ECE3050
Homework Set 2
1. A diode has the parameters IS = 10 fA, n = 2, and VT = 25 mV.
(a) Calculate rd for VD = 0.6 V.
rd =
nVT
nVT e−VD /nVT
=
= 30.72 MΩ
ID + IS
IS
(b) Calculate rd for VD = 0 V.
rd =
nVT
= 5 × 1012 Ω
IS
(c) At what voltage does rd exceed 1015 Ω?
µ
¶
nVT
VD = nVT ln
= −0.265 V
IS rd
2. A diode current-controlled attenuator circuit is shown. It is given that R = 20 kΩ. The diode
parameters are n = 2 and VT = 0.025 V.
(a) Calculate the bias current which will provide a small-signal attenuation of 20 dB, i.e.
vo /vs = 0.1.
#
"µ ¶
vo −1
2nVT
I=
− 1 = 45 µA
R
vs
(b) If the current is halved, what is the new attenuation? (−14.8 dB). (c) If the current is
doubled, what is the new attenuation? (−25.6 dB)
3. A diode has the current ID1 = 1 mA for VD1 = 0.55 V and ID2 = 2 mA for VD2 = 0.58 V. If
IS ¿ ID1 , determine the ideality factor or emission coefficient n and the saturation current
IS .
n =
IS =
VD2 − VD1
= 1.73
VT ln (ID2 /ID1 )
ID2
ID1
=
= 3.03 nA
exp (VD1 /nVT )
exp (VD2 /nVT )
4. The diagram shows a zener diode regulator. It is given that V1 = 35 V. The diode has the
zener voltage VZ = 24 V. The load resistance varies between the limits 500 Ω ≤ RL ≤ 10 kΩ.
1
(a) Calculate R1 if IZ is to have a value that is no smaller than 10 mA. Note that I1 is a
constant once R1 is determined and I1 = IZ + IL . Thus the minimum value of IZ occurs
when IL is a maximum (when RL is a minimum) because this makes IZ have the smallest
value.
35 V − 24 V
= 190 Ω
R1 =
0.01 A + 24 V/500 Ω
(b) What is the power dissipation in R1 and the maximum power dissipation in the zener
diode? Note that I1 is a constant and I1 = IZ + IL . Thus the maximum dissipation in the
zener diode occurs when IL is its smallest value because this makes IZ have the largest value.
¶
µ
24 V
(35 V − 24 V)2
35 V − 24 V
= 0.637 W
PZ max = 24 V
−
= 1.33 W
P1 =
190 Ω
190 Ω
10 kΩ
5. Calculate the values of β and IS for the transistor shown if VCB = VBE = 0.7 V, IB = 0.2 mA,
and IE = 10 mA.
Figure 1:
β=
10 mA − 0.2 mA
= 49
0.2 mA
IS =
9.8 × 10−3
= 6.78 × 10−15 A
exp (0.7/0.025)
6. Calculate the values of β and IS for the transistor shown if VEB = VBC = 0.7 V, IB = 50 µA,
and IC = 2.5 mA.
β=
2.5 mA
= 50
50 µA
IS =
2.5 × 10−3
= 1.73 × 10−15 A
exp (0.7/0.025)
7. Calculate the collector, emitter, and base currents if V + = 3.3 V, VEE = −3.3 V, VBE = 0.7 V,
RE = 47 kΩ, and β = 90.
IE =
−0.7 V − (−3.3 V)
= 55.3 µA
47 kΩ
IB =
IC = IE − IB = 54.7 µA
2
55.3 µA
= 0.608 µA
91
8. An npn transistor is operated in the active mode with a base current of 3 µA. It is found that
IC = 240 µA for VCE = 5 V and IC = 265 µA for VCE = 10 V. What are the values of β 0 and
VA for this transistor? [β 0 = 71.7, VA = 43.1 V]
9. A BJT has the parameters β 0 = 75, VA = 100 V, and VCE = 10 V.
(a) Calculate IC for rπ = 10 kΩ.
µ
¶
VCE
VT
IB = 0.2063 mA
= 2.5 µA
IC = β 0 1 +
IB =
rπ
VA
(b) Calculate the values of gm and r0 .
gm =
IC
1
=
VT
121.2
r0 =
VA + VCE
= 533.3 kΩ
IC
(c) Calculate α and re .
µ
¶
VCE
β0 1 +
β
VA
µ
¶ = 0.9880
=
α=
VCE
1+β
1 + β0 1 +
VA
re =
VT or
VT
or
=
=
IE
(1 + β) IB
r
µπ
¶ = 119.8 Ω
VCE
1 + β0 1 +
VA
10. The output characteristics of a BJT are shown. (a) Determine β 0 and VA . [β 0 = 120,
VA = 30 V] (b) Calculate β at iB = 4 µA and VCE = 5 V. [135] (c) Calculate β at iB = 8 µA
and VCE = 15 V. [225]
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