1. Subject : Analog Electronics
Branch : Electrical – A
Semestar : 3rd
Guide by :
Dhanush Sir
Hardik Sir

2. Inverting & non- Inverting
amplifier

3. Prepared by : 1. Sandip Baldha. 130940109001
2. Keval Chandarana
3. Vishal Domadiya
4. Ronak Kakadiya
5. Chetan katariya
5. Chetan katariya

4. 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.

5. 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
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, input-offset current, and input-offset voltage.

6. 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.

7. Inverting Amplifier:Voltage Gain
The negative voltage gain implies that there is a 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
The inverting input of the op amp is at ground potential (although it is not connected directly to ground) and is said to be at virtual ground.
But is= i2 and v- = 0 (since vid= v+ - v-= 0)
and

8. 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
But i1=i2
Since v- = 0, i1=0. Therefore vx = 0 irrespective of the value of ix .

9. 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.
and Av = -100
A minus sign is added since the amplifier is inverting.

10. 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 .

11. Non-inverting Amplifier: Voltage Gain, Input Resistance and Output Resistance
Since i-=0 and
But vid =0
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.

12. The Difference Amplifier
Since v-= v+
For R2= R1
This circuit is also called a differential amplifier, since it amplifies the difference between the input signals.
Rin2 is series combination of R1 and R2 because i+ is zero.
For v2=0, Rin1= R1, as the circuit reduces to an inverting amplifier.
For general case, i1 is a function of both v1 and v2.
Also,

13. 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,
Here Adm is called the “differential mode voltage gain” of the difference amplifier.