disconnecting vI , RI , and R3 from circuit.



All capacitors in original amplifier circuits are replaced by open circuits, disconnecting vI , RI , and R3 from circuit.


AC Equivalent for BJT Amplifier

Make ac equivalent circuit by replacing all- capacitors by short circuits - inductors by open circuits - dc voltage sources by ground connections - dc current sources by open circuits



21

3

160k 300k

22k 100kB

L C

R R R

R R R


Small-Signal Modeling

• For AC analysis (or general time varying analysis) we would need to use phasor , time domain, Laplace Transform (or similar) methods.

• Those are working better with linear systems.

• Need to make the linear model – small signal model.

• Assume that the time varying components are small signals and construct the two port model, which is linear.

• The concept of small signal is device dependent.

• We start from developing small signal model for a diode and then switch to transistor


Small-Signal Operation of Diode• The slope of the diode characteristic at the Q-

point is called the diode conductance and is given by:

ID and VD represent the DC bias point Q-point, vD and iD are small changes around Q-point




• gd is small but non-zero for ID = 0 because slope of diode equation is nonzero at the origin.

exp 1

exp 1 1

exp

40 for 0.025V

D D S DSd

T T TDD Qp

DS D

TT

S

D S

T

D Dd D

T

vI

Vv

v

V

i I vg

V Vv

I iV V

ig

I

Ii IV

i






• Diode resistance is given by:

exp 1

exp 1 1

exp

40 for 0.025V

D D S DSd

T T TDD Qp

DS D

TT

S

D S

T

D Dd D

T

vI

Vv

v

V

i I vg

V Vv

I iV V

ig

I

Ii IV

i

rd 1

gd


• The slope of the diode characteristic at the Q-point is called the diode conductance and is given by:


• Diode resistance is given by:

• Thus, linear model in the vicinity of ID :

when


Small-Signal Operation of Diode

exp 1

exp 1 1

exp

40 for 0.025V

D D S DSd

T T TDD Qp

DS D

TT

S

D S

T

D Dd D

T

vI

Vv

v

V

i I vg

V Vv

I iV V

ig

I

Ii IV

i

rd 1

gd


, small signal model D dd dD d

i g vi I i

vd2V

T0.05V or v

d5 mV


• BJT is a three terminal device and to build similar model to a diode, we would need to use 2-port y-parameter network.

Small-Signal Model of BJT

Using 2-port y-parameter network,

The port variables can represent either time-varying part of total voltages and currents or small changes in them away from Q-point values.

ib y

11v

be y

12v

ce

ic y21vbe

y22

vce

y12

ib

vce v

be0

i

B

vCE Q point

0

y21

ic

vbe v

ce0

i

C

vBE Q point

IC

VT

y22

ic

vce v

be0

i

C

vCE Q point

IC

VAV

CE

y11

ib

vbe v

ce0

i

B

vBE Q point

I

C

oV

T


Small-Signal Model of BJT

y12

ib

vce v

be0

i

B

vCE Q point

0

y21

ic

vbe v

ce0

i

C

vBE Q point

IC

VT

y22

ic

vce v

be0

i

C

vCE Q point

IC

VAV

CE

y11

ib

vbe v

ce0

i

B

vBE Q point

I

C

oV

T

• BJT is a three terminal device and to build similar model to a diode, we would need to use 2-port y-parameter network.



ib y

11v

be y

12v

ce

ic y21vbe

y22

vce


Small Signal Hybrid-Pi Model of BJT

• The hybrid-pi small-signal model is most widely accepted model for BJT amplifier.

• Small-signal parameters are controlled by the Q-point





• VA - Early Voltage

Trans-conductance:

21

1 40Cm C

Ty

Ig IV

Input resistance:

11

1 o o T

m Cy

Vrg I

Output resistance:

ro 1

y22

V

AV

CE

IC

V

A

IC



Trans-conductance:

| 40CCm Q C

BE T

i

v

Ig IV

Input resistance:

o o T

m C

Vrg I

Output resistance:

ro 1

y22

V

AV

CE

IC

V

A

IC

bo is the small-signal common-emitter current gain of the BJT.






Trans-conductance:

| 40CCm Q C

BE T

i

v

Ig IV

Input resistance:

o o T

m C

Vrg I

Output resistance:

ro 1

y22

V

AV

CE

IC

V

A

IC

2 0.005be T be

v V or v V

Definition of small signal for BJT:

bo is the small-signal common-emitter current gain of the BJT.





Small-Signal Current Gain and Voltage Gain of BJT

The important small signal parameter is trans-conductance, or current-voltage (iC/vBE ) gain. It shows how the collector current changes in response to base-emitter voltage.




Other two important parameters are: - small signal current gain

Small signal current gain (iC/iB = iC/(vBE /r) - gain) :

In practice, however, bF and bo are often assumed to be equal.

og

mr

F

1 IC

1

F

F

iC

Q point




Other two important parameters are: - small signal current gain - intrinsic voltage gain

Small signal current gain (iC/iB = iC/(vBE /r) - gain) :

In practice, however, bF and bo are often assumed to be equal.

