ECE 342 Problem Set #9 Due: 5 P.M. Wednesday, October 28, 2015 Fall Semester 2015 Prof. E. Rosenbaum Prof. T. Trick Reading Assignment: Sections 6.6.3, 6.6.4, 6.6.6, 5.6.6 1. Consider the circuit shown below, (a) Find Gv (vout/vsig) for IBIAS values of 0.1, 0.2, 0.5, 1 and 1.25 mA. (b) Why isn’t Gv a linear function of IBIAS? (c) For IBIAS = 1mA, what is the maximum allowed value of Vsig, where vsig = Vsig*sin( t)? The circuit must provide linear amplification. vsig ∞ VCC vout ∞ IBIA S RB=480kΩ -VEE RL=100kΩ ∞ Rsig=10kΩ RC=10kΩ ß = 100 VA=25V 2. Consider the circuit shown below, (a) Choose the value of Re such that Gv is maximized, subject to the constraint that vbe ≤ 10mV and the transistor stays in the active mode. What is the value of Gv? (b) If β drops by 10%, what is the new value of Gv? All other design variables are unchanged. vsig=0.05sin(ωt) ∞ +5V vout ∞ 0.2mA 100kΩ 20kΩ ∞ 20kΩ 20kΩ ß = 100 VA=∞ Re -5V 10V -10V 3. Find Rin, Rout, Gv, and the overall current gain io/isig (“Gi”). You are given that β=100 and VA=∞. VBE,ON = 0.7V. 100kΩ vsig ∞ 5V -5V 3.3kΩ 2kΩ vo io ii 4. Choose the value of IBIAS such that vout is a sinusoidal signal with amplitude 200 mV or larger. 5V 10kΩ vout 75Ω ∞ vsig=0.5sin(ωt) VIN=3V IBIAS ß = 100 VA=∞ 5. Consider the circuit shown below, (a) Find the dc bias point and small-signal model parameters. Assume λ = 0. (b) Find Rin, Rout, Avo and Gv. (c) Repeat part (b) for the case that λ = 0.03. You will have to recalculate ro. (d) Finally, you will consider the body effect; i.e., you will no longer assume that V B = VS but, instead, that VB = -VSS, which is -5V in this circuit. You are given γ = 0.4V 1/2 and 2ϕF = 0.6V. You may assume λ = 0 for simplicity. First, you need to recalculate the dc bias point. You may iteratively solve for VS and Vt, using ( ) and (√ √ ). Do not iterate more than 2 or 3 times, as this should be sufficient to obtain the value of V t with less than 10mV error. Then, redraw the small-signal model, now including the gmb current source shown in Fig. 5.62 of the textbook. The value of gmb can be found using equations (5.110) and (5.111). You are to calculate the value of Gv. 1MΩ vsig ∞ 5V 5kΩ vout 4.7MΩ -5V ∞ 0.5mA Vto=0.75V k=2mA/V 2 ENSC 324 HOMEWORK #2 Fall 2015 DUE: Monday October 19, 2015 at 2 PM (note new time!) Please note that unless you show work in the derivations and solutions you will get no credit for the answers. Obviously copied answers from study partners or other sources, etc., will also receive no credit. Please do all parts of all eight problems. It is suggested that you make a copy of your homework before turning it in in case it cannot be returned before Exam 2 (solution key will be provided). Problem #1 In a particular sample of n-type sil ...

1. ECE 342
Problem Set #9
Due: 5 P.M. Wednesday, October 28, 2015
Fall Semester 2015 Prof. E. Rosenbaum
Prof. T. Trick
Reading Assignment: Sections 6.6.3, 6.6.4, 6.6.6, 5.6.6
1. Consider the circuit shown below,
(a) Find Gv (vout/vsig) for IBIAS values of 0.1, 0.2, 0.5, 1 and
1.25 mA.
(b) Why isn’t Gv a linear function of IBIAS?
(c) For IBIAS = 1mA, what is the maximum allowed value of
Vsig, where vsig = Vsig*sin( t)? The circuit
must provide linear amplification.
vsig
∞
VCC
vout
∞
IBIA S
RB=480kΩ

2. -VEE
RL=100kΩ
∞
Rsig=10kΩ
RC=10kΩ
ß = 100
VA=25V
2. Consider the circuit shown below,
(a) Choose the value of Re such that Gv is maximized, subject
to the constraint that vbe ≤ 10mV and the
transistor stays in the active mode. What is the value of Gv?
(b) If β drops by 10%, what is the new value of Gv? All other
design variables are unchanged.
vsig=0.05sin(ωt) ∞
+5V
vout
∞
0.2mA
100kΩ

3. 20kΩ
∞
20kΩ
20kΩ
ß = 100
VA=∞
Re
-5V
10V
-10V
3. Find Rin, Rout, Gv, and the overall current gain io/isig
(“Gi”). You are given that β=100 and VA=∞.
VBE,ON = 0.7V.
100kΩ
vsig
∞
5V
-5V
3.3kΩ 2kΩ

4. vo
io
ii
4. Choose the value of IBIAS such that vout is a sinusoidal
signal with amplitude 200 mV or larger.
5V
10kΩ
vout
75Ω
∞
vsig=0.5sin(ωt)
VIN=3V
IBIAS
ß = 100
VA=∞
5. Consider the circuit shown below,
(a) Find the dc bias point and small-signal model parameters.
Assume λ = 0.

5. (b) Find Rin, Rout, Avo and Gv.
(c) Repeat part (b) for the case that λ = 0.03. You will have to
recalculate ro.
(d) Finally, you will consider the body effect; i.e., you will no
longer assume that V B = VS but, instead,
that VB = -VSS, which is -5V in this circuit. You are given γ =
0.4V
1/2
and 2ϕF = 0.6V. You may
assume λ = 0 for simplicity. First, you need to recalculate the dc
bias point. You may iteratively
solve for VS and Vt, using
( )
and (√ √ ).
Do not iterate more than 2 or 3 times, as this should be
sufficient to obtain the value of V t with less
than 10mV error. Then, redraw the small-signal model, now
including the gmb current source shown
in Fig. 5.62 of the textbook. The value of gmb can be found
using equations (5.110) and (5.111). You
are to calculate the value of Gv.
1MΩ

6. vsig
∞
5V
5kΩ
vout
4.7MΩ
-5V
∞
0.5mA
Vto=0.75V
k=2mA/V
2
ENSC 324 HOMEWORK #2 Fall 2015
DUE: Monday October 19, 2015 at 2 PM (note new time!)
Please note that unless you show work in the derivations and
solutions you will get no credit for the
answers. Obviously copied answers from study partners or other

7. sources, etc., will also receive no credit.
Please do all parts of all eight problems. It is suggested that you
make a copy of your homework before
turning it in in case it cannot be returned before Exam 2
(solution key will be provided).
Problem #1
In a particular sample of n-type silicon, the Fermi energy level
varies linearly with distance over a short
range. At x=0, the difference between the Fermi level and the
intrinsic Fermi level is 0.4 eV. At x=10
-3
cm, the same difference is only 0.15eV. The electron diffusion
coefficient is 25 cm
2
/s.
a) Write an expression for the Fermi level with respect to the
intrinsic fermi level versus distance.
b) What is the electron concentration versus distance?
c) What is the electron diffusion current density at x=0, and at
x=5 x 10
−4
cm?

8. Problem #2
A rectangular bar shaped silicon semiconductor resistor with a
cross-sectional area of 100 µm
2
,
and a length of 0.1 cm, is doped with a concentration of 5 x 10
16
cm
−3
arsenic atoms (and no
acceptors). Let T = 300 K. A bias of 5V is applied across the
length of the silicon resistor.
a) Calculate the current in the resistor.
b) Repeat part a) if the length is reduced by 0.01 cm.
c) What is the drift current in parts a) and b)?
Problem #3
Assume we have silicon with typical mid-range doping levels of
donors (no acceptors) at 300K. Assume
that the mobility for electrons is limited by lattice scattering
and has a temperature dependence of
approximately T
-3/2

9. . Determine the electron mobility at: a) T=200K; b) 400K.
Problem #4
A silicon bar has a length of L = 0.1 cm and a cross-sectional
area of A = 10
−4
cm
2
. The
semiconductor is uniformly doped with phosphorous at a level 5
x 10
16
cm
−3
(no other doping).
A voltage of 5 V is applied across the length of the material
which is at T=300K. The minority
carrier lifetime is 3 x 10
−7
s. For t < 0, the semiconductor has been uniformly illuminated
with
light, producing a uniform excess carrier generation rate of g' =
5 x 10
21
cm

10. −3
s
−1
. At t = 0, the
light source is turned off. Determine the current in the
semiconductor as a function of time for -∞
≥ t ≥ ∞.
Problem #5
An n-type GaAs semiconductor at 300 K is uniformly doped
with donors (no acceptors) at a
level of 5 x 10
15
cm
−3
(mid level doping). The minority carrier lifetime is 5 x 10
−8
s. A light
source is turned on at t = 0 generating excess carriers uniformly
at a rate of g' = 4 x 10
21
cm
−3

11. s
−1
.
There is no external electric field.
a) Determine the excess carrier concentrations versus time over
the range 0 ≤ t ≤ ∞.
b) Calculate the conductivity of the semiconductor versus time
over the same time period as
part (a).
Problem #6
Consider a bar of p-type silicon material at T=300K that is
homogeneously doped to a value of 3
x 10
15
cm
−3
(ND=0), which may be considered mid level doping. The
applied electric field is
zero. A light source is incident on the end of the semiconductor
on one end (where x = 0, with x
increasing into the silicon bar). The excess carrier concentration
generated at x = 0 is 10

12. 13
cm
−3
.
The lifetimes for electrons and holes are 0.5 µs and 0.1 µs,
respectively.
a) Calculate the steady-state excess electron and hole
concentrations as a function of
distance into the semiconductor.
b) Calculate the electron diffusion current density as a function
of x.
Problem #7
Calculate the position of the quasi-Fermi level with respect to
the intrinsic level for the following silicon
crystal that is steadily illuminated with an excess carrier
generation rate of 10
21
/cm
3
S:
NA = 10
16
/cm
3

13. , ND = 0
τn0 = τp0 = 1 µS
Problem #8
A silicon sample at 300K has the following impurity
concentrations: ND = 10
15
/cm
3
and NA = 0 . The
equilibrium recombination rate is Rp0 = 10
11
cm
−3
s
−1
. A uniform generation rate produces an excess
carrier concentration of 10
14
/cm
3
.
a) What is the excess carrier lifetime?
b) By what factor does the total recombination rate increase?