EE 321 Analog Electronics, Fall 2013
Exam 4 December 6, 2013
solution
This is a closed-book exam. You are only allowed pen, paper, brain (your own
only), and calculator. The exam is designed for conceptual understanding not
long derivations. You MUST box your answers. When there are multiple values
in an answer summarize them in a single box. To earn full points you need
the correct solution, the correct derivation, and all of it organized to be easily
readable.
Clamped capacitor
1. Draw a clamped capacitor circuit which clamps the minimum voltage of a
bipolar signal to ground.
Vin
Vout
2. Plot the input and output of this circuit when the input is a square wave
between −10 V and +10 V.
+10 V
−10 V
+20 V
0V
Diode circuit analysis
3. Carefully plot Vout as a function of Vin for this circuit, for −10 V < Vin <
10 V.
Vin
2R
Vout
R
4. Using two parallel diodes and a summing inverter design a circuit which
limits the input voltage to the ±0.7 V range and shifts/amplifies that range
to 0 − 2.8 V range appropriate for inputing into a microprocessor ADC.
MOSFET amplifier
2
A NMOS has kn′ W
=
1
mA/V
, and Vt = 1 V. VDD = 15 V, VSS = 0 V.
L
5. If it is used to build a common-source amplifier with VG = 3 V pick RD to
produce a gain of -10.
Avo = −gm RD = −
2ID RD
VGS − Vt
We have VGS = 3 V, and assuming saturation mode
ID =
1
2
kn′
W
L
(VGS − Vt )2 = 0.5 × 22 = 2 mA
Then re-arranging the expression for the gain
RD = −
Avo (VGS − Vt )
2ID
=
10 × 2
2×2
= 5 kΩ
MOSFETs at DC
In the following assume k′ W
=
1 mA/V2 , Vt = 1 V, voltage at top of 15 V, and
L
voltage and bottom of 0 V. In each case obtain iD , all voltages, and mode of
operation.
6. Do the analysis for this PMOS circuit
5K
We see that vSG = 5 V. Assuming saturation mode we
get
1
(5 − 4)2 = 8 mA
2
which produces vD = 8 V which is less than the gate
voltage. Thus saturation mode is confirmed. Answer is
then
iD =
10K
1K
vG = 10 V
vD = 8 V
vS = 15 V
iD = 8 mA
7. Do the analysis for this NMOS circuit
We have
10K
1K
vGS = VG − iD RS
and then, guessing saturation mode,
iD =
1
2
kn′
W
L
(vGS − Vt )2 =
1
2
kn′
W
L
(VG − iD RS − Vt )2
Inserting numbers in units of kΩ and V will resulting in
a solution for iD in mA.
2iD = (4 − iD )2
5K
1K
I guessed iD = 2 mA. Now the values are
iD = 2 mA
VG = 5 V
VS = 2 V
VD = 13 V