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Op-Amps
Microprocessor Interface
Operational Amplifier
(Op-Amp)
+Vcc
 Very high differential
Input 1
gain
+
Vo
 High input impedance
Vd
Output

 Low output impedance
Input 2
 Provide voltage changes
-V
cc Rout~0
R
in~inf
(amplitude and polarity)
Vo  GdVd
 Used in oscillator, filter
and instrumentation
Gd : differenti al gain normally
 Accumulate a very high
ver y large, say 105
gain by multiple stages
IC Product
OFFSET
NULL
-IN
1
8
N.C.
2
7
V+
6
OUTPUT
5
OFFSET
NULL
+IN
3
V
4

+
DIP-741
OUTPUT A
1
-IN A
2
+IN A
3
V
4
8 V+

+
7 OUTPUT B

6
+
5 +IN B
-IN B
Dual op-amp 1458 device
Distortion
+V =+5V
cc
+
V
V
+5V
o
d

0
5V
V =5V
cc
The output voltage never excess the DC voltage
supply of the Op-Amp
Op-Amp Properties
(1) Infinite Open Loop gain
-
The gain without feedback
Equal to differential gain
Zero common-mode gain
Pratically, Gd = 20,000 to 200,000
V1
+
V2

i1~0
+
i2~0

(2) Infinite Input impedance
-
Input current ii ~0A
T- in high-grade op-amp
m-A input current in low-grade op-amp
(3) Zero Output Impedance
-
act as perfect internal voltage source
No internal resistance
Output impedance in series with load
Reducing output voltage to the load
Practically, Rout ~ 20-100 
Vo
Vo
Rout
Vo' +
Vload
Rload
Rload
 Vo
Rload  Rout
Frequency-Gain Relation
•
•
•
•
•
Ideally, signals are amplified
from DC to the highest AC
(Voltage Gain)
frequency
Gd
Practically, bandwidth is limited 0.707Gd
741 family op-amp have an limit
bandwidth of few KHz.
Unity Gain frequency f1: the
gain at unity
Cutoff frequency fc: the gain
drop by 3dB from dc gain Gd
20log(0.707)=3dB
1
0
fc
f1
(frequency)
GB Product : f1 = Gd fc
GainBandwidth Product
Example: Determine the cutoff frequency of an op-amp having a unit gain
frequency f1 = 10 MHz and voltage differential gain Gd = 20V/mV
(Voltage Gain)
Sol:
Since f1 = 10 MHz
Gd
0.707Gd
? Hz
By using GB production equation
f1 = Gd fc
fc = f1 / Gd = 10 MHz / 20 V/mV
= 10  106 / 20  103
= 500 Hz
10MHz
1
0
fc
f1
(frequency)
Ideal Op-Amp
Applications
Analysis Method :
Two ideal Op-Amp Properties:
(1) The voltage between V+ and V is zero V+ = V
(2) The current into both V+ and V termainals is zero
For ideal Op-Amp circuit:
(1) Write the kirchhoff node equation at the
noninverting terminal V+
(2) Write the kirchhoff node eqaution at the inverting
terminal V
(3) Set V+ = V and solve for the desired closed-loop
gain
Noninverting Amplifier
(1) Kirchhoff node equation at V+
yields, V  V

V
+
in
V
o
i

(2) Kirchhoff node equation at V
yields, V  0 V  Vo
Ra

Rf
0
(3) Setting V+ = V– yields
Vi Vi  Vo

0
Ra
Rf
Rf
Vo
 1
or
Vi
Ra
Ra
Rf
v+
vi
v-
+
v+
vo
R
1
Rf
v+
v-
Rf
Ra
2
v-
)vi

Rf
Voltage follower
o
R
f
Noninverting input with voltage divider
Rf
R2
vo  (1  )(
)vi
Ra R1  R2
v
v+
i
vo
v

a
+
vo  vi
R
R
Noninverting amplifier
vo  (1 
+
i

Ra
vi
v
R
1
R
2
v-
+
v
o

R
f
Less than unity gain
R2
vo 
vi
R1  R2
Inverting Amplifier
Rf
(1) Kirchhoff node equation at V+
yields, V  0
Ra

(2) Kirchhoff node equation at V
yields, Vin  V_ V  V
 o  0
Ra
Rf
(3) Setting V+ = V– yields
Vo  R f

Vin
Ra
V ~
in

V
o
+
Notice: The closed-loop gain Vo/Vin is
dependent upon the ratio of two
resistors, and is independent of the
open-loop gain. This is caused by the
use of feedback output voltage to
subtract from the input voltage.
Multiple Inputs
(1) Kirchhoff node equation at V+
yields, V  0

Va
Vb
(2) Kirchhoff node equation at V Vc
yields,
V_  Vo
Rf

V  Va V  Vb V  Vc


0
Ra
Rb
Rc
(3) Setting V+ = V– yields
c V
 Va Vb Vc 
j
Vo   R f       R f 
j a R j
 Ra Rb Rc 
Rf
Ra
Rb
Rc

V
o
+
Inverting Integrator
Now replace resistors Ra and Rf by complex
components Za and Zf, respectively, therefore
Zf
Vo 
Vin
Supposing
Za
(i) The feedback component
is a capacitor C, in
i.e., Z  1
f
jC
(ii) The input component is a resistor R, Za = R
Therefore, the closed-loop gain (Vo/Vin) become:
1
vo (t ) 
vi (t )dt

RC
vi (t )  Vi e jt
where
Zf
Za
V ~
in
V
o
V ~
What happens if Za = 1/jC whereas, Zf = R
Inverting differentiator

+
C
R

V
o
+
Op-Amp Integrator
Example:
(a) Determine the rate of change
of the output voltage.
C
R
+5V
0
100s
(b) Draw the output waveform.
V
i
10 k
0.01F

V
o
+
Vo(max)=10 V
Solution:
(a) Rate of change of the output voltage
Vo
V
5V
 i 
t
RC (10 k)(0.01 F)
 50 mV/ s
+5V
0
V
i
0
(b) In 100 s, the voltage decrease
Vo  (50 mV/ s)(100μs)  5V
-5V
-10V
V
o
Op-Amp Differentiator
R
C
0
to
t1
t2
V

i
V
o
0
+
to
 dV 
vo   i  RC
 dt 
t1
t2
Slew Rate
The maximum possible rate at which an amplifier’s
output voltage can change, in volts per second, is called
its slew rate.
FIGURE 10-17 The rate of change of a linear, or ramp, signal is the change in voltage
divided by the change in time