ee101_opamp_1

5/27/2018 ee101_opamp_1

1/85

EE101: Op Amp circuits (Part 1)

M. B. [email protected]

Department of Electrical EngineeringIndian Institute of Technology Bombay

M. B. Patil, IIT Bombay
http://find/

5/27/2018 ee101_opamp_1

2/85

Op Amps: introduction

* The Operational Amplifier (Op Amp) is a versatile building block that can beused for realizing several electronic circuits.

http://find/http://goback/

5/27/2018 ee101_opamp_1

3/85



* The use of Op Amps frees the user from cumbersome details such as transistorbiasing and coupling capacitors.

http://find/

5/27/2018 ee101_opamp_1

4/85




* The characteristics of an Op Amp are nearly ideal Op Amp circuits can be

expected to perform as per theoretical design in most cases.

http://find/

5/27/2018 ee101_opamp_1

5/85




* The characteristics of an Op Amp are nearly ideal Op Amp circuits can be

expected to perform as per theoretical design in most cases.* Amplifiers built with Op Amps work with DC input voltages as well useful in

sensor applications (e.g., temperature, pressure)

http://find/

5/27/2018 ee101_opamp_1

6/85




* The characteristics of an Op Amp are nearly ideal Op Amp circuits can beexpected to perform as per theoretical design in most cases.

* Amplifiers built with Op Amps work with DC input voltages as well useful insensor applications (e.g., temperature, pressure)

* The user can generally carry out circuit design without a thorough knowledge ofthe intricate details (next slide) of an Op Amp. This makes the design processsimple.

http://find/

5/27/2018 ee101_opamp_1

7/85




* The characteristics of an Op Amp are nearly ideal Op Amp circuits can beexpected to perform as per theoretical design in most cases.

* Amplifiers built with Op Amps work with DC input voltages as well useful insensor applications (e.g., temperature, pressure)

* The user can generally carry out circuit design without a thorough knowledge ofthe intricate details (next slide) of an Op Amp. This makes the design processsimple.

* However, as Einstein has said, we should make everything as simple as possible,but not simpler. need to know where the ideal world ends, and the real onebegins.

http://find/

5/27/2018 ee101_opamp_1

8/85

Op Amp 741

OUT

offset adjust

Q2Q1

Q3 Q4

Q5 Q6

Q7

Q8

Q9

Q10

Q11

Q12

Q13 Q14

Q15

Q16

Q17

Q18

Q19

Q20

Q21

Q23

Q22

Q24R1 R2R3

R4

R6

R7

R8R9

R10

R5

OUT

CC

SymbolActual circuit

V

CC

VEE

VEE

VCC

+

http://find/

5/27/2018 ee101_opamp_1

9/85

Op Amp: equivalent circuit

OUT

Vo VoVi ViAV Vi AV Vi

Ro

Ri

VEE

VCC


5/27/2018 ee101_opamp_1

10/85


OUT


Ro

Ri

VEE

VCC

* The external resistances ( a few k) are generally much larger than Ro andmuch smaller than Ri we can assume Ri , Ro 0 without significantlyaffecting the analysis.

http://find/

5/27/2018 ee101_opamp_1

11/85


OUT


Ro

Ri

VEE

VCC


* VCC andVEE ( 5 V to 15 V) must be supplied; an Op Amp will not workwithout them!

http://find/

5/27/2018 ee101_opamp_1

12/85


OUT


Ro

Ri

VEE

VCC



In Op Amp circuits, the supply voltages are often not shown explicitly.

http://find/

5/27/2018 ee101_opamp_1

13/85


OUT


Ro

Ri

VEE

VCC



In Op Amp circuits, the supply voltages are often not shown explicitly.

*Parameter Ideal Op Amp 741

AV 105 (100 dB)

Ri 2 M

Ro 0 75


5/27/2018 ee101_opamp_1

14/85


OUT

10

10

5

0

5

10

10

5

0

5

linearsaturation saturation

saturation

linear

saturation

05 50.2 0.1 0 0.1 0.2

Vo VoVi ViAVVi AVVi

Ro

RiVEE

VCC

Vo

(V)

Vi(V)Vi(mV)

Vo

(V)

Vsat

Vsat

slope= AV


O A i l i i
http://find/

5/27/2018 ee101_opamp_1

15/85


OUT

10

10

5

0

5

10

10

5

0

5


saturation

linear

saturation

05 50.2 0.1 0 0.1 0.2

Vo VoVi ViAVVi AVVi

Ro

RiVEE

VCC

Vo

(V)

Vi(V)Vi(mV)

Vo

(V)

Vsat

Vsat

slope= AV

* The output voltage Vo is limited to Vsat, where Vsat 1.5 V less than VCC.