Intrinsic voltage gain (vCE/vBE – gain) :

For VCE << VA,

mF represents maximum voltage gain individual BJT can provide, doesn’t change with operating point, and ranges from 1000 to 4000.

Small signal model for pnp-BJT is similar

F

VA

VT

40VA

og

mr

F

1 IC

1

F

F

iC

Q point

F gmro

I

C

VT

VA V

CEIC

V

A V

CE

VT




ig y

11v

gs y

12v

ds

id y

21v

gs y

22v

ds

y11

ig

vgs v

ds0

i

G

vGS Q point

0

y12

ig

vds v

gs0

i

G

vDS Q point

0

y21

id

vgs v

ds0

i

D

vGS Q point

2ID

VGS V

TN

y22

id

vds v

gs0

i

D

vDS Q point

ID

1V

DS

Small-Signal Model of MOSFET (nmos)




ig y

11v

gs y

12v

ds

id y

21v

gs y

22v

ds

y11

ig

vgs v

ds0

i

G

vGS Q point

0

y12

ig

vds v

gs0

i

G

vDS Q point

0

y21

id

vgs v

ds0

i

D

vGS Q point

2ID

VGS V

TN

y22

id

vds v

gs0

i

D

vDS Q point

ID

1V

DS




• Since gate is insulated from channel by gate-oxide input resistance of transistor is infinite.

• Small-signal parameters are controlled by the Q-point.

• For same operating point, MOSFET has higher transconductance and lower output resistance that BJT.

Trans-conductance:

21

2 2Dn Dm

GS TN

Ig y K IV V

Output resistance:

22

1/1 1DSo

D D

Vry I I

Amplification factor (intrinsic voltage gain) for lVDS<<1:

Where l – channel-length modulation parameter.

)/2

21/ 1(

DS nf m o

DGS TNV

KVg rIV

Definition of small signal for MOSFET:

0.2 GS TNgsv V V


Small-Signal Model of JFET

1r

g

y11i

G

vGS Q point

I

GI

SG

VT

0 for IG0

gmy

21i

D

vGS Q point

ID

VGS V

P2

gm2 I

DSS

VP2

(VGS V

P)

1r

o

y22i

D

vDS Q point

ID

1V

DS

iDI

DSS1

vGS

VP

2

1vDS

for vDSv

GS V

P

iGI

SGexp

vGS

VT

1


Small-Signal Model of JFET

Since JFET is normally operated with gate junction reverse-biased,

IG ISG

rg

For small-signal operation, the input signal limit is:

The amplification factor is given by:

vgs 0.2 V

GS V

P

f g

mr

o 2

1 V

DS

VGS V

P

2V

P

IDSSI

D


Summary of FET and BJT Small-Signal Models


Common-Emitter (C-E) Amplifier

• Construct AC equivalent circuit assuming Q-point is known.

Input is applied to Base

Output appears at Collector

Emitter is common (through RE) to both input and output signal - Common-Emitter (CE) Amplifier.

Developing a small signal model:




• Use the small signal hybrid model for BJT




• Use the small signal hybrid model for BJT3

|| ||L o C

R r R R• Simplify:




3|| ||

L o CR r R R • Use the small signal hybrid model for BJT• Simplify:


C-E Amplifier – Voltage Gain

• Goal – develop the overall voltage gain from vi to vo - Av

CE .




CE .

•

where - terminal gain -

gain between transistor terminals.

v v v vCE CEo o b bA Av vtv v v v

i b i i

vCE oAvt v

b




CE .

•



• To find AvtCE , we connect a test source vb

to the base terminal (right circuit):


i b i i

vCE oAvt v

b

or CEv g R v A g Ro m L b vt m L




CE .

•






i b i i

vCE oAvt v

b

• For the input resistance RiB :


b

b

vR r

iB i




CE .

•






i b i i

vCE oAvt v

b



b

b

vR r

iB i

|| Now, since ,

||

we have

||

||CE CE b Bv vt m L

i

B ibb i

I B ib

I B

v R rA A g

R Rv v

Rv R

R R R

R r




CE .

•






i b i i

vCE oAvt v

b


• We see that the overall voltage gain is the terminal gain reduced by the voltage division between R1 and equivalent resistance at the base.

• Thus AvtCE is the upper limit of voltage

gain.or CEv g R v A g R

o m L b vt m L

b

b

vR r

iB i

|| Now, since ,

||

we have

||

||CE CE b Bv vt m L

i

B ibb i

I B ib

I B

v R rA A g

R Rv v

Rv R

R R R

R r


C-E Amplifier - Simplifications and Limits

• If we obtain the max gain: I B

R R r

AvCE g

mR

L g

mro R

CR

3

||

||CE CE b Bv vt m L

i I B

v R rA A g R

v R R r



• If we obtain the max gain:

• For maximum gain we set R3 >> RC and RC << rO . . If we assume IC RC = VCC with 0 < < 1:

I B

R R r

AvCE g

mR

L g

mro R

CR

3

40Cv vt

CCm

TC C

Ig RA A RV

V

cRRRor C

3

||

||CE CE b Bv vt m L

i I B

v R rA A g R

v R R r




• For maximum gain we set R3 >> RC and RC << ro . . If we assume IC RC = VCC with 0 < < 1:

Typically, = 1/3, since common design allocates one-third power supply across RC. To further account for other approximations leading to this result, we use:

I B

R R r

AvCE g

mR

L g

mro R

CR

3

40Cv vt

CCm

TC C

Ig RA A RV

V

10v CC

A V

cRRRor C

3




• For maximum gain we set R3 >> RC and RC << ro . . If we assume IC RC = VCC with 0 < < 1:

Typically, = 1/3, since common design allocates one-third power supply across RC. To further account for other approximations leading to this result, we use:

• Also, if the load resistor approaches ro (RC and R3 are very large), voltage gain is limited by amplification factor, mf , of BJT itself.

• For large RE, voltage gain can be approximated as:

I B

R R r

AvCE g

mR

L g

mro R

CR

3

40Cv vt

CCm

TC C

Ig RA A RV

V

10v CC

A V

AvtCE

gmRL

1gmRE

R

LR

E

cRRRor C

3


C-E Amplifier – Example

• Problem: Find voltage gain, input and output resistances.

• Given data: bF = 65, VA = 50 V• Assumptions: Active-region

operation, VBE = 0.7 V, small signal operating conditions.




• Given data: bF = 65, VA = 50 V• Assumptions: Active-region

operation, VBE = 0.7 V, small signal operating conditions.

We start from finding the Q-point. DC equivalent circuit:

The KVL for the input loop :

The KVL for the output loop (blue):

5 410 ( 1) (1.6 10 ) 5 0B BE F B

I V I

3.71

65 241

66 245

B

C B

E B

I A

I I A

I I A

4 45 10 (1.6 10 ) 5 0

3.67

C CE E

CE

I V I

V V



Next we construct the ac equivalent and simplify it.

gm 40I

C 9.64 10 3 S

r

oV

T

IC

6.64 k

ro

VA V

CE

IC

223 k

Rin R

Br 6.23 k

RoutR

Cr

o9.57 k

Av

vo

vi

gm(R

outR

3)

Rin

RIR

in

84.0


Common-Source Amplifier

• Construct AC equivalent circuit.• Assume that Q-point is known.




• Use the small signal hybrid model for BJT




• Use the small signal hybrid model for MOSFET3

|| ||L o C





• Use the small signal model3

|| ||L o C



C-S Amplifier - Terminal Voltage Gain

Terminal voltage gain between gate and drain:

Overall voltage gain from source vi to output voltage across R3

Avtv

dv

gv

ov

gs g

mR

L

gs gso ov vt

i gs i i

Gv m L

I G

v vv vA Av v v v

RA g R

R R

Voltage division


C-S Amplifier - Simplifications

• If we assume RI << RG

• Generally R3 >> RD and RD << ro. Hence, total load resistance on drain is RD. For this case, common design allocates half the power supply for voltage drop across RD and (VGS - VTN ) = 1V

• Also, if load resistor approaches ro, (RD and R3 are very large), voltage gain is limited by amplification factor, mf of the MOSFET itself.

AvA

vt g

mR

L g

mr

oR

DR

3

This implies that total signal voltage at input appears across gate-source terminals.

Av g

mR

D

IDR

D

VGS V

TN

2

VDD


C-S Amplifier - Input and Output Resistance

• Input resistance of C-S amplifier is much larger than that of corresponding C-E amplifier.

vxi

xR

G

RinR

G

• For comparable bias points, output resistances of C-S and C-E amplifiers are similar.

for xoout D D o D

x

vR R r R r Ri

In this case, vgs= 0.


C-S Amplifier - Example


• Given data: Kn = 500 mA/V2, VTN = 1V, l = 0.0167 V-1

Do example on the board




• Given data: Kn = 500 mA/V2, VTN = 1V, l = 0.0167 V-1

Construct dc equivalent circuit

1 6,

Since 0,

5 10

G

DS

IVI

41

2

10 2 10 ( ) 0

(0.4 )2

5 , 2 ,

250

DS D

nD DS TN

DS GS

D

V I IK

I V V

V V V V

I A

KVL:


Next we construct the ac equivalent and simplify it.

RinR

G1R

G21 M

3( ) 7.93ino

v m outI ini

RvA g R RR Rv

gm 2K

nI

DS(1V

DS)5.20104 S

ro

1VDS

ID

260 k


3 18.2 o Gout DR r R R k

GIR R


Summary of CE / CS Amplifiers

NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 1 Topic 6 Analog Circuits Ideal Operational Amplifiers One...

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Transcript of NJIT ECE 271 Dr. Serhiy Levkov Topic 6 - 1 Topic 6 Analog Circuits Ideal Operational Amplifiers One...