ECE 3042 SƵŵŵĞƌϮϬϭϮ Homework No. 1 Due Week of DĂLJϮϴ Section L01, T12‐3 Tom Brewer GTNUM: gt54321 ECE 3042
Homework Assignment No. 1
Fall 2008 Homework Problem Set for Experiment No. 1
Due Week of September 1
1. Shown below is a single stage common emitter amplifier using an 2N3904 NPN BJT as the active device.
Design the circuit so that it clips symmetrically, viz. pick RB1 and RB2 so that the collector current is that which
yields symmetric clipping. The design specification for the magnitude of the small signal midband voltage gain is 25.
For the design calculations assume the the base-to-emitter dc voltage drop is 0.65 V, that the current in RB1 is 9IB ,
β = 416, the base spreading resistance is 10 Ω, and the Early voltage is infinity.
2. For the circuit parameters indicated in the diagram, use both Cadence and National Instruments SPICE to
determine:
• the dc operating point of the circuit, viz. the dc voltage at each terminal of the transistor and the current flowing
into the collector and base leads and out of the emitter.
• the small signal ac voltage gain, viz. a plot of the gain, Av versus frequency where the frequency range is from
10 Hz to 100 MHz.
• the positive and negative clipping levels, viz. the maximum and minimum possible values of the output voltage,
vo .
• plot of the output voltage versus time for 2 cycles of the input for an input signal a sine wave with a frequency of
1 kHz and a peak value 0.7 V.
• plot of the output voltage versus frequency for a frequency span from dc to 10 kHz, viz. the spectra of the output
with an input signal a sine wave with a frequency of 1 kHz and a peak value 0.7 V. Plot the output voltage on a
log scale.
Assume that the SPICE parameters for the 2N3904 NPN BJT are: saturation current, 6.734 f A; forward
beta, 416; Early voltage, 74.03 V, zero-bias base collector capacitance, 3.638 pF; forward transit time, 301.2 ps, and
base spreading resistance, 10 Ω. (These are the values given in the model statement on page 7 of Experiment 1.)
3. Verify the SPICE solution for the above with a hand calculation using the parameters given for the SPICE
simulation. Calculate the dc operating point, mid-band small signal voltage gain, and positive and negative clipping
levels.
V + = 30 V
V+
6.8k Ω
R
R
vi
v
G
B2
C1
C
C2
10 μ F
RL
0.22 μ F
RB1
vo
22k Ω
R
E1
220 Ω
R E2
C
E
100 μ F
0
1
2
3
4
5
6
7
8
A
A
VCC
VCC
30V
VCC
30V
VCC
RC
B
B
RB2
496.4kΩ
6.8kΩ
C2
6 RL
4
C
Q1
C1
1
10uF
22kΩ
2
0
C
3.585kΩ
D
V10.22uF
0.7 Vpk
1kHz
0°
D
CE
RB1
22.63kΩ
0
0
3
RE1
100uF
220 Ω
0
5
RE2
0
E
E
Fall 2008
F
F
BJT_NPN_VIRTUAL*
G
0
1
2
3
4
5
G
6
7
8
Common Emitter Fall 2008
Printing Time:August 27, 2008 19:02:26
AC Analysis
100
Magnitude
10
1
100m
1
100
10k
1M
100M
10G
100M
10G
Frequency (Hz)
Traces:
V(6),
V(6):
x1
21.5307k
y1
23.5643
x2
1.0000
y2
496.6534m
min x
1.0000
