ANU
ENGN 2211
AUSTRALIAN NATIONAL UNIVERSITY Department of Engineering
ENGN 2211 Electronic Circuits and Devices Problem Problem Set #8 BJT CE Amplifier Circuits Circuits Q1 Consider the common-emitter BJT amplifier circuit shown in Figure 1. Assume V CC = 15 V, β = 150, V BE = 0.7 V, R E = 1 k Ω, RC = 4.7 k Ω, R 1 = 47 k Ω, R 2 = 10 k Ω, R L = 47 k Ω, R s = 100 Ω. +V CC CC
RC
R1
C 2
C 1
Rs
vo R L vs
vin
R2
R E
C E
Figure 1: The circuit for Question 1. (a) Determine the Q-point. (b) Sketch the DC load-line. load-line. What is the maximum (peak to peak) output voltage voltage swing availab available le in this amplifier. (c) Draw the AC equivalent circuit and determine the AC model parameters. (d) Derive expressions for R in , R out , A voc , A v , A i , G . (e) Find R in , R out , A voc , A v , A i , G . v s = 10 × 10−3 sin(2π5000t ). Sketch (f) Find the output voltage waveform if v Sketch the source and output voltage voltage waveforms. (g) Determine whether clipping will take place if v s = 25 × 10−3 sin(2π5000t ).
Problem Set #8
page 1
ANU
ENGN 2211
Q2 Consider the common-emitter BJT amplifier circuit shown in Figure 2. Assume V CC = 15 V, β = 150, V BE = 0.7 V, R E = 2.7 k Ω, RC = 4.7 k Ω, R 1 = 47 k Ω, R 2 = 10 k Ω, R L = 47 k Ω, R s = 100 Ω. +V CC
RC
R1
C 2
C 1
Rs
vo R L vs
vin
R2
R E
C E
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 R in , R out , A voc , A v , A i , G . (e) Find R in , R out , A voc , A v , A i , G . (f) Find the output voltage waveform if v s = 10 × 10−3 sin(2π5000t ). Sketch the source and output voltage waveforms.
Problem Set #8
page 2
ANU
ENGN 2211
AUSTRALIAN NATIONAL UNIVERSITY Department of Engineering
ENGN 2211 Electronic Circuits and Devices Problem Set #8 Solution Q1 Complete Solution Given that V CC = 15 V, β = 150, V BE = 0.7 V, R E = 1 k Ω, RC = 4.7 k Ω, R 1 = 47 k Ω, R 2 = 10 k Ω, R L = 47 k Ω, R s = 100 Ω. +V CC
RC
R1
C 2
C 1
Rs
vo R L vs
vin
R2
R E
C E
(a) Analyzing the DC Voltage-divider bias circuit, we have V T H
= =
RT H
= =
I B
= =
I C I E V E V C V CE
R2 R1 + R2 10k
V CC
(15) = 2.63 V 10k + 47k R2 R1 R1 + R2 (10k )(47k ) = 8.2456 k Ω 10k + 47k V T H − V BE RT H + (β + 1) R E 2.63 − 0.7 = 12.12 µA 8.2456k + (151)(1k )
= β I B = (150)(12.12 µ) = 1.8179 mA = (β + 1) I B = (151)(12.12 µ) = 1.83 mA = I E R E = (1.83m)(1k ) = 1.83 V = V CC − I C R C = 15 − (1.8179m)(4.7k ) = 6.456 V = V CC − I C R C − I E R E = 15 − (1.8179m)(4.7k ) − (1.83m)(1k ) = 4.626 V
Problem Set #8
page 3
ANU
ENGN 2211
As I B > 0 and V CE > 0.2 V, the transistor is in active region of operation. The Q-point lies at I CQ V CEQ
= 1.8179 mA = 4.626 V
(b) For ideal cut-off
= V CC = 15 V
V CE (o f f )
For ideal saturation I C (sat )
=
V CC RC + R E
=
15 = 2.63 mA 5.7k
The plot of DC load line is shown in figure below 3 Ideal Saturation
2.5 ) A m (
2
t n e r r u C
1.5
I C
Q−point
1 0.5 Ideal Cut−off 0
0
5
10
15
Voltage V (V) CE
We see that the Q-point lies closer to saturation (V CE = 0.2 V) than cut-off (V CE = 15 V). Hence the maximum available peak to peak output voltage swing = 2(V CEQ − 0.2) = 8.852 V.
(c) Replacing the capacitors by short circuits and V CC by virtual AC ground, the AC equivalent circuit is Rs vo
vs
Problem Set #8
RC vin
R1
R L
R2
page 4
ANU
ENGN 2211
Replacing the transistor by the small-signal AC equivalent circuit, we have iin
Rs
ib
ic
B vs
v in
R B vbe
E
io
C r π
β i b
R C
vo
R L
E
The AC Model parameters are r e
= =
r π R B
R L
26 mV I EQ 26 = 14.207 Ω 1.83 (β + 1)r e
= = (151)(14.207) = 2.1453 k Ω = R1 || R2 = 8.2456 k Ω = R L || RC = (47 k )||(4.7 k ) = 4.27 k Ω
(d) For derivations, please see Lecture 13.
