This discussion focuses on Amplifiers, Operational Amplifiers in particular.

1. • The Ideal Op Amp
• The Inverting and Non-Inverting
Configurations
• The Voltage follower
For aid and reference only

2. INTRODUCTION
• This discussion focuses on Amplifiers, Operational
Amplifiers in particular.
• Signal Amplification- A fundamental signal
processing task is employed in some form in
almost every electronic system.
• Need for amplification arises because transducers
provide signals that are said to be “weak,” that is,
in the microvolt (µV) or millivolt (mV) range and
possessing little energy.

3. INTRODUCTION
• Such signals are too small for normal processing.
Processing can be made easier if signal
magnitude is increased. The functional block that
accomplishes this task is the amplifier.
• It is equally important to understand the need for
linearity in amplifiers. While amplifying, one must
keep in mind that the information contained in
the signal is not changed.
• Thus when feeding the signal/waveform to an
amplifier, we want the output waveform to be an
exact replica of that of the input except having
larger magnitude.

4. THE IDEAL OP AMP
The Op-Amp terminals
• The Op-Amp has three terminals: Terminal 1 and 2 are
input terminals and terminal 3 is the output terminal.
1
3
2
• Two terminals 4 and 5 are brought out of the op-amp and
connected to a positive voltage VCC and a negative voltage
–VEE ,respectively. We assume that these two terminals are
implicitly present in the op-amp device.

5. THE IDEAL OP AMP
Function of the Op-Amp
• The Op-Amp senses the difference between the
voltage signals applied at its two input terminals 1 & 2,
multiplies this by a value A , and causes the resulting
voltage A(v2 – v1) to appear at output terminal 3.
Key features of an ideal op-amp
1. The input impedance of an ideal op amp is supposed
to be infinite. The ideal op amp is not supposed to
draw any current through its input terminals.
2. The output impedance of an ideal op amp is supposed
to be zero. Voltage between terminal 3 and ground is
independent of the current drawn from terminal 3.

6. THE IDEAL OP AMP
3. From the expression of the
output, note that the output is in
phase with v2 and is out of phase
with v1.Hence,terminal 1 is called
inverting input terminal and
terminal 2 is called non-inverting
input terminal .
4. Common-mode rejection if
v1=v2,the output will ideally be
zero. From this, we conclude that
an ideal op amp has zero
common-mode gain or infinite
common-mode rejection.
-
+

7. THE IDEAL OP AMP
5.Ideal op amps will amplify signals of any
frequency with equal gain A, and thus are said
to have infinite bandwidth.
6.The ideal op amp should have a gain A whose
value is very large and ideally infinite. Why so?
This will be justified in the later sections.
Characteristics of the Ideal Op Amp (in short)
Infinite input impedance
Zero output impedance
Zero common-mode gain or infinite common-mode rejection
Infinite open-loop gain A
Infinite bandwidth

8. THE INVERTING CONFIRUGATION
• Op amps are not used alone, rather, the op
amp is connected to passive components
(resistors) in a feedback circuit. There are two
such op amp circuit configurations employing
2 resistorsinverting
& non-inverting .

9. THE INVERTING CONFIRUGATION
• Figure below depicts the inverting configuration.
It consists of two resistorsResistor R2 is connected from the output
terminal 3,back to the inverting or negative input
terminal, terminal 1.R2 is seen as applying
negative feedback because it is connected to the
negative terminal.
Terminal 2 is grounded and R1 is connected
between terminal 1 and input voltage source v1.

10. THE INVERTING CONFIGURATION
• Closed loop gain G, is defined asG= vo /vi
Assuming the op amp to be ideal and gain A very
large(infinite), then by definition,
v2 - v1 = vo /A = 0
vo being the output voltage.
• From the result we may conclude that because A
is very large, voltage v1 approaches and ideally
equals v2 .Hence a virtual short circuit appears
between the terminals 1 & 2.
• Since terminal 2 is grounded thus, v1 =0 & v2 =0.

11. THE INVERTING CONFIGURATION
• On applying ohm’s law across R1 ,we geti1 = (vi -v1)/R1 = (vi -0)/R1 =vi /R1
• This current cannot flow through the op amp
because an ideal op amp draws zero current. It
follows that i1 will have to flow through R2 to
low-impedance terminal 3. Thus,
vo =v1 - i1R2 = 0-(vi /R1)R2
vo /vi = - R2 / R1
which is the closed loop gain of the inverting
configuration.(Refer to the figure in the next slide).

12. THE INVERTING CONFIGURATION

13. THE NON-INVERTING CONFIGURATION
vi
In the non-inverting configuration, the input
signal is applied directly to the positive input
terminal of the op amp while one terminal of R1
is connected to the ground.

14. THE NON-INVERTING CONFIGURATION
• The closed-loop gain –
Assuming that the op
amp has infinite gain
A, a virtual short
circuit exists between
its two input terminals.
Hence the difference
input signal isvid =v2 - v1 = vo /A = 0
• The current through R1
can be determined as
v1 /R1 .

15. THE NON-INVERTING CONFIGURATION
• Because of the infinite input impedance of the
op amp, the current vi /R1 will flow through R2
as shown in previous figure. Now the output
voltage can be determined from
vo = vi +(vi /R1)R2
which yields
vo /vi =1+ R2/R1
which is the open loop gain of the noninverting configuration.

16. THE VOLTAGE FOLLOWER
• The property of high input impedance is a very
desirable feature of the non-inverting configuration.
• We can make R1 =∞ and R2 =0 to obtain the unity gain
amplifier shown in the figure. This circuit is known as
the voltage follower, since the output “follows” the
input. In the ideal case, vo =vi ,Rin =∞,Rout =0. The
equivalent circuit of the follower is also shown above.

17. Thanks.
For feedback, drop a mail at –
f2011263@hyderabad.bits-pilani.ac.in
Mahesh Naidu
B.E. Electrical & Electronics,
BITS-Pilani Hyderabad Campus
For aid and reference only