Characteristics of an Ideal Op-Amp
• Infinite input impedance
• Zero output impedance
• Zero common-mode gain, or, infinite common-mode
rejection
• Infinite open-loop gain A
• Infinite bandwidth.
1
Difference Amplifier
•
A difference amplifier is one that responds to the difference between the two
signals applied at its input and ideally rejects signals that are common to the two
inputs.
v Id = v I 2 − v I1
vIcm
1
= (v I 1 + v I 2 )
2
v Icm
vo = Ad v Id + Acm v Icm
Ad
CMRR = 20 log
Acm
+
-
v Id
v I1 = v Icm −
2
+
v Id
2
+
v Id
2
v Id
v I 2 = v Icm +
2
2
More Characteristics of Op-Amp
• Since the ideal op-amp responds only to the difference between the
two input signals, the ideal op-amp maintains a zero output signal
when the two input signals are equal.
• When the two input signals are unequal, there is what is called a
common-mode input signal.
• For the ideal op-amp, the common-mode output signal is zero. This
characteristic is referred to as common-mode rejection.
• Another characteristic, because op-amp is biased by both positive and
negative power supplies, most op-amps are direct coupled devices (no
coupling capacitors are required on the input). Accordingly, the two
input voltages can be DC.
• Because the OP is composed of transistors biased in the active region
by the DC power supply, the output voltage is limited.
3
Difference Amplifier
R2
vI1
vI2
R1
-
R3
+
vo
R4
4
A Single Difference or Differential Amplifier
vout
R2
(v2 − v1 )
=
R1
R2
vout =
v Id
R1
Figure
8.10
R2
Ad =
R1
5
Instrumentation Amplifier
Input (a) and output (b) stages of
Instrumentation amplifier
Figure
8.14,
8.15
vout
RF  2 R2 
1 +

AV =
=
v1 − v2
R 
R1 
6
Design Example: Determine the range required for resistor R1 in the
instrumentation amplifier to realize a differential gain adjustable from 5 to
500. Assume RF = 2R, so that the difference amplifier gain is 2.
•
Assume R1 is a combination of a fixed resistor R1f and a variable resistor R1v.
Assume R1v = 100 kΩ
vout
RF  2 R2 
1 +

AV =
=
v1 − v2
R 
R1 
R1f
 2R 
500 = 21 + 2  and the minimum differential gain is
 R1 f 


R1v

2 R2 

5 = 2 1+
and from the maximum gain expression
 R1 f + 100 


2 R2 = 249 R1 f
249 R1 f
2 R2
=
1.5 =
R1 f + 100 R1 f + 100
R1 f = 0.606 kΩ and R2 = 75.5 kΩ
7
Op-amp Differentiator
dvS (t )
vout (t ) = − RF CS
dt
Figure 8.35
8
Large Signal Operation of Op Amp
• Like other amplifiers, op amps operate linearly over a limited range of
output voltages.
• Another limitation of the operation of op amps is that their output current
is limited to a specified maximum. For example, the op amp 741 is
specified to have a maximum output current of ±20 mA.
• Read Example 2.5.
• Another phenomenon that can cause nonlinear distortion when large
output signals are present is that of slew-rate limiting. This means there is
a specific maximum rate of change possible at the output of a real op amp.
This maximum is known as the slew rate (SR) of the op amp and is
defined as
dv
SR = o
dt max
9
Design Example. Design a difference amplifier with a
specified gain and minimum differential input resistance.
Design the circuit such that the differential gain is 30 and the
minimum differential input resistance is Ri = 50 kΩ
Ri = 2 R1 = 50 kΩ
R1 = R3 = 25 kΩ
R2
= 30,
Since the differential gain is
R1
we must have R2 = R4 = 750 kΩ
10
Design Example: Calculate the common-mode rejection ratio of a
differential amplifier. Consider the difference amplifier shown in page 2.
Let R2/R1 = 10 and R4/R3 = 11. Determine CMRR (dB)
vo = vo1 + vo 2
 R4 


R
R
3 v − ( R 2 )v
vo = (1 + 2 )
I1
R4  I 2
R1 
R1
 1+

R
3


11
vo = (1 + 10)(
)v I 2 − (10)v I1 = 10.0833v I 2 − 10v I1
1 + 11
v
v
v +v
vd = v I 2 − v I1; vcm = I1 I 2 ; v I1 = vcm − d ; v 2 = vcm + d
2
2
2
v
v
vo = 10.0833(vcm + d ) − 10(vcm − d )
2
2
vo = 10.042vd + 0.0833vcm ; vo = Ad vd + Acm vcm
Ad = 10.042; Acm = 0.0833
 10.042 
CMRR(dB) = 20log10 
 = 41.6 dB
0.0833


11
Op-amp Circuits Employing Complex Impedances
Vout
ZF
( jω ) = −
VS
ZS
Figure
8.20
Vout
ZF
( jω ) = 1 +
VS
ZS
12
Active Low-Pass Filter
Figure 8.21,
8.23
Normalized Response of Active Low-pass Filter
13
Active High-Pass Filter
Figure
8.24,
8.25
Normalized Response of Active High-pass Filter
14
Active Band-Pass Filter
Figure
8.26,
8.27
Normalized Amplitude Response of Active Band-pass Filter
15
Op-amp Integrator
1
vout (t ) = −
RS C F
t
∫ vS (t )dt
Figure
8.30
−∞
16
Op-amp Differentiator
dvS (t )
vout (t ) = − RF CS
dt
Figure 8.35
17