O A i l i i
http://find/

5/27/2018 ee101_opamp_1

16/85


OUT

10

10

5

0

5

10

10

5

0

5


saturation

linear

saturation

05 50.2 0.1 0 0.1 0.2

Vo VoVi ViAVVi AVVi

Ro

RiVEE

VCC

Vo

(V)

Vi(V)Vi(mV)

Vo

(V)

Vsat

Vsat

slope= AV

* The output voltage Vo is limited to Vsat, where Vsat 1.5 V less than VCC.

* For Vsat

5/27/2018 ee101_opamp_1

17/85

Op Amp circuits

OUT

10

10

5

0

5 saturation

linear

saturation

05 5

VoViAVVi

Ro

Ri

VEE

VCC

Vsat

Vsat

Vo

(V)

Vi(V)


Op Amp circ its
http://find/

5/27/2018 ee101_opamp_1

18/85

Op Amp circuits

OUT

10

10

5

0

5 saturation

linear

saturation

05 5

VoViAVVi

Ro

Ri

VEE

VCC

Vsat

Vsat

Vo

(V)

Vi(V)

* Broadly, Op Amp circuits can be divided into two categories:


Op Amp circuits
http://find/

5/27/2018 ee101_opamp_1

19/85

Op Amp circuits

OUT

10

10

5

0

5 saturation

linear

saturation

05 5

VoViAVVi

Ro

Ri

VEE

VCC

Vsat

Vsat

Vo

(V)

Vi(V)


- Op Amp operating in the linear region


Op Amp circuits
http://find/

5/27/2018 ee101_opamp_1

20/85

Op Amp circuits

OUT

10

10

5

0

5 saturation

linear

saturation

05 5

VoViAVVi

Ro

Ri

VEE

VCC

Vsat

Vsat

Vo

(V)

Vi(V)



- Op Amp operating in the saturation region


Op Amp circuits
http://find/

5/27/2018 ee101_opamp_1

21/85

Op Amp circuits

OUT

10

10

5

0

5 saturation

linear

saturation

05 5

VoViAVVi

Ro

Ri

VEE

VCC

Vsat

Vsat

Vo

(V)

Vi(V)




* Whether an Op Amp in a given circuit will operate in linear or saturation region

depends on


Op Amp circuits

5/27/2018 ee101_opamp_1

22/85

Op Amp circuits

OUT

10

10

5

0

5 saturation

linear

saturation

05 5

VoViAVVi

Ro

Ri

VEE

VCC

Vsat

Vsat

Vo

(V)

Vi(V)





depends on

- input voltage magnitude


Op Amp circuits
http://find/

5/27/2018 ee101_opamp_1

23/85

Op Amp circuits

OUT

10

10

5

0

5

saturation

linear

saturation

05 5

VoViAVVi

Ro

Ri

VEE

VCC

Vsat

Vsat

Vo

(V)

Vi(V)





depends on

- input voltage magnitude

- type of feedback (negative or positive)

(We will take a qualitative look at feedback later.)


Op Amp circuits (linear region)

5/27/2018 ee101_opamp_1

24/85


OUT

10

10

5

0

5

saturation

linear

saturation

05 5

VoVi

iin

AVVi

Ro

VEE

VCC

Ri

Vsat

Vsat

Vo

(V)

Vi(V)


http://find/

5/27/2018 ee101_opamp_1

25/85


OUT

10

10

5

0

5

saturation

linear

saturation

05 5

VoVi

iin

AVVi

Ro

VEE

VCC

Ri

Vsat

Vsat

Vo

(V)

Vi(V)

In the linear region,

* V+ V =Vo/AV, which is very small

V+ V


http://find/

5/27/2018 ee101_opamp_1

26/85

p p ( g )

OUT

10

10

5

0

5

saturation

linear

saturation

05 5

VoVi

iin

AVVi

Ro

VEE

VCC

Ri

V

sat

Vsat

Vo

(V)

Vi(V)



V+ V

* Since Ri is typically much larger than other resistances in the circuit,we can assume Ri .

