OPERATIONAL AMPLIFIERS

BASIC OP-AMP

Symbol and Terminals

• A standard operational amplifier (op-amp) has;• Vout is the output voltage, • V+ is the non-inverting input voltage,• V- is the inverting input voltage.

• Typical op-amp operates with 2 dc supply voltages, • +ve supply.• –ve supply.

Figure 1a: Symbol Figure 1b: Symbol with dc supply connections

An op amp is an active

circuit element designed

to perform mathematical

Operations of addition,

subtraction,

multiplication, division,

differentiation, and

integration.741 general purpose op-amp made by Fairchild Semiconductor

The Ideal Op-Amp

• The ideal op-amp has;• Infinite voltage gain.• Infinite bandwidth.• Infinite input impedance• zero output impedance.

• The input voltage, Vin appears between the two input terminal.

• The output voltage is AvVin as indicated by the internal voltage source symbol.

_

+

AvVin

Zout=0

Zin=∞Vin

Av=∞

Figure 2a: Ideal op-amp representation

The Practical Op-Amp• Characteristic of a practical op-amp are;

• Very high voltage gain.• Very high input impedance.• Very low output impedance.• Wide bandwidth.

_

+

AvVin

Zout

Zin

Vin

Figure 2b: Practical op-amp representation

OP-AMP INPUT MODES AND PARAMETERS

Input Signal Modes

A) Single-Ended Input• Operation mode;

• One input is grounded.• The signal voltage is applied only to the other input.

• When the signal voltage is applied to the inverting input, • an inverted amplified signal voltage appears at the

output. (figure 3a)

.

V out

.V in

_

+

Figure 3a

• When the signal voltage is applied to the noninverting input with the inverting input grounded,

• a noninverted amplified signal voltage appears at the output. (figure 3b)

.

V out

V in

.

_

+

Figure 3b

B) Differential Input• Operation mode;

• Two opposite-polarity (out-of-phase) signals are applied to the inputs

• This type of operation is also referred to as double-ended.• The amplified difference between the two inputs appears

on the output.

.

V out

V in1

V in2

.

_

+

Figure 3c

C) Common-Mode Input• Operation mode

• Two signal voltages of the same phase, frequency and amplitude are applied to the two inputs. (figure 3d)

• When equal input signals are applied to both inputs, they cancel, resulting in a zero output voltage.

• This action is called common-mode rejection.• Means that this unwanted signal will not appear on the

output and distort the desired signal.

.

V out

V in

V in

.

_

+

Figure 3d

Common-Mode Rejection Ratio

• Desired signals can appear on only • one input or • with opposite polarities on both input lines.

• These desired signals are • amplified and appear on the output.

• Unwanted signals (noise) appearing with the same polarity on both input lines are • essentially cancelled by the op-amp and do not appear

on the output.• The measure of an amplifier’s ability to reject common-

mode signal is called • CMRR (common-mode rejection ration).

• Ideally, op-amp provides • a very high gain for desired signal (single-ended or

differential) • zero gain for common-mode signal.

• The higher the open-loop gain with respect to the common-mode gain, • the better the performance of the op-amp in terms of

rejection of common-mode signals.• Therefore;

where Aol = open-loop voltage gain

Acm = common-mode gain• The higher the CMRR, the better.• A very high value of CMRR means that

• the open-loop gain, Aol is high and • the common-mode gain, Acm is low.

• The CMRR expressed in decibels (dB) is

ol

cm

ACMRR

A

20log ol

cm

ACMRR

A

Open-Loop Voltage Gain

• Open-loop voltage gain, Aol of an op-amp • is the internal voltage gain of the device• represents the ration of output voltage to input

voltage when there are no external components.

• The open-loop voltage gain is set entirely by the internal design.

• Open-loop voltage gain can range up to • 200,000 and is not a well-controlled parameter.

• Data sheet often refer to the open-loop voltage gain as • the large-signal voltage gain.

Example 1

A certain op-amp has an open-loop voltage gain of 100,000 and a common-mode gain of 0.2.

Determine the CMRR and express it in decibels.

Answer: a) 500,000 b) 114dB

Common-Mode Input Voltage Range

• All op-amp have limitation on the range of voltages over which they will operate.

• The common-mode input voltage range is

• the range of input voltages which when applied to both inputs will cause clipping or other output distortion.

• Many op-amp have common-mode input ranges of

• ±10V with dc supply voltages of ±15V.

Input Bias Current

• The input bias current is

• the dc current required by the inputs of the amplifier to properly operate the first stage.

• By definition, the input bias current is

• the average of both input currents and is calculated as;

1 2

2BIAS

I II

Figure 4a: Input bias current is the average of the two op-amp input currents.

V2

V1 _

+

I2 Vout

I1

Input Impedance• Two basic ways of specifying the input impedance of an

op-amp are• Differential.• Common-mode.

