ECE 3410 – Homework 5
For all problems in this set, assume VDD = 5V. You are allowed to use SPICE to check your
answers (in fact it is encouraged), however you must also show that you can get the answer through
manual analysis. Here is an example SPICE file for a simple MOSFET circuit:
∗ Example SPICE s i m u l a t i o n f o r homework problems
∗ G e n e r i c MOSFET models :
. model nmos nmos ( l e v e l =2 KP=250u VT0=0.5 lambda =0.01)
. model pmos pmos ( l e v e l =2 KP=75u VT0=−0.75 lambda =0.01)
∗
∗
∗
∗
Nodes :
1 −− VDD
2 −− Drain o f NMOS
3 −− Gate o f NMOS
VDD 1 0 DC 5V
VG 3 0 DC 0 . 6V
R1 1 2 1k
∗ MOSFET s t a t e m e n t : MX Drain Gate S o u r c e S u b s t r a t e ModelName
M1 2 3 0 0 nmos
∗ O p e ra t i n g p o i n t s i m u l a t i o n :
. op
. end
Also note that SPICE analysis should not be necessary for logic problems.
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ECE 3410 – Homework 5
Problem 1. The MOSFET circuit shown below uses a MOSFET with the indicated characteristics.
For each of the cases listed, solve MOSFET’s device current and drain voltage, and
indicate whether the device is operating in saturation or triode. [Hint: To perform
this analysis, first assume the device is in saturation, and solve accordingly. Then check for
consistency. If the result is inconsistent, then repeat the analysis using the triode equation.]
VDD
R
Kn = 250µA/ V2
iD =?
VThN = 0.5V
vD =?
vG = 0.6V
(A) R = 100Ω
(B) R = 1kΩ
(C) R = 10kΩ
Solution
If the device is in saturation, then the device’s current and drain voltage are:
1
2
iD = kn vOV
2
vD = VDD − iD R
where vOV = vG − VTh = 0.1 V.
If the device is in triode, then the solution is more complex:
1 2
iD = kn vOV vD − vD
2
vD = VDD − iD R
1
2
⇒ iD = kn vOV (VDD − iD R) − (VDD − iD R)
2
⇒ 0 = Ai2D + BiD + C
Since the triode solution is more complex, it’s easier to first assume saturation and
see if the result is consistent with that assumption. In saturation, the device current
should always be 2.5 µA.
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ECE 3410 – Homework 5
Solution (cont.)
(A) R = 100Ω — In this case, we find
iD = 2.5 µA
vD = 4.999 75 V
This result is consistent with being in saturation since vD > vOV .
(B) R = 1kΩ — In this case, we find
iD = 2.5 µA
vD = 4.9975 V
This result is consistent with being in saturation since vD > vOV .
(C) R = 10kΩ— In this case, we find
iD = 2.5 µA
vD = 4.975 V
This result is consistent with being in saturation since vD > vOV .
Problem 2. The MOSFET circuit shown below uses a MOSFET with the indicated characteristics.
For each of the cases listed, solve MOSFET’s device current and drain voltage, and
indicate whether the device is operating in saturation or triode.
VDD
R
iD =?
Kn = 250µA/ V2
VThN = 0.5V
vD =?
(A) R = 100Ω
(B) R = 1kΩ
(C) R = 10kΩ
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ECE 3410 – Homework 5
Solution
In this problem, the device is always in saturation due the diode connection, which
guarantees vDS > vOV in all cases. The solution requires solving a quadratic:
vOV = vD − VTh
1
2
iD = kn vOV
2
vD = VDD − iD R
It is perhaps quickest to solve for vOV :
1
2
vOV = VDD − kn RvOV
− VTh
2
1
2
+ vOV + (VTh − VDD )
⇒ 0 = kn RvOV
2
Now we can use the quadratic formula to solve:
p
−1 ± 1 + 2kn R (VDD − VTh )
vOV =
kn R
vD = vD + VTh
VDD − vD
iD =
R
In the numerator, we choose ‘+’ from the ‘±’ in order to obtain a positive-valued
solution (a negative valued solution doesn’t make physical sense).
(A) R = 100Ω — In this case, we find
vOV = 4.27 V
vD = 4.77 V
iD = 2.28 mA
(B) R = 1kΩ — In this case, we find
vOV = 3.21 V
vD = 3.71 V
iD = 1.29 mA
(C) R = 10kΩ — In this case, we find
vOV = 1.54 V
vD = 2.04 V
iD = 296 µA
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ECE 3410 – Homework 5
Problem 3. The MOSFET circuit shown below uses a MOSFET with the indicated characteristics.
For each of the cases listed, solve MOSFET’s device current and drain voltage, and
indicate whether the device is operating in saturation or triode.
VDD
R
iD =?
Kn = 250µA/ V2
vD =?