iin 0



5/27/2018 ee101_opamp_1

27/85

p p ( g )

OUT

10

10

5

0

5

saturation

linear

saturation

05 5

VoVi

iin

AVVi

Ro

VEE

VCC

Ri

Vsat

Vsat

Vo

(V)

Vi(V)



V+ V

* Since Ri is typically much larger than other resistances in the circuit,we can assume Ri .

iin 0

These two golden rules enable us to understand several Op Amp circuits.


http://find/

5/27/2018 ee101_opamp_1

28/85

( g )

ii

RL

R2

R1Vi

Vo

http://find/

5/27/2018 ee101_opamp_1

29/85

ii

RL

R2

R1Vi

Vo

i1

Since V+ V

, V

0 V i1 = (Vi 0)/R=Vi/R.(The non-inverting input is at realground here, and the inverting input is atvirtualground.)


5/27/2018 ee101_opamp_1

30/85

ii

RL

R2

R1Vi

Vo

i1

Since V+ V

, V


Since ii(current entering the Op Amp) is zero, i1 goes through R2.

http://find/

5/27/2018 ee101_opamp_1

31/85

ii

RL

R2

R1Vi

Vo

i1

Since V+ V

, V



http://find/

5/27/2018 ee101_opamp_1

32/85

ii

RL

R2

R1Vi

Vo

i1

SinceV

+ V

,V

0V

i1 = (

Vi 0)/

R=

Vi/

R.

(The non-inverting input is at realground here, and the inverting input is atvirtualground.)


Vo= V i1R2 = 0

Vi

R1

R2 =

R2

R1

Vi.

http://find/

5/27/2018 ee101_opamp_1

33/85

ii

RL

R2

R1Vi

Vo

i1

SinceV

+ V

,V

0V

i1 = (

Vi 0)/

R=

Vi/

R.



Vo= V i1R2 = 0

Vi

R1

R2 =

R2

R1

Vi.

The circuit is called an inverting amplifier.

http://find/

5/27/2018 ee101_opamp_1

34/85

ii

RL

R2

R1Vi

Vo

i1

Since V+ V, V 0 V i

1= (V

i 0)/R=V

i/R.



Vo= V i1R2 = 0

Vi

R1

R2 =

R2

R1

Vi.

The circuit is called an inverting amplifier.Where does the current go?

http://find/

5/27/2018 ee101_opamp_1

35/85

ii

RL

R2

R1Vi

Vo

i1

ii

RL

R2

R1

i1

Vi

Vo

Since V+ V, V 0 V i

1= (V

i 0)/R=V

i/R.



Vo= V i1R2 = 0

Vi

R1

R2 =

R2

R1

Vi.

The circuit is called an inverting amplifier.Where does the current go?


Op Amp circuits: inverting amplifier
http://find/

5/27/2018 ee101_opamp_1

36/85

0

5

5

1 k

10 k

0 0.5 1 1.5 2

t (msec)

RL

R2

R1

Vm = 0.5V

f= 1kHz

Vi

Vo Vi

Vo

Vi,

Vo

(Volts)


http://find/

5/27/2018 ee101_opamp_1

37/85

0

5

5

1 k

10 k

0 0.5 1 1.5 2

t (msec)

RL

R2

R1

Vm = 0.5V

f= 1kHz

Vi

Vo Vi

Vo

Vi,

Vo

(Volts)

* The gain of the inverting amplifier is R2/R1. It is called the closed-loop gain(to distinguish it from the open-loop gain of the Op Amp which is 105).


http://find/

5/27/2018 ee101_opamp_1

38/85

0

5

5

1 k

10 k

0 0.5 1 1.5 2

t (msec)

RL

R2

R1

Vm = 0.5V

f= 1kHz

Vi

Vo Vi

Vo

Vi,

Vo

(Volts)


* The gain can be adjusted simply by changing R1 orR2!



5/27/2018 ee101_opamp_1

39/85

0

5

5

1 k

10 k

0 0.5 1 1.5 2t (msec)

RL

R2

R1

Vm = 0.5V

f= 1kHz

Vi

Vo Vi

Vo

Vi,

Vo

(Volts)



* For the common-emitter amplifier, on the other hand, the gain gm(RC RL)depends on how the BJT is biased (since gm depends on IC).


http://find/

5/27/2018 ee101_opamp_1

40/85

0

5

5

1 k

10 k

0 0.5 1 1.5 2t (msec)

RL

R2

R1

Vm = 0.5V

f= 1kHz

Vi

Vo Vi

Vo

Vi,

Vo

(Volts)



* For the common-emitter amplifier, on the other hand, the gain gm(RC RL)depends on how the BJT is biased (since gm depends on IC).

