ENGN 2211 Electronic Circuits and Devices Problem Set #8 ...
[Pages:9]ANU
AUSTRALIAN NATIONAL UNIVERSITY Department of Engineering
ENGN 2211 Electronic Circuits and Devices Problem Set #8 BJT CE Amplifier Circuits
Q1
Consider the common-emitter BJT amplifier circuit shown in Figure 1.
Assume VCC = 15 V, = 150, VBE = 0.7 V, RE = 1 k, RC = 4.7 k, R1 = 47 k, R2 = 10 k, RL = 47 k, Rs = 100 .
+VCC
ENGN 2211
R1
Rs
C1
vs
vin
R2
RC C2
RE CE
vo RL
Figure 1: The circuit for Question 1.
(a) Determine the Q-point. (b) Sketch the DC load-line. What is the maximum (peak to peak) output voltage swing available in this amplifier. (c) Draw the AC equivalent circuit and determine the AC model parameters. (d) Derive expressions for Rin, Rout , Avoc, Av, Ai, G. (e) Find Rin, Rout , Avoc, Av, Ai, G. (f) Find the output voltage waveform if vs = 10 ? 10-3 sin(25000t). Sketch the source and output voltage waveforms. (g) Determine whether clipping will take place if vs = 25 ? 10-3 sin(25000t).
Problem Set #8
page 1
ANU
Q2
Consider the common-emitter BJT amplifier circuit shown in Figure 2.
Assume VCC = 15 V, = 150, VBE = 0.7 V, RE = 2.7 k, RC = 4.7 k, R1 = 47 k, R2 = 10 k, RL = 47 k, Rs = 100 .
+VCC
ENGN 2211
R1
Rs
C1
vs
vin
R2
RC C2
RE CE
vo RL
Figure 2: The circuit for Question 2.
(a) Determine the Q-point. (b) Sketch the DC load-line. What is the maximum (peak to peak) output voltage swing available in this amplifier. (c) Draw the AC equivalent circuit and determine the AC model parameters. (d) Derive expressions for Rin, Rout , Avoc, Av, Ai, G. (e) Find Rin, Rout , Avoc, Av, Ai, G. (f) Find the output voltage waveform if vs = 10 ? 10-3 sin(25000t). Sketch the source and output voltage waveforms.
Problem Set #8
page 2
ANU
AUSTRALIAN NATIONAL UNIVERSITY Department of Engineering
ENGN 2211 Electronic Circuits and Devices Problem Set #8 Solution
Q1
Complete Solution
Given that VCC = 15 V, = 150, VBE = 0.7 V, RE = 1 k, RC = 4.7 k, R1 = 47 k, R2 = 10 k, RL = 47 k, Rs = 100 .
+VCC
R1
Rs
C1
vs
vin
R2
RC C2
RE CE
vo RL
ENGN 2211
(a)
Analyzing the DC Voltage-divider bias circuit, we have
VT H
=
R1
R2 + R2
VCC
10k
=
(15) = 2.63 V
10k + 47k
RT H
=
R2R1 R1 + R2
= (10k)(47k) = 8.2456 k 10k + 47k
IB
=
VT H - VBE RT H + ( + 1)RE
2.63 - 0.7
=
= 12.12 ?A
8.2456k + (151)(1k)
IC = IB
= (150)(12.12?) = 1.8179 mA
IE = ( + 1)IB = (151)(12.12?) = 1.83 mA
VE = IE RE = (1.83m)(1k) = 1.83 V
VC = VCC - ICRC = 15 - (1.8179m)(4.7k) = 6.456 V
VCE = VCC - ICRC - IE RE = 15 - (1.8179m)(4.7k) - (1.83m)(1k) = 4.626 V
Problem Set #8
page 3
ANU As IB > 0 and VCE > 0.2 V, the transistor is in active region of operation.
The Q-point lies at ICQ = 1.8179 mA
VCEQ = 4.626 V
ENGN 2211
(b)
For ideal cut-off
VCE(o f f ) = VCC = 15 V
For ideal saturation
IC(sat)
=
VCC
15 = = 2.63 mA
RC + RE 5.7k
The plot of DC load line is shown in figure below
3 Ideal Saturation
2.5
Current IC (mA)
2 Q-point
1.5
1
0.5 0 0
5
10
Voltage VCE (V)
Ideal Cut-off 15
We see that the Q-point lies closer to saturation (VCE = 0.2 V) than cut-off (VCE = 15 V). Hence the maximum available peak to peak output voltage swing = 2(VCEQ - 0.2) = 8.852 V.
