MH 02N 2

MH-03: Operational Amplifiers and
Circuits
EG1008 Principles of Electronics
Dr Masood Hajian
2015-16
The OpAmp
Operational
amplifiers (OpAmps) are the basic building

blocks of electronic systems,
It is made of transistors, diodes, resistors, capacitors,
It is fabricated as one integrated circuit (IC),
It is only necessary to add power, and any required external
components,
At the moment we need only consider the OpAmp as a black-
box with a defined set of inputs and outputs,

We will ignore its internal structure.
Differential OpAmp
+VS
vO
v-
v+
-VS
Op amp operated in a wide range of modes depending

on external components,
Many functions are possible using standard opamps:
Non-inverting & inverting amps, adders, subtractors,
integrators, differentiator, filters.
3
The Differential OpAmp

Basic opamp based on the differential amplifier
Two inputs, -ve and +ve
Output voltage a function of the two inputs difference,
The output voltage is vo = Av(v+ - v-)
Av is the internal (open loop) voltage gain of the
components)
-ve I/P is inverting
+ve I/P is non-inverting
Some work with single supply,
Some need positive and negative
power supply
amplifier (ie gain with no external
+VS
_
+
v-
-VS
v+
vO
OpAmp complete circuit would be fabricated onto a single slice
(chip) of silicon, integrated circuit (IC)

In practice, it may have many similar stages on a single chip,
All you see are inputs, outputs and power supply connection.
A typical real OpAmp looks like this
This is a 14 pin IC (integrated circuit) but many other
configurations are available,

The pins are numerically identified with reference to a notch or
dot on an IC chip as shown.
OpAmp Properies
Properties common to all OpAmps:
an inverting input (marked -) & a non-inverting input (+),
a high input impedance (at both inputs) & low output
impedance,
a very large voltage gain (with no feedback),
constant voltage gain over a wide frequency range,
free of temperature drift (relatively),
Real opamps
Practical opamps are often based on the design below:
But it may have many stages on a single chip of Si,
Avof 104 -106 typical in real opamps,

input impedance - of 10k up to many M (with FETs),
It has nonzero output impedance - usually quite low 10.
Has a finite gain
Feedback in an Engineering System

Many engineering systems use of the concept of feedback (closed loop
system) ,
Feedback: portion of output signal returned & fed back to input,

This controls and modifies the output often improving behavior,
A regulator controls the temperature in a central heating boiler unit,
The temp in car is controlled by the climate control unit,
Controlling the temp in a car can be done manually
If temp drops you turn up the heater - When it gets too warm you turn it
down,
Reacting manually to input change - an open-loop control system,
In a climate control unit you set desired T to a specific level

As T drops then unit automatically turns up the heater until ref temp is
reached,
If T increases, the heat input is reduced back - closed loop control,
This is an example of feedback - a portion of the output is fedback and
compared at the input
Feedback in an Electronic System
The concept of feedback is very important in amplifier
design,
Consider the op amp open loop (internal) voltage gain of Av,
Since the output is vout, input to the amp must be vout/Av,
Suppose some fraction B of the output voltage is obtained,
and added to, or subtracted from, the voltage applied to
the input,
this is feedback
vout/Av
vIN
Av
-
B
8
vOUT
Feedback in an Electronic System

The output voltage is :
vout = Av(vin Bvout)

Thus :
vout = [Av/(1 + BAv)]vin.
The new gain of the system, the closed loop gain, is :
vOUT
Av
A = ----- = ----------vIN
1 + B Av
where Av is the open-loop gain , and A is called closed loop
gain.
9
Feedback Types
Positive feedback
If either B or Av is negative - then BAv becomes negative:
(1 + BAo) < 1, then A > Av,
Gain of system increases with positive feedback (Av was
already large!),
Special case: if BAv = -1 :
Closed-loop gain becomes infinite and the system becomes
unstable oscillates
Negative feedback
If Av and B are either both positive or both negative:
then BAv is positive.
Closed-loop gain decreases with feedback A < Av ,
BAv 1, then A Av /BAv 1/B

Closed-loop gain only depends on feedback path & independent
of Av.
Special case (but usual) - if
10
Negative Feedback in an OpAmp Circuit

Very large (open loop) gain Av of opamp reduced negative FB,
also raises the input impedance and reduces the output
impedance,
Forces inverting inverting input to follow non-inverting input,