VThN = 0.5V
vG = 1V
100
(A) R = 100Ω
(B) R = 1kΩ
(C) R = 10kΩ
Solution
In this problem, we can use the same procedure from Problem 1. Since RS is small,
it should have a minor effect on the DC solution. We can evaluate that effect by
iterating as follows:
vS = 0 (initial assumption)
vOV = vG − vS − VTh
1
2
iD = kn vOV
2
vD = VDD − iD R
vS ← iD RS
If we initially assume vS = 0, then we can solve for a new vS , then repeat the
equations until the results stabilizes. From this procedure, we find the following
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ECE 3410 – Homework 5
Solution (cont.)
results:
vS = 3.09 mV
iD = 30.9 µA
vOV = 4.97 V
As long as the device stays in saturation, these results should not depend on R and
will be the same for each case. Then the solutions are:
(A) R = 100Ω:
vD = 4.9969 V
vDS = 4.994 V > vOV
⇒ saturationX
(B) R = 1kΩ:
vD = 4.969 V
vDS = 4.966 V > vOV
⇒ saturationX
(C) R = 10kΩ:
vD = 4.691 V
vDS = 4.688 V > vOV
⇒ saturationX
Problem 4. The MOSFET circuit shown below uses a pair of identical MOSFETs with the indicated
characteristics. For each of the cases listed, solve MOSFETs’ device currents and drain
voltages, and indicate whether each device is operating in saturation or triode.
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ECE 3410 – Homework 5
VDD
VDD
R
R
iD1 =?
vD1 =?
M1
iD2 =?
Kn = 250µA/ V2
VThN = 0.5V
vD2 =?
M2
(A) R = 100Ω
(B) R = 1kΩ
(C) R = 10kΩ
Solution
Notice that the left-hand side of this circuit is identical to the one in Problem
2. The solution should be exactly the same. On the right hand side, there is an
identical MOSFET and an identical resistor. Furthermore the two MOSFETs share
identical gate and source voltages, and should therefore have identical currents
(iD2 = iD1 in all cases). With all other things being equal, the only remaining
unknown (vD2 ) should also be identical to its twin on the right-hand side. This is
called a symmetry argument, and is a very useful method for analyzing current
mirrors and differential circuits.
Problem 5. The MOSFET circuit shown below uses a MOSFET with the indicated characteristics.
For each of the cases listed, solve MOSFET’s device current and drain voltage, and
indicate whether the device is operating in saturation or triode.
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ECE 3410 – Homework 5
VDD
vD =?
Kp = 75µA/ V2
iD =?
VThP = 0.75V
R
(A) R = 100Ω
(B) R = 1kΩ
(C) R = 10kΩ
Solution
This problem is nearly identical to Problem 2, except that the PMOS device has
a different k value and the voltages are “upside down”. By modifying the solution
for Problem 2, we obtain:
vOV = VDD − vD − VThP
1
2
iD = kp vOV
2
vD = i D R
1
2
⇒ vOV = VDD − VThP − kp RvOV
2
1
2
⇒ 0 = kp RvOV
+ vOV + (VThP − VDD )
2
Notice that this is exactly the same equation as in Problem 2, except VTh and k
are numerically different. So we just need to repeat the quadratic equation from
Problem 2 to get the results:
(A) R = 100Ω:
vOV = 4.184 V
vD = 65.7 mV
iD = 656.7 µA
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ECE 3410 – Homework 5
Solution (cont.)
(B) R = 1kΩ:
vOV = 3.729 V
vD = 521.4 mV
iD = 521.4 µA
(C) R = 10kΩ:
vOV = 2.288 V
vD = 1.962 V
iD = 196.2 µA
Problem 6. The MOSFET circuit shown below uses a MOSFET with the indicated characteristics.
For each of the cases listed, solve MOSFET’s device current and drain voltage, and
indicate whether the device is operating in saturation or triode.
VDD
vG = 4.2V
vD =?
iD =?
Kp = 75µA/ V2
VThP = 0.75V
R
(A) R = 100Ω
(B) R = 1kΩ
(C) R = 10kΩ
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ECE 3410 – Homework 5
Solution
This problem is very similar to Problem 1. We begin by assuming the device is in
saturation, then solve for iD , then for vD , and finally verify that the device is in
saturation.
vOV = VDD − vG − VThP = 50 mV
1
2
iD = kp vOV
= 93.7 nA
2
vD = i D R
Applying these solutions to each of the cases:
(A) R = 100Ω:
vD = 9.37 µV
(B) R = 1kΩ:
vD = 93.7 µV
(C) R = 10kΩ:
vD = 937 µV
Problem 7. The MOSFET circuit shown below uses a MOSFET with the indicated characteristics.
For each of the cases listed, solve MOSFET’s device current and drain voltage, and
indicate whether the device is operating in saturation or triode.
VDD
vG
vD =?
iD =?
Kp = 75µA/ V2
VThP = 0.75V
R = 1kΩ
R
(A) vG = 4.5V
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ECE 3410 – Homework 5
(B) vG = 4.0V
(C) vG = 3.0V
(D) Assuming the device is in deep triode (i.e. it is in the linear region), solve for the
case when vG = 0V by approximating the device as a resistor with resistance RON .
