EE 321 Analog Electronics, Fall 2013
Exam 2 November 4, 2013
solution
200K
1K
200K
10K
200K
1K
200K
This is a closed-book exam. Calculators allowed. The exam is designed for conceptual understanding not long derivations. You MUST box your answer. When
there are multiple values in an answer summarize them in a single box. Correct
answer boxed and derivation (where necessary) gives you 10 points on each question. No boxed answer or no derivation (except where not needed) is 5 points.
BJT DC analysis
For each of these circuit assume VCC = 10 V (that is the voltage at the top) and
β = 100.
(a)
(b)
(c)
(d)
1. Determine all voltages and currents for (a) and state the mode of operation
VB = 0.7 V, VE = 0 V. Assume active mode. Then
IB =
10 − 0.7
VCC − VBE
=
= 46.5 µA
RB
200 × 103
IC = βIB = 4.65 mA
VC = VCC − IC RC = 10 − 4.65 = 5.35 V
Active mode verified.
IE = (β + 1) IB = 4.7 mA
2. Determine all voltages and currents for (b) and state the mode of operation
VE = 0 V, VB = 0.7 V, IB = 46.5 µA (as before). VC = 0.3 V
IC =
VCC − VC
10 − 0.3
=
= 0.97 mA
RC
10
Since IC < βIB saturation mode is verified.
IE = IC + IB = 0.97 + 46.5 × 10−3 = 1.02 mA
3. Determine all voltages and currents for (c) and state the mode of operation
Assume cutoff mode. IB = IC = IE = 0. Then VE = VB = VCC and cutoff mode is
verified. And VC = 0 V.
4. Determine all voltages and currents for (d) and state the mode of operation
VE = VCC , VC = 0 V. VB = VCC − VBE = 10 − 0.7 = 9.3 V. This BJT is in active mode.
IB =
VB
9.3
=
= 46.5 µA
RB
200
IC = βIB = 4.65 mA
IE = IB (β + 1) = 4.70 mA
BJT Biasing
Consider these two biasing schemes
(a)
(b)
5. Which of those two biasing schemes is best for biasing (in active mode, of
course)? Do the calculation with equation and then explain the result.
(a) In this case IB =
VCC −VBE
RB
and IC = βIB so that
IC = β
VCC − VBE
RB
The bias point is linearly dependent on β.
(b) In this case we have
IB =
VCC − VBE − IE RE
VCC − VBE − (β + 1) IB RE
=
RB
RB
Substitute in IB =
IC
β
βVCC − βVBE − (β + 1) IC RE
RB
VCC − VBE
(β + 1) RE
=β
IC 1 +
RB
RB
IC =
IC =
BE
β VCCR−V
B
1+
(β+1)RE
RB
=
VCC − VBE
RB
+ RαE
β
In the second case there is no proportionality to β. We just need to make sure that RE
dominates over RB /β.
Common-emitter amplifier
Consider this common-emitter amplifier which has β = 100 and VCC = 15 V. Ignore
the Early effect.
Rb
Vs
Rc
Rs
Vbe
6. Select RB and RC such that IC = 1 mA and VC = 5 V (assuming no current
from the source)
To make IC = 1 mA we need IB = 10 µA, and
RB =
VCC − VBE
15 − 0.7
=
= 1.43 MΩ
IB
10 × 10−6
To make VC = 5 V we need to make
RC =
VCC − VC
15 − 5
=
= 10 kΩ
IC
1
7. Draw the small-signal circuit
8. Determine Avo , Gvo , Rin and Rout , when RS = 1 kΩ.
We have gm =
IC
VT
=
1
25
= 0.04 Ω−1 . Then
Rin = RB ||rπ = RB ||
beta
100
= 1.43 × 106 ||
= 2.5 kΩ
gm
0.04
Rout = RC = 10 kΩ
Avo = −gm RC = 0.04 × 10 × 103 = −400
Gvo =
Rin
2.5
Avo = −400 ×
= −286
Rin + RS
1 + 2.5