Brings the two I/Ps very close together in voltage,
If not, non-zero voltage difference produces infinite Vo,
The variation in system gain (closed-loop) is very insensitive
to the variability of the open-loop gain if AvB >>1.
11
Frequency response
Ideal amp amplifies all signals equally
independent from frequency,

In practice drops off at low and high
frequencies,
Frequencies where Ap drops to halfpeak - 3 dB points,
low & high frequency cut-off
points, fL & fH,
Bandwidth of amp : BW = fH - fL
Voltage gain
1 Hz
100 Hz
1 kHz
10 kHz
100 kHz 1 MHz
100 Hz
1 kHz
10 kHz
100 kHz 1 MHz

fH
100 Hz
1 kHz
10 kHz
100 kHz 1 MHz

fH
Voltage gain
1
0.707
For an amplifier with equal Rin & Rout
Ap(dB) = 10log(Po/Pi) =
10log[(vo2/Ro)/(vi2/Ri)] =
10log(vo2/vi2) = 20log(vo/vi) =
20log(Av)
1 Hz
fL
Power gain
1
- 3 dB
A 3 dB drop in Ap is an Av drop to
0.707 max
12
1 Hz
fL
Effect of Feedback on Frequency

Response
The closed-loop gain (A) will also fall at
high and low frequencies,
But since the open-loop gain >> closed-loop gain,
Av falls by a considerable amount before this has

an effect on A,
The closed-loop gain (A) will be stable over
a wider frequency range - bandwidth will

increase,
In some cases, the increase in bandwidth is
proportional to decrease in gain
Gain x bandwidth = constant
13
Non-inverting Amplifier
+
_
vIN
RF
Ri
14
v OUT
Non-inverting Amp.
Differential Amp. functions as a non-inverting Amp,
Input signal is applied to +ve,
Feedback network output is applied to ve input
15
(negative feedback),
Feedback gain is obtained as:
[B = Ri/(RF+Ri)]
Ideal opamp non-inverting input has infinite impedance,
Assuming ideal opamp:
The open-loop gain is infinite,
Thus the differential input to the opamp must tend
towards zero.
Gain of Non-Inverting OpAmp

So v+ - v- = 0 and v+ = v- = vin,
Because the op amp differential input is zero,
this is known as a virtual short circuit,

It follows that if the voltage is zero then input
current is zero too,
So v- = vo [Ri/(Ri+RF)] = vin
Where the feedback fraction is given by B = R i/(Ri+RF)
Thus Gain of amp is

A = 1/B = (Ri+RF)/Ri
RF
A = 1 + ----Ri
16
Worked Example
In a non-inverting amplifier, RF = 22 k and Ri
= 2 k, what is the closed-loop gain? Calculate

the feedback factor.
17
Special Case: Unity Gain Op-Amp

+
_
vIN
vOUT
Unity gain amplifier is special case of non-inverting amp,

RF is zero and Ri is infinity,
A = 1 + RF/Ri = 1 + 0/ = 1,
B=1,
Has a very high input resistance and a very low output resistance,
Often used as a buffer between amplifier stages to help match
impedances,
Also known as a voltage follower.
This is a negative feedback system with
18
Inverting Amplifier
RF
IF
Rin
iin
Ia
Va
vin
One of the most common and useful opamp
configurations,
Utilizes RF as a feedback resistor.
19
Gain of an Inverting OpAmp

IF
Assume ideal opamp,

so Av = -
iin = iF + ia and
Rin
iin
Ia
Va
vin
RF
(vin-va)/Rin = (va-vout)/RF + ia,

But for an ideal opamp, the input impedance is
infinite, so:
ia = 0 and va = vout/Av
Since Ao = -, va 0.
20
Thus input current is nearly zero and the
differential input is also nearly zero,

it is a virtual earth - negative feedback
forces the two inputs to be equal:
v- = v+ = 0 & vin/Rin = -vo/RF.
So
Av = vout/vin = -RF/Rin
Voltage Gain of Inverting OpAmp
Since Rout of ideal opamp is zero, Rout of full
circuit is also zero,

Input resistance is Rin = vin/iin, This is a drawback of
inverting amplifier!
21
Inverting Amp. Example

An inverting OpAmp has circuit values of Rin = 10 k
and Rf = 100 k. What is the gain of the circuit? A
sinusoidal wave of 1 kHz frequency and 2 V pp (peakto-peak) amplitude is input. Sketch the input and
output waveforms.
22
Current to Voltage Converter