(SEQUEL file: ee101 inv amp 1.sqproj)



5/27/2018 ee101_opamp_1

41/85

15

0

15

1 k

10 k

0 0.5 1 1.5 2

t (msec)

R

L

R2

R1Vi

Vm = 2V

f= 1kHz

Vo

Vo

ViVi,

Vo

(Volts)


http://find/

5/27/2018 ee101_opamp_1

42/85

15

0

15

1 k

10 k

0 0.5 1 1.5 2

t (msec)

R

L

R2

R1Vi

Vm = 2V

f= 1kHz

Vo

Vo

ViVi,

Vo

(Volts)

* The output voltage is limited to Vsat.


http://find/

5/27/2018 ee101_opamp_1

43/85

15

0

15

1 k

10 k

0 0.5 1 1.5 2

t (msec)

R

L

R2

R1Vi

Vm = 2V

f= 1kHz

Vo

Vo

ViVi,

Vo

(Volts)

* The output voltage is limited to Vsat.

* Vsat

is 1.5 V less than the supply voltageVCC

.


http://find/

5/27/2018 ee101_opamp_1

44/85

10 k

1 k

0 40 806020

0

10

10

RL

R2

R1Vi

Vm = 1 V

f= 25kHz

Vo

t (sec)

Vi

Vo

Vo(expected)

Vi,

Vo

(Volts)


http://find/

5/27/2018 ee101_opamp_1

45/85

10 k

1 k

0 40 806020

0

10

10

RL

R2

R1Vi

Vm = 1 V

f= 25kHz

Vo

t (sec)

Vi

Vo

Vo(expected)

Vi,

Vo

(Volts)

* If the signal frequency is too high, a practical Op Amp cannot keep up with theinput due to its slew rate limitation.



5/27/2018 ee101_opamp_1

46/85

10 k

1 k

0 40 806020

0

10

10

RL

R2

R1Vi

Vm = 1 V

f= 25kHz

Vo

t (sec)

Vi

Vo

Vo(expected)

Vi,

Vo

(Volts)


* The slew rate of an Op Amp is the maximum rate at which the Op Amp outputcan rise (or fall).



5/27/2018 ee101_opamp_1

47/85

10 k

1 k

0 40 806020

0

10

10

RL

R2

R1Vi

Vm = 1 V

f= 25kHz

Vo

t (sec)

Vi

Vo

Vo(expected)

Vi,

Vo

(Volts)



* For the 741, the slew rate is 0.5 V/sec.


http://find/

5/27/2018 ee101_opamp_1

48/85

10 k

1 k

0 40 806020

0

10

10

RL

R2

R1Vi

Vm = 1 V

f= 25kHz

Vo

t (sec)

Vi

Vo

Vo(expected)

Vi,

Vo

(Volts)



* For the 741, the slew rate is 0.5 V/sec.

(SEQUEL file: ee101 inv amp 2.sqproj)



5/27/2018 ee101_opamp_1

49/85

Circuit 1 Circuit 2

RL

RL

R2 R2

R1 R1Vi Vi

Vo Vo

What if the + (non-inverting) and (inverting) inputs of the Op Amp are

interchanged?


http://find/

5/27/2018 ee101_opamp_1

50/85

Circuit 1 Circuit 2

RL

RL

R2 R2

R1 R1Vi Vi

Vo Vo


interchanged?

Our previous analysis would once again give us Vo= R2

R1Vi.


http://find/

5/27/2018 ee101_opamp_1

51/85

Circuit 1 Circuit 2

RL

RL

R2 R2

R1 R1Vi Vi

Vo Vo


interchanged?


R1Vi.

However, from Circuit 1 to Circuit 2, the nature of the feedback changes fromnegative to positive.

Our assumption that the Op Amp is working in the linear region does not hold for

Circuit 2, and Vo= R2

R1Vidoes not apply any more.


http://find/

5/27/2018 ee101_opamp_1

52/85

Circuit 1 Circuit 2

RL

RL

R2 R2

R1 R1Vi Vi

Vo Vo


interchanged?