(c)
Replacing the capacitors by short circuits and VCC by virtual AC ground, the AC equivalent circuit is
Rs
vs
vin
R1
R2
vo RC RL
Problem Set #8
page 4
ANU Replacing the transistor by the small-signal AC equivalent circuit, we have
Rs
iin
ib
ic
B
C
vs
vin
RB vbe r ib
RC
ENGN 2211
io
vo
RL
E
E
The AC Model parameters are
26 mV
re =
IEQ
= 26 = 14.207 1.83
r = ( + 1)re
= (151)(14.207) = 2.1453 k
RB = R1||R2 = 8.2456 k
RL = RL||RC = (47 k)||(4.7 k) = 4.27 k
(d)
For derivations, please see Lecture 13.
(e)
The BJT CE amplifier parameters are
Rin = RB||r = (8.2456 k)||(2.1453 k) = 1.7024 k
Ro = RC
= 4.7 k
Avoc
=
- RC r
= (4.7 k)(150) = -328.62 2.1453 k
Av
=
- RL r
(4.27 k)(150)
=
= -298.56
2.1453 k
Ai
=
Av
Rin RL
1.7024 k
= (-298.56)
= -10.81
47 k
G = AiAv
= (-10.81)(-298.56) = 3228.69
Problem Set #8
page 5
ANU
(f)
Finding the equation for output voltage with load, we have
vs = 10 ? 10-3 sin(25000t)
vin
=
Rs
Rin + Rin
vs
1.7024 k = 100 + 1.7024 k vs
= 0.9445vs
vo = Avvin
= (-298.56)(0.9445 vs) = (-298.56)(0.9445)(10 ? 10-3 sin(25000t))
= -2.82 sin(25000t)
The required peak to peak output voltage swing = 2(2.82)=5.64 V. The maximum available peak to peak output voltage swing = 8.852 V > 5.64 V. Hence no clipping will take place.
ENGN 2211
The -ve sign indicates that output voltage is 180 out of phase with input voltage (inverting amplifier).
The
time
period
is
T
=
1 5000
=
0.2
ms.
The sketch of input and output voltages is shown in figures below:- (Note y-axis has different units in the two
figures)
Voltage vs (mV) Voltage vo (V)
10
5
0
-5
-10 0
0.1
0.2
0.3
0.4
Time t (ms)
Figure 3: Source voltage vs(t).
3
X: 0.00015
2
Y: 2.824
1
0
-1
-2
-3
0
0.1
0.2
0.3
0.4
Time t (ms)
Figure 4: Output voltage vo(t).
(g)
For vs = 25 ? 10-3 sin(25000t), we have
vo = Avvin = (-298.56)(0.9445 vs) = (-298.56)(0.9445)(25 ? 10-3 sin(25000t)) = -7.05 sin(25000t)
The required peak to peak output voltage swing = 2(7.05)=14.1 V. However the maximum available peak to peak output voltage swing = 8.852 V < 14.1 V. Hence clipping will take place.
See ProbSet08_Q1.sch.
Problem Set #8
page 6
ANU
Q2
Solution
Given that VCC = 15 V, = 150, VBE = 0.7 V, RE = 2.7 k, RC = 4.7 k, R1 = 47 k, R2 = 10 k, RL = 47 k, Rs = 100 .
+VCC
R1
Rs
C1
vs
vin
R2
RC C2
RE CE
vo RL
ENGN 2211
(a)
Analyzing the DC Voltage-divider bias circuit, we have
VTH = 2.63 V RTH = 8.2456 k
IB = 4.64 ?A IC = 0.696 mA IE = 0.7006 mA VC = 11.72 V VE = 1.89 V VCE = 9.837 V
As IB > 0 and VCE > 0.2 V, the transistor is in active region of operation.
The Q-point lies at
ICQ = 0.696 mA VCEQ = 9.837 V
Problem Set #8
page 7
ANU
(b)
For ideal saturation and cut-off
VCE(o f f ) = 15 V IC(sat) = 2.027 mA
The plot of DC load line is shown in figure below
ENGN 2211
3
2.5
Current IC (mA)
2 Ideal Saturation
1.5
1 0.5
0 0
Q-point
5
10
Voltage VCE (V)
Ideal Cut-off 15
We see that the Q-point lies closer to cut-off (VCE = 15 V) than saturation (VCE = 0.2 V). Hence the maximum available peak to peak output voltage swing = 2(VCC -VCEQ) = 10.34 V.
(c)
The AC Model parameters are re = 37.11 r = 5.6037 k RB = 8.2456 k RL = 4.27 k
(d)
For derivations, please see Lecture 13.
(e)
The BJT CE amplifier parameters are
Rin = 3.336 k Ro = 4.7 k Avoc = -125.81 Av = -114.3 Ai = -8.11 G = 926.97
Problem Set #8
page 8
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