A special case of the inverting amplifier
sometimes called a trans-impedance amplifier.
With a suitable opamp (very low input current) and
high-value feedback resistor:

this is valuable for measuring small currents,
e.g. from low-light photodetectors and similar
probes.
iIN
_
+
23
RF
vOUT iIN RF
Current to Voltage Converter

Trans-impedance
RF
Rin
ID
Va
Ia
+
VO = -IDRF
24
Current-to-Voltage Converter Example

The response of a typical photodiode is 0.2
A/W (at a given wavelength) for a constant

reverse bias. Design a circuit to display the
output on a voltmeter.
Select, for example, RF = 10 k. If the radiant
power on the photodiode is 100 W, then
current developed through diode is
iD = S = 100 W x 0.2 A/W = 20 A
The voltage on meter is
Vm = -RFiD = -10x103 x 20 A = -200 mV
25
Summing Amplifier
R1
I1
I2
v1
v2
26
Iout
R3
R2
+
R3
R3
v2
v OUT v1
R1
R2
Summing Amplifier
Using the virtual earth concept, Rin converts vin
into a current (iin) which is fed to virtual earth

and back to vout,
Currents from other inputs can be fed in and summed,
Current into inverting input node from input 1:

i1 = v1/R1 and i2 = v2/R2,
And current in R3:
iout = i1 + i2 = (v1/R1) + (v2/R2),
vout/R3 = -iout = -[(v1/R1)+(v2/R2)],

vout = -[v1(R3/R1)+v2(R3/R2)].
27
Example
If you have two input voltages of 2 V and 3 V
respectively, design a circuit to add the two

inputs together:
Hence: V0 = -[V1R3/R1 + V2R3/R2]
Set R1 = R2:
V0 = - (R3/R1) [V1 + V2],
V0 = - (R3/R1) [2 V + 3 V] = -5 V
So R3/R1 = 5 V/5 V
So suitable values could be R3 = R1 = 1 k.
28
Difference Amplifier
_
R1
v1
v2
RF
R1
RF
vout
A common operation in electronics is to be able to subtract one

signal from another,
For ease of implementation it is usual to set the two input resistors
equal to each other, & set the feedback resistor and the ground
resistor in the divider chain equal to each other,
No currents flow into either input, so respective voltages are
determined by voltage divider.
29
Difference Amplifier
The voltage at the non-inverting input is:
v+ = v2[RF/(R1 + RF)], And at the inverting input:

v- = v1RF/(R1+RF) + voutR1/(R1+RF)
o (This is an application of the principle of superposition,
which states that the effect of two sources is the sum of

the effects of the sources separately),
So since the two inputs are forced to be equal
v- = v+:
and so
v1RF/(R1+RF) + voutR1/(R1+RF) = v2[RF/(R1 + RF)]
Thus:
voutR1/(R1+RF) = v2[RF/(R1 + RF)] - v1RF/(R1+RF)
And:
vout = v2[RF/(R1 + RF)] - v1RF/(R1+RF) [(R1+RF) /R1]
Finally
30
vout = (v2 - v1)RF/R1
Common Mode and Differential Mode

In an ideal differential amp if v- = v+ then vo will be zero,
This a common-mode (CM) signal and is the same at both inputs,
If v- v+ output would be proportional to the difference, differential-
mode (DM) signal,

A Diff Amp should reject a CM signal and amplify a DM signal,
This is often used to remove noise from a signal,
If an input signal is connected to an amp by a long co-axial conductor
cable it may be affected by noise by electromagnetic interference,
Both noise and signal will be amplified,
If two cables are connected to the inputs of a diff amp:
Then noise is common to both inputs and will be rejected,
The signal is only connected to one and hence amplified,
Common Mode Rejection Ratio (CMRR) is defined as Differential
Gain/Common Mode Gain,
How good is a typical amp at rejecting a CM signal, the larger the
better.
In dB: CMRR(dB) = 20 log(CMRR) For a 741, CMRR is about 90 dB
31
So the linear CMRR = [antilog(90 dB/20)] = 31620
Power Supplies & Slew Rate