R1Vi.

However, from Circuit 1 to Circuit 2, the nature of the feedback changes fromnegative to positive.

Our assumption that the Op Amp is working in the linear region does not hold for

Circuit 2, and Vo= R2

R1Vidoes not apply any more.

(Circuit 2 is also useful, and we will discuss it later.)


http://find/

5/27/2018 ee101_opamp_1

53/85

RL

R2

R1

i1

Vi

Vo

* V+ V =Vi


http://find/

5/27/2018 ee101_opamp_1

54/85

RL

R2

R1

i1

Vi

Vo

* V+ V =Vi

i1 = (0 Vi)/R1 = Vi/R1.


http://find/

5/27/2018 ee101_opamp_1

55/85

RL

R2

R1

i1

Vi

Vo

* V+ V =Vi

i1 = (0 Vi)/R1 = Vi/R1.

* Vo =V+ i1R2 = Vi

Vi

R1

R2 =Vi

1 +

R2

R1

.



5/27/2018 ee101_opamp_1

56/85

RL

R2

R1

i1

Vi

Vo

* V+ V =Vi

i1 = (0 Vi)/R1 = Vi/R1.

* Vo =V+ i1R2 = Vi

Vi

R1

R2 =Vi

1 +

R2

R1

.

* This circuit is known as the non-inverting amplifier.


http://find/

5/27/2018 ee101_opamp_1

57/85

RL

R2

R1

i1

Vi

Vo

* V+ V =Vi

i1 = (0 Vi)/R1 = Vi/R1.

* Vo =V+ i1R2 = Vi

Vi

R1

R2 =Vi

1 +

R2

R1

.

* This circuit is known as the non-inverting amplifier.* Again, interchanging + and changes the nature of the feedback from negative

to positive, and the circuit operation becomes completely different.


Inverting or non-inverting?

5/27/2018 ee101_opamp_1

58/85

Inverting amplifier

Noninverting amplifier

RL

RL

R2

R2

R1

R1

Vs

Vs

RL

RL

Vs

Vs

R1

R2i1

R1

R2

Vo

Vo

Vi

Vi

AVVi

AVVi

Ro

Ro

Ri

Ri

Vo = R2

R1

Vs

Vo =

1+

R2

R1

Vs

* If the sign of the output voltage is not a concern, which configuration should bepreferred?


http://find/

5/27/2018 ee101_opamp_1

59/85

Inverting amplifier


RL

RL

R2

R2

R1

R1

Vs

Vs

RL

RL

Vs

Vs

R1

R2i1

R1

R2

Vo

Vo

Vi

Vi

AVVi

AVVi

Ro

Ro

Ri

Ri

Vo = R2

R1

Vs

Vo =

1+

R2

R1

Vs


* For the inverting amplifier, since V 0 V, i1 = Vs/R1 Rin =Vs/i1 =R1.


http://find/

5/27/2018 ee101_opamp_1

60/85

Inverting amplifier


RL

RL

R2

R2

R1

R1

Vs

Vs

RL

RL

Vs

Vs

R1

R2i1

R1

R2

Vo

Vo

Vi

Vi

AVVi

AVVi

Ro

Ro

Ri

Ri

Vo = R2

R1

Vs

Vo =

1+

R2

R1

Vs


* For the inverting amplifier, since V 0 V, i1 = Vs/R1 Rin =Vs/i1 =R1.

* For the non-inverting amplifier, Rin Riof the Op Amp, which is a few M.

Non-inverting amplifier is better if a large Rin is required.


Non-inverting amplifier
http://find/

5/27/2018 ee101_opamp_1

61/85

RL

RL

R2

R1

Vi

Vo

Vi

Vo

ConsiderR1 , R2 0 .


http://find/

5/27/2018 ee101_opamp_1

62/85

RL

RL

R2

R1

Vi

Vo

Vi

Vo

ConsiderR1 , R2 0 .

Vo

Vi 1 +

R2

R1 1 , i.e.,Vo=Vi.


http://goforward/http://find/http://goback/

5/27/2018 ee101_opamp_1

63/85

RL

RL

R2

R1

Vi

Vo

Vi

Vo

ConsiderR1 , R2 0 .

Vo

Vi 1 +

R2

R1 1 , i.e.,Vo=Vi.