Solution
Again, begin by assuming the device is in saturation. Then we have:
vOV = VDD − vG − VThP
1
2
iD = kp vOV
2
vD = iD R
saturation:VDD − vD > vOV
Applying this to the four cases:
(A) vG = 4.5V:
vOV = −0.25 V
⇒cutoff, iD = 0
vD = 0
(B) vG = 4.0V:
vOV = 0.25 V
iD = 2.34 µA
vD = 2.34 mV
|vDS | = 4.9977 V > vOV
⇒ saturationX
(C) vG = 3.0V:
vOV = 1.25 V
iD = 58.6 µA
vD = 58.6 mV
|vDS | = 4.941 V > vOV
⇒ saturationX
(D) vG = 0V: Here the device is obviously in triode. As an initial guess, we can
assume |vDS | = 0, so that vD = VDD , but then some current must be flowing
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ECE 3410 – Homework 5
Solution (cont.)
through R, so |vDS | must be greater than zero. To estimate the value of |vDS |,
we can use a Taylor approximation:
"
vD ≈ VDD − iD
diD
dvDS
#−1
vDS =0
= VDD − iD RON
By solving the derivative in the above equation, we find that the “on resistance” is given by
1
k (VDD − VTh )
= 3.137 kΩ
RON =
Then we can use a voltage-divider to predict vD :
R
vD ≈ VDD
R + RON
= 1.2 V
This result can be used as an initial guess for iterating with the triode formula.
Upon iteration, we obtain a more precise answer of vD = 0.677 V. These
results indicate that this device is not very good at pulling up the output
voltage across a 1 kΩ load (either the load resistance needs to be larger or we
need a MOSFET with a bigger k).
Problem 8. The MOSFET circuit shown below uses a MOSFET with the indicated characteristics.
For each of the cases listed, solve MOSFET’s device current and drain voltage, and
indicate whether the device is operating in saturation or triode. [Hint: To perform
this analysis, the usual procedure is to assume the device is in saturation, and solve
accordingly. Then check for consistency. If the result is inconsistent, then you must
repeat the analysis using the triode equation.]
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ECE 3410 – Homework 5
VDD
Kn = 250µA/ V2
R
iD =?
VThN = 0.5V
vD =?
R = 1kΩ
vG
(A)
(B)
(C)
(D)
vG = 0.25V
vG = 0.74V
vG = 3V
Assuming the device is in deep triode (i.e. it is in the linear region), solve for the
case when vG = 5V by approximating the device as a resistor with resistance RON .
Solution
This problem is very similar to Problem 7. We begin by assuming saturation:
vOV = vG − VThN
1
2
iD = kn vOV
2
vD = VDD − iD R
saturation:vD > vOV
Applying this to the four cases:
(A) vG = 0.25V:
vOV = −0.25 V
⇒cutoff, iD = 0
vD = VDD
(B) vG = 0.74V:
vOV = 0.24 V
iD = 7.2 µA
vD = 4.9928 V > vOV
⇒ saturationX
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ECE 3410 – Homework 5
Solution (cont.)
(C) vG = 3.0V:
vOV = 2.5 V
iD = 781 µA
vD = 4.219 V > vOV
⇒ saturationX
(D) vG = VDD : Here the device is obviously in triode. As an initial guess, we can
assume vD = 0 and use a Taylor approximation as before:
1
k (VDD − VTh )
= 889 Ω
RON =
Then we can use a voltage-divider to predict vD :
RON
vD ≈ VDD
R + RON
= 2.35 V
This result can be used as an initial guess for iterating with the triode formula.
Upon iteration, we obtain a more precise answer of vD = 2.82 V (so the RON
voltage divider prediction was pretty close). These results indicate that this
device is not very good at pulling down the output voltage across a 1 kΩ load
(either the load resistance needs to be larger or we need a MOSFET with a
bigger k).
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ECE 3410 – Homework 5
Problem 9. The logic circuits shown below use both NMOS and PMOS devices as switches. Analyze
each circuit and complete the truth tables by entering “L” or “H” in each position within
the table. From your analysis, state the logic operation implemented by each circuit.
VDD
A
VDD
Truth table:
B
Q
(A)
A
A
L
H
L
H
B
Q
L
L
H
H
What kind of logic gate is this?
B
VDD
Truth table:
B
A
(B)
Q
A
B
A
L
H
L
H
B
Q
L
L
H
H
What kind of logic gate is this?
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ECE 3410 – Homework 5
(C) For this circuit, recall that A refers to the logical inverse of A, i.e. if A = H then
A = L. For this problem, first identify standard logic gates and then analyze the
circuit using traditional logic-gate methods.
VDD
B
VDD
A
VDD
A
B
Q
VDD
B
A
B
A
Truth table:
A
L
H
L
H
B
Q
L
L
H
H
What kind of logic gate is this?
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ECE 3410 – Homework 5
Solution
(A) NAND gate:
A
L
H
L
H
B
L
L
H
H
B
(B) NOR gate:
A
L
H
L
H
B
(C) XOR gate:
A
L
H
L
H
L
L
H
H
L
L
H
H
Q
H
H
H
L
Q
H
L
L
L
Q
L
H
H
L
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