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EE 321 Analog Electronics, Fall 2008
Exam #3, 2008/11/
Solutions
Rules You may use your calculator, your text book, and your notes. I cannot answer clarification questions. Answer what you think the question asks. Each sub-question counts equally. In- clude units in all numbers that you compute. Do your calculations in symbolic form and only insert numbers at the end. None of these problems require long derivations. If you find yourself doing long derivations you may be on the wrong track.
Question 1: BJT DC analysis
In each of the following cases, con- sider a circuit like the one to the right, with values as specified in each of the problems below. Determine the operating mode of the BJT and calculate IB, IC, IE, VB, VC, and VE. Assume that β = 100.
V
R
V
R
VBB
CC
B
R (^) C
E
EE (a) VBB = VCC = 10 V, VEE = 0, RB = 1 MΩ, RC = 5 kΩ, and REE = 0. (b) VBB = VEE = 0 V, VCC = 10 V, RB = 0, RE = 1 kΩ, and RC = 1 kΩ. (c) VBB = 3 V, VCC = 10 V, VEE = 0, RB = 0, RE = 1 kΩ, and RC = 10 kΩ. (d) VBB = 5 V, VCC = 10 V, VEE = 0, RB = 100 kΩ, RE = 1 kΩ, RC = 2 kΩ. (a) Since RE = 0, VE = 0. Since VBB > VEE + 0.7 the BJT will be conducting, and
VB = VBE = 0.7 V
IB =
VBB − VBE
RB
1 × 106
= 9. 3 μA
Assume active mode we get
IC = β IB = 0.93 mA
VC = VCC − IC RC = 10 − 0. 93 × 5 × 103 = 5.35 V
Since VCE = VC > 0 .3 V, the assumption of active mode operation is correct. Finally we get
IE = IC + IB = 0.93 mA + 9. 3 muA = 0.9393 mA
(b) The BE junction is in cutoff, so we get IB = IC = IC = 0, and thus VB = VBB = 10, V, VE = VEE = 0 V, and VC = VCC = 10, V. (c) VB = 3 V. Since VBB > VEE +0.7 V, the BJT is conducting, and VE = VB − 0 .7 V = 2.3 V. Then we have
IE =
VE − VEE
RE
1 × 103
= 2.3 mA
and, assuming active mode
IB =
IE
β + 1
2. 3 × 10 −^3
= 22. 8 μA
IC = βIB = 100 × 22. 8 μA = 2.28 mA
VC = VCC − RC IC = 10 − 10 × 103 × 2 .28 = − 12 .8 V
This does not agree with our assumption of active mode operation, which is that VCE > 0 .3 V, so we must be in saturation mode where we assume that VCE = 0.2 V. That gives us then that
VC = VE + VCE = 2.3 + 2.5 V
IC =
VCC − VC
RC
10 × 103
= 0.75 mA
and finally
IB = IE − IC = 2. 3 − 0 .75 = 1.55 mA
(d) Since VBB > VEE + 0.7 V, the BJT is conducting. We then have
VBB = IB RB + VBE + IE RE
If we assume active mode then
VBB = IB [RB + (β + 1) RE ] + VBE
IB =
VBB − VBE
RB + (β + 1) RE
100 × 103 + 101 × 1 × 103
= 21. 4 μA
VB = VBB − IB RB = 2.86 V
VE = VB − VBE = 2. 86 − 0 .7 = 2.16 V
RC =
1 × 10 −^3
RE = RC × 0 .392 = 2450 Ω
(d) VB = VE + 0.7 V = 2.45 + 0.7 = 3.15 V. If β = 100 we get
IB =
IC
β
1 × 10 −^3
= 10 μA
There are two ways to achieve this bias. One way is to connect a resistor between VCC and the base such that it produces the right voltage drop at the right current. That is,
IB RB = VCC − VB
RB =
VCC − VB
IB
10 × 10 −^6
= 685 kΩ
A second way to achieve this is to make a voltage divider whose middle connects to the base. We should then make sure that the current through the voltage divider is much larger than the base current such that the base current does not disturb the bias point. We could for example choose a 1 mA current throug the voltage divider as that is much larger than the base current. In that case we have
I = (R 1 + R 2 ) VCC
R 1 + R 2 =
VCC
I
1 × 10 −^3
= 10 kΩ
It is also true that
R 2 R 1 + R 2
VB
VCC
R 2 =
VB
VCC
(R 1 + R 2 ) =
× 10 × 103 = 3.15 kΩ
R 1 = (R 1 + R 2 ) − R 2 = 10 × 103 − 3. 15 × 103 = 6.85 kΩ
