Total Points Possible: 60
For the circuit shown, Vin=2 mV Ri=R3=30 k12, R2=R4=120 k12, VDDP= 12 V and VDDN=-12 V and the OPAMP is ideal and you notice
there is negative feedback.
VOUT
V
R4
R1
M
{R1}
R2
M
{R2}
R3
M
{R1}
{R2}
VDDP.
VDDP
OUT
U1
U2
IN
V3
VDDNA
VDDN
+
{VIN}
The relative tolerance for this problem is 1 %.
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Question 1. Points possible: 10
Unlimited attempts.
For the circuit shown, Vin=2 mV R1=R3=5 k12, R2=R4= 60 k12, VDDP=9 V and VDDN=-9 V and the OPAMP is ideal and you notice there is
negative feedback.
VOUT=
V
R2
R3
R4
IN
R1
M
{R1}
{R2}
{Ř1}
{Ř2)
V3
U1
U2
{VIN}
The relative tolerance for this problem is 1 %.
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Question 2. Points possible: 10
Unlimited attempts.
For the circuit shown, Vin=1 mV R1=R3=15 kN2, R2=R4= 60 k12, VDDP= 24 V and VDDN=-24 V and the OPAMP is ideal and you notice
there is negative feedback.
VOUT
V
R1
R2
R3
R4
IN
{R1}
{R2}
{R1}
{R2}
V3
U1
U2
{VIN}
The relative tolerance for this problem is 1%.
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Question 3. Points possible: 10
Unlimited attempts.
R1
R2
{R1}
{R2}
-VDDP
+VDDP
U1
U3
R3
out
in 1
V1
+
{R3}
VDDN
VDDP
VDDN
U2
R4
{V1}
in 2
V2
+
{R4}
VDDN
+ VDDP
VDDN
{V2}
VSP
VSN
{VDDP)
{VDDN)
For the circuit shown assume the OPAMP is ideal.
In terms of V1, V2, Vx, R1, R2, R3, or R4, write the equation for the node voltage at X, due to V1 only
Preview
In terms of V1, V2, Vx,R1, R2, R3, or R4, write the equation for the node voltage at X, due to V2 only
Preview
In terms of Vx, R1, and R2 write the equation for the node voltage at out
Preview
In terms of V1, V2, R1, R2, R3, or R4, write the equation for the node out, due to V1 and V2
Preview
If Ri=1000 32, R2= 20000 12, R3=5000 12, R4=5000 12, V1=-0.01 V, V2= -0.02 V, VDDP=24 V, and VdDn=-24 V
Vout=
V.
The relative tolerance for this problem is 1 %.
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Question 4. Points possible: 10
Unlimited attempts.
R1
R2
R5
R6
in 2
Vb
{R1}
{R2}
{R5}
{R6}
{Vb}
U2
U1
x
R3
in1
M
Va {R3}
R4
{R4}
{Va}
{
+
For the circuit shown assume the OPAMP is ideal.
In terms of Va, Vb, VX, R1, R2, R3, R4, R5, Ro, write the equation for the node voltage at X, due to Va only
Preview
In terms of Va, Vb, Vx, R1, R2, R3, R4, R5, and Ro, write the equation for the node voltage at X, due to Vh only
Preview
In terms of Vx, R5, and Ro write the equation for the node voltage at out
Preview
In terms of Va, Vb, R1, R2, R3, R4, R5, and Ro, write the equation for the node out, due to Va and Vb
Preview
If Ri=10000 12, R2=100000 S2, R3=10000 12, R4=100000 12, R3=20000 12, R6=180000 12, Va=0.01 V, Vb=-.02 V, VDDP=24 V, and VDDN=-24
V
Vout=
V.
The relative tolerance for this problem is 1 %.
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Question 5. Points possible: 10
Unlimited attempts.
For the circuit shown, 11=5 uA R1=500 k32, C1= 200 pF, VDDP= 12 V and VDDN=-12 V and the OPAMP is ideal and you notice there is
negative feedback.
VOUT due to I1 at DC (frequency=0Hz)
V
C1
HE
{C1}
R1
M
{R1}
11
8
U1
d
OUT
{11}
VDDN
C2
R2
{R1}
{C1}
The relative tolerance for this problem is 1 %.
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Question 6. Points possible: 10
Unlimited attempts.
Total Points Possible: 90
R1
R2
{R}
{R}
out
U1
VDDNA ZVDDP
;op
Х
R4
M
{R}
R3
in
M
{R}
V1
MV
D1
For the circuit shown, derive an equation for Vout using nodal analysis in terms of VIN, N, VT, Io, R3 if the OPAMP is ideal
VOUT=
Preview
VD
Note: ID = I. . envy
Note: R3-R1=R2=R4
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Question 1. Points possible: 10
Unlimited attempts.
R2
R1
M
{R}
M
{
R}
out
U1
VDDNÆVDDP
;op
Х
in
R3
M
{R}
R4
M
{R}
V1
D1
D
For the circuit shown, derive an equation for Vout using nodal analysis in terms of VIN, N, VT, Io, R3 if the OPAMP is ideal . All
resistances are equal.
