4 Operational Amplifier 1

7/30/2019 4 Operational Amplifier 1

1/15

-1-

Operational Amplifier

An operational amplifier is a high gain, differential,

voltage amplifier. It is a voltage amplifier. The input is a voltage and

the output is a voltage.

The gain is high. Typically, the gain is over100 000

It is a differential amplifier. It actually amplifies the

difference between two voltages It has a very high input resistance It has a very low output resistance

power supply voltage


2/15

-2-

Op-amps can be used:

To amplify signalsTo isolate and buffer signals (Some signal sources

e.g. many sensors - cannot provide enough current to

"drive" circuits (loading!). Opamps provide close to

an exact copy of the signal, but the opamp can

provide more current.)

To integrate, add, subtract or filter signals (and othermathematical operations)

Characteristics of Ideal OpAmp

Rule #1: Have a gain so high that consider it to beinfinite.

the Infinite Gain Assumption(V+ - V-) 0 i.e. V+ = V-

Rule #2: Have such a high input resistance(impedance) that they draw no current at the input

terminals (inverting and non-inverting).

the Infinite Input Resistance Assumption

I+ = I- = 0 Rule #3: Have a very low output resistance, so in

most cases there are no loading effects the Zero

Output Resistance Assumption


3/15

-3-

Amplifies signals of all frequencies equallyInfiniteBandwidth Assumption

Some Basic Amplifying Circuits

1) the invertingconfiguration 2) the noninverting

configuration 3) the differentialconfiguration 4) the

summing Inverterconfiguration 5) theIntegrator

configuration 6) theDifferentiatorconfiguration

7) the Voltage Followerconfiguration


4/15

-4-

Inverting Amplifier

An Inverting simply reverses the polarity of the input

signal.


5/15

-5-

Calculating the gain of an inverting amplifier

I1 + I2 = I- = 0

Vin/R1 = -Vout/R2

Vout = -(R2/R1)*Vin

Gain = Vout/Vin= -(R2/R1)

The closed loop gain is dependent only on the two

resistances and not on the amplifier gain.

The Voltage Follower (or Unity Gain Buffer Amplifier)

V- = Vout

Since V- = V+ = Vin

Vout = Vin


6/15

-6-

This circuit is often used when a voltage source with

high output impedance is used, and want to draw more

current than the source can deliver. The Vout appears at

the output of the op amp can deliver more current

without lowering the output voltage because theinternal resistance of the operational amplifier is lower

than the internal resistance of the original source i.e. no

voltage amplification but current amplification

Summing Amplifier

2

3

2

1

1

2V

R

RV

R

RV

out


7/15

-7-

Non-Inverting Amplifier.

I1+I2 = 0

V+ = VinVin/R1+ (Vin-Vout)/R2 =0

Vout = [1+R2/R1] Vin

Gain = Vout/Vin

= 1+R2/R1

Differential Amplifier


8/15

-8-

1 2 3 4

1 2 3 4

1 2

1 2 3 4

42

3 4

4 1 22

1 3 4

Apply KCL at a and b

0 0

Rule 1 0

0 and 0

by Ohms Law

0 (1) 0 (2)

Rule 2

from (2)

substitute in (1)

a a out b b

a b

b a

out

I I I I I I

I I

I I I I

V V V V V V V

R R R R

V V

RV V V

R R

R R RV V

R R R

21

1

3 1 4 2

22 1

1

stipulate and

out

RV

R

R R R R

RV V V

R

The output voltage is proportional to the voltage

difference and the closed-loop gain is R2/R1


9/15

-9-

Voltage-to-Current Converter

1 2 3 5 5 4

1 1 2 2 1 1 1

1 2 3 5 5 4

L L

41

4 5

Apply KCL b and Design Rule 1 0

0, 0 and 0

by Ohms Law

0 (1) 0 (2) 0 (3)

Substitute IR for V in (2) and (3)

then (4)

So

in L L L

L

I I

I I I I I I I

V V V V V V V V V V VI

R R R R R R

R RV I

R R

2 1

2 4 1 22

1 1 4 5

2 4 1 3 5 2

1 3 4 5 1 3

2 4 1 3 5

2

1 3

lve (1) for V and substitute for V from (4)

(5)

substitute (4) and (5) in (1)

I 1

select

I

L

a

L in

in

R R R R RV V I

R R R R

R R R R R RR V

R R R R R R

R R R R R

RV

R R


10/15

-10-

Current-to-Voltage Converter

The Integrator

The integrator integrates - in the calculus sense - the

input signal to produce the output signal

sum currents at

summing point

dtV

RC

V

dt

dVC

R

V

inou t

ou tin

1

0


11/15

-11-

Differentiator

dt

dVRCV

R

V

dt

dVC

in

ou t

ou tin

0

By combining the elementary units discussed, we can

perform many mathematical operations, hence the term

operational amplifier. Only a rudimentary knowledge

of electronics is required to design operational amplifier

circuits.


12/15

-12-

The performance of the circuit depends only on the

circuit elements and is independent of the gain of the

operational amplifier, as long as it remains constant.

Linearisation

Place a non-linear element in the feedback loop

out

out

0

where I(V ) nonlinear variation of current with voltage

where nonlinear function of the input voltage

the inverse function of I(V

in

out

in

out

in

VI V

R

VV G

R

VG

R

)


13/15

-13-

Logarithmic Amplifier

exp

1 1log log

out o out

out e in e o

I V I V

V V I R


14/15

-14-

Bias (zero drift) Removal

Sometimes a bias exists in the output signal

y = Kx + Cwhere x is the physical quantity

y is the measurement signal

C is the bias in the output signal that needs

to be removed

Vo = (R3/R1) (VpVi)

where Vi=y=Kx+C and Vrefis set such that Vp=C

substituting Vo=-K(R3/R1)x

now for x=0 the output signal is zero (no bias)


15/15

-15-

Design Guidelines

Parameter: Nature of output. Usually voltage.

Range: Desired range of output parameter. 0 to 5 V,4 to 20 mA

Input Impedance: Presented to input signal source.

Prevent loading.

Output Impedance: Offered to output load circuit.

If input is resistance change, consider nonlinearityand current through sensor.

For opamp design develop equation for input vs.

output.

Always consider loading.

Documents

4 Operational Amplifier 1