EE 321 Analog Electronics, Fall 2013
Homework #6 solution
5.21. Measurements on the circuits of Fig. P5.21 produce labeled voltages as
indicated. find the value of β for each transistor.
4.3
= 21.5 µA, IC = RVCC = 12 = 2 mA. We can verify that the transistor is
(a) IB = RVBB = 200
in the active mode since VC < VB . The we get β = IIBC = 2000
= 93.
21.5
2.3
= 10 mA. It is IE because IB and IC are merged.
(b) In this case we have IE = RVCC = 0.23
VB −VC
4.3−2.3
We also have IB = RB = 20 = 0.1 mA. We verify active mode since VC < VB ,
10
− 1 = 99.
and we get β = IIBE − 1 = 0.1
E
= 10−7
= 3 mA. Then it must also be that VC = 0 + 3 V = 3 V since
(c) IE = VEER−V
1
E
C
RC = RE and we have IE flowing through RC . Then we have IB = VBR−V
= 6.3−3
=
100
B
IE
3
33 µA. We verify active mode operation and get β = IB − 1 = 0.033 − 1 = 90.
5.24. for each of the circuits shown in Fig. P5.24, find the emiter, base, and collector voltages and currents. Use β = 30, but assume |VBE | = 0.7 V independent
of current level.
1
(a) First, VB = 0 V. The transistor must be on, so VE = VB − VBE = 0 − 0.7 V = −0.7 V.
−VEE
= −0.7+3
= 1.0 mA. Next assume active mode and we get IC =
Then IE = VE R
2.2
E
β
30
αIE = β+1 IE = 30+1
×1.0 = 1.01 mA. Then VC = VCC −IC RC = 3−1.01×2.2 = 0.78 V.
= 33 µA.
Active mode is verified. Then finally, IB = IβC = 1.01
30
E
(b) VB = 0 V. The transistor is on, so VE = 0.7V. Then IE = VEER−V
= 3−0.7
= 2.3 mA.
1
E
30
If we assume active mode then IC = αIE = 30+1 × 2.3 = 2.2 mA. Then VC = VCC +
IC RC = −3 + 2.2 = −0.8 V, and active mode is verified. Finally, IB = IβC = 2.2
=
30
73, µA.
(c) VB = 3 V, and the transistor is on, so VE = VB + VEB = 3 + 0.7 = 3.7 V. Then
β
E
= 9−3.7
= 4.8 mA. Assuming active mode we get IC = αIE = β+1
IE =
IE = VEER−V
1.1
E
30
× 4.8 = 4.6 mA. Then VC = RC IC = 0.56 ∗ 4.6 = 2.6 V. Since VC < VB active
30+1
= 155 µA.
mode is verified. Finally, IB = IβC = 4.6
30
(d) VB = 3 V, and the transistor is on so VE = VB −VBE = 3−0.7 = 2.3 V. Then IE = RVEE =
β
30
2.3
= 4.9 mA. Then, assuming active mode, IC = αIE = β+1
IE = 30+1
× 4.9 = 4.7 mA.
0.47
Then VC = VCC − IC RC = 9 − 4.7 × 1 = 4.3 V. Since VC > VB active mode is verified.
Then finally IB = IβC = 4.3
= 143 µA.
30
5.26. For the circuit shown in Fig. P5.26, measurement indicates that VB =
−1.5 V. Assuming VBE = 0.7 V, calculate VE , α, β, and VC . If a transistor with
β = ∞ is used, what values of VB , VE , and VC result?
2
First, we have the base current,
−VB
1.5
=
= 0.15 mA
RB
10 × 103
= −1.5 − 0.7 = −2.2 V, and thus the emitter current
IB =
We also have VE = VB − VBE
−2.2 + 9
VE − VEE
=
= 0.68 mA
RE
10 × 103
Assuming active mode operation the collector current is
IE =
IC = IE − IB = 0.68 − 0.15 = 0.53 mA
The collector voltage is the
VC = VCC − IC RC = 9 − 0.53 × 10 = 3.7 V
which is larger than the base voltage so it is operating in active mode. Then
α=
IC
0.53
=
= 0.78
IE
0.68
and
0.53
IC
=
= 3.53
IB
0.15
If instead we assume that β = ∞, then IB = 0, and VB = 0. Then VE = VB −VBE = 0−0.7 =
−0.7 V. Then, because IE = IC ,
β=
VCC − VC
VE − VEE
=
RC
RE
VC = VCC − (VE − VEE ) = 9 − (−0.7 + 9) = 0.7 V
5.39. For a BJT having an Early voltage of 200 mA, what is its output resistance
at 1 mA? At 100 µA?
200
The output resistance is ro = VICA . So for IC = 1 mA, we get ro = VICA = 1×10
−3 = 200 kΩ. For
VA
200
IC = 100 µA, we get ro = IC = 100×10−6 = 2 MA.
3
5.54. A common-emitter amplifier circuit operated with VCC = +10 V is biased
at VCE = +1 V. Find the voltage gain, the maximum allowed output negative
swing without the transistor entering saturation, and the corresponding maximum input signal permitted.
The voltage gain is
RC IC
VCC − VCE
10 − 1
vc
=−
=−
=−
= −360
vb
VT
VT
25 × 10−3
The maximum negative output voltage swing without entering saturation is that which makes
vCE = 0.3. Since vCE = VCE + vc we get
vc = VCE − vCE,sat = 0.3 − 1 = −0.7 V
The corresponding maximum input signal is
vb = vc
vb
−0.7
=−
= 1.94 mV
vc
360
5.55. For the common-emitter amplifier circuit in Fig 5.26(a) with VCC = +10 V
and RC = 1 kΩ, find VCE and the voltage gain at the following dc collector bias
current: 1 mA, 2 mA, 5 mA, 8 mA, and 9 mA. For each, give maximum possible
positive- and negative-output signal swing as determined by the need to keep
the transistor in the active region. Present your result in a table.
The value of VCE is
VCE = VCC − IC RC
This is tabulated in second column. The gain, A, is
RC IC
VT
This is tabulated in the third column. The minimum output voltage swing is determined by
the edge of saturation, vCE,sat = 0.3 V. It is
A=−
vo,min = vCE,sat − VCE
4
This is tabulated in the fourth column. The maximum output voltage swing is determined
by cutoff, iC = 0, so that
vo,max = VCC − VCE
This is tabulated in the fifth column.
IC (mA) VCE (V)
1
9
2
8
5
5
8
2
9
1
A vo,min (V)
40
-8.7
80
-7.7
200
-4.7
320
-1.7
360
-0.7
5
vo,max (V)
1
2
5
8
9