1 of 29

Operational Amplifiers

2 of 29

Op-Amp Introduction

  • Op-amps (amplifiers/buffers in general) are drawn as a triangle in a circuit schematic
  • There are two inputs
    • inverting and non-inverting
  • And one output
  • Also power connections (note no explicit ground)

2

−

+

2

3

4

7

6

divot on pin-1 end

inverting input

non-inverting input

V+

V−

output

741

3 of 29

The ideal op-amp

  • Infinite voltage gain
    • a voltage difference at the two inputs is magnified infinitely
    • in truth, something like 200,000
    • means difference between + terminal and − terminal is amplified by 200,000!
  • Infinite input impedance
    • no current flows into inputs
    • in truth, about 1012 Ω for FET input op-amps
  • Zero output impedance
    • rock-solid independent of load
    • roughly true up to current maximum (usually 5–25 mA)
  • Infinitely fast (infinite bandwidth)
    • in truth, limited to few MHz range
    • slew rate limited to 0.5–20 V/μs

3

4 of 29

Op-amp without feedback

  • The internal op-amp formula is:

Vout = gain×(V+ − V−)

  • So if V+ is greater than V−, the output goes positive
  • If V− is greater than V+, the output goes negative

​

​

​

​

​

  • A gain of 200,000 makes this device (as illustrated here) practically useless

4

−

+

V−

V+

Vout

5 of 29

Infinite Gain in negative feedback

  • Infinite gain would be useless except in the self-regulated negative feedback regime
    • negative feedback seems bad, and positive good—but in electronics positive feedback means runaway or oscillation, and negative feedback leads to stability
  • Imagine hooking the output to the inverting terminal:
  • If the output is less than Vin, it shoots positive
  • If the output is greater than Vin, it shoots negative
    • result is that output quickly forces itself to be exactly Vin

5

−

+

Vin

negative feedback loop

6 of 29

Even under load

  • Even if we load the output (which as pictured wants to drag the output to ground)…
    • the op-amp will do everything it can within its current limitations to drive the output until the inverting input reaches Vin
    • negative feedback makes it self-correcting
    • in this case, the op-amp drives (or pulls, if Vin is negative) a current through the load until the output equals Vin
    • so what we have here is a buffer: can apply Vin to a load without burdening the source of Vin with any current!

6

−

+

Vin

Important note: op-amp output terminal

sources/sinks current at will: not like

inputs that have no current flow

7 of 29

Positive feedback pathology

  • In the configuration below, if the + input is even a smidge higher than Vin, the output goes way positive
  • This makes the + terminal even more positive than Vin, making the situation worse
  • This system will immediately “rail” at the supply voltage
    • could rail either direction, depending on initial offset

7

−

+

Vin

positive feedback: BAD

8 of 29

Op-Amp “Golden Rules”

  • When an op-amp is configured in any negative-feedback arrangement, it will obey the following two rules:

​

    • The inputs to the op-amp draw or source no current (true whether negative feedback or not)

​

    • The op-amp output will do whatever it can (within its limitations) to make the voltage difference between the two inputs zero

8

9 of 29

Inverting amplifier example

  • Applying the rules: − terminal at “virtual ground”
    • so current through R1 is If = Vin/R1
  • Current does not flow into op-amp (one of our rules)
    • so the current through R1 must go through R2
    • voltage drop across R2 is then IfR2 = Vin×(R2/R1)
  • So Vout = 0 − Vin×(R2/R1) = −Vin×(R2/R1)
  • Thus we amplify Vin by factor −R2/R1
    • negative sign earns title “inverting” amplifier
  • Current is drawn into op-amp output terminal

9

−

+

Vin

Vout

R1

Rf

3

2

6

10 of 29

OPAMP Amplifier as inverting amplifier with different Gain

  • Aim: To study the use of operational Amplifier as Inverting Amplifier
  • Apparatus: OPAMP 741, batteries, dual power supply(±15), voltmeter, wires etc.
  • Formula:

Vout = −Vin×(Rf/R1)

Av= Vout /Vin= − (Rf/R1)=-(100K/10K)=-10

  • Circuit Diagram:

R1

​

​

R2

10

−

+

Vin

Rf

11 of 29

Winter 2012

UCSD: Physics 121; 2012

11

Observation Table 1

S.R. No.

Vin(volt)

Vout(volt)

Gain=Vout/Vin

1.

0.2

-2

G=-2/0.2=-10

2.

0.3

-3

G=-3/0.3=-10

3.

0.4

-4

G=-4/0.4=-10

4.

0.5

-5

G=-5/0.5=-10

R1=R2=10K ohm

Rf=100K ohm

12 of 29

Winter 2012

UCSD: Physics 121; 2012

12

Observation Table 2

S.R. No.

Vin(volt)

Vout(volt)

Gain=Vout/Vin

1.

0.2

-1

G= -1/0.2=-5

2.

0.3

-1.5

G= -1.5 /0.3=-5

3.

0.4

-2

G= -2/0.4= -5

4.

