Operational Amplifiers

Dr. Deepti Gola

Assistant Professor
BIT Mesra, Ranchi
 Op-Amp – The Surface
▪ An Operational Amplifier (Op-Amp) is an integrated circuit that uses external
voltage to amplify the input through a very high gain.

▪ We recognize an Op-Amp as a mass-produced component found in

countless electronics.

Fig (1) OPAMP IC and Pin Configuration

What is an Op-Amp? – The Inside
 Op-amps (amplifiers/buffers in general) are drawn as a triangle in a circuit schematic

 The actual count varies, but an Op-Amp contains several Transistors, Resistors, and a
few Capacitors and Diodes.

 For simplicity, an Op-Amp is often depicted as follows which has minimum of five
terminals such as inverting input, non inverting input ,power supply and an output.

Positive Power Supply divot on pin-1 end

Inverting Input

Output

Non-Inverting Input

Negative Power Supply

Basic Block Diagram of Op-Amp

 An Op-Amp can be conveniently divided in to four main blocks

1. An Input Stage or Input Diff. Amp.

2. The Gain Stage
3. The Level Translator
4. An Out put Stage

V1 Input Stage Gain Stage (C Level Out put

I/P Shifter Stage VO
(Diff. Amp.) E Amp.) (Buffer)
V2

Op-Amp
IC
Basic Block Diagram of Op-Amp

• A Differential Amplifier i.e, it is the input stage for the op-amp. It

provides the amplification of the difference voltage b/w the two
inputs.

• A Voltage Amplifier(s) i.e, it is the second stage of op-amp

Basic Block Diagram of Op-Amp

 The internal op-amp formula is:

Vout = gain(V+ − V−)
 So if V+ is greater than V−, the output goes positive
 If V− is greater than V+, the output goes negative

V− −
Vout
V+ +

 A gain of 200,000 makes this device (as illustrated here)

practically useless
Ideal Op-Amp
The Ideal Op-Amp
•It has infinite voltage gain and infinite bandwidth.
•It has infinite input impedance (open) so that it does not load the driving source.
•It has zero output impedance.
•These characteristics are illustrated in figure. The input voltage ,Vin , appears b/w the
two terminals, and the output voltage is AvVin , as indicated by the internal voltage
source symbol.
Op-Amp Parameters
Common Mode

 Two signal voltages of the same phase, frequency and amplitude are
applied to the two inputs as shown is given in below figure.
 This results in a zero output voltage (as difference is 0V).
 This action is called common-mode rejection.
Op-Amp Parameters
CMRR (COMMON MODE REJECTION RATIO):

 Desired signals can appear on only one input or with opposite polarities on both input
lines. These desired signals are amplified and appear on the output.
 Unwanted signals (noise) appearing with the same polarity on both input lines are
essentially cancelled by the op-amp and don’t appear on the output. The measure of an
amplifier’s ability to reject common-mode signals is a parameter called the CMRR
(Common-Mode Rejection Ratio).

• The CMRR is often expresses in decibels (dB) as

Op-Amp Parameters

INPUT OFFSET VOLTAGE

• Ideal op-amp produces zero volts out for zero volts in.

• However, in op-amp a small dc voltage, VOUT(error) , appears at the output when no

differential input voltage is applied. Its primary cause is a slight mismatch of the base-
emitter voltages of the differential amplifier input stage of an op-amp.

• Input offset voltage, VOS ,is the differential dc voltage required between the inputs to
force the output to zero volts.

• Typical values of input offset voltage are in the range of 2mV or less. It is 0V in ideal case.
Op-Amp Parameters

INPUT OFFSET VOLTAGE

Op-Amp Parameters

INPUT BIAS CURRENT

• Input terminals of a differential amplifier are the transistor bases and therefore , the
input currents are the base currents as shown in the figure.
Op-Amp Parameters

INPUT OFFSET CURRENT

 Ideally, the two input bias currents are equal and thus their difference is zero.

 However, in a practical op-amp, the bias currents are not exactly equal.

 The input offset current , IOS is the difference of the input bias currents expressed as
an absolute value.
Op-Amp Parameters
INPUT IMPEDANCE

 The Differential input impedance is the total resistance between the inverting and
non-inverting terminals as illustrated in figure (a). Differential impedance is measured
by determining the change in bias current for a given change in differential input
voltage.

 The common-mode input impedance is the resistance between each input and
ground and is measured by determining the change in bias current for a given change
in common-mode input voltage shown in figure.
Op-Amp Parameters

OUTPUT IMPEDANCE

• The output impedance is the resistance viewed from the output terminal of the
op-amp as shown in figure below
Op-Amp Parameters
Slew Rate
 The slew rate (SR) of an op-amp is the maximum rate at which the output voltage can
change in response to input voltage.
 When the SR is too slow for the input, distortion results.
 For example when an input sine wave is applied to a voltage follower it produces a
triangular output waveform.
 The triangular waveform results because the op-amp simply cannot move fast enough
to follow the sine wave input.
 This happens because voltage change in the second stage ( Voltage Amplifier(s) ) is
limited by the charging and discharging of capacitors.
 The slew rate is expressed as :
SR = ∆VO/∆t
 The unit of SR is volts per microseconds.
 Inverting amplifier example
R2

R1
Vin −
Vout
+

• Applying the rules: − terminal at “virtual ground”

– so current through R1 is If = Vin/R1
• Current does not flow into op-amp (one of our rules)
– so the current through R1 must go through R2
– voltage drop across R2 is then IfR2 = Vin(R2/R1)
• So Vout = 0 − Vin(R2/R1) = −Vin(R2/R1)
• Thus we amplify Vin by factor −R2/R1
– negative sign earns title “inverting” amplifier
• Current is drawn into op-amp output terminal
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 Non-inverting Amplifier
R2

R1
−
Vout
Vin +

• Now neg. terminal held at Vin

– so current through R1 is If = Vin/R1 (to left, into ground)
• This current cannot come from op-amp input
– so comes through R2 (delivered from op-amp output)
– voltage drop across R2 is IfR2 = Vin(R2/R1)
– so that output is higher than neg. input terminal by Vin(R2/R1)
– Vout = Vin + Vin(R2/R1) = Vin(1 + R2/R1)
– thus gain is (1 + R2/R1), and is positive
• Current is sourced from op-amp output in this example

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 Summing Amplifier
Rf
R1
V1

−
R2
Vout
V2 +

• Much like the inverting amplifier, but with two input

voltages
– inverting input still held at virtual ground
– I1 and I2 are added together to run through Rf
– so we get the (inverted) sum: Vout = −Rf(V1/R1 + V2/R2)
• if R2 = R1, we get a sum proportional to (V1 + V2)
• Can have any number of summing inputs
– we’ll make our D/A converter this way

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 UCSD: Physics 121; 2012

Differencing Amplifier
R2

R1
V− −
Vout
V+ +
R1
R2

• The non-inverting input is a simple voltage divider:

– Vnode = V+R2/(R1 + R2)
• So If = (V− − Vnode)/R1
– Vout = Vnode − IfR2 = V+(1 + R2/R1)(R2/(R1 + R2)) − V−(R2/R1)
– so Vout = (R2/R1)(V+ − V−)
– therefore we difference V+ and V−

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 Differentiator (high-pass)
R

C
Vin −
Vout
+

• For a capacitor, Q = CV, so Icap = dQ/dt = C·dV/dt

– Thus Vout = −IcapR = −RC·dV/dt
• So we have a differentiator, or high-pass filter
– if signal is V0sint, Vout = −V0RCcost
– the -dependence means higher frequencies amplified more

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 Low-pass filter (integrator)
C

R
Vin −
Vout
+

• If = Vin/R, so C·dVcap/dt = Vin/R

– and since left side of capacitor is at virtual ground:
−dVout/dt = Vin/RC
– so

– and therefore we have an integrator (low pass)

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