6
6
22
THE OPERATIONAL AMPLIFIER
The Operational Amplifier
Operational Amplifiers
OpAmp is an active component that can be used for amplification, addition, buffering, ... etc of signals. Its symbol has three
terminals, two terminals are for input and labeled n (inverting input) and p (noninverting input) and one output labeled Out.
Operational Amplifiers-Ideal
n
p
−
+
Out
opampsymbol
Figure 5: OPAmp Symbol
The following are the basic characteristics of an ideal OpAmp:
1. in = ip = 0
2. vp = vn
6
23
THE OPERATIONAL AMPLIFIER
Examples-Ideal OpAmp
3. Buffer Amplifier (Voltage Follower):
1. Inverting Amplifier:
i2 Rf
R1
−
vs +
−
i1
−
+
+vo
+
opamp006-inverting
vs +
−
−
vs
, and since in = 0, then
here, vp = 0 = vn , hence i1 = R
1
i1 = i2 .
Start from OpAmp.n node through Rf then vo we write
:
vs
kvl: i2 ∗ Rf + vo = 0, vo = −Rf ∗ i2 = −Rf ∗ R
1
−Rf
R1
vs
2. Noninverting amplifier:
−
1
+
Rf
if
vs +
−
+
v
2 o
i1
R1 −
opamp006-noninverting
Figure 7: Noninverting Amplifier
i1 R1
if Rf
−
+
+vo
vs
opamp006-noninvertingx
−
vs = vp = vn
and R1 i1 + vs = 0
vs
, also if = i1 .
i1 = − R
1
−vs + Rf if + vo = 0, then
vs
vo = vs − Rf if = vs − Rf (i1 ) = vs − Rf (− R
)
1
⇒ vo = vs (1 +
vo = vs (1 +
Rf
R1
Rf
R1
)
)
−
opampbuffer.m4
Figure 6: Inverting Amplifier
and vo =
+
v0
vs = vp = vn = vo , i.e
vo = vs
.......................................................
6
24
THE OPERATIONAL AMPLIFIER
Important Use of Buffer Amplifier
−
+
v0
+
vs +
−
−
opampbuffer.m4
The following example shows an important use of the Buffer Amplifier: Buffer amplifiers are used to isolate the load from
the circuit.
Consider the circuit:
20kΩ
+
vs +
−
60kΩ
vo
−
buffer-1.m4
60k
= 0.75vs
the voltage vo = vs 60k+20k
Consider a load of 30kΩ to be connected as shown below:
20kΩ
+
vs +
−
60kΩ
vo
30kΩ
−
buffer-2.m4
60k||30k
vs (60k||30k)+20k
the voltage vo =
= 0.5vs .
Notice that the voltage across the load dropped from 0.75vs to 0.5vs .
To keep the voltage across the load constant with a value =0.75vs , we insert a buffer amplifier between the circuit and the
load as shown below:
20kΩ
−
+
+
vs +
−
60kΩ
30kΩ
vo
−
buffer-3.m4
Here vp = 0.75vs as in the first circuit, then vn = vp = vo = 0.75vs .
Notice that the voltage across the load =0.75vs irrespective of the value of the load resistor.
6
25
THE OPERATIONAL AMPLIFIER
Operational Amplifiers-(Simplified)Linear
Approximation
In this model we have:
1. Very high input resistance Rin ≈ 10M egΩ
2. small output resistance Rout ≈ 75Ω)
3. High gain A ≈ 10000
VOLTAGE SOURCE CURRENTS
NAME
CURRENT
vs
-1.000E-04
TOTAL POWER DISSIPATION
n
Rout
Output
Rin
****
+
−
p
n
p
A(Vp − Vn )
−
+
Output
Op AMP-Circuit Model
Example: The following values are for an OpAmp. A =
200k, Rin = 2M eg, Rout = 50. Find Vout in the following circuit:
First replace the OpAmp by its equivalent circuit, the circuit becomes:
10kΩ
20kΩ
vs + 1V
−
−
+vo
+
−
1V
10kΩ
+ v
− s
2
20kΩ
−
4
2M Ω vd
+ +
−
50
3
+
Vout
−
(2e5)vd
opampgain
(a) Using PSpice
OpAmp.gain
vs
1
r12
1
r20
2
r23
2
r34
3
e40
4
.tf
v(3,0)
.end
0
2
0
3
4
0
vs
1.00E-04
WATTS
SMALL-SIGNAL CHARACTERISTICS
V(3,0)/vs = -2.000E+00
INPUT RESISTANCE AT vs = 1.000E+04
OUTPUT RESISTANCE AT V(3,0) = 7.525E-04
JOB CONCLUDED
TOTAL JOB TIME
.02
(b) Using Maxima
OpAmp tf.m4
1
****
SMALL SIGNAL BIAS SOLUTION
NODE
VOLTAGE
(
1)
1.0000
(
2) 10.02E-06
(
3)
-2.0000
(
4)
-2.0050
dc
10k
2Meg
20k
50
0
1
2
200k
(%i2)
(%o2)
globalsolve : true
true
- (v2 - vo)
v2
-(v2 - vs)
(%i3) eq1 : ----------- - --- + ----------- = 0
R2
Rin
R1
vo - v2
vs - v2
v2
(%o3)------- + ------- - --- = 0
R2
R1
Rin
-(vo - v2)
vo - A vd
(%i4)eq2 : ----------- - --------- = 0
R2
Ro
v2 - vo
vo - vd A
(%o4) ------- - --------- = 0
R2
Ro
(%i5) eq3 : v2 + vd = 0
(%o5) vd + v2 = 0
(%i6) solve([eq1, eq2, eq3], [vd, v2, vo])
Rin vs R2 + Rin Ro vs
(%o6) [[vd : - ----------------------------------,
(R1+Rin)R2+(Rin (A+1)+Ro)R1+Rin Ro
Rin vs R2 + Rin Ro vs
v2 : --------------------------------------------,
(R1 + Rin)R2 +(Rin (A + 1) + Ro) R1 + Rin Ro
Rin vs A R2 - Rin Ro vs
vo : - ------------------------------------------]]
(R1+Rin) R2 + (Rin (A+1) + Ro) R1 + Rin Ro
(%i7) ev([vd,v2,vo],FLOAT,[A=10000,Rin=2000000,
R1=10000, R2=20000,Ro=50])
(%o7)[- 2.00939E-4 vs, 2.00939E-4 vs, - 1.99939 vs]
6
26
THE OPERATIONAL AMPLIFIER
6.1
Difference Amplifier
Rf
R
v1 +
−
−
+
R
v2
+
−
RL
Rf
+
vo
−
opampdifference.amplifier.m4
For the loop from ground, v2 , R, OP.p, Rf , ground: −v2 + i2 ∗ (R + Rf ) = 0 (ideal opamp), hence vp = vn = i2 Rf =
−v1
1
o
− vnR−v
= 0 ⇒ −vn ( R
+ R1f ) + vR1 + Rvof = 0
KCL at OP.n: − vnR
f
Rf +R
v2 Rf Rf +R
vo
1
= vn ( RR
) − vR1 = R+R
( RRf ) − vR1 = vR2 − vR1 = v2 −v
, hence
Rf
R
f
f
vo =
6.2
Summing Amplifier
if Rf
+ vf −
R
v1 +
−
i1
Rf
(v2 − v1 )
R
−
R
v2
+
R
− + v
3
−
i2
i3
+
RL
+
vo
−
opampsumming.amplifier.m4
vp = vn = 0, and i1 = vR1 , i2 = vR2 , i3 = vR3
KCL at OP.n: i1 + i2 + i3 − if = 0, then if = v1 +vR2 +v3
R
KVL: +vf + vo = 0, then if Rf + vo = 0, ⇒ vo = −if Rf = − Rf (v1 + v2 + v3 )
hence
Rf
vo = −
(v1 + v2 + v3 )
R
v2 Rf
R+Rf
.