Guided By :
Dhanush Patel
or Op Amps for short
 An
Operational Amplifier (Op-Amp) is an
integrated circuit that uses external
voltage to amplify the input through a
very high gain.
 We recognize an Op-Amp as a massproduced component found in countless
electronics.
OpAmp
LM324
Positive power supply
(Positive rail)
Non-inverting
Input terminal
Output terminal
Inverting input
terminal
Negative power supply
(Negative rail)
v2
v1
vd = v2 – v1
A is the open-loop voltage gain
Represented by:
A = open-circuit voltage gain
vid = (v+-v-) = differential input signal
voltage
Rid = amplifier input resistance
Ro = amplifier output resistance
The signal developed at the amplifier
output is in phase with the voltage
applied at the + input (non-inverting)
terminal and 180° out of phase with
that applied at the - input (inverting)
terminal.
R
Since v-= v+ vo   2 (v1  v2 )
R
1
For R2= R1 vo  (v1  v2)
 This circuit is also called a
differential amplifier, since it
amplifies the difference
between the input signals.
 Rin2 is series combination of R1
v o  v-  i R  v-  i R
2 2
1 2
and R2 because i+ is zero.


R R 
R
R
 For v2=0, Rin1= R1, as the


2
1
2
2
 v- 
( v  v- )  
v
circuit reduces to an inverting
 v- 
1
1


R
R
R
amplifier.

1
1 
1

 For general case, i1 is a
R
function of both v1 and v2.
2 v
Also, v 

R R 2
1 2




Problem: Determine vo
Given Data: R1= 10kW, R2 =100kW, v1=5 V, v2=3 V
Assumptions: Ideal op amp. Hence, v-= v+ and i-= i+= 0.
Analysis: Using dc values,
R
100kW
A  2 
 10
dm
R
10kW
1


Vo  A V V  10(5 3)
dm 1 2 
Vo  20.0 V
Here Adm is called the“differential mode voltage gain” of the difference
amplifier


The “ideal” op amp is a special case of the ideal differential
amplifier with infinite gain, infinite Rid and zero Ro .
v
v  o and lim vid  0
id A
A 
 If A is infinite, vid is zero for any finite output voltage.
 Infinite input resistance Rid forces input currents i+ and i- to be
zero.
The ideal op amp operates with the following assumptions:
 It has infinite common-mode rejection, power supply rejection,
open-loop bandwidth, output voltage range, output current
capability and slew rate
 It also has zero output resistance, input-bias currents, inputoffset current, and input-offset voltage.


The positive input is grounded.
A “feedback network” composed of resistors R1 and R2 is
connected between the inverting input, signal source and
amplifier output node, respectively.
The negative voltage gain
implies that there is a 1800
phase shift between both dc
and sinusoidal input and
output signals.
 The gain magnitude can be
greater than 1 if R2 > R1
 The gain magnitude can be
less than 1 if R1 > R2
vs  isR  i R  vo  0
 The inverting input of the op
1 2 2
amp is at ground potential
But is= i2 and v- = 0 (since vid= v+ - v-= (although it is not connected
0)
R
directly to ground) and is said
vs
vo
2
is 
and Av   
to be at virtual ground.
R
vs
R
1
1

Rout is found by applying a test
current (or voltage) source to the
amplifier output and determining
the voltage (or current) after
turning off all independent
sources. Hence, vs = 0
vx  i R  i R
2 2 11
But i1=i2
vx  i ( R  R )
1 2 1
vs
R   R since v  0
in i
1
s

Since v- = 0, i1=0. Therefore vx
= 0 irrespective of the value of
ix .
Rout  0





Problem: Design an inverting amplifier
Given Data: Av= 20 dB, Rin = 20kW,
Assumptions: Ideal op amp
Analysis: Input resistance is controlled by R1 and voltage gain
is set by R2 / R1.
AvdB 20log Av , Av 1040dB/20dB 100
and
10 
Av = -100
R since
R  the
20kW
A minus sign is added
amplifier is inverting.
1 in
R
Av  2  R 100R  2MW
2
1
R
1

• The input signal is applied to the non-inverting input terminal.
• A portion of the output signal is fed back to the negative input
terminal.
• Analysis is done by relating the voltage at v1 to input voltage vs
and output voltage vo .
R
vs  v  v
1
and
v  vo
id 1
1
R R
1 2
vs  v
But vid =0
1
R R
vo  vs 1 2
R
1
R
v o R1  R2
 Av 

 1 2
R
vs
R
1
1
vs
R  
in i
Since i+=0

Rout is found by applying a test current source to the amplifier
output after setting vs = 0. It is identical to the output resistance
of the inverting amplifier i.e. Rout = 0.
Since i-=0




Problem: Determine the output voltage and current for the
given non-inverting amplifier.
Given Data: R1= 3kW, R2 = 43kW, vs= +0.1 V
Assumptions: Ideal op amp
R
Analysis:
43kW
Av 1 2 1
15.3
R
3kW
1
vo  Av vs  (15.3)(0.1V)1.53V
Since i-=0,
vo
1.53V
io 

 33.3A
R  R 43kW  3kW
2 1