• Differential input impedance is • the total resistance between the inverting and the

noninverting input. • Measured by determining the change in bias current

for a given change in differential input voltage.

+

.

_

ZIN(d)

Figure 4b: Differential input impedance

• Common-mode input impedance is

• the resistance between each input and ground.

• Measured by determining the change in bias current for a given change in common-mode input voltage.

+

.

_

ZIN(cm)

Figure 4c: Common-mode impedance

Output Impedance

• The output impedance is

• the resistance viewed from the output terminal of the op-amp as indicated in figure 4d

.

+

_

Zout

Figure 4d: Op-amp output impedance

Slew Rate• What is slew rate?

• The maximum rate of change of the output voltage in response to a step input voltage.

• Is dependent upon the high-frequency response of the amplifier stages within the op-amp.

• Is measured with an op-amp connected as shown in figure 4e

R

+

V out

V in

_

Figure 4e: Test circuit

• A pulse is applied to the input, the output voltage is measured as indicated in figure 4f.

• The width of the input pulse must be sufficient • to allow the output to slew

from its lower limit to its upper limit.

• A certain time interval ∆t, is required for the output voltage • to go from its lower limit

–Vmax to its upper limit +Vmax, once the input step is applied.

Figure 4f: Step input voltage and the resulting output voltage

-Vmax

∆t

Vout

+Vmax

Vin

0

• The slew rate is expressed as

Where ∆Vout = +Vmax-(-Vmax).

• The unit is volts per microsecond (V/μs).

outV

t

Example 2The output voltage of a certain op-amp appears as shown in figure below in response to a step input.

Determine the slew rate.

t

2μs

12μs

-9

-10

9

10

0

Vout(V)

Answer: 1.8 V/us

OP-AMPS WITH NEGATIVE FEEDBACK

Closed-Loop Voltage Gain, Acl

• The closed-loop voltage gain is • the voltage gain of an op-amp with external

feedback.• The amplifier configuration consists of

• the op-amp • an external negative feedback circuit that

connects the output to the inverting input.• The closed-loop voltage gain is determined by

• the external component values and can be precisely controlled by them.

Noninverting Amplifier

• Noninverting amplifier is • an op-amp connected in a closed-loop with a controlled

amount of voltage gain is shown in figure 5.• The input signal is applied to

• the noninverting (+) input.• The output is applied back to

• the inverting (-) input through the feedback circuit (closed loop) formed by the input resistor Ri and the feedback resistor Rf.

V in

V f

.

Ri

_Rf

+

V out

Figure 5: Noninverting amplifier

Feedback network

• This creates negative feedback as follows.• Resistor Ri and Rf form a voltage divider circuit, which

reduces Vout and connects the reduced voltage Vf to the inverting input.

• The feedback voltage is expressed as

• The closed-loop gain of the noninverting (NI) amplifier is

• Where

• Therefore;

if out

i f

RV V

R R

( )

1 i foutcl NI

in i

R RVA

V B R

( ) 1 fcl NI

i

RA

R

i

i f

RB

R R

Example 3

Determine the gain of the amplifier in figure below. The open-loop voltage gain of the op-amp is 100,000.

Ri

_

V in

Rf

+

V out

4.7kΩ

100kΩ

Answer: 22.3

Voltage-Follower

• The voltage-follower configuration is a special case of the noninverting amplifier • where all the output voltage is fed back to the inverting

(-) input by a straight connection. (figure 6)• The straight feedback connection has a voltage gain of 1

(no gain).• The closed-loop voltage gain of a noninverting amplifier is

1/B.

Vout

Vin

_

+

Figure 6: Op-amp voltage-follower

• Since B=1, for a voltage-follower, • the closed-loop voltage gain of the voltage follower is

Acl(VF)=1

• The most important features of the voltage-follower configuration are • very high input impedance• very low output impedance.

• These features make it a nearly ideal buffer amplifier for the • interfacing high-impedance sources• low-impedance loads.

Inverting Amplifier

• Inverting amplifier

• An op-amp connected with a controlled amount of voltage gain. (figure 7)

• The input signal is applied through a series input resistor Ri to the inverting (-) input.

• The output is fed back through Rf to the same input.

• The noninverting (+) input is grounded.

AolVin

Ri

Rf

Vout

+

_

Figure 7: Inverting Amplifier

• For inverting amplifier

• The closed-loop voltage gain is the ratio of the feedback resistance (Rf) to the input resistance (Ri).

• This gain is independent of the op-amp’s internal open-loop gain.

• Thus, the negative feedback stabilizes the voltage gain.• The negative sign indicates inversion. Therefore;

fout

in i

RV

V R

( )f

cl Ii

RA

R

( )f

cl Ii

RA

R

Vin

Ri

Rf

Vout

+

_

Example 4Given the op-amp configuration in figure below, determine the value of Rf required to produce a closed-loop voltage gain of -100.

Aol

2.2kΩ

Answer: 220 kΩ