0.5

-2.5

G= -2.5/0.5=-5

R1=R2=10K ohm

Rf=50K ohm

13 of 29

OPAMP Amplifier as non-inverting amplifier with different Gain

  • Now neg. terminal held at Vin
    • so current through R1 is If = Vin/R1 (to left, into ground)
  • This current cannot come from op-amp input
    • so comes through R2 (delivered from op-amp output)
    • voltage drop across R2 is IfR2 = Vin×(R2/R1)
    • so that output is higher than neg. input terminal by Vin×(R2/R1)
    • Vout = Vin + Vin×(R2/R1) = Vin×(1 + R2/R1)
    • thus gain is (1 + R2/R1), and is positive
  • Current is sourced from op-amp output in this example

13

−

+

Vin

Vout

R1

R2

14 of 29

Winter 2012

UCSD: Physics 121; 2012

14

  • Aim: To study the use of operational Amplifier as Non Inverting Amplifier
  • Apparatus: OPAMP 741, batteries, dual power supply(±15), voltmeter, wires etc.
  • Formula:

Vout = Vin×(1+Rf/R1)

Av= Vout /Vin= (1+Rf/R1)

  • Circuit Diagram:

R1

​

​

R2

15 of 29

15

Observation Table 1

S.R. No.

Vin(volt)

Vout(volt)

Gain=Vout/Vin

1.

0.2

2.2

G=2.2/0.2=11

2.

0.3

3.3

G=3.3/0.3=11

3.

0.4

4.4

G=4.4/0.4=11

4.

0.5

5.5

G=5.5/0.5=11

5.

0.6

6.6

G=6.6/0.6=11

R1=R2=10K ohm

Rf=100K ohm

16 of 29

16

Observation Table 2

S.R. No.

Vin(volt)

Vout(volt)

Gain=Vout/Vin

1.

0.2

0.8

G=0.8/0.2=4

2.

0.3

1.2

G=1.2/0.3=4

3.

0.4

1.6

G=1.6/0.4=4

4.

0.5

2.0

G=2.0/0.5=4

5.

0.6

2.4

G=2.4/0.6=4

R1=R2=10K ohm

Rf=30K ohm

17 of 29

Voltage Follower

17

  • Aim: To study the use of operational Amplifier as voltage follower
  • Apparatus: OPAMP 741, batteries, dual power supply(±15), voltmeter, wires etc.
  • Formula:

Vout = Vin

Av= Vout /Vin= 1

  • Circuit Diagram:

​

​

18 of 29

18

Observation Table 1

S.R. No.

Vin(volt)

Vout(volt)

Gain=Vout/Vin

1.

0.2

0.2

G=0.2/0.2=1

2.

0.3

0.3

G=0.3/0.3=1

3.

0.4

0.4

G=0.4/0.4=1

4.

0.5

0.5

G=0.5/0.5=1

5.

0.6

0.6

G=0.6/0.6=1

19 of 29

Summing Amplifier

  • Much like the inverting amplifier, but with two input voltages
    • inverting input still held at virtual ground
    • I1 and I2 are added together to run through Rf
    • so we get the (inverted) sum: Vout = −Rf×(V1/R1 + V2/R2)
      • if R2 = R1, we get a sum proportional to (V1 + V2)
  • Can have any number of summing inputs
    • we’ll make our D/A converter this way

19

−

+

V1

Vout

R1

Rf

V2

R2

20 of 29

20

21 of 29

21

22 of 29

22

23 of 29

23

24 of 29

24

25 of 29

25

1. Set up the integrator circuit as shown in figure. Give a rectangular wave of ±5V (10V pp) and 1 kHz frequency at the input and observe the input and output simultaneously on CRO.

​

2. Vary the dc offset of the square wave input and observe the difference in the output waveform.

​

3. Repeat the experiment by feeding triangular wave and sine wave at the input and observe the output.

Integrator

Procedure:

Differnerator

1. Set up the differentiator circuit as shown in figure. Give a rectangular wave of ±5V (10V pp) and 1 kHz frequency at the input and observe the input and output

​

2. Repeat the experiment by feeding triangular wave and sine wave at the input and observe the output.

26 of 29

26

For Procedure:

​

​

👉🏻Please Follow the link

Procedure

27 of 29

27

Sr. No.

C(micro farad)

R.C

Time

Voltage

1.

0.1

​

0.63 ms

1.2V

2.

0.01

​

1 ms

3.6V

3.

0.001

​

1.1 ms

4.5V

Vi(p-p)=1Volt Frequency=1Khz R=10K ohm Rd=100K ohm

Integrator

Sr. no.

Input Frequency

Vo(p-p)

Time

1/(Vo(p-p)/Time)

1.

500Hz

3v

1.6ms

​

2.

1000Hz

1.2v

0.63ms

​

3.

1500Hz

1.1V

0.4ms

​

Vi(p-p)=1Volt C=0.1micro farad R=10K ohm Rd=100K ohm

28 of 29

28

Differnerator

Sr. no.

Frequency

Vo(p-p)

Time

1/(Vo(p-p)/Time)

1.

50Hz

8.93v

1.6ms

​

2.

100Hz

13.2v

0.63ms

​

3.

150Hz

13V

0.4ms

​

Vi(p-p)=1Volt Frequency=1Khz R=10K ohm Rd=100K ohm

Sr. No.

C(micro farad)

Rc

Time

Voltage

1.

0.1

1000

0.63 ms

1.2V

2.

0.01

100

1 ms

3.6V

3.

0.001

10

1.1 ms

4.5V

Vi(p-p)=1Volt C=0.1micro farad R=10K ohm Rd=100K ohm

29 of 29

Differencing Amplifier

  • The non-inverting input is a simple voltage divider:
    • Vnode = V+R2/(R1 + R2)
  • So If = (V− − Vnode)/R1
    • Vout = Vnode − IfR2 = V+(1 + R2/R1)(R2/(R1 + R2)) − V−(R2/R1)
    • so Vout = (R2/R1)(V+ − V−)
    • therefore we difference V+ and V−

29

−

+

V−

Vout

R1

R2

V+

R1

R2