(e) The BJT CE amplifier parameters are Rin Ro
Avoc
= R B ||r π = (8.2456 k )||(2.1453 k ) = 1.7024 k Ω = RC = 4.7 k Ω RC β = − =
r π (4.7 k )(150)
2.1453 k
= −328.62
Av
R L β
= − =
Ai
= =
G
r π (4.27 k )(150)
= −298.56 2.1453 k Rin Av R L 1.7024 k (−298.56) = −10.81 47 k Ai Av
= = (−10.81)(−298.56) = 3228.69
Problem Set #8
page 5
ANU
ENGN 2211
(f) Finding the equation for output voltage with load, we have vs vin
= 10 × 10−3 sin(2π5000t ) = =
vo
Rin
vs Rs + Rin 1.7024 k
vs 100 + 1.7024 k 0.9445vs
= = Av vin = (−298.56)(0.9445 vs ) = (−298.56)(0.9445)(10 × 10−3 sin(2π5000t )) = −2.82sin(2π5000t )
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. The -ve sign indicates that output voltage is 180 ◦ out of phase with input voltage (inverting amplifier). 1 = 0.2 ms. The time period is T = 5000 The sketch of input and output voltages is shown in figures below:- (Note y-axis has different units in the two figures)
10
3 X: 0.00015 Y: 2.824
2 5 ) V m (
) V ( o
v
s
v
e g a t l o V
1
e g a t l o V
0
0 −1
−5 −2 −10
0
0.1
0.2 Time t (ms)
0.3
0.4
−3
Figure 3: Source voltage v s (t ).
0
0.1
0.2 Time t (ms)
0.3
0.4
Figure 4: Output voltage v o (t ).
(g) For v s = 25 × 10−3 sin(2π5000t ), we have vo
= Av vin = (−298.56)(0.9445 vs ) = (−298.56)(0.9445)(25 × 10−3 sin(2π5000t )) = −7.05sin(2π5000t )
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
ENGN 2211
Q2 Solution Given that V CC = 15 V, β = 150, V BE = 0.7 V, R E = 2.7 k Ω, RC = 4.7 k Ω, R 1 = 47 k Ω, R 2 = 10 k Ω, R L = 47 k Ω, R s = 100 Ω. +V CC
RC
R1
C 2
C 1
Rs
vo R L vs
vin
R2
R E
C E
(a) Analyzing the DC Voltage-divider bias circuit, we have V T H RT H I B I C I E V C V E V CE
= 2.63 V = 8.2456 k Ω = 4.64 µA = 0.696 mA = 0.7006 mA = 11.72 V = 1.89 V = 9.837 V
As I B > 0 and V CE > 0.2 V, the transistor is in active region of operation. The Q-point lies at I CQ V CEQ
= 0.696 mA = 9.837 V
Problem Set #8
page 7
ANU
ENGN 2211
(b) For ideal saturation and cut-off V CE (o f f ) I C (sat )
= 15 V = 2.027 mA
The plot of DC load line is shown in figure below 3 2.5 ) A m (
I C
t n e r r u C
2 Ideal Saturation 1.5 1 Q−point 0.5 Ideal Cut−off 0
0
5
10
15
Voltage V (V) CE
We see that the Q-point lies closer to cut-off (V CE = 15 V) than saturation (V CE = 0.2 V). Hence the maximum available peak to peak output voltage swing = 2(V CC − V CEQ ) = 10.34 V.
(c) The AC Model parameters are r e r π R B
R L
= = = =
37.11 Ω 5.6037 k Ω 8.2456 k Ω 4.27 k Ω
(d) For derivations, please see Lecture 13.
(e) The BJT CE amplifier parameters are Rin Ro Avoc Av Ai G
= = = = =
3.336 k Ω 4.7 k Ω
−125.81 −114.3 −8.11
= 926.97
Problem Set #8
page 8
ANU
ENGN 2211
(f) vs vin vo
= 10 × 10−3 sin(2π5000t ) = 0.97089vs = −1.11sin(2π5000t )
The required peak to peak output voltage swing = 2(1.11)=2.2 V. The maximum available peak to peak output voltage swing = 10.34 V Hence no clipping will take place.
>
2.2 V.
The sketch of input and output voltages is shown in figures below:- (Note y-axis has different units in the two figures) 10
2 X: 0.00015 Y: 1.111
1.5 5
1
) V m (
) V (
v
e g a t l o V
0.5
o
v
s
e g a t l o V
0
−5
0 −0.5 −1 −1.5
−10
0
0.1
0.2 Time t (ms)
0.3
Figure 5: Source voltage v s (t ).
0.4
−2
0
0.1
0.2 Time t (ms)
0.3
0.4
Figure 6: Output voltage v o (t ).
See ProbSet08_Q2.sch.
Problem Set #8
page 9
