UNIT II CHARACTERISTICS OF OPAMP

Ideal OP-AMP characteristics, DC characteristics, AC

characteristics, differential amplifier; frequency response of OP-
AMP; Basic applications of op-amp – Inverting and Non-inverting
Amplifiers, summer, differentiator and integrator-V/I & I/V
converters.

2.1 INTRODUCTION TO IDEAL OP-AMP CIRCUIT

CHARACTERISTICS

2.1.1 Introduction
Operational amplifier (op amp for short) is basically a voltage
amplifying device designed to be used with components like
capacitors and resistors, between its in/out terminals, or is
simply a linear Integrated Circuit (IC) having multiple-terminals.
In electronics, the open-loop voltage gain of the actual
operational amplifier is very large, which can be seen a
differential amplifier with infinite open loop gain, infinite input
resistance and zero output resistance. In addition, it has positive
and negative inputs which allow circuits that use feedback to
achieve a wide range of functions. And meanwhile, it can be
further simplified into an ideal op amp model, referred to as an
ideal op amp (also called ideal OPAMP).
2.1.2 Ideal Op Amp Characteristics

When analyzing various application circuits of operational

amplifiers, the integrated operational amplifier is often regarded
as an ideal operational amplifier. The so-called ideal op amp is to
idealize various technical indicators of op amps, and it must have
the following characteristics.

2.1.3 Infinite Input Resistance

The input terminal of an ideal operational amplifier does not
have any current to flow in. In electronics, op amps are voltage
gain devices. They amplify a voltage fed into the op amp and
give out the same signal as output with a much larger gain. In
order for an op amp to receive the voltage signal as its input, the
voltage signal must be dropped across the op amp. If you know
the concept of a voltage divider, voltage drops primarily across
components with high impedances, proportionally according to
ohm’s law by the formula V=IR. So the greater the resistance (or
impedance) of a device, the greater the voltage drop across that
device is. To make sure that the voltage signal drops fully on the
op amp, it must have a very high input impedance, so that the
voltage drops fully across it. If it had a low input impedance, the
voltage may not drop across it and it would not receive the
signal. This is why op amps must have high-input impedances.
It’s also easy to make the input impedance lower (put a resistor
in parallel) or the source impedance higher (put a resistor in
series).
Figure 2.1. Ideal Op Amp Symbol and Transfer Characteristic
Curve

2.1.4 Zero Output Impedance

The output of an ideal op amp is a perfect voltage source, no
matter how the current flowing to the amplifier load changes, the
output voltage of the amplifier is always a certain value, that is,
the output impedance is zero. In practice, zero output impedance
is actually a distinct property from infinite input impedance, but
for a very long-time infinite input impedance was approached
only with compromises in offset voltage and noise.

2.1.5 Infinite Open-loop Gain

In an open-loop state, the differential signal at the input has an
infinite voltage gain. This feature makes the operational amplifier
very suitable for practical applications with upper negative
feedback configuration.
2.1.6 Infinite Common-mode Rejection Ratio
An ideal operational amplifier can only respond to the difference
between the voltages at both ends of V+ and V-. In addition, the
same part of the two input signals (ie common mode signal) will
be completely ignored. What’s more, a high CMRR is required
when a differential signal must be amplified in the presence of a
possibly large common-mode input, such as strong
electromagnetic interference (EMI). An example is audio
transmission over balanced line in sound reinforcement or
recording.

2.1.7 Infinite Bandwidth

The ideal operational amplifier will amplify the input signal of any
frequency with the same differential gain, which will not change
with the change of signal frequency.

2.2 ASSUMPTIONS OF IDEAL OP AMP

The op amp can be considered a voltage controlled current
source, or it is an integrated circuit that can amplify weak
electric signals. Based on it, for an ideal OPAMP, what is the
relationship between it and these electrical signals?

First, assume that the current flowing into the input of the op
amp is zero. This assumption is almost completely correct for
FET op amps, because the input current for FET op amps is below
1pA. But for dual high-speed op amps, this assumption is not
always correct, because the input current of it can sometimes
reach tens of microamperes.

Second, assume that the gain of the op amp is infinite, so the op

amp can swing the output voltage to any value to meet the input
requirements. It means that the output voltage of the op amp
can reach any value. In fact, when the output voltage is close to
the power supply voltage, the op amp will saturate. Maybe this
hypothesis does exit, but needs a limit in practical. For example,
at higher frequencies, the internal junction capacitors of
transistor come into play, thus reducing the output and therefore
the gain of amplifier. The capacitor reactance decreases with
increase in frequency bypassing the majority of output. The op-
amp is in saturation state.

Figure 2.2. Op Amp Saturation

For example, as per datasheet of LM741, large signal voltage

gain is 200V/mv. It means an open loop gain of 200,000. If you
operate an op-amp in open-loop condition (i.e. without negative
feedback), even microvolts of input voltage (input offset voltage
of LM741 is 3mv) will drive the output to saturation.

In most of the amplifier circuits op-amp is configured to use

negative feedback which greatly reduces the voltage gain (i.e.
closed loop gain). In oscillators and schmit triggers, Op-amp is
configured to use positive feedback. Comparator circuit is an
example of the circuit which utilizes open-loop gain of op-amp.
Its output will be always at saturation either positive or negative
saturation. In an integrator circuit, the DC gain should be limited
by adding a feedback resistor in parallel with capacitor; else the
output will get saturated.

Even in amplifier circuits, the amplitude of the input signal and

the voltage gain of the circuit should be balanced so that the
output voltage does not exceed power supply voltage. For
example, for a non-inverting amplifier with a voltage gain of 100,
the maximum permissible input voltage will be 150 mv if the VCC
is 15 Volts. If you apply a signal of 200 mv, the op-amp output
will go to saturation as the required output will be 20 volts which
exceeds the VCC of 15 Volts.

Third, the assumption of infinite gain also means that the input
signal must be zero. The gain of the op amp will drive the output
voltage until the voltage (error voltage) between the two input
terminals is zero. The voltage between the two input terminals is
zero. The zero voltage between two input terminals means that if
one input terminal is connected to a hard voltage source like
ground, the other input terminal will also be at the same
potential. In addition, since the current flowing into the input
terminal is zero, the input impedance of the op amp is infinite.
Fourth, of course, the output resistance of an ideal op amp is
zero. An ideal op amp can drive any load without any voltage
drop due to its output impedance. At low currents, the output
impedance of most op amps is in the range of a few tenths an
ohm, so this assumption is true in most cases.

2.3 WORKING CHARACTERISTICS OF IDEAL OPERATIONAL

AMPLIFIERS

2.3.1 Work in Linear Region

When the ideal op amp works in the linear region, the output and
the input voltage show a linear relationship. Where u0 is the
output voltage of the integrated op amp; u+ and u- are the
voltages at the non-inverting input terminal and the inverting
input terminal, respectively. Auo is the open loop differential
voltage magnification. According to the characteristics of the
ideal op amp, two important characteristics of the ideal op amp
in the linear region.

1. Zero differential input voltage

Since the open-loop differential voltage magnification of an ideal
op amp is equal to infinity, and the output voltage is a certain
value, the voltage values at the non-inverting input terminal and
the inverting input terminal are approximately equal. Just like
short circuit between input and output, but it is fake. Because it
is an equivalent short circuit, not a real short circuit, so this
phenomenon is called "virtual short".

2. Zero input current

Since the open-loop input resistance of an ideal op amp is
infinite, no current flows into the op amp at either input. At this
time, the current at the non-inverting input terminal and the
inverting input terminal are both equal to zero. Like an
disconnection, but an equivalent disconnection, so this
phenomenon is called "virtual break". Virtual short and virtual
break are two important concepts for analyzing the ideal op amp
working in the linear region.

The ideal operational amplifier has the characteristics of "virtual

short" and "virtual break". These two characteristics are very
useful for analyzing linear amplifier circuits. The necessary
condition for virtual short is negative feedback. When negative
feedback is introduced, at this time, if the forward terminal
voltage is slightly higher than the reverse terminal voltage, the
output terminal will output a high voltage equivalent to the
power supply voltage after the amplification of the op amp.
 In fact, the op amp has a respond time changing from the
original output state to the high-level state (the golden rule of
analyzing analog circuits: the change of the signal is a
continuous change process). Due to the feedback resistance of
the reverse end change will inevitably affect its voltage, when
the reverse end voltage infinitely close to the forward end
voltage, the circuit reaches a balanced state. The output voltage
does not change anymore, that is, the voltage at the forward end
and the reverse end is always close. (Note: The analysis method
is the same when the voltage decreases.)

2.3.2 Work in Nonlinear Region

When the op-amp operates in the nonlinear region, the output
voltage no longer increases linearly with the input voltage, but
saturates. The ideal op amp also has two important
characteristics when operating in the nonlinear region.

1) When u+ ≠ u-, the output voltage of the ideal op amp reaches

the saturation value. When u+ > u-, the op-amp operates works
in positive saturation region with a positive output voltage. When
u+ < u-, the op-amp operates works in negative saturation
region with a negative output voltage. Ideal op amp operates in
the nonlinear region, u+ ≠ u-, there is no “virtual short”.

2)The input current is equal to zero.

Although the input voltage u+ ≠ u- above, the input current is
considered to be zero.
2.3.3 Characteristics Analysis of Ideal Operational
Amplifier

As for Op-amp, there's probably a description like this: three-

terminal element (circuit structure with double-ended input,
single-ended output), ideal transistor, high-gain DC amplifier.

1. High input resistance:

Under this situation, the current flowing into the input
terminal is close to 0, almost no signal source current is
used, which is close to the voltage control characteristic.
And virtual break is derived from this.
2. Lower output resistance:
It has the characteristics of adapting to any load. And the
impedance of the subsequent load circuit will not affect the
output voltage.

3. Infinite voltage amplification: Under a certain supply

voltage condition, the amplifier can only work in closed-
loop (negative feedback) mode, and the actual
amplification is limited. Because op-amps themselves don't
have a 0V connection but their design assumes the typical
signals will be more towards the center of their positive
and negative supplies.
Thus, if your input voltage is right at one extreme or forces the
output toward one supply, chances are it won't work properly.
Working in open-loop mode is the like a comparator, and the
output is high level or low level.
In the closed-loop (limited amplification) state, the amplifier is
randomly comparing the potentials of the two input terminals.
The output stage makes immediate adjustments when they are
not equal. So, the final purpose of amplification is to make the
potentials of the two input terminals equal. And virtual short is
derived from this.

2.4 BALANCED RESISTANCE PRESETS

2.4.1 The Role of Balanced Resistance

1. A suitable resistance is generally required to ensure that

the input impedance is matched.
2. In order to reduce the input current imbalance, the in-
phase resistor should be equal to the parallel value of the
two resistors at the reverse end in theory. In practice, as a
result of the closed loop, especially in deep negative
feedback conditions, the misalignment is not obvious at
the output. And there is no need of in-phase grounding
resistor when the misalignment is not the main problem.
Because a balanced resistor is the starting point for an
ideal op amp. In-phase grounding resistance is useful for
bipolar op amps, and has no meanings for MOS-type op
amps.
 3. Ground input termination resistance: it is necessary for
impedance matching and high frequency setting.
4. Bias current and offset current.
For operational amplifiers with bias current greater than
offset current, input resistance matching can be reduced,
and precision circuits can compensate bias current to a
minimum. If the bias current and offset current are similar,
the matching resistance will increase the error.
5. Set for the bias current at the input, the purpose of which
is to equalize the impedance of the inverting and non-
inverting inputs, so that two inputs with equal bias currents
are assumed to have equal voltage drops, thereby
counteraction can be made.

2.4.2 Input Balancing Resistor Explanation

A op-amp is connected to an inverting amplifier:
Set the input resistance for R1, feedback resistance for Rfi,
Assume that the non-inverting end is not connected to a
balanced resistor, but grounded directly.
Set the input bias current for the op-amp IB (same voltage in
inverting and non-inverting end).

The current flows through R1 and Rf are represented by I1 and If.

Inverting voltage is V-, The op-amp gain is A.
Use KCL in the inverting end (set the input signal to 0).

Where
 (0-V-)/R1- (A+1)V- /Rf=IB

From the above equation, it follows that

V-=-(IB×R1×Rf/(Rf+(A+1)R1))

At this time, the output voltage of the op-amp is

Vo=A×(IB×R1×Rf/(Rf+(A+1)R1))

The above formula can be approximated as

Vo=IB×((A×R1)/Rf)

If the in-phase terminal passes through a resistor R2 to ground

and R2=R1/Rf, then the voltage at the in-phase terminal is V+=-
IB×R2 KCL is applied to the inverted terminal, where (0-V-)/R1+
(A×(V+-V-)-V-)/Rf=IB
>At this time the output voltage of the op-amp is Vo=0.

2.5 IDEAL OP AMP EQUATIONS

Understanding the basic conditions of an ideal op amp, and
combining it with the Kirchhoff's current law (KCL) node voltage
method and the superposition theorem of the node, is an
effective method to analyze the ideal op amp circuit.
As shown below, find the output voltage u o,
1) Equation based on KCL
From the concept of virtual break, i+=i-=0, then i1=i2, i3=i4, so

Based on virtual break, u+=u-, then

R2
u0 = ( u −u ) … … … … … … .(b)
R1 2 1

1) Node voltage method

List the node voltage equations for node 1 and node 2, and get

Note: Because the output current of the op amp is unknown at 1)

and 2), it is not possible to list the KCL equation or node voltage
equation at the output of the op amp. In addition, the op amp
output uo in 2) should be treated as an independent voltage
source.

3) Superposition theorem
When there are multiple signal inputs, choosing the
superposition theorem to solve can simplify the analysis and
calculation process. The size of the output signal u 0 can be
regarded as the superposition of the output signal obtained by
the independent action of u 1 and u2. When u1 acts alone, the u 2
terminal is grounded, and the op amp output is:
−R2
u01= u
R1 1
R2
u02= u … … … … … … .(d)
R1 2

Therefore, the final output of the operational amplifier is:

R2
u0 =u01+u 02= ( u −u ) … … (e)
R1 2 1

2.6 SEVERAL COMMON OP AMP CIRCUITS

Non-inverting Amplifier Circuit

A non-inverting amplifier is an op-amp circuit configuration which
produces an amplified output signal. It provides a high input
impedance along with all the advantages gained from using an
operational amplifier.
RG
V ¿ =V OUT
RG+ RF
V OUT R G + R F RF
= =1+
V¿ RG RG
 Figure 2. 3. Non-inverting Amplifier Circuit

Inverting Amplifier Circuit

An inverting amplifier (also known as an inverting operational
amplifier or an inverting op-amp) is a type of operational
amplifier circuit which produces an output which is out of phase
with respect to its input by 180 degrees out of phase with
respect to input signal. In the following figure, two external
resistors to create feedback circuit and make a closed loop
circuit across the amplifier.
V¿ −V OUT
I 1= =−I 2=
RG RF
V OUT −R F
=
V¿ RG

Figure 2.4. Inverting Amplifier Circuit

Op-amp as Adder
An adder circuit can be made by connecting more inputs to the
inverting op amp. The circuit diagram of a summing amplifier is
as shown in the following figure.
−R F
V OUTN = V
RN N
−R F
V OUT 1= V
R1 1
−R F
V OUT 2= V
R2 2

(
V OUT =−
RF
R1
RF
V 1+ V 2+
R2
RF
V
RN N )

2.Figure 2.5. Op-amp as Adder

Differential Amplifier
Differential amplifier is an analog circuit with two inputs and one
output in which the output is ideally proportional to the
difference between the two voltages. It is a very useful op-amp
circuit and by adding more resistors in parallel with the input
resistors as shown in the following figure.
 V OUT =V 1
R2
(
R1 + R2
R3 + R 4
R3 )
−V 2
R4
R3

Figure 2.6. Differential Amplifier

Composite Amplifier
The composite amplifier is termed as a combination of multiple
operational amplifiers that are cascaded together with a
negative-feedback loop around the entire network.
R 2 R3
R 2+ R 3 +
−V OUT R4
=
V¿ R1

Figure 2.7. Composite Amplifier

The resistance in the circuit is generally selected at the K ohm

level, the ratio of the resistance affects the gain and bias, in
addition, the supply current, frequency response and capacitive
load driving capability of the op amp determine their specific
values in circuits. If it is used in a high-frequency circuit, the
resistance needs to be reduced to obtain a better high-frequency
response, but it will increase the input bias current, thereby
increasing the current of the power supply.

2.6.1 Difference Between Ideal Op-Amp and Practical Op-

Amp
Ideal op amps use no power, have infinite input impedance,
unlimited gain-bandwidth and slew rate, no input bias current,
and no input offset. They have unlimited voltage compliance.
Practical op amps consume some power, have very high input
impedance have limited gain-bandwidth and limited slew rate,
have some input bias current and input offset voltage. Voltage
compliance is limited by the power supply rail, or frequently even
less.
Still practical op amps are very useful because most of the
limitations listed above are way better than what your circuit
needs.
For an ideal amplifier, it does not draw any current at all from its
input. Assuming a two-input amplifier the signal current in both
input probes is zero. In other words, the input impedance must
be infinite. The output, should operate as the output of an ideal
voltage source. This means that the potential between the
output and the ground must be A(v2−v1), no matter how much
current would a load connected to the output would draw. In
other words, the output impedance must be zero.
For a real amplifier, the input impedance must be as large as
possible while the output impedance must be as low as possible.
In fact, an op-amp in real life, however, cannot operate with zero
current flow.

2.7 DC CHARACTERISTICS OF OP-AMP

Current is taken from the source into the op-amp inputs respond
differently to current and voltage due to mismatch in transistor.

DC output voltages are,

1. Input bias current
2. Input offset current
3. Input offset voltage
4. Thermal drift
1. Input bias current:

The op-amp‘s input is differential amplifier, which may be made

of BJT or FET.

 In an ideal op-amp, we assumed that no current is drawn

from the input terminals.

 The base currents entering into the inverting and non-

inverting terminals (IB-& IB+ respectively)
  Even though both the transistors are identical, I B- and
IB+ are not exactly equal due to internal imbalance between
the two inputs.

 Manufacturers specify the input bias current I B

Fig.2.8 Input bias current

If input voltage V i=0V . The output voltage V 0 should also be (

V 0=0), I B=500 nA .
Op-amp with a 1M feedback resistor
V0 = 5000nA X 1M = 500mV
The output is driven to 500mV with zero input, because of the
bias currents.

In application where the signal levels are measured in mV, this is

totally unacceptable. This can be compensated. Where a
compensation resistor Rcomp has been added between the non-
inverting input terminal and ground as shown in the figure below.
Fig.2.9 Bias current compensation

Current IB+ flowing through the compensating resistor R comp, then

by KVL we
-V1+0+V2-Vo = 0 (or)
Vo = V2 – V1 ——>(3)

By selecting proper value of R comp, V2 can be cancelled with

V1 and the Vo = 0. The value of Rcomp is derived a

V1 =I=+Rcomp (or)

IB+ = V1/Rcomp ——>(4)

The node =a‘ is at voltage (-V1). Because the voltage at the non-
inverting input terminal is (-V1). So with Vi = 0 we get,

I1 = V1/R1 ——>(5)
 I2 = V2/Rf ——>(6)

For compensation, Vo should equal to zero (V o = 0, Vi = 0). i.e.

from equation (3) V2 = V1. So that,
I2 = V1/Rf ——>(7)

KCL at node =a‘ gives,

IB-= I2 + I1
IB- = RfR1
Assume IB- = IB+ and using equation (4) & (8) we get
Rcomp = R1 + Rf
R comp = R1 || Rf --->(9)
i.e. to compensate for bias current, the compensating resistor,
Rcomp combination of resistor R1 and Rf.

2. Input offset current:

Bias current compensation will work if both bias currents IB + and
IB- are equal.Since the input transistor cannot be made identical.
There will always be some small difference between I B+ and IB-.
This difference is called the offset current
|Ios| = IB+ -IB- -à(10)

Offset current Ios for BJT op-amp is 200nA and for FET op-amp is
10pA. Even with bias current compensation, offset current will
produce an output voltage when Vi = 0.
 +¿ Rcomp …… …… … .(11)¿
V 1=I B
I 1=V 1 / R1 … … … … … .(12)
KCL at node a gives
Again
V 0=I 2 R f −V 1
+¿ R comp ¿
V 0=I 2 R f −I B

V 0=1 MΩ X 200 nA
V 0=200 mV with V i=0

Equation (16) the offset current can be minimized by keeping

feedback resistance small.

 Unfortunately to obtain high input impedance, R1 must be

kept large.

 R1 large, the feedback resistor R f must also be high. So as

to obtain reasonable gain.

 The T-feedback network is a good solution. This will allow

large feedback resistance, while keeping the resistance to
ground low (in dotted line).

 The T-network provides a feedback signal as if the network

were a single feedback resistor.
By T to Π conversion,to design T- network first pick R t<<Rf/2
Then calculate
R s=Rf 2 Rt

Fig. 2.10 Input offset current

3. Input offset voltage:

Inspite of the use of the above compensating techniques, it is
found that the output voltage may still not be zero with zero
input voltage [Vo ≠ 0 with Vi = 0]. This is due to unavoidable
imbalances inside the op-amp and one may have to apply a
small voltage at the input terminal to make output (V o) = 0.
This voltage is called input offset voltage V os. This is the voltage
required to be applied at the input for making output voltage to
zero (Vo = 0).
Fig.2.11 Inverting amplifier and its equivalent circuit for V i=0V

Let us determine the Vos on the output of inverting and non-

inverting amplifier. If Vi = 0 become the same as in figure 2.11.

4. Total output offset voltage:

The total output offset voltage VOT could be either more or less
than the offset voltage produced at the output due to input bias
current (IB) or input offset voltage alone(Vos).

This is because IB and Vos could be either positive or negative

with respect to ground. Therefore the maximum offset voltage at
the output of an inverting and non-inverting amplifier (figure
2.11) without any compensation technique used is given by
many op-amp provide offset compensation pins to nullify the
offset voltage.

 10K potentiometer is placed across offset null pins 1&5.

The wipes connected to the negative supply at pin 4.
  The position of the wipes is adjusted to nullify the offset
voltage

Fig. 2.12 Output offset voltage

When the given (below) op-amps does not have these offset
null pins, external balancing techniques are used.

5. Thermal drift:

 Bias current, offset current, and offset voltage change with

temperature.

 A circuit carefully nulled at 25ºC may not remain. So when

the temperature rises to 35ºC. This is called drift.

 Offset current drift is expressed in nA/ºC.

  These indicate the change in offset for each degree Celsius
change in temperature.

2.8 BALANCING CIRCUIT

Inverting amplifier:
V R4
V os=±
R3 + R4

Fig 2.13

Non-inverting amplifier:
V R4
V os=±
R3 + R4
Fig. 2.14

2.9 AC CHARACTERISTICS
For small signal sinusoidal (AC) application one has to know the
ac characteristics such as frequency response and slew-rate.

2.9.1 Frequency Response

The variation in operating frequency will cause variations in gain
magnitude and its phase angle. The manner in which the gain of
the op-amp responds to different frequencies is called the
frequency response. Op-amp should have an infinite bandwidth
BW =∞ (i.e.) if its open loop gain in 90dB with dc signal its gain
should remain the same 90 dB through audio and onto high radio
frequency. The op-amp gain decreases (roll-off) at higher
frequency what reasons to decrease gain after a certain
frequency reached. There must be a capacitive component in the
equivalent circuit of the op-amp. For an op-amp with only one
break (corner) frequency all the capacitors effects can be
represented by a single capacitor C. Below fig is a modified
variation of the low frequency model with capacitor C at the
output.
 Fig.2.15 Equivalent circuit of practical circuit

There is one pole due to R 0C and one -20dB/decade. The open

loop voltage gain of an op-amp with only one corner frequency is
obtained from above figure 2.15 f1 is the corner frequency or the
upper 3 dB frequency of the op-amp. The magnitude and phase
angle of the open loop volt gain are f1 of frequency can be
written as, The magnitude and phase angle characteristics:

 For frequency f<< f1 the magnitude of the gain is 20 log

AOL in db.

 At frequency f = f1 the gain in 3 dB down from the dc value

of AOL in db. This frequency f1 is called corner frequency.

 For f>> f1 the fain roll-off at the rate off -20dB/decade or -

6dB/decade.
 Fig.2.16 Frequency response of op-amp

From the phase characteristics that the phase angle is zero at

frequency f = 0. At the corner frequency f1 the phase angle is -
45º (lagging and an infinite frequency the phase angle is -90 .
It shows that a maximum of 90 phase change can occur in an
op-amp with a single capacitor C. Zero frequency is taken as the
decade below the corner frequency and infinite frequency is one
decade above the corner frequency.

Fig.2.17 Roll off rate of op-amp gain

2.9.2 Circuit Stability

A circuit or a group of circuit connected together as a system is
said to be stable, if its o/p reaches a fixed value in a finite time. A
system is said to be unstable, if its o/p increases with time
instead of achieving a fixed value. In fact the o/p of an unstable
sys keeps on increasing until the system break down. The
unstable system is impractical and need be made stable. The
criterion gn for stability is used when the system is to be tested
practically. In theoretically, always used to test system for
stability, ex: Bode plots.

Bode plots are compared of magnitude Vs Frequency and phase

angle Vs frequency. Any system whose stability is to be
determined can represented by the block diagram.

Fig. 2.18 Feedback loop system

The block between the output and input is referred to as forward

block and the block between the output signal and f/b signal is
referred to as feedback block. The content of each block is
referred as transfer frequency. From fig. 2.18 we represented it
by AOL (f) which is given by

AOL (f) = V0/Vin if Vf = 0 ----- (1)

where AOL (f) = open loop volt gain.
The closed loop gain Af is given by AF = V0/Vin
= AOL / (1+(AOL ) (B) ----(2)
B = gain of feedback circuit.B is a constant if the feedback circuit
uses only resistive components.
Once the magnitude Vs frequency and phase angle Vs frequency
plots are drawn, system stability may be determined as follows

Method 1:
Determine the phase angle when the magnitude of (AOL) (B) is
0dB (or) 1.
If phase angle is >-180 , the system is stable. However, the
some systems the magnitude may never be 0, in that cases
method 2, must be used.

Method 2:
Determine the phase angle when the magnitude of (AOL) (B) is
0dB (or) 1.
If phase angle is > - 180 , If the magnitude is –ve decibels then
the system is stable. However, the some systems the phase
angle of a system may reach -1800, under such conditions
method 1 must be used to determine the system stability.

2.10 DC CHARACTERISTICS OF OP-AMP

Current is taken from the source into the op-amp inputs respond
differently to current and voltage due to mismatch in transistor.

DC output voltages are,

 Input bias current
 Input offset current
 Input offset voltage
 Thermal drift

Input bias current

The op-amp‘s input is differential amplifier, which may be made
of BJT or FET. In an ideal op-amp, we assumed that no current is
drawn from the input terminals the base currents entering into
the inverting and non-inverting terminals (IB- & IB+ respectively).

Even though both the transistors are identical, I B- and IB+ are not
exactly equal due to internal imbalance between the two inputs.
Manufacturers specify the input bias current I B

Input Bias Current

−¿
IB
+¿+ ¿
2
I B=I B ¿
If input voltage Vi = 0V. The output Voltage Vo should also be
(Vo = 0) but for IB = 500nA We find that the output voltage is
offset by Op-amp with a 1M feedback resistor

Vo = 500nA X 1M = 500mV

The output is driven to 500mV with zero input, because of the

bias currents.
In application where the signal levels are measured in mV, this is
totally unacceptable. This can be compensated by a
compensation resistor Rcomp has been added between the non-
inverting input terminal and ground as shown in the figure below.

Fig.1.23 Bias Compensated circuit

Current IB+ flowing through the compensating resistor R comp, then

by KVL we get,
- V1+0+V2-Vo = 0 (or)
Vo = V2 – V1 --------- (1)
By selecting proper value of R comp, V2 can be cancelled with
V1 and the Vo = 0. The value of Rcomp is derived as
V1 = IB+Rcomp (or)
IB+ = V1/Rcomp ------------------------ (2)

The node ‘a’ is at voltage (-V 1). Because the voltage at the non-
inverting input terminal is (-V1). So with Vi = 0 we get,
I1 = V1/R1 ------------------------ (3)
I2 = V2/Rf ------------------------ (4)

For compensation, Vo should equal to zero (Vo = 0, Vi = 0). i.e.

from equation (3) V2 = V1. So that,

I2 = V1/Rf ——> (5)

KCL at node ‘a’ gives,
I B- = I2 + I1 =( V1/Rf ) +(V1/R1) =
V1(R1+Rf)/R1Rf ------------------------ (5)

Assume IB- = IB+ and using equation (2) & (5) we get
V1 (R1+Rf)/R1Rf = V1/Rcomp
R comp = R1 || Rf
------------------------ (6)

i.e. to compensate for bias current, the compensating resistor,

Rcomp should be equal to the parallel combination of resistor
R1 and Rf.
Input offset current:
 Bias current compensation will work if both bias currents
IB+ and IB- are equal.
 Since the input transistor cannot be made identical. There
will always be some small difference between I B+ and IB-. This
difference is called the offset current
|Ios| = IB+-IB------------------------- (7)
Offset current Ios for BJT op-amp is 200nA and for FET op-amp is
10pA. Even with bias current compensation, offset current will
produce an output voltage when Vi = 0.

+¿ Rcomp ……(11)¿
V 1=I B
And
I 1=V 1 / R1 … … .(12)
KCL at node a gives
I 2=¿
Again V0 = I2 Rf – V1
Vo = I2 Rf - IB+ Rcomp
Vo = 1M Ω X 200nA
Vo = 200mV with Vi = 0
Equation (16) the offset current can be minimized by keeping
feedback resistance small.
 Unfortunately to obtain high input impedance, R1 must be
kept large.
 R1 large, the feedback resistor Rf must also be high. So as
to obtain reasonable gain.
The T-feedback network is a good solution. This will allow large
feedback resistance, while keeping the resistance to ground low
(in dotted line).
 The T-network provides a feedback signal as if the network
were a single feedback resistor.
By T to Π conversion,
2
Rt +2 R t Rs
Rf=
Rs

To design T- network first pick Rt<<Rf/2 and=calculate

2
Rt
R s=
R f −2 R t

Input offset voltage:

In spite of the use of the above compensating techniques, it is
found that the output voltage may still not be zero with zero
input voltage [Vo ≠ 0 with Vi= 0]. This is due to
unavoidable imbalances inside the op-amp and one may have to
apply a small voltage at the input terminal to make output (Vo) =
0.

This voltage is called input offset voltage Vos. This is the voltage
required to be applied at the input for making output voltage to
zero (Vo = 0).
Let us determine the Vos on the output of inverting and non-
inverting amplifier. If Vi = 0 (Fig (b) and (c)) become the same as
in figure (d).

Total output offset voltage:

The total output offset voltage V OT could be either more or less
than the offset voltage produced at the output due to input bias
current (IB) or input offset voltage alone(Vos). This is because IB
and Vos could be either positive or negative with respect to
ground. Therefore the maximum offset voltage at the output of
an inverting and non-inverting amplifier (figure b, c) without any
compensation technique used is given by many op amps provide
offset compensation pins to nullify the offset voltage. A 10K
potentiometer is placed across offset null pins 1&5. The wipes
connected to the negative supply at pin 4. The position of the
wipes is adjusted to nullify the offset voltage.

Fig. 1.23. Compensation circuit for offset voltage

When the given (below) op-amps does not have these offset null
pins, external balancing techniques are used.

( )
V OT = 1+
Rf
V +R f
R1 OS f B

With Rcomp, the total output offset

( )
V OT = 1+
Rf
V +R I
R1 OS f OS
Thermal drift:
Bias current, offset current, and offset voltage change with
temperature. A circuit carefully nulled at 25ºC may not remain.
So when the temperature rises to 35ºC. This is called drift. Offset
current drift is expressed in nA/ºC. These indicate the change in
offset for each degree Celsius change in temperature.

2.11 DIFFERENTIAL AMPLIFIER

Thus far we have used only one of the operational amplifiers
inputs to connect to the amplifier, using either the "inverting" or
the "non-inverting" input terminal to amplify a single input signal
with the other input being connected to ground. But we can also
connect signals to both of the inputs at the same time producing
another common type of operational amplifier circuit called a
Differential Amplifier.

Basically, as we saw in the first tutorial about operational

amplifiers, all op-amps are "Differential Amplifiers" due to their
input configuration. But by connecting one voltage signal onto
one input terminal and another voltage signal onto the other
input terminal the resultant output voltage will be proportional to
the "Difference" between the two input voltage signals of V1 and
V2.

Then differential amplifiers amplify the difference between two

voltages making this type of operational amplifier circuit a
Subtractor unlike a summing amplifier which adds or sums
together the input voltages. This type of operational amplifier
circuit is commonly known as a Differential Amplifier
configuration and is shown below:

By connecting each input inturn to 0v ground we can use

superposition to solve for the output voltage Vout. Then the
transfer function for a Differential Amplifier circuit is given as:
 V 1−V a V 2−V b V a −( V out )
I 1= , I 2= , If =
R1 R2 R3
Summing point V a −V b
And

V b =V 2
( R4
R 2+ R 4 )
If

V b =0 then; V out (a)=−V 1

( )
R3
R1
If

V a =0 then; V out (b)=V 2

( R4
R2 + R4 )( R 1+ R 3
R1 )
V out =V out(a )+V out(b)

∴ V out =−V 1
( ) (
R3
R1
+V 2
R4
R 2+ R 4 )( R1 + R3
R1 )

When resistors, R1 = R2 and R3 = R4 the above transfer function

for the differential amplifier can be simplified to the following
expression:
Differential Amplifier Equation
R3
V out = ( V −V 1 )
R1 2

If all the resistors are all of the same ohmic value, that is: R1 =
R2 = R3 = R4 then the circuit will become a Unity Gain
Differential Amplifier and the voltage gain of the amplifier will
be exactly one or unity. Then the output expression would simply
be Vout = V2 - V1. Also note that if input V1 is higher than input
V2 the ouput voltage sum will be negative, and if V2 is higher
than V1, the output voltage sum will be positive.
The Differential Amplifier circuit is a very useful op-amp circuit
and by adding more resistors in parallel with the input resistors
R1 and R3, the resultant circuit can be made to either "Add" or
"Subtract" the voltages applied to their respective inputs.

2.12 SUMMING AMPLIFIER

Op-amp may be used to design a circuit whose output is the sum
of several input signals. Such a circuit is called a summing
amplifier or a summer.

The Summing Amplifier is a very flexible circuit based upon the

standard Inverting Operational Amplifier configuration. As its
name suggests, the ―summing amplifier‖ can be used for
combining the voltage present on multiple inputs into a single
output voltage.
We saw previously in the Inverting Operational Amplifier that the
inverting amplifier has a single input voltage,( Vin ) applied to
the inverting input terminal. If we add more input resistors to the
input, each equal in value to the original input resistor, Rin we
end up with another operational amplifier circuit called a
Summing Amplifier,―summing inverter‖ or even a ―voltage adder
circuit as shown below.

2.12.1 Summing Amplifier Circuit

The output voltage, ( Vout ) now becomes proportional to the

sum of the input voltages, V1, V2, V3 etc. Then we can modify
the original equation for the inverting amplifier to take account
of these new inputs thus:

I F =I 1+ I 2 + I 3=−[ V1 V2 V3
+ +
R¿ R¿ R¿ ]
Inverting equation:
−R f
V out = XV¿
R¿
 [
−V out =
Rf
R¿
Rf Rf
V 1+ V 2 + V 3
R¿ R¿ ]

However, if all the input impedances, (Rin ) are equal in value,

we can simplify the above equation to give an output voltage of:
 RF
−V out = ( V + V + V … . etc )2.12.2 Inverting Summing Amplifier
R¿ 1 2 3

A typical summing amplifier with three input voltages V1, V2 and

V3 three input resistors R1, R2, R3 and a feedback resistor Rf is
shown in figure 2.

The following analysis is carried out assuming that the op-amp is

an ideal one, that is, AOL = ∞. Since the input bias current is
assumed to be zero, there is no voltage drop across the resistor
Rcomp and hence the non- inverting input terminal is at ground
potential.

Thus the output is the average of the input signals (inverted). In

a practical circuit, input bias current compensating resistor Rcomp
should be provided.

To find Rcomp, make all inputs V1 = V2 = V3 = 0. So the

effective input resistance Ri = R1 || R2 || R3.
Therefore, Rcomp = Ri || Rf = R1 || R2 || R3 || R,f.
2.12.3 Non-Inverting Summing Amplifier

A summer that gives a non-inverted sum is the non-inverting

summing amplifier of figure 3. Let the voltage at the (-) input
teriminal be Va.

The voltage at (+) input terminal will also be Va. Let R1 = R2 = R3

= R = Rf/2, then Vo = V1+V2+V3

The electronic circuits which perform the mathematical operations such as

differentiation and integration are called as differentiator and integrator,
respectively.
This chapter discusses in detail about op-amp based differentiator and
integrator. Please note that these also come under linear applications of
op-amp.

2.13 DIFFERENTIATOR
A differentiator is an electronic circuit that produces an output
equal to the first derivative of its input. This section discusses
about the op-amp based differentiator in detail.
An op-amp based differentiator produces an output, which is
equal to the differential of input voltage that is applied to its
inverting terminal. The circuit diagram of an op-amp based
differentiator is shown in the following figure −

In the above circuit, the non-inverting input terminal of the op-

amp is connected to ground. That means zero volts is applied to
its non-inverting input terminal.
According to the virtual short concept, the voltage at the
inverting input terminal of opamp will be equal to the voltage
present at its non-inverting input terminal. So, the voltage at the
inverting input terminal of op-amp will be zero volts.
The nodal equation at the inverting input terminal's node is −
Cd(0−Vi)dt+0−V0R=0Cd(0−Vi)dt+0−V0R=0
=>−CdVidt=V0R=>−CdVidt=V0R
=>V0=−RCdVidt=>V0=−RCdVidt
If RC=1sec RC=1sec, then the output voltage V0 will be −
V0=−dVidtV0=−dVidt
Thus, the op-amp based differentiator circuit shown above will
produce an output, which is the differential of input voltage ViVi,
when the magnitudes of impedances of resistor and capacitor
are reciprocal to each other.
Note that the output voltage V0V0 is having a negative sign,
which indicates that there exists a 1800 phase difference
between the input and the output.
2.14 INTEGRATOR
An integrator is an electronic circuit that produces an output
that is the integration of the applied input. This section discusses
about the op-amp based integrator.
An op-amp based integrator produces an output, which is an
integral of the input voltage applied to its inverting terminal.
The circuit diagram of an op-amp based integrator is shown in
the following figure −
In the circuit shown above, the non-inverting input terminal of
the op-amp is connected to ground. That means zero volts is
applied to its non-inverting input terminal.
According to virtual short concept, the voltage at the inverting
input terminal of op-amp will be equal to the voltage present at
its non-inverting input terminal. So, the voltage at the inverting
input terminal of op-amp will be zero volts.
The nodal equation at the inverting input terminal is −
0−ViR+Cd(0−V0)dt=00−ViR+Cd(0−V0)dt=0
=>−ViR=CdV0dt=>−ViR=CdV0dt
=>dV0dt=−ViRC=>dV0dt=−ViRC
=>dV0=(−ViRC)dt=>dV0=(−ViRC)dt
Integrating both sides of the equation shown above, we get −
∫dV0=∫(−ViRC)dt∫dV0=∫(−ViRC)dt
=>V0=−1RC∫Vtdt=>V0=−1RC∫Vtdt
If RC=1secRC=1sec, then the output voltage, V0V0 will be −
V0=−∫VidtV0=−∫Vidt
So, the op-amp based integrator circuit discussed above will
produce an output, which is the integral of input voltage ViVi,
when the magnitude of impedances of resistor and capacitor are
reciprocal to each other.
Note − The output voltage, V0V0 is having a negative sign,
which indicates that there exists 180 0 phase difference between
the input and the output.
2.15 VOLTAGE TO CURRENT CONVERTER WITH FLOATING
LOADS (V/I)

 Voltage to current converter in which load resistor RL

is floating (not connected to ground).
 Vin is applied to the non inverting input terminal,
and the feedback voltage across R1 devices the inverting input
terminal.
 This circuit is also called as a current – series
negative feedback amplifier.
 Because the feedback voltage across R1 (applied
Non-inverting terminal) depends on the output current i0 and is
in series with the input difference voltage Vid .
From the fig input voltage Vin is converted into output current of
Vin/R1 [Vin -> i0 ] . In other words, input volt appears across R 1. If R1
is a precision resistor, the output current (i 0 = Vin/R1 ) will be
precisely fixed.

Applications
 Low voltage ac and dc voltmeters
 Diode match finders
 LED
 Zener diode testers.

Fig. 3, shows a voltage to current converter in which load

resistor RL is floating (not connected to ground). The input
voltage is applied to the non-inverting input terminal and the
feedback voltage across R drives the inverting input terminal.
This circuit is also called a current series negative feedback,
amplifier because the feedback voltage across R depends on the
output current iL and is in series with the input difference voltage
vd.
FIG.3

Writing the voltage equation for the input loop.

V ¿ =V d +V f

But vd » since A is very large,therefore,

V ¿ =V f V ¿ =R¿
V¿
i ¿=
R

and since input current is zero.

V¿
i L =i ¿ =
R

The value of load resistance does not appear in this equation.

Therefore, the output current is independent of the value of load
resistance. Thus the input voltage is converted into current, the
source must be capable of supplying this load current.

2.15.1 Voltage – to current converter with Grounded load

This is the other type V – I converter, in which one terminal of the

load is connected to ground.
Grounded Load

If the load has to be grounded, then the above circuit cannot be

used. The modified circuit is shown in fig. 4.

FIG.4

Since the collector and emitter currents are equal to a close

approximation and the input impedance of OPAMP is very high,
the load current also flows through the feedback resistor R. On
account of this, there is still current feedback, which means that
the load current is stabilized.

Since
V d =0 V 2=V 1=V ¿
i out =( V cc −V ¿ ) /R

Thus the load current becomes nearly equal to iout. There is a

limit to the output current that the
circuit can supply. The base current in the transistor equals iout /
bdc. Since the op-amp has to supply this base current i out / bdc
must be less than Iout (max) of the op-amp, typically 10 to 15mA.

There is also a limit on the output voltage, as the load resistance

increases, the load voltage increases and then the transistor
goes into saturation. Since the emitter is at V in w. r. t. ground, the
maximum load voltage is slightly less than Vin.

In this circuit, because of negative feedback V BEis automatically

adjusted. For instance, if the load resistance decreases the load
current tries to increase. This means that more voltage is
feedback to the inverting input, which decreases VBE just enough
to almost completely nullify the attempted increase in load
current. From the output current expression it is clear that as V in
increases the load current decreases.

2.16 CURRENT TO VOLTAGE CONVERTER (I –V)

Sensitivity of the I – V converter
 The output voltage V0 = -RF Iin.
 Hence the gain of this converter is equal to -RF. The
magnitude of the gain (i.e) is also called as sensitivity of I to V
converter.
 The amount of change in output volt ∆V0 for a
given change in the input current ∆Iin is decide by the
sensitivity of I-V converter.
 By keeping RF variable, it is possible to vary the
sensitivity as per the requirements.