max x
10.0000G
min y
496.6534m
max y
23.5643
offset x
0.0000
offset y
0.0000
200
Phase (deg)
100
0
-100
-200
1
100
10k
1M
Frequency (Hz)
Traces:
V(6),
Common Emitter Fall 2008
Printing Time:January 07, 2009 18:19:08
AC Analysis
30
Magnitude
20
10
0
-10
1
100
10k
1M
100M
10G
100M
10G
Frequency (Hz)
Traces:
V(6),
V(6):
x1
31.0771k
y1
27.4451
x2
35.4549
y2
23.7409
min x
1.0000
max x
10.0000G
min y
496.6534m
max y
23.5643
offset x
0.0000
offset y
0.0000
200
Phase (deg)
100
0
-100
-200
1
100
10k
1M
Frequency (Hz)
Traces:
V(6):
V(6),
x1
7.7598M
y1
135.3429
x2
42.1959
y2
-134.9463
min x
1.0000
max x
10.0000G
min y
-179.9967
max y
179.9973
offset x
0.0000
offset y
0.0000
Common Emitter Fall 2008
Printing Time:August 27, 2008 19:06:30
DC Operating Point
1
2
3
4
DC Operating Point
V(4)
V(2)
V(3)
(V(vcc)-V(4))/6800
13.96126
1.20334
519.96452 m
2.35864 m
Common Emitter Fall 2008
Printing Time:August 27, 2008 19:07:52
Transient Analysis
15
10
Voltage (V)
5
0
-5
-10
-15
0
500.0µ
1.0m
1.5m
Time (s)
Traces:
V(6),
V(6):
x1
211.8227µ
y1
-12.7878
x2
743.8424µ
y2
12.2569
min x
0.0000
max x
2.0000m
min y
-12.8052
max y
12.2595
offset x
0.0000
offset y
0.0000
2.0m
Common Emitter Fall 2008
Printing Time:August 27, 2008 19:09:49
1 Fourier analysis fo
V(6)
1.54492
2 DC component:
9
3 No. Harmonics:
THD:
16.1151 %
4
256
5 Gridsize:
6 Interpolation Degr 1
7
Frequency
8 Harmonic
1000
9 1
2000
10 2
3000
11 3
4
4000
12
5000
13 5
6000
14 6
7000
15 7
8
8000
16
9000
17 9
Magnitude
13.6459
1.90322
1.06229
0.18746
0.0159872
0.201528
0.0327852
0.0680078
0.0587069
Phase
-176.7
99.6903
-175.41
-74.752
-20.661
-74.953
-1.2864
130.178
44.7043
Norm. Mag
1
0.139471
0.0778464
0.0137374
0.00117157
0.0147683
0.00240256
0.00498374
0.00430215
Norm. Phase
0
276.395
1.29969
101.953
156.044
101.752
175.418
306.883
221.409
18
Fourier Analysis
15.0
12.5
Voltage (V)
10.0
7.5
5.0
2.5
0
-2.5
0
2k
4k
6k
Frequency (Hz)
Traces:
V(6),
8k
10k
rx := 10Ω
β := 416
Homework 1 Fall 2008
Rp ( A , B) :=
(
A⋅ B
A+ B
Rtc := Rp RC , RL
α :=
)
RC := 6.8kΩ
RL := 22kΩ
VBE := 0.65V
Av := 25
β
β+1
Rte :=
(
α⋅ Rp RC , RL
RE1 := 220Ω
Vplus := 30V
)
RE2 :=
Av
For Symmetric Clipping
IC
IC
IB :=
IE :=
β
α
−6
VC := Vplus − IC⋅ RC
IB2 := IB1 + IB
4
A
RB1 :=
VB
IB1
VCB := VC − VB
)
ro + Rp rie , Rte
Rte
(
−3
IC = 2.415 × 10
VE := IE⋅ RE1
VC = 13.577 V
RB2 = 4.964 × 10 Ω
rx
rie :=
+ re
1+β
(
−3
5
RB1 = 2.263 × 10 Ω
1 − α⋅
Vplus
IC :=
Rtc + RC + RE1 + Rte
A IE = 2.421 × 10
IB1 := 9 ⋅ IB
ric :=
ro :=
)
Rp ric , Rtc
Av := α⋅
rie + Rte
VE = 0.533 V
RB2 :=
Vplus − VB
IB2
VT := 25.9mV
VA + VCB
v oplus = 12.545 V
)
v ominus = −12.545 V
A
VB = 1.183 V
VA := 74.03V
VT
re :=
IE
4
Av = 23.594
(
RE1
ro = 3.578 × 10 Ω
IC
v ominus := −⎡Vplus − IC⋅ RC + RE1 ⎤ ⋅
⎣
⎦ R
1
VB := VE + VBE
rie + Rte
v oplus := IC⋅ Rtc
−
Rte
3
RE2 = 3.585 × 10 Ω
IB = 5.806 × 10
1
1
Rtc
tc + Rte