Most op-amps operate between 15 V power rails with some variations,
No matter what its Av, or input voltage, can never get more than 15 V at
the output,
Driving the amp harder causes output to saturate and signal will clip at top &
bottom,
Another important OpAmp parameter is its slew rate,
If a square-wave is input to OpAmp, it takes a finite time to reach its max
at output,
The 741 has a slew rate of about 0.5 V/s,
To reach a voltage of 5 V, will take about 10 s this is quite slow,
The slew rate tells what the maximum signal frequency the amp can respond,
fmax = slew rate/(2Vpk),
Vpk = 10 V, fmax = (0.5 V/s)/( 2 x 10 V) = (0.5/10-6 s)/(6.28x10 V) = 7.96 kHz
This is not very fast for a modern OpAmp,
However, note that fmax depends on output voltage; if the amp output is 100
mV, then max frequency rises to 796 kHz,
32
Typical OpAmp: The 741
Characteristic
Conditions
Power rails
input offset voltage
15 V
RS 10 k
2 mV
Input resistance
2 M
output resistance
75
input voltage range
13 V
Slew rate
0.5 V/s
CMRR
RS 10 k
Large signal Av
o/p voltage swing
90 dB
200,000
RL 10 k
Power consumption
33
Typical Value
14 V
50 mW
Rise time
Vin = 20 mV, RL = 2 k
0.3 s
Bandwidth
BW = 0.35/rise time (s)
1 MHz
Non-inverting vs. Inverting OpAmps

No feedback
Closed-loop
Gain (A)
Phase
(referred to
i/p)
Input
impedance
(Zin)
Output
impedance
(Zout)
Max frequency
CMRR
34
Very high OL
gain
Non-inverting
Amp
1 + RF/Ri
Voltagefollower
Unity
Inverting Amp
In phase
In phase
180 out of
phase
-RF/Ri
Very high
> Zin of op-amp > Zin of op-amp Low; Ri
Very low
< Zout of op-
< Zout of op-
< Zout of op-
amp
amp
amp
Slew rate/
(2Vpk)
A/ACM
Slew rate/
(2Vpk)
A/ACM
Slew rate/
(2Vpk)
A/ACM
Some Other Useful OpAmp Applications

Integrator
Differentiator
Low pass filter
High pass filter
Waveform generator
35
Worked Example
741 opamp parameters: Circuit & input parameters:
ACM = 0.001
RF = 100 k
AOL = 200,000
Ri = 10 k
Zin = 2 M RL = 10 k
Zout = 75 vin = 1 V pp
Slew rate = 0.5 V/s
Power supplies: 15 V
Using the circuit parameters given, compare and
contrast the performance of an inverting and noninverting amplifier in terms of voltage gain, input
resistance, output resistance, CMRR and maximum
operating frequency.
36

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MH-03: Operational Amplifiers and

amplifiers (OpAmps) are the basic building

It is only necessary to add power, and any required external

At the moment we need only consider the OpAmp as a black-

box with a defined set of inputs and outputs,

Op amp operated in a wide range of modes depending

The Differential OpAmp

amplifier (ie gain with no external

OpAmp complete circuit would be fabricated onto a single slice

(chip) of silicon, integrated circuit (IC)

This is a 14 pin IC (integrated circuit) but many other

configurations are available,

Avof 104 -106 typical in real opamps,

Feedback in an Engineering System

Feedback: portion of output signal returned & fed back to input,

In a climate control unit you set desired T to a specific level

This is an example of feedback - a portion of the output is fedback and

compared at the input

Feedback in an Electronic System

The concept of feedback is very important in amplifier

Feedback in an Electronic System

vout = Av(vin Bvout)

vout = [Av/(1 + BAv)]vin.

The new gain of the system, the closed loop gain, is :

BAv 1, then A Av /BAv 1/B

Special case (but usual) - if

Negative Feedback in an OpAmp Circuit

Forces inverting inverting input to follow non-inverting input,

to the variability of the open-loop gain if AvB >>1.

Ideal amp amplifies all signals equally

independent from frequency,

100 kHz 1 MHz

100 kHz 1 MHz

100 kHz 1 MHz

For an amplifier with equal Rin & Rout

Effect of Feedback on Frequency

high and low frequencies,

But since the open-loop gain >> closed-loop gain,

Av falls by a considerable amount before this has

The closed-loop gain (A) will be stable over

a wider frequency range - bandwidth will

Gain of Non-Inverting OpAmp

this is known as a virtual short circuit,

Where the feedback fraction is given by B = R i/(Ri+RF)

Thus Gain of amp is

= 2 k, what is the closed-loop gain? Calculate

Special Case: Unity Gain Op-Amp

Unity gain amplifier is special case of non-inverting amp,

One of the most common and useful opamp

Gain of an Inverting OpAmp

Assume ideal opamp,

(vin-va)/Rin = (va-vout)/RF + ia,

Thus input current is nearly zero and the

differential input is also nearly zero,

circuit is also zero,

Inverting Amp. Example

Current to Voltage Converter

high-value feedback resistor: