The Operational Amplifier
Why Study Op-Amps?
• Understand the “magic” of negative feedback and the characteristics of
ideal op amps.
• Understand the conditions for non-ideal op amp behavior so they can be
avoided in circuit design.
• Demonstrate circuit analysis techniques for ideal op amps.
• Characterize inverting, non-inverting, summing and instrumentation
amplifiers, voltage follower and first order filters.
• Learn the factors involved in circuit design using op amps.
• Find the gain characteristics of cascaded amplifiers.
• Special Applications: The inverted ladder DAC and successive
approximation ADC
Differential Amplifier Basics
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.
Ideal Operational Amplifier
• The “ideal” op amp is a special case of the ideal differential amplifier with infinite
gain, infinite Rid and zero Ro .
vo
v 
id A
and
lim vid  0
A 
• 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.
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.
Inverting Amplifier: Voltage Gain
• The negative voltage gain
implies that there is a 1800
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
vs  isR  i R  vo  0
1 2 2
• The inverting input of the op
But is= i2 and v- = 0 (since vid= v+ - v-= 0) amp is at ground potential
(although it is not connected
R
vs
vo
directly to ground) and is
is 
 2
and Av 
R
said to be at virtual ground.
vs
R
1
1
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
vx  i R  i R
2 2 11
But i1=i2
v
R  s  R since v  0
in i
1
s

vx  i ( R  R )
1 2 1
Since v- = 0, i1=0. Therefore vx = 0
irrespective of the value of ix .
Rout  0
The 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
AvdB 20log Av ,  Av 1040dB/20dB 100
10







A minus sign is added since the amplifier is inverting.
R  R  20k
1 in
R
Av   2  R 100R  2M
2
1
R
1

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 .
Non-inverting Amplifier: Voltage Gain, Input
Resistance and Output Resistance
Since i-=0
But vid =0
Since i+=0
Rand
1
v  vo
1
R R
1 2
vs  v  v
id 1
vs  v
1
R R
vo  vs 1 2
R
1
R
v o R1  R2
 Av 

 1 2
R
vs
R
1
1
vs
R  
in i

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.
Non-inverting Amplifier: Example
• Problem: Determine the output voltage and current for the given non-inverting
amplifier.
• Given Data: R1= 3kW, R2 = 43kW, vs= +0.1 V
• Assumptions: Ideal op amp
• Analysis:
R
43k
Av 1 2 1
15.3
R
3k
1
vo  Av vs  (15.3)(0.1V)1.53V
Since i-=0,
vo
1.53V
io 

 33.3A
R  R 43k  3k
2 1
Finite Open-loop Gain and Gain Error
vo  Av  A(vs  v )  A(vs  vo )
1
id
vo
A
Av 

v s 1  A
A is called loop gain.
R
1 v  v
v 
o
1 R R o
1 2
R
1
is called the

R R
feedback factor.
1 2
For A >>1,
R
1
Av  1 2

R
1

This is the “ideal” voltage gain of
the amplifier. If Ab is not >>1,
there will be “Gain Error”.
Gain Error
• Gain Error is given by
GE = (ideal gain) - (actual gain)
For the non-inverting amplifier,
GE 
1
A
1

 1 A  (1 A )

• Gain error is also expressed
as a fractional or percentage
1
A

error.
1
1
 1 A
FGE 
1


PGE 

1
100%
A

1 A A
Gain Error: Example
• Problem: Find ideal and actual gain and gain error in percent
• Given data: Closed-loop gain of 100,000, open-loop gain of 1,000,000.
• Approach: The amplifier is designed to give ideal gain and deviations from the
ideal case have to be determined. Hence,
.
 1
5
10
Note: R1 and R2 aren’t designed to compensate for the finite open-loop gain of
the amplifier.
• Analysis:
Av 
A

1 A
10 6
10 6
1
10 5

 9.09x104
10 5  9.09x10 4
PGE
100% 9.09%
5
10

Output Voltage and Current Limits
Practical op amps have limited output
voltage and current ranges.
Voltage: Usually limited to a few volts less
than power supply span.
Current: Limited by additional circuits (to
limit power dissipation or protect against
accidental short circuits).
The current limit is frequently specified in
terms of the minimum load resistance that
the amplifier can drive with a given output
voltage swing. Eg:
5V
io 
10mA
500
v
vo
v
io  i  i  o 
 o
L F R
R R R
L
2 1
EQ
R
 R (R  R )
L 1 2
EQ
For the inverting amplifier,
R
R R
L 2
EQ
The Summing Amplifier

Since the negative amplifier
input is at virtual ground,
v
v
1
i 
i  2 i   vo
1 R
2 R
R
1
2 3
3
Since i-=0, i3= i1 + i2,
R
R
vo   3 v  3 v
R 1 R 2
1
2
• Scale factors for the 2 inputs can
be independently adjusted by the
proper choice of R2 and R1.
• Any number of inputs can be
connected to a summing junction
through extra resistors.
• This circuit can be used as a simple
digital-to-analog converter. This
will be illustrated in more detail,
later.
The Difference Amplifier
R
Since v-= v+ vo   2 (v  v )
R 1 2
1
For R2= R1 vo  (v1  v2)
• 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
v o  v-  i R  v-  i R
because i+ is zero.
2 2
1 2


R R 
R
R
• For v2=0, Rin1= R1, as the circuit


2
1
2
2
 v- 
( v  v- )  
v
 v- 
reduces to an inverting amplifier.
1
1


R
R
R

1
1 
1

• For general case, i1 is a function of
R
both v1 and v2.
2 v
Also, v 

R R 2
1 2
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,
R
100k
A  2 
 10
dm
R
10k
1


Vo  A V V  10(5 3)
dm 1 2 
Vo  20.0 V
Here Adm is called the“differential mode voltage gain” of the difference amplifier.
The Common-Mode Rejection Ratio (CMRR)
A(or Adm) = differential-mode gain
Acm = common-mode gain
vid = differential-mode input voltage
vic = common-mode input voltage
A real amplifier responds to signal
common to both inputs, called the
common-mode input voltage (vic). In
general,
v  v 
vo  A (v  v ) Acm 1 2 
dm 1 2

2 

vo  A (v ) Acm(v )
ic
dm id

v
v  v  id
1 ic 2
v
v  v  id
2 ic 2
An ideal amplifier has Acm = 0, but for a
real amplifier,




Acm v 
v

ic  A v  ic 
vo  A v 

dm id
dm id CMRR 
A

dm 
A
CMRR  dm
Acm





and CMRR(dB) 20log (CMRR)
10
Instrumentation Amplifier
R
vo   4 (v a  v )
b
R
3
va  iR  i(2R )  iR  v
2
1
2 b
v v
i 1 2
2R
1
R  R 
vo   4 1 2 (v  v )
R  R  1 2
3
1
NOTE
Combines 2 non-inverting
amplifiers with the difference
amplifier to provide higher gain and
higher input resistance.
Ideal input resistance is infinite
because input current to both op
amps is zero. The CMRR is
determined only by Op Amp 3.
Instrumentation Amplifier: Example
•
•
•
•
Problem: Determine Vo
Given Data: R1 = 15 kW, R2 = 150 kW, R3 = 15 kW, R4 = 30 kW V1 = 2.5 V, V2 = 2.25 V
Assumptions: Ideal op amp. Hence, v-= v+ and i-= i+= 0.
Analysis: Using dc values,
R







R  30k  150k 
4
1
 22
A 
1 2  


dm
R
R  15k  15k 
3
1 
Vo  A (V V ) 22(2.5 2.25) 5.50V
dm 1 2

The SlewRate