VOUT=
Preview
VD
Note: Ip = 1. envy
Note: R3=R1=R2=R4
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Question 2. Points possible: 10
Unlimited attempts.
For the circuit shown below complete all steps to derive the input output time based relationship using nodal analysis. Assume the OPAMP is
ideal.
Note: This question has derivatives and integrals which are part of the answers. Myopenmath does not allow these functions to be used in the
question answers. To enter in a function that has an integral in it use the variable Int (these answers do not need limits.) and dt. For example
if the answer is supposed to be A rdt then enter in to the answer box as A*Int*VIN*dt. If you need to enter in a derivative such as
d(Vx – VOUT)
Ici C1
then enter into the answer box as C1*d*(VX-VOUT)/dt.
dt
Afrinatt
=
To be mathematically precise when integrating over time one should change the integration variable to T
[ f(-)dr. Given that this would be to much overhead to program, just use dt. Vin is assume to be Vivo).
For the circuit shown, derive an equation for Vout using nodal analysis in terms of Int, dt VIN, R3 and C1 if the OPAMP is ideal.
VOUT
Preview
R1
M
{R}
R2
M
{R}
VDDP
out
U1
;op
VDDN,
X
R3
R4
in
{R}
{R}
V1
C1
{C1}
Note: R3=R1=R2=R4
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Question 3. Points possible: 10
Unlimited attempts.
R1
R2
{R}
{R}
out
U1
VDDNA EVDDP
;op
х
R3
in
M
{R}
V1
R4
M
{R}
B1
I={gm}*V(X)*V(X)
For the circuit shown, derive an equation for Vout using nodal analysis in terms of VIN, R3 and gm if the OPAMP is ideal.
VOUT
Preview
Note: R3=R1 =R2=R4
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Question 4. Points possible: 10
Unlimited attempts.
R1
D1
in
Х
V1
{R}
ox
VDDNA AVDDP
U1
out
For the circuit shown, derive an equation for V out using nodal analysis in terms of VIN, R1 and lo, N, and VT if the OPAMP is ideal.
VD
Note: Id = 1, · enVT
VOUT=
Preview
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Question 5. Points possible: 10
Unlimited attempts.
R1
D1
in
Х
V1
{R}
out
VDDNA VDDP
up
For the circuit shown, derive an equation for V Tout using nodal analysis in terms of VIN, R1 and lo, N, and VT if the OPAMP is ideal.
VD
Note: Id = 1, · envy
VOUT
Preview
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Question 6. Points possible: 10
Unlimited attempts.
D1
in
X
R1
M
{R}
V1
out
VDDN AVDDP
For the circuit shown, derive an equation for Vout using nodal analysis in terms of VIN, R1 and lo, N, and VT if the OPAMP is ideal. Note:
Use e^(x) instead of exp(x).
VD
Note: ID = I envy
VOUT=
Preview
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Question 7. Points possible: 10
Unlimited attempts.
D1
in
X
计
R1
N
{R}
V1
D
out
U1
VDDNA VDDP
For the circuit shown, derive an equation for Vout using nodal analysis in terms of VIN, R1 and Io, N, and VT if the OPAMP is ideal. Note:
Use e^(x) instead of exp(x).
VD
Note: ID = I. enVT
VOUT=
Preview
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Question 8. Points possible: 10
Unlimited attempts.
For the circuit shown below complete all steps to derive the input output time based relationship using nodal analysis. Assume the OPAMP is
ideal.
Note: This question has derivatives and integrals which are part of the answers. Myopenmath does not allow these functions to be used in the
question answers. To enter in a function that has an integral in it use the variable Int (these answers do not need limits.) and dt. For example
if the answer is supposed to be A then enter in to the answer box as A*Int*VIN*dt. If you need to enter in a derivative such as
d(Vx – VouT)
Ici = C1
then enter into the answer box as C1*d*(VX-VOUT)/dt.
dt
ſvindt
To be mathematically precise when integrating over time one should change the integration variable to t
[* $(7)dr. Given that this would be to much overhead to program, just use dt. Vin is assume to be VINCO.
Write the nodal equation at Node X in terms of Iri and Ici. based on the drawn current directions and assigned voltage drops. Use the
convention that currents into the node are on the left hand side of the equals sign and the current leaving the node are entered in on the
right hand side of the equals sign:
Preview
Preview
(Equation 1)
Iri in terms of VIN, RVOUT and/or Vx is
Preview
Ici in terms of Vin, VOUT, VX, C1, d and dt is is
Preview
Based on the facts that the OPAMP is ideal and Rį provides negative feedback form Vout to Vx, Vx=1
Based on the correct answer for the last question:
Iri in terms of VIN, R, VOUT and/or Vx is
Preview
(Equation 2)
Ici in terms of VIN, VOUT, VX, C1, d and dt is is
Preview
(Equation 3)
Substitute equation 2, and 3 into equation 1 and solve for Vout in terms of R1, C1, Int, d, dt, Vin, and Vx:
VouT=
Preview
+ C1
+ R1
IN
누
С
R
V1
4VDDP
ç
+
X
OUT
V
VDDN
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Question 9. Points possible: 10
Unlimited attempts.
