Download Inverting amplifier

Subject
: Analog Electronics
Branch : Electrical – A
Semestar : 3rd
Guide by :
Dhanush Sir
Hardik Sir
Inverting & non- Inverting
amplifier
Prepared by :
1. Sandip Baldha.
2. Keval Chandarana.
3. Vishal Domadiya.
4. Ronak Kakadiya.
5. Chetan katariya.
130940109001
130940109008
130940109018
130940109033
130940109038
Differential Amplifier Model: Basic
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.
Ideal Operational Amplifier

The “ideal” op amp is a special case of the ideal differential amplifier
with infinite gain, infinite Rid and zero Ro .
and
lim vid  0
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, openloop bandwidth, output voltage range, output current capability and
slew rate
 It also has zero output resistance, input-bias currents, input-offset
current, and input-offset voltage.


v
v  o
id A
The Inverting Amplifier: Configuration


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.
Inverting Amplifier:Voltage Gain
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 amp
1 2 2
is at ground potential (although it
But is= i2 and v- = 0 (since vid= v+ - v-= is not connected directly to
0)
R
vs
ground) and is said to be at
vo
2
is 
A



and
v
virtual ground.
R
vs
R
1
1

Inverting Amplifier: Input and Output
Resistances
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
v x  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
Inverting Amplifier: Example

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 Av =
10
-100

 R the amplifier
20kW
A minus sign is addedRsince
is inverting.
1 in
R
Av  2  R 100R  2MW
2
1
R
1

The Non-inverting Amplifier:
Configuration
•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 .
Non-inverting Amplifier: Voltage Gain,
Input Resistance and Output Resistance
R
1
Since i-=0 andv  vo
1
R R
1 2
vs  v  v
id 1
vs  v
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.
But vid =0
The Difference Amplifier
R
Since v-= v+ vo   2 (v  v )
R 1 2
1
For R2= R1 vo  (v1  v2 )
 This circuit is also called a
differential amplifier, since it
amplifies the difference between
the input signals.
v o  v-  i R  v-  i R
 Rin2 is series combination of R1
2 2
1 2


and R2 because i+ is zero.
R R 
R
R

2 v-  2 v  For v2=0, Rin1= R1, as the circuit
 v-  2 ( v  v- )   1

R 1
R 
R 1

reduces to an inverting amplifier.
1
1 
1

 For general case, i1 is a function
R
2 v
Also, v 
of both v1 and v2.
 R R 2
1 2
Difference Amplifier: Example

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
 “differential mode voltage gain” of the difference amplifier.
Here Adm is called the