ee2301: basic electronic circuit amplifiers - unit 1: amplifier model1 block d: amplifiers

EE2301: Basic Electronic Circuit

Amplifiers - Unit 1: Amplifier Model 1

Block D: Amplifiers

Some Applications

EE2301: Block D Unit 1 2

Audio Amplifier

Kalaok Amplifier

Power Amplifier

WiFi Repeater

Some Applications


Servo Amplifier

Some Applications


TV and Satellite Antenna Amplifier

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Some Applications


Motion Sensor

Speed Sensor

Temperature Sensor

Location Sensor

Alcohol Sensor

Heart Beat Rate Sensor

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Amplifiers Need Everywhere



Block D Unit 1 Outline Concepts in amplifiers

> Gain

> Input resistance

> Output resistance Introduction to the Operational Amplifier

> Characteristics of the Ideal Op Amp

> Negative feedback


Gain: Inverting and Non-inverting The gain of an amplifier describes the ratio of the output amplitude to that

of the input If the output is an inverted version of the input with a larger amplitude, we

say the amplifier is INVERTING

> Gain is represented by a negative sign If the output is simply an amplified version of the input but with no shift in

phase, we say the amplifier is NON-INVERTING

Non-invertingInverting


The decibel The gain of amplifiers are commonly expressed in decibels (written as dB) The decibel is a logarithmic unit related to the power gain:

Gain in dB = 10 log10(Pout/Pin)

Gain in dB = 20 log10(Vout/Vin)

The dB is extremely useful in finding the overall gain when we cascade amplifiers (connecting the output of one stage to the input of another) like in the figure below:

> Note that the overall gain = multiplication of individual gains

> Therefore, overall gain (in dB) = sum of individual gains (in dB)

A1 A2Total gain (in dB) = 20 log10(A1A2)

= 20 log10(A1) + 20 log10(A2)


Ideal Amplifier

We will first consider the amplifier as a black box and see how it connects with the rest of the system. We can see right away that the amplifier is really two port network (REF Block A Unit 3).

Input terminals:The input terminals of the amplifier are connected to the source, modeled as a Thevenin circuit.

Output terminals:The output terminals of the amplifier are connected to the load.


Ideal Amplifier

INPUT port:

Amplifier acts as an equivalent load with respect to the source

OUTPUT port:

Amplifier acts as an equivalent source with respect to the load

Now we take a look at how the amplifier looks like on the inside

1) Input resistance Rin: This is seen across the input terminals

2) Dependent source Avin: In the circuit model above, the voltage of the source depends on the voltage drop across Rin. A is known as the open loop gain.

3) Output resistance Rout: This is seen in series with the dependent source and the positive output terminal.


Effect of finite resistances

From the source to the input terminals, there will be some voltage drop across RS due to the finite value of Rin. Hence Vin < VS since:

SinS

inin v

RR

Rv

vin is then amplified by a factor A through the dependent source Avin

From the dependent source to the output terminals, there will be some voltage drop across Rout since Rout is non-zero. Hence VL < Avin since:

inLout

LL Av

RR

Rv


Impedance Dependence

Amplification now dependent on both source and load impedances Also dependent on input & output resistance of the amplifier Therefore, different performance using different load/source for

same amplifier

SLout

L

inS

inL Av

RR

R

RR

Rv

Rin should be very large (ideally infinite), so that vin ≈ vS

Rout should be very small (ideally zero), so that vL ≈ Avin

inLout

LL Av

RR

Rv

SinS

inin v

RR

Rv


Impedance: Example 1

RS = Rin = Rout = RL = 50Ω, A = 50

Find vL, (1) without RL, (2) with RL

S

SSin

inin

V

VRR

RV

5.0

S

S

inOL

LL

V

V

AVRR

RV

5.12

5.0502

1

With RL:

S

S

inL

V

V

AVV

25

5.050

Without RL:


Impedance: Example 2RS = Rin = Rout = RL = 50Ω, A = 40

If two of the above amplifiers are cascaded, find VL


Impedance: Example 2

Given: RS = Rin = Rout = RL = 50Ω, A = 40

+-

vs

Rs

Rin Rin RL

RoutRout

+-

+-

s

s

outL

L

outin

in

Sin

insL

v200

2

140

2

140

2

1v

RR

RA

RR

RA

RR

Rvv

Av Av


Impedance: Example 3RS = Rin = Rout = RL = 50Ω, A = 20

If it is required for vL = 40vS, how many amplifier stages are needed?


Impedance: Example 3 solution

From previous example 2:

sn

s

n

s

n

outin

in

outL

LL

v105.0

v2

120

2

1

vRR

RA

RR

Rv

To achieve VL = 40VS

9.1n

40105.0 n

Need an integer number of stages, so therefore 2 stages required.



> Gain

> Input resistance



> Negative feedback

Let’s con’t in

next lecture

Much Simpler to Use



> Gain

> Input resistance



> Negative feedback


Operational Amplifier

vS+

vS-

vout

+

_

v+

v-

v+ Non-inverting input

v- Inverting input

vS+ Positive power supply

vS- Negative power supply

vout Output

The operational amplifier (or op-amp for short) behaves much like an ideal difference amplifier

It amplifies the difference between two input voltages v+ and v-

vout = AV(OL) (v+ - v-)


Op-amp Model

v+

v-

General amplifier model Op-amp model

+

vin

-

Rout

Rin +

-

iin

Avin

+

vin

-

RoutRin

AV(OL)vin

+

-

iin

+

vout

-


Ideal Op-amp Characteristics

+

vin

-

RoutRin

v+

v-

AV(OL)vin

+

-

iin

+

vout

-

There are 3 main assumptions we make for an ideal op amp:

1)Infinite input resistance (i.e. Rin → ∞)

2)Zero output resistance (i.e. Rout = 0)

3)Infinite open loop gain (i.e. AV(OL) → ∞)

Given that Rin → ∞, this means that no current flows into or out of any of the input terminals (inverting as well as non-inverting)

Given that Rout = 0, this means that vout = Avin

But what about the effect of infinite open loop gain?

Negative feedback


In this course, we will focus on the op amps being used in negative feedback. What this means is that we introduce a direct electrical path between the output and the inverting input terminals.

What this does is to take some of the output and feed it back to the input in the opposite sense. This provides stability to the circuit and is commonly used to build amplifiers and filters.

+

-

Vin

Vout

We can see that:Vout = A(Vin – Vout)

Re-arranging the terms:

inout VA

AV

1

Now if A is infinite, this will then force Vout to equal Vin.In other words, an infinite A has the effect of forcing the input terminals to be at the same voltage.V+ = V-

Before we continue,let’s go through some op-amp

applications


Amplifiers - Unit 2: The operational amplifier 26

Inside the Op Amp

Model 741

Made up of many transistors and circuits…

Circuit Example 1 - Timer


Circuit Example 2 - Ramp Generator


Circuit Example 3 – Optical Receiver


Circuit Example 5 – Active Filter


Circuit Example 6 – Dual Power Regulator


Circuit Example 7 – Radio Receiver



If you want to do these designs, you have to start from the basic first

Feel Confusion!



Block D Unit 2 Outline

Op-amp circuits with resistors only

> Inverting amplifier

> Non-inverting amplifier

> Summing amplifier

> Differential amplifier

> Instrumentation amplifier Op-amp circuits with reactive components

> Active filters (Low pass, High pass, Band pass)

> Differentiator & Integrator Physical limits of practical op-amps


Source follower

+

-

VinVout

Rs

RL

Find the gain of the above circuit

The key features of the source follower are:1)Large input resistance2)Small output resistance3)Unity gain (i.e. gain of close to one)It is therefore commonly used as a buffer between a load and source where the impedances are not well matched

How to prove it?


Inverting op amp

Note that the inverting input (node X) is at ground

R1 and R2 are in series

+

-VinVout

R2

R1

Applying KCL at X

021

R

V

R

V outin

1

2

R

R

V

V

in

out

Closed-loop gain

Negative sign indicates a 1800 shift in the phase

Gain is set by the resistor values


Non-inverting amplifier

R+

_+_ +

vout

-RF

RS

vin

+

vRS-

iin

iin

Note that since iin = 0:

1) v+ = vin (no voltage drop across R)

2) We can apply voltage divider rule to RS & RF

outFS

SRS v

RR

Rv

But we also note that A is infinite, so:

1) vin = vRS

2) Hence we then obtain:

S

F

in

out

R

R

v

vG 1


Summing Amplifier

+

-

+

vout

-+_

+_vS1

vS2

RS1

RS2

RFApply NVA at the inverting input terminal:

02

2

1

1 F

out

S

S

S

S

R

v

R

v

R

v

2

2

1

1

S

S

S

S

F

out

R

v

R

v

R

v

Hence the form of the gain relation can be described by: Vout = -(A1vS1 + A2vS2)

A1 and A2 are set by the resistor values chosen. Sn

N

n Sn

Fout v

R

Rv

1

We can extend this result to write a general expression for the gain:


Differential amplifier

+

-V1Vout

R1

V2

R2

R1

R2

One way of analyzing this circuit is to apply superposition (find Vout for V1 or V2 only)

If we short V2 first, we obtain:

+

-V1Vout1

R1 R2

11

21 v

R

Rvout

If we short V1 now, we obtain:

+

-

[R2/(R1+R2)]V2

R2R1

Vout2

21

22 v

R

Rvout

Hence, finally: Vout = (R2/R1)(v2 – v1)Output is the difference between the inputs amplified by a factor set by the resistor values.


Difference amplifier

11

22

43

4

1

21 vR

Rv

RR

R

R

Rvout

243

4

1

22 1 v

RR

R

R

Rvout

In the case whereby all the resistors are different, as shown in the circuit below, while vout1 remains unchanged, vout2 now becomes:

Hence the overall gain expression is given by:

This is similar to the form of the summing amplifier except that we take the difference between the two inputs:

Vout = A2V2 – A1V1

A2 and A1 are set by the resistor values

+

-V1Vout

R1

V2

R2

R3

R4


Diff Amp: Example 1

Find vout if R2 = 10kΩ and R1 = 250Ω

+

-V1Vout

R1

V2

R2

R1

R2


Diff Amp: Example 2

For the same circuit in the previous example, given v2 = v1:

Find vout if v1 and v2 have internal resistances of 500Ω and 250Ω respectively

+

-V1Vout

R1

V2

R2

R1

R2

500Ω

250Ω

11

22

43

4

1

21 vR

Rv

RR

R

R

Rvout

Now: R2 = 10kΩ, R1 = 750Ω, R3 = 500Ω, R4 = 10kΩ

vout = 13.65 v2 – 13.33v1

We can see that as a result of the source resistances:

1)Differential gain has changed

2)vout is not zero for v2 = v1


Recap in last lecture

EE2301: Block B Unit 2 43



Ideal Op-Amp Characteristic

> Three basic assumptions1. Infinitive input impedance

2. Zero output impedance

3. Infinitive open-loop gain V+ = V-


> Source Follower



> Summing amplifier

> Differential amplifier (Difference amplifier)

> Instrumentation amplifier


Source follower

+

-

VinVout

Rs

RL

Find the gain of the above circuit

The key features of the source follower are:1)Large input resistance2)Small output resistance3)Unity gain (i.e. gain of close to one)It is therefore commonly used as a buffer between a load and source where the impedances are not well matched


Inverting op amp

+

-VinVout

R2

R1

1

2

R

R

V

V

in

out

Negative sign indicates a 1800 shift in the phase

Gain is set by the resistor values


Non-inverting amplifier

R+

_+_ +

vout

-RF

RS

vin

+

vRS-

iin

iin

outFS

SRS v

RR

Rv

S

F

in

out

R

R

v

vG 1


Summing Amplifier

+

-

+

vout

-+_

+_vS1

vS2

RS1

RS2

RF

2

2

1

1

S

S

S

S

F

out

R

v

R

v

R

v

Sn

N

n Sn

Fout v

R

Rv

1

We can extend this result to write a general expression for the gain:


Differential amplifier

+

-V1Vout

R1

V2

R2

R1

R2

+

-V1

R1

V2

R2

R3

R4

11

22

43

4

1

21 vR

Rv

RR

R

R

Rvout

121

2 vvR

Rvout


Diff Amp: Example 2

For the same circuit in the previous example, given v2 = v1:

Find vout if v1 and v2 have internal resistances of 500Ω and 250Ω respectively

+

-V1Vout

R1

V2

R2

R1

R2

500Ω

250Ω

11

22

43

4

1

21 vR

Rv

RR

R

R

Rvout

Now: R2 = 10kΩ, R1 = 750Ω, R3 = 500Ω, R4 = 10kΩ

vout = 13.65 v2 – 13.33v1

We can see that as a result of the source resistances:

1)Differential gain has changed

2)vout is not zero for v2 = v1

121

2 vvR

Rvout


Instrumentation Amplifier

+

-

R2

RX

R3

R1

R3

vo1

vo2

The instrumentation amplifier takes care of this problem by including a non-inverting amplifier (which possesses an infinite input resistance) between the differential amplifier an the inputs.

-

+

R2

R1

+

-

V1

Vout

V2


Instrumentation Amplifier

2nd stage is a differential amplifier:

Vout = (R3/R2) (Vo2 - Vo1)

R3

R3

+

- Vout

Vo1

Vo2

+

-

RX/2

R1

V1

1st stage comprises a pair of non-inverting amplifiers:

Vo1 = (2R1/Rx +1)V1

Vo1

Closed-loop gain:

Vout = (R3/R2)(2R1/Rx + 1)(V2 - V1)


Op amp example 1Problem 8.5

Find v1 in the following 2 circuits

What is the function of the source follower?


Op amp example 1 solution

This slide is meant to be blank

Fig (a):

6Ω || 3Ω = 2Ω

v1 = {2 / (2+6)}*Vg = 0.25Vg

Fig (b):

Voltage at the non-inverting input = 0.5Vg

Voltage at output = Voltage at non-inverting input = 0.5Vg


Op amp example 2

Problem 8.7

Find the voltage v0 in the following circuit

Transform to Thevenin


Op amp example 2 solution

This slide is meant to be blank

Thevenin equivalent circuit:

Rth = 6 + 2||4 = 22/3 kΩ

Vth = {4/(2+4)}*11 = 22/3 V

Closed loop gain expression:

A = - 12kΩ / Rth (With Vth as input source)

Vout = 12/(22/3) * (22/3) = 12V


Op amp example 3

+

-

+ vout-

R2

R1

vin

RB

Find the closed-loop gain

Consider:

Current through RB = 0A (Infinite input resistance of op amp)

Voltage across RB = 0V

Voltage at inverting input = 0V

Same as inverting op amp:

A = -R2 / R1


Op amp example 4

+

-

+ vout-

R2

Find the closed-loop gain

+ vin-

Consider:

Current through R2 = 0A (Infinite input resistance of op amp)

Voltage across R2 = 0V

Vout = V-

Vout = Vin (Infinite open loop gain of op amp)

Common and differential mode


A differential amplifier should ideally amplify ONLY DIFFERENCES between the inputs. That is to say identical inputs should give an output of zero.

Ideal differential amplifier: Vout = A(V2 – V1)

In reality this is not the case as we have seen in a previous example. The output of the amplifier is more accurately described by:

Vout = A2V2 + A1V1

A2: Gain when V1 = 0 (V2 is the only input)A1: Gain when V2 = 0 (V1 is the only input)

It is then useful to describe the performance of a differential amplifier in terms of the gain when both input are identical (common mode) and when the inputs are equal and out-of-phase (differential)

We would like to express A1 and A2 by Acm and Adm

Common mode rejection ratio


Common mode gain (Acm): Gain when both inputs are exactly the same (V1 = V2 = Vin) Vocm = (A2 + A1)Vin Voltage output for common input

12 AAVVA inocmcm

Vodm = (A2 - A1)Vin Voltage output for differential input

Differential mode gain (Adm): Gain when both inputs are equal but out-of-phase (V2 = -V1 = Vin)

22 12 AAVVA inodmdm

The common mode rejection ratio (CMRR) is simply the ratio of the differential mode gain (Adm) over the common mode gain (Acm):CMRR = Adm/Acm

A large CMRR is therefore desirable for a differential amplifier

Divide by 2 since the difference of a pair of differential inputs is twice that of each input

2

1212

vvAvvAv cmdmout






> Summing amplifier

> Differential amplifier

> Instrumentation amplifier Op-amp circuits with reactive components

> Active filters (Low pass, High pass, Band pass)

> Differentiator & Integrator Physical limits of practical op-amps


Let’s con’t in this lecture

EE2301: Block B Unit 2 62

Active filters


Range of applications is greatly expanded if reactive components are used

Addition of reactive components allows us to shape the frequency response

Active filters: Op-amp provides amplification (gain) in addition to filtering effects

Substitute R with Z now for our analysis:

S

F

S

out

Z

Zj

V

V

S

F

S

out

Z

Zj

V

V1

ZF and ZS can be arbitrary (i.e. any) complex impedance

Active low pass filter


S

F

S

out

Z

Zj

V

V

FF

F

FF

F

F

FFF

RCj

R

CjR

CjR

CRZ

1

1

||

ZS = RS

FF

SF

S

out

RCj

RRj

V

V

1Shaping of frequency response

Amplification

Active high pass filter


S

F

S

out

Z

Zj

V

V

SSS

SSS

CjRCj

CjRZ

1

1

ZF = RF

SS

Fs

S

out

RCj

RCjj

V

V

1

Shaping of frequency response

Amplification

ω→∞, Vout/VS → -RF/RS

Active band pass filter


S

F

S

out

Z

Zj

V

V

SSS

SSS

CjRCj

CjRZ

1

1

FFSS

Fs

S

out

RCjRCj

RCjj

V

V

11

FF

F

FFF

RCj

R

CRZ

1

||

ZS: CS blocks low frequency inputs but lets high frequency inputs throughHigh pass filter

ZF: CF shorts RF at high frequency (reducing the gain), but otherwise looks just like a high pass filter at lower frequencies Low pass filter

Active band pass filter


FFSS

Fs

S

out

RCjRCj

RCjj

V

V

11

ωHP = 1/(CSRS) – Lower cut-off frequency ωLP = 1/(CFRF) – Upper cut-off frequency

For ωHP < ωLP: Frequency response curve is shown below

Second-order Low Pass Filter


S

F

S

out

Z

Zj

V

V

LjRZS 1

2

2

2

1

||

CRj

R

CRZF

LjRCRj

Rj

V

V

S

out

12

2

1

R1, R2, C and L are specially chosen so that: ω0 = 1/(CR2) = R1/L.

The frequency response function then simplifies to:

2

01

2

1

jR

Rj

V

V

S

out

Second-order Low Pass Filter


R1, R2, C and L are specially chosen so that: ω0 = 1/(CR2) = R1/L.

2

01

2

1

jR

Rj

V

VjH

S

outv

Above ω0, Hv is reduced by a factor of 100 for a ten fold increase in ω (40dB drop per decade)

First-order filter: Hv is reduced by a factor of 10 for a ten fold increase in ω (20dB drop per decade)

Ideal integrator


KCL at inverting input: iS = -iF

Virtual ground at inverting input: iS = VS/RS

For a capacitor: iF = CF[dVout/dt]

dt

dVC

R

V out

S

S

dttVRC

tV

RCF

V

dt

dV

SSF

out

S

Sout

)(1

)(

Output is the integral of the input

Ideal differentiator


KCL at inverting input: iS = -iF

Virtual ground at inverting input: iF = Vout/RF

For a capacitor: iS = CS[dVS/dt]

dt

dVCRV

dt

dVC

R

V

SSFout

SS

F

out

Output is the time-differential of the input

Physical limit: Voltage supply limit


The effect of limiting supply voltages is that amplifiers are capable of amplifying signals only within the range of their supply voltages.

RS = 1kΩ, RF = 10kΩ, RL = 1kΩ; VS

+ = 15V, VS- = -15V; VS(t) = 2sin(1000t)

Gain = -RF/RS = -10Vout(t) = 10*2sin(1000t) = 20sin(1000t)

But since the supply is limited to +15V and -15V, the op-amp output voltage will saturate before reaching the theoretical peak of 20V.

Physical limit: Frequency response limit


So far we have assumed in our ideal op-amp model that the open loop gain AV(OL) is infinite or at most a large constant value. In reality, AV(OL) varies with a frequency response like a low pass filter:

ω0: frequency when the response starts to drop off

The consequence of a finite bandwidth is a fixed gain-bandwidth productIf closed loop gain is increased, -3dB bandwidth is reducedIncreasing the closed loop gain further results in a bandwidth reduction till the gain-bandwidth produce equals the open-loop gainGain bandwidth product = A0ω0

ee2301: basic electronic circuit amplifiers - unit 1: amplifier model1 block d: amplifiers

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