EE 321 Analog Electronics, Fall 2013
Homework #7 solution
5.56. Consider the CE amplifier circuit of Fig 5.26(a) when operated with a dc
supply VCC = +5 V. It is required to find the point at which the transistor
should be biased; that is, find the value of VCE so that the output sine-wave
signal vce resulting from an input sine-wave signal vbe of 5 mV peak amplitude
has the maximum possible magnitude. What is the peak amplitude of the output
sine wave and the vlaue of the gain obtained? Assume linear operation around
the bias point. (Hint: To obtain the maximum possible output amplitude for a
given input, you need to bias the transistor as close to the edge of saturation as
possible without entering saturation at any time, that is without, vCE decreasing
below 0.3 V.
The maximum negative output swing is −Vo = −gm RC Vi , which would make the bias point
equal to
VCE = VCE,sat + Vo = VCE,sat + gm RC Vi = VCE,sat +
IC RC
Vi
VT
Note also that
VCE = VCC − IC RC
Eliminating IC RC we get
VCE = VCE,sat +
VCC − VCE
Vi
VT
VCC
Vi
= VCE,sat +
Vi
VCE 1 +
VT
VT
VCE =
VCC
Vi
VT
Vi
VT
VCE,sat +
1+
=
5
0.3 + 25
×5
= 1.13 V
5
1 + 25
5.58. Sketch and label the voltage transfer characteristics of the pnp commonemitter amplifiers shown in Fig. P5.58.
1
(a)
(b)
2
5.65. For the circuit in Fig. P5.56 select a value for RB so that the transistor
saturates with an overdrive factor of 10. The BJT is specified to have a minimum
β of 20 and VCE sat = 0.2 V. What is the value of forced β archieved?
When the transistor is saturated then VCE = 0.3 V, and thus
IC =
5 − 0.2
VCC − VCE sat
=
= 4.8 mA
RC
1
β
To get a overdrive factor of 10 means that βbeta
= 10. This result in βforced = 10
= 2. The
forced
IC
4.8
base current is then IB = βforced = 2 = 2.4 mA, and the required value of the base resistor
is
VCC − VBE
5 − 0.7
=
= 1.8 kΩ
IB
2.4
5.87. Find approximate values for the collector voltages in the circuits of Fig.
P5.87. Also, calculate forced β for each of hte transistor. (Hint: Initially, assume
all transistors are operating in saturation, and verify the assumption).
RB =
3
(a) Assuming saturation mode we have VCE = 0.2 V. We can write the following equations
IC =
VCC − VC
RC
IE =
VC − VCE
RE
IB =
VCC − VC − VBC
RB
(VCE = 0.2 V, VBC = 0.5 V). Inserting thes three into
IC = IE − IB
we get
VC − VCE VCC − VC − VBC
VCC − VC
=
−
RC
RE
RB
VC
1
1
1
+
+
RB RC RE
=
VCC
RC
+
VCE
RE
VC =
=
VCC VCE VCC
VBC
+
+
−
RC
RE
RB
RB
+
VCC
RB
−
VBC
RB
1
+ R1C + R1E
RB
10
+ 0.2
+ 10
− 0.5
10
1
20
20
1
1
1
+ 10 + 1
20
=1.46 V
Now we verify. The verification is that the βforced = IC /IB < β.
βforced
IC
=
=
IB
VCC −VC
RC
VCC −VC −VBC
RB
=
10−1.46
10
10−1.46−0.5
20
Guessing that β ≫ 2.12 we have verified saturation mode.
4
= 2.12
(b) In this case we can write
IC =
VC
RC
IE =
VCC − VC − VEC
RE
IB =
VC − VCB
RB
(VEC = 0.2 V, VCB = 0.5 V). Inserting these into IE = IC + IB we get
VCC − VC − VEC
VC
VC − VCB
=
+
RE
RC
RB
VC
1
1
1
+
+
RC
RB RE
VC =
=
=
VCC
VEC
VCB
−
−
RE
RE
RB
VCC
− VREC
RE
E
1
1
+ RB
RC
10
− 0.2
−
1
1
1
1
+ 10 +
1
−
+
VCB
RB
1
RE
0.5
10
1
1
=4.64 V
Next we verify. The verification is that βforced =
βforced
IC
=
=
IB
VC
RC
VC −VCB
RB
=
IC
IB
< β.