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Total Points Possible: 80
For the circuit shown I1=1 uA, R1 = 5000 12, Ri= 200 12, VDDP=3.3 V and VDDN=-3.3 V and the OPAMP is ideal and you notice there is
negative feedback.
Vout=5
✓mV
IR1 = 1
UA
IRL=2.5x10^10
* uA
11
R1
M
{R1}
{11}
4VDDP
OUT
U1
.op
RL
VDDN
{RL}
The relative tolerance for this problem is 1 %.
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Question 1. Points possible: 10
Unlimited attempts.
Score on last attempt: (3.33, 3.33, 0), Score in gradebook: (3.33, 3.33, 0), Out of: (3.33, 3.33, 3.34)
For the circuit shown I1=10 uA, R1 = 1000 S2, RL= 2000 12, VDDP= 24 V and VDDN=-24 V and the OPAMP is ideal and you notice there is
negative feedback.
VOUT = -10
✓ mV
IR1 =10
UA
IRL=-5000000000
|x pA
{11}
R1
M
{R1}
11
OUT
VDDP
ş
VDDN
.op
RL
{RL}
The relative tolerance for this problem is 1 %.
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Question 2. Points possible: 10
Unlimited attempts.
Score on last attempt: (3.33, 3.33, 0), Score in gradebook: (3.33, 3.33, 0), Out of: (3.33, 3.33, 3.34)
For the circuit shown 11=500 uA, 12=200 uA, R1 = 2000 12, Ri= 200 2, VDDP=3.3 V and VDDN=-3.3 V and the OPAMP is ideal and you
notice there is negative feedback.
VOUT due to 11
* mV
VOUT due to 12
* mV
VOUT due to both I1 and 127
* mV
11
R1
V
{R1}
{11}
{12}
ÆVDDP
OUT
A
U1
12
VDDNA
RL
{RL}
The relative tolerance for this problem is 1 %.
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Question 3. Points possible: 10
Unlimited attempts.
Score on last attempt: (0, 0, 0), Score in gradebook: (0, 0, 0), Out of: (3.33, 3.33, 3.34)
For the circuit shown 11=200 uA, 12=500 uA, R1 = 2000 12, R2 = 2000 12, RL= 5000 S2, VDDP= 24 V and VDDN=-24 V and the OPAMP is
ideal and you notice there is negative feedback.
VOUT due to 11
* MV
VOUT due to 12
* mv
VOUT due to both 11 and 12
* mv
11
R1
M
{R1}
{11}
OUT
U1
VDDNA VDDP
RL
R2
{zi}
9
{RL}
12
R
The relative tolerance for this problem is 1 %.
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Question 4. Points possible: 10
Unlimited attempts.
Score on last attempt: (0, 0, 0), Score in gradebook: (0, 0, 0), Out of: (3.33, 3.33, 3.34)
For the circuit shown, VINI=20 mV, VIN2=10 mV, R1 = 10000 12, R2 = 2000 12 , R3 = 5000 12 Ri= 200 S2, VDDP=5 V and VDDN=-5 V
and the OPAMP is ideal and you notice there is negative feedback.
VOUT due to VIN1
= 0
* mV
VOUT due to VIN2
0
* mV
VOUT due to both VIN1 and VIN27
* mV
IN2
VIN2
R1
M
{R1}
R2
M
{R2}
R3
M
{R3}
IN1
{V2}
OUT
U1
VIN1
VDDNA ÆVDDP
RL
{V1}
{RL}
The relative tolerance for this problem is 1 %.
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Question 5. Points possible: 10
Unlimited attempts.
Score on last attempt: (0, 0, 0), Score in gradebook: (0, 0, 0), Out of: (3.33, 3.33, 3.34)
For the circuit shown, VIN1=20 mV, VIN2=10 mV, R1 = 20000 12, R2 = 5000 12 , R3 = 1000 12 Ru= 5000 12, VDDP=5 V and VDDN=-5 V
and the OPAMP is ideal and you notice there is negative feedback.
VOUT due to VIN1
* mV
VOUT due to VIN2
* mV
VOUT due to both VIN1 and VIN27
* mV
IN2
VIN2
R2
M
{R2}
R1
M
{R1}
+ 1
{V2}
OUT
U1
R3
IN1
VDDNA ÆVDDP
{R3}
RL
{RL}
VIN1
+
{V1}
The relative tolerance for this problem is 1 %.
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Question 6. Points possible: 10
Unlimited attempts.
Score on last attempt: (0, 0, 0), Score in gradebook: (0, 0, 0), Out of: (3.33, 3.33, 3.34)
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