Chapter 2 - Linear Circuit Analysis
Recommended problems to study:
Warmups
Resistance of a wire
Transients
Transformers/mutual inductance
Linear circuit models
Voltage regulator
i-v diode models
Photodetector circuit
Temperature dependent diode modeling
Hