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Chapter 6: The Operational Amplifier
Exercises
Ex. 6.4–
Ex. 6.4-
Ex. 6.5-
Ex. 6.6–
Ex. 6.7-
v =
R
R R
v
v R
v R
a 2 1
s
a a 0
2
3 4
v
v R
v R
v v
= R
R R
s
0 s
2 1
0 2
0 1
2
v
in out in f f 1 out 1 in f 1 out^3 in^3
v v (^) R R R 0 = 0^ v^ =^ 1+^ v R when R = 100k and R 25k then v (^10010) v =^ 1+^ = 5 2510
− v ^
- ⇒ Ω = Ω (^) ⋅ ⋅
2 2 0 4 (^2 1 1 2 2) s 3
R R v R When R >>R then (^) R + R − (^) R = 1 and (^) v − 1 +R
v R
v R
v v
= 1+ R
R
s s 0
0 s
2 1
1 2
a)
0 4 0 2 4 a 3 s (^1 2 )
v R v (^) R R v 1 R v 1 R R R
= + ⇒ = + +
b)
3 os b os b 3 3 3 os b
Analysis of the circuit in Section 6.7 showed that output offset voltage = 6 v (50.10 )i For a A741 opamp, v 1mV and i 80nA so output offset voltage 6 v (50.10 )i 6 (10 ) (50.10 )(
μ −
≤ ≤ = + ≤ + 80.10 −^9 ) =10mV
Ex. 6.7-
Ex. 6.7-
Ex. 6.7-
(^2 ) 0 1 in 1 os 2 b
2 1 os bl 3 3 9 3
v R^ v 1 R v R i R (^) R When R 10k , R 2k , v 5mV and i 500nA then output offset voltage 6 5.10 −^ 10.10 500.10 −^ 35.10 − 35mV
= − + + + = Ω = Ω ≤ ≤ ≤ + ≤ =
3 os b os b 3 3 9
Analysis of this circuit in Section 6.7 showed that output offset voltage = 6v 50 10 i For a typical OPA1O1AM, v 0.1mV and i 0.012nA so output offset voltage 6 0.1 10 −^ 50 10 0.012 10−
- × = = ≤ ^ × ^ + × × 0.6 10 3 0.6 10 6 0.6 10^3 0.6mV
− − −
≤ × + × − × ≤
V V
R
V V
R
V
R R
V A
R
R R
V
R
V V
R
After some algebra
A
V
V
R R R AR R
R R R R R R R R R AR R
For the given values, A
s a
o b i s
0 i i s 0
0 b
v 0 s
0 i s i f f 0 i s a f 0 i s i a v
− − −
− −
Ex. 6.8-
Spice deck V1 1 0 200mV V2 2 0 125mV V3 3 0 250mV R1 1 5 5k R2 2 5 2.5k R3 5 4 10k XOA1 5 0 4 UA741_OP_AMP R4 4 6 2.5k R5 3 6 1k R6 6 7 5k R7 7 0 10k XOA2 6 0 7 UA741_OP_AMP SUBCKT UA741_OP_AMP 1 2 5 IB1 1 0 70nA IB2 2 0 90nA VOS 3 2 1mV RI 1 3 2MEG E 4 0 1 3 - RO 4 5 75 .ENDS UA741_OP_AMP .END NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE ( 1) .2000 ( 2) .1250 ( 3). ( 4) -.8958 ( 5) .0010 ( 6) 997.2E- ( 7) .5429 (XOA1.3) .0010 (XOA1.4) -. (XOA2.3) .0010 (XOA2.4).
Problems
Section 6-4: The Ideal Operational Amplifier
P6.4-
P6.4-
P6.4-
v V v 12
V
i 8000
mA
a 0
0 0 0
v v .
v 0 = v a − i 01 4000 6 = − 2 4. − 4 2 4 .× 10 −^391 4000 6 = − 12 V
KVL 1:
i+0+2000i = 0
i = 12 5000
mA i i = 2.4mA i i = 2.4mA v i = V
0 1 a 1
a − +
⇒ =
= = − + −
P6.4-
P6.4-
v Ri v i
R
0 s 0 s
^
^
^
^
^
^
^
^
^
v R
v R
v R R
v
i R R
R R
R R
v
R R
R R
v 0 R
v 0 R v
R
R
v
R R
R R
v
s 1
a 2 a 2 1 s
0 a 2
a 3
2 3 2 3
a 2 3 1 3
s
0 4
a 3
0 4 3
a 2 4 1 3
s
v v v
P6.4-
P 6.4-
KCL at node a:
( 18 )^
0 0 12 V
a (^) a a
v v
v
The node voltages at the input nodes of ideal op amps are equal so vb = va.
Voltage division:
8 V
v o = vb = −
P6.5-
P6.5-
P6.5-
KCL at node b:
a c c a
v v
v v
e e
KCL at node a:
0 V
a a a (^) a a a
v v
v v v
v
e e e e
− ^ −
− − + ^
+ + + ^ = ⇒ = −
so
c 4 a 16
v = − v = −.
i = v^ v R v R R R i
=
R R R
R
v v
1 2 2 0 1 2 3 1 2 3 2
1 2
1 a 1 2 1 1 1 7 a^71 2 b (^1 2) b 2 2 2 1 (^2 7 7 )
v v (^) R R R R 0 0 1 v^ R v R v v v v R R 0 0 v 1 v v R R R (^) R
− v − v (^) v ^
- = ⇒ = + − − − ^ − + = ⇒ = (^) +
b c c^6 4 6 c^4 6 b a c c (^0 0 5) a 5 c 3 5 3 3
v v v^0 0 0 v R v R R R R v v v v (^0 0) v R (^) v (1+ R )v R R R R
(^) − (^) − − + + = ⇒ = +
− −^ + − + = ⇒ = − +
5 1 6 3 5 2 5 1 6 3 5 2 0 3 7 3 4 6 7 2 3 7 3 4 6 7 1
c 0 0
R R R^ R^ R^ R R R R^ R^ R^ R v (1+ ) v (1+ ) v R R R R R R R R R R R R v v i (^) R
(^) + (^) + = + − + +^ + − = = "
P6.5-
P6.5-
P6.5-
KCL at node b:
0 4 V
b b b
v v
v
e e
vc = vb = − 4 V because the node voltages at the input nodes of an ideal op amp are equal.
vd = vc + 0 10 3⋅ e = −4 V because the currents into the inputs of an ideal op amp are zero.
KCL at node g:
f g g g f
v v v
v v
e e
KCL at node d:
0 V
d f^ d^ f f d
v v v^ v
v v
e e
+ = ⇒ = = − so
V
g 3 f 5
v = v = −.
V
v c = vg = − because the node voltages at the input nodes of an ideal op amp are equal
^
^
^
^
− + + − − + +^ −^ =
v v
V
v v (^) + v v^0
a a
a a a a b a
v v
i i i =^1 6
mA
v = 10000 i =^10 6
V
v 10000
v 20000
0 0 0
a 0 a a 0
v
+ −^ − −
^
^
^
+ + −^ = ⇒ = −
3v v 6 + v v 3 v 6 v 0 v 2v 6 3 V v 10000
v v 30000
v v 30000
v 6 v 10000
3v v v v v 3 v 6 v 0 v 8v 4v 18 12 V
i v 30000
v v 30000
0 i 0.7mA
a a a b a b b a b b 0 a b a^ b
b b 0 a b a b 0 b a
0 0 0 b 0
⇒ v 0 = 3v (^) a = 5 V
P6.6-
P6.6-
v
R
R R
v
v v R
v v R
a 4 3 4
2
a 1 1
a 0 2
(a)
^
^
^
^ +
v 0 2 v 1
a 2 1
1
2 1
4 3 4 2 2 1 1
2 1 3 4
2 2 1 1
R
R
R
R
v
= R
R
R
R R
v R R
v
R
R
R
R
v R R
v
(b) (^) 11 = R R
and 4 =
R
R
R
R
R
R
R
R
For example: R k , R k , R k & R k
2 1
2 1 3 4
3 4
3 4
1 2 3 4
Using superposition, vo = v 1 + v 2 + v 3 = − 9 − 16 + 32 =7 V
- P6.6-
- P6.6-
- P6.6-
- P6.6-
- P6.6-
- P6.6-
- R 1 6 12 24 6||12 6||24 12||12 12||24 6||12||12 6||12||24 12||12||
- R 2 12||12||24 6||12||24 6||12||12 12||24 12||12 6||24 6||12
- -v o /v s 0.8 0.286 0.125 2 1.25 0.8 0.5 8 3.5 1.
P6.6-
P 6.6-
( )
( ) ( ( ) )
c a
o b c b a b a
v v
v v v v v v v
v k
v k
v 20k
v 80k +4 (v +10v )+4v v k v
a 1 a 2 a a 0
a 1 a 2 a a 0 a 0 1 2
v v v
v v v v
v v
Section 6-7: Characteristics of the Practical Operational Amplifier
P6.7-
P6.7-
so v = 1+100k 10k
v +100k i = 11v k i
= 11 0.03mV +100k 1.2nA = 0.45mV
out os b1 os b
^
^
The node equation at node a is
so v + 90k 10k v k i + 90k i
mV + 90k (.05nA) = 50.0045mV mV
0 =^ os b1 os b
^
^ +^ =
v
( ) ~
The node equation at node a is
v k
v 10k
out os os (^) + ib −v 100
v 10k
os (^) b1 = 0 −vos 90 Ω
P6.7-
P6.7-
P6.7-
v R R +R
v R
v +v ) R
a 3 2 3 cm p
a 0 a cm n 1
v v
v v 4
v R
v R
v R v +Av R
v R
v v
R (R AR )
(R +R )(R +R )+R R (1+A)
1 in 1
1 in
1 0 2 0 1 0
0 1 2
0 in
in 0 2 1 in 0 2 1 in
KK
K
K
v v
v
v v
= 2M^ (75^ k^ )) (5.1k +2M )(75+49k )+(5.1k )(2M )(1+200,000) =
0 in
v v
= R
R
= 49k 5.1k
in
2 1
a)
v v
2M 75 k (5k +2M )(75+50k )+(5k )(2M )(1+200,000) =
0 in
b)
c)
v = R R
=
R
R
(R +R R
R (R +R
0 4 1
cm n 4 1 1
a
4 1
cm n 4 1 3 1 2 3
cm p
v v
v v v
when R R
R
R
then (R^ +R^ R R (R +R
R
R
R
R
R
R
R
R
so v R R
) = R
R
4 1
3 2
4 1 3 1 2 3
4 1 3 2
3 2
4 1
0 4 1
cm n 4 1
cm p 4 1
n p
× =
v v v v
