Linear integrated circuits are being used in a number of electronic applications, such as in the fields like communication, medical electronics, instrumentation control etc. An important linear IC is an operational amplifier.

**Operational amplifier (OP - AMP)**

Linear integrated circuits are
being used in a number of electronic applications, such as in the fields like
communication, medical electronics, instrumentation control etc. An important
linear IC is an operational amplifier.

OP-AMP is a solid state device capable of sensing and amplifying
dc and ac input signals. OP-AMP is an amplifier with two inputs (differential
inputs) and a single output. OP-AMP consists of 20 transistors, 11 resistors
and one capacitor. It usually requires a positive and negative power supply
(dual power supply). This allows the output voltage to swing positive and
negative with respect to ground.

The most important
characteristics of OP-AMP are : (i) very high input impedance or even infinity
which produces negligible current at the inputs, (ii) very high gain, (iii)
very low output impedance or even zero, so as not to affect the output of the
amplifier by loading.

An OP-AMP is so named, because it
was originally designed to perform mathematical operations such as addition,
subtraction, multiplication, division, integration, differentiation etc in
analog computer. Nowdays OP-AMPs are used in analog computer operations and in
timing circuits.

*Circuit symbol and Pin-out
configuration of an OP-AMP*

The OP - AMP is represented by a
triangular symbol as shown in Fig. It has two input terminals and one output
terminal. The terminal with *negative*
sign is called as the inverting input and the terminal with *positive* sign is called as the
non-inverting input. The input terminals are at the base of the triangle. The
output terminal is shown at the apex of the triangle.

The widely used very popular type
Op-Amp IC 741, which is available in DIP. Referring to the top view of the
dual-in-package, the pin configuration of IC 741 can be described (Fig) as
follows. The top pin on the left side of the notch indicates Pin 1. The pin
number 2 is inverting input terminal and 3 is non-inverting input terminal. Pin
6 is the output terminal. A d.c. voltage or a.c signal placed on the inverting
input will be 180^{o} out of phase at the output. A d.c. voltage or
a.c. signal placed on the non-inverting input will be inphase at the output.
Pins 7 and 4 are the power supply terminals. Terminals 1 and 5 are used for
null adjustment. Null adjustment pins are used to null the output voltage when
equal voltages are applied to the input terminals for perfect balance. Pin 8
indicates no connection.

*Basic OP-AMP circuits*

This section concentrates on the
principles involved with basic OP-AMP circuit viz, (i) inverting and (ii)
non-inverting amplifiers.

*(i) Inverting amplifier*

The basic OP-AMP inverting
amplifier is shown in Fig. The input voltage V_{in} is applied to the
inverting input through the input resistor R_{in}. The non inverting
input is grounded. The feedback resistor R_{f} is connected between the
output and the inverting input.

Since the input impedance of an
op-amp is considered very high, no current can flow into or out of the input
terminals. Therefore I_{in} must flow through R_{f} and is
indicated by I_{f} (the feedback current). Since R_{in} and R_{f}
are in series, then I_{in} = I_{f}. The voltage between
inverting and non-inverting inputs

is essentially equal to zero
volt. Therefore, the inverting input terminal is also at 0 volt. For this
reason the inverting input is said to be at virtual ground. The output voltage
(Vout) is taken across Rf.

It can be proved that

I_{f}=V_{out}/R_{f}

Since I_{in} =I_{f },
then

V_{in}/R_{in} =
-V_{out}/R_{f}

Rearranging the equation, we
obtain

-V_{out}/V_{in} =
R_{f}/R_{in}

∴ The voltage gain of an inverting amplifier can be expressed as

A_{v} = -R_{f}/R_{in}

The amplifier gain is the ratio
of R_{f} to R_{in}

Finally, the output voltage can
be found by

V_{out} = -R_{f}/R_{in}
x V_{in}

The output voltage is out of
phase with the input voltage.

*(ii) Non-inverting amplifier*

The basic OP-AMP non-inverting
amplifier is shown in Fig. The input signal V_{in} is applied to the
non-inverting input terminal. The resistor R_{in} is connected from the
inverting input to ground. The feedback resistor R_{f} is connected between
the output and the inverting input.

Resistors R_{f} and R_{in}
form a resistive ratio network to produce the feedback voltage (V_{A})
needed at the inverting input. Feedback voltage (V_{A}) is developed
across R_{in}. Since the potential at the inverting input tends to be
the same as the non-inverting input (as pointed out with the description of
virtual ground), V_{in} = V_{A}.

Since V_{A} = V_{in},
the gain of the amplifer can be expressed as

_{Av}_{= }^{V}*ou t*

*V _{A}*

However, V_{A} is
determined by the resistance ratio of R_{in} and R_{f} ; thus,

V_{A}=[R_{in}/(R_{f}/R_{in})]
V_{out}

A_{v}=1+(R_{f}/R_{in})

Finally, the output voltage can
be found by, Vout = (1+R_{f}/R_{in})V_{in}

It is seen that the input and
output voltages are in phase

**(iii) Summing amplifier**

The summing amplifier provides an
output voltage equal to the algebraic sum of the input voltages.

Fig shows an inverting amplifier,
used to sum two input voltages. The input voltages v_{1} and v_{2}
are applied through the resistors R_{1} and R_{2} to the
summing junction (P) and Rf is the feedback resistor. At the point P,

i_{1} + i_{2} =i_{f}

Since the voltage at the point P
is ideally 0

(V_{1}/R_{1}) +
(V_{2}/R_{2}) =(V_{out}/R_{f})

Hence the output voltage,

V_{out}=-[(R_{f}V_{1}/R_{1})
+(R_{f}V_{2}/R_{2}) ]

If R_{1} = R_{2}
= R_{f} = R, then vout = - (v_{1}
+ v_{2})

Hence the output voltage is equal
to the sum of the input voltages and the circuit acts as a summing amplifier.
The negative sign indicates that OP-AMP is used in the inverting mode.

**(iv)Difference amplifier**

The difference amplifier is shown
in Fig. The output voltage can be obtained by using superposition principle. To
find the output voltage v_{01} due to v_{1} alone, assume that
v_{2} is shorted to ground. Then

V^{+} = R_{2}V_{1
}/ R_{1}+R_{2}

Therefore, with both inputs
present, the output is

V_{0} = V_{01} +
V_{02}

= (R_{3}+R_{4 / }R_{3})
(R_{2}/R_{1}+R_{2}) V_{1
} - (R_{4}/R_{3})V_{2}

If R_{1} = R_{2}
= R_{3} = R_{4} = R

then vo = v_{1} - v_{2}

If all the external resistors are
equal, the voltage difference amplifier functions as a voltage subtractor.

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