This circuit is known as unity-gain amplifier/voltage follower/buffer.


http://find/

5/27/2018 ee101_opamp_1

64/85

RL

RL

R2

R1

Vi

Vo

Vi

Vo

ConsiderR1 , R2 0 .

Vo

Vi 1 +

R2

R1 1 , i.e.,Vo=Vi.

This circuit is known as unity-gain amplifier/voltage follower/buffer.

What has been achieved?


Loading effects
http://find/

5/27/2018 ee101_opamp_1

65/85

Vs RL

Rs

VoViAV Vi

Ro

Ri

Consider an amplifier of gain AV. We would like to have Vo=AV Vs.


Loading effects

5/27/2018 ee101_opamp_1

66/85

Vs RL

Rs

VoViAV Vi

Ro

Ri


However, the actual output voltage is,

Vo= RL

Ro+RL AV Vi = AV

RL

Ro+RL

Ri

Ri+RsVs.


Loading effects
http://find/

5/27/2018 ee101_opamp_1

67/85

Vs RL

Rs

VoViAV Vi

Ro

Ri



Vo= RL

Ro+RL AV Vi = AV

RL

Ro+RL

Ri

Ri+RsVs.

To obtain the desired Vo, we need Ri and Ro 0 .


Loading effects
http://find/

5/27/2018 ee101_opamp_1

68/85

Vs RL

Rs

VoViAV Vi

Ro

Ri



Vo= RL

Ro+RL AV Vi = AV

RL

Ro+RL

Ri

Ri+RsVs.

To obtain the desired Vo, we need Ri and Ro 0 .

The buffer (voltage follower) provides this feature (next slide).


Op Amp buffer
http://find/

5/27/2018 ee101_opamp_1

69/85

A

B

Op Amp

RL

Vs

RL

Vs

VoViAV Vi

Ro

R

RiVo

* The current drawn from the source (Vs) is small (since Riof the Op Amp islarge) the buffer has a large input resistance.


Op Amp buffer
http://find/

5/27/2018 ee101_opamp_1

70/85

A

B

Op Amp

RL

Vs

RL

Vs

VoViAV Vi

Ro

R

RiVo


* As we have seen earlier, AV is large Vi 0 V VA = VB =Vs.


Op Amp buffer
http://find/

5/27/2018 ee101_opamp_1

71/85

A

B

Op Amp

RL

Vs

RL

Vs

VoViAV Vi

Ro

R

RiVo


* As we have seen earlier, AV is large Vi 0 V VA = VB =Vs.

* The resistance seen by RL is R

Ro, which is small the buffer has a smalloutput resistance. (To findR, deactivate the input voltage source (Vs) AVVi= 0 V.)


Op Amp buffer

5/27/2018 ee101_opamp_1

72/85

amplifier

buffer

buffer

source

load

RL

Vs

Rs

ViAVVi

Ro

Ri

Vo2

i1

i2

VoVo1


Op Amp buffer

5/27/2018 ee101_opamp_1

73/85

amplifier

buffer

buffer

source

load

RL

Vs

Rs

ViAVVi

Ro

Ri

Vo2

i1

i2

VoVo1

Since the buffer has a large input resistance, i1 0 A,

and V+ (on the source side) = Vs Vo1 = Vs.


Op Amp buffer

5/27/2018 ee101_opamp_1

74/85

amplifier

buffer

buffer

source

load

RL

Vs

Rs

ViAVVi

Ro

Ri

Vo2

i1

i2

VoVo1



Similarly, i2 0 A, and Vo2 =AVVs.


Op Amp buffer
http://find/

5/27/2018 ee101_opamp_1

75/85

amplifier

buffer

buffer

source

load

RL

Vs

Rs

ViAVVi

Ro

Ri

Vo2

i1

i2

VoVo1



Similarly, i2 0 A, and Vo2 =AVVs.

Finally, Vo= Vo2 =AVVs, as desired, irresepectiveofRS and RL.


http://find/

5/27/2018 ee101_opamp_1

76/85

RL

Vo

if

iii

Vi1

Vi2

Vi3

Rf

R3 i3

R2

i2

R1 i1



5/27/2018 ee101_opamp_1

77/85

RL

Vo

if

iii

Vi1

Vi2

Vi3

Rf

R3 i3

R2

i2

R1 i1

V V+ = 0 V i1 = Vi1/R1, i1 = Vi2/R2, i1 =Vi3/R3.