4.64
1
4.64−0.5
10
= 11.2
Guessing that β > 11.2 we have verified saturation mode operation.
(c) We can write the equations
IC3 =
VC3
+ IB4
RC3
IC4 =
IE3 =
V+ − VC3 − VEC
RE3
V+ − VC3 + VBC
RC4
IE4 =
IE3 = IC3 + IB3
IB3 =
VC3 − VCB
RB3
VC3 − VBE
RE4
IE4 = IC4 + IB4
where VEC = 0.2 V, VCB = 0.5 V, VBC = 0.5 V, VBE = 0.7 V. If we insert into the last
two current continuity equations we end up with expressions for IB4 in terms of VC3 .
Then we can eliminate IB4 between them to obtain an expression for VC3 .
VC3
VC3 − VCB
V+ − VC3 − VEC
=
+ IB4 +
RE3
RC3
RB3
5
VC3 − VBE
V+ − VC3 + VBC
=
+ IB4
RE4
RC4
Subtract
VC3 − VBE
VC3
VC3 − VCB V+ − VC3 − VBC
V+ − VC3 − VEC
−
=
+
−
RE3
RE4
RC3
RB3
RC4
VC3
VC3
VC3
VC3
VC3
V+
VEC VBE VCB
V+
VBC
+
+
+
+
=
−
+
+
+
−
RC3 RB3 RC4 RE3 RE4
RE3 RE3 RE4 RB3 RC4 RC4
VC3 =
V+
RE3
−
VEC
RE3
+
VBE
RE4
+
VCB
RB3
+
V+
RC4
−
VBC
RC4
1
+ R1B3 + R1C4 + R1E3 + R1E4
RC3
10
− 0.2 + 0.7 + 0.5 + 10 − 0.5
= 10 1 10 1 10 1 10 1 30 1 30
+ 10 + 30 + 10 + 10
10
=3.27 V
Next we see that
VC4 = VC3 − VBC = 3.27 − 0.5 = 2.77 V
Now we should verify the assumptions of saturation mode. Begin with Q4
!−1
−1
VE4
IC4
IC4
IE4
RE4
βforced,4 =
=
=
−1
= VCC −VC4 − 1
IB4
IE4 − IC4
IC4
RC4
−1 −1
2.77 − 0.2 30
VC4 − VCE RC4
=
−1
= 15.1
=
VCC − VC4 RE4
10 − 2.77 10
Next for Q3 we get
βforced,3
IC3
IE3 − IB3
IE3
=
=
=
−1=
IB3
IB3
IB3
=
VCC −VE3
RE3
VB3
RB3
−1=
VCC − VE3 RB3
−1
VB3
RE3
VCC − VC3 − VEC RB3
10 − 3.27 − 0.2 10
−1 =
− 1 = 1.9
VC3 − VCB
RE3
2.77 − 0.5 10
NOTE: This problem is a good example of how relatively minor simplifications can
make an enormous difference in the complexitiy of a calculation, as well as the likelyhood of mistakes. Sedra and Smith apparently assumed that VB = VC = VE (i.e.
VBE = 0 and VCE = 0) for each of the transistors, got an approximate solution and
then verified that both transistors were indeed in saturation.
6
5.96.
(a) Using a 3-V power supply, design the feedback cicuit of Fig. 5.46 to provide
IC = 3 mA and VC = VCC /2 for β = 90.
(b) Select standard 5% resistor values, and reevaluate VC and IC for β = 90.
(c) Find VC and IC for β = ∞
(d) To improve the situation that obtains when high-β transistors are used, we
have to arrange for an additional current to flow through RB . This can be
achieved by connecting a resistor between base and emitter, as shown in
Fig. P5.96. Design this circuit for β = 90. Use a current through RB2 equal
to the base current. Now, what values of VC and IC result with β = ∞?
(a) We have VCC /2 = IC RC , or RC =
VCC
2IC
=
To find RB use
7
3
2×3×10−3
= 500 Ω.
VCC
IC
− VBE = RB
2
β
RB =
VCC
− VBE
2
β
=
IC
3
− 0.7
2
90
= 24 kΩ
3
(b) Select RC = 510 Ω and RB = 24 kΩ. In that case we have
VCC − VC = IC RC
and
VC − VBE =
IC
RB
β
Eliminate IC to get
VC − VBE
VCC − VC
=β
RC
RB
VC
β
1
+
RB RC
VC =
=
=
VCC
VBE
+β
RC
RB
VCC
+ β VRBE
RC
B
β
1
+
RB
RC
3
0.7
+ 90 24×10
3
510
1
90
+ 510
24×103
=1.49 V
The collector current is
IC =
VCC − VC
3 − 1.49
=
= 2.96 mA
RC
510
(c) For β = ∞ the collector and base are at the same voltage because there is zero current
through RB . Thus, VC = VBE = 0.7 V. The collector current is then
IC =
3 − 0.7
VCC − VC
=
= 4.5 mA
RC
510
8
2IC
β
(d) In this case we have 2IB =
RB2 =
RB1 =
flowing through RB1 and half that through RB2 . So
VBE
βVBE
90 × 0.7
=
=
= 21 kΩ
IB
IC
3 × 10−3
VC − VBE
1.5 − 0.7
VC − VBE
=β
= 90
= 12 kΩ
2IB
2IC
2 × 3 × 10−3
RC will have the same value as before. Now, if β = ∞ there is the same current through
RB1 and RB2 . That current is
IB =
VBE
0.7
=
= 0.0333 mA
RB2
21
and the collector voltage is
VC = IB (RB1 + RB2 ) = 0.0333 × (21 + 12) = 1.1 V
and the collector current is
IC =
VCC − VC
3 − 1.1
=
= 3.8 mA
RC
500
5.103. An npn BJT with grounded emitter is operated with VBE = 0.700 V, at
which the collector current is 1 mA. A 10 kΩ resistor connects the collector to
+15 V supply. What is the resulting collector voltage VC ? Now, if a signal applied
to the base raised vBE to 705 mV, find the resulting total collector current iC
and total collector voltage vC using the exponential iC -vBE relationship. For this
situation, what are vbe and vc ? Calculate the voltage gan vc /vbe . Compare with
the value obtained using the small-signal approximation, that is −gm RC .
The output voltage is VC = VCC − IC RC = 15 − 1 × 10 = 5 V. Next we rais the base voltage.
First we want to find the value of IC from the bias point, and
iC = IS e
IS =
iC
e
vBE
VT
=
1 × 10−3
e
0.7
25×10−3
vBE
VT
= 6.91 × 10−16 A
Next for vBE = 0.705 V:
iC = IS e
vBE
VT
0.705
= 6.91 × 10−16 × e 25×10−3 = 1.22 mA
vC = VCC − iC RC = 15 − 1.22 × 10 = 2.8 V
For this situation vbe = 5 × 10−3 , and vc = 2.8 − 5 = −2.2 V, and the gain is vc /vbe = −440.
The gain using the linear approximation is
9
IC RC
1 × 10
=−
= 400
VT
25 × 10−3
5.104. A transitor with β = 120 is biased to operate at a DC collector current of
1.2 mA. Find the values of gm , rπ , and re . Repeat for a bias current of 120 µA.
Avo = −
re =
gm =
1.2
IC
=
= 0.048 Ω−1
VT
25
rπ =
120
β
=
= 2500 Ω
gm
0.048
β 1
120 1
α
=
=
= 20.7 Ω
gm
β + 1 gm
121 0.048
If instead the transistor is biased at IC = 120 µA, we get
gm =
IC
0.12
=
= 0.0048 Ω−1
VT
25
rπ =
re =
120
β
=
= 25 kΩ
gm
0.0048
β 1
120 1
α
=
=
= 207 Ω
gm
β + 1 gm
121 0.0048
10