V
http://find/

5/27/2018 ee101_opamp_1

78/85

RL

Vo

if

iii

Vi1

Vi2

Vi3

Rf

R3 i3

R

2

i2

R1 i1


i=i1+ i2+ i3 =

Vi1

R1+

Vi2

R2+

Vi3

R3

.



V
http://find/

5/27/2018 ee101_opamp_1

79/85

RL

Vo

if

iii

Vi1

Vi2

Vi3

Rf

R3 i3

R

2

i2

R1 i1


i=i1+ i2+ i3 =

Vi1

R1+

Vi2

R2+

Vi3

R3

.

Because of the large input resistance of the Op Amp, ii 0 if = i, which gives,



Vi3
http://find/

5/27/2018 ee101_opamp_1

80/85

RL

Vo

if

iii

Vi1

Vi2

Vi3

Rf

R3 i3

R

2

i2

R1 i1


i=i1+ i2+ i3 =

Vi1

R1+

Vi2

R2+

Vi3

R3

.


Vo=V if Rf = 0

Vi1

R1

+ Vi2

R2

+ Vi3

R3

Rf =

Rf

R1

Vi1+ Rf

R2

Vi2+ Rf

R3

Vi3

,

i.e., Vo is a weighted sum ofVi1, Vi2, Vi3.



Vi3
http://find/

5/27/2018 ee101_opamp_1

81/85

RL

Vo

if

iii

Vi1

Vi2

Vi3

Rf

R3 i3

R

2

i2

R1 i1


i=i1+ i2+ i3 =

Vi1

R1+

Vi2

R2+

Vi3

R3

.


Vo=V if Rf = 0

Vi1

R1

+ Vi2

R2

+ Vi3

R3

Rf =

Rf

R1

Vi1+ Rf

R2

Vi2+ Rf

R3

Vi3

,

i.e., Vo is a weighted sum ofVi1, Vi2, Vi3.

IfR1 = R2 = R3 =R , the circuit acts as a summer, giving

Vo= K(Vi1+ Vi2+ Vi3) with K =Rf/R.


Summer example

1.2

V
http://find/

5/27/2018 ee101_opamp_1

82/85

1

2

3

0.6

0.6

0

t (msec)

0 1 2 3 4

RL

Vo

if

iii

Vi1

Vi2

Vi3

Rf

R3 i3

R2 i2

R1 i1

R1= R2= R3= 1 k

Rf= 2 k

Vo= 2 (Vi1 + Vi2 + Vi3)

Vo

Vi2

Vi1

Vi3


Summer example

1.2

Vi1
http://find/

5/27/2018 ee101_opamp_1

83/85

1

2

3

0.6

0.6

0

t (msec)

0 1 2 3 4

RL

Vo

if

iii

Vi1

Vi2

Vi3

Rf

R3 i3

R2 i2

R1 i1

R1= R2= R3= 1 k

Rf= 2 k

Vo= 2 (Vi1 + Vi2 + Vi3)

Vo

Vi2

Vi1

Vi3

* Note that the summer also works with DC inputs. This is true about theinverting and non-inverting amplifiers as well.


Summer example

1.2

Vi1
http://find/

5/27/2018 ee101_opamp_1

84/85

1

2

3

0.6

0.6

0

t (msec)

0 1 2 3 4

RL

Vo

if

iii

Vi1

Vi2

Vi3

Rf

R3 i3

R2 i2

R1 i1

R1= R2= R3= 1 k

Rf= 2 k

Vo= 2 (Vi1 + Vi2 + Vi3)

Vo

Vi2

Vi1

Vi3


* Op Amps make life simpler! Think of adding voltages in any other way.


Summer example

1.2

Vi1
http://find/

5/27/2018 ee101_opamp_1

85/85

1

2

3

0.6

0.6

0

t (msec)

0 1 2 3 4

RL

Vo

if

iii

Vi1

Vi2

Vi3

Rf

R3 i3

R2 i2

R1 i1

R1= R2= R3= 1 k

Rf= 2 k

Vo= 2 (Vi1 + Vi2 + Vi3)

Vo

Vi2

Vi1

Vi3


* Op Amps make life simpler! Think of adding voltages in any other way.

(SEQUEL file: ee101 summer.sqproj)

http://find/

ee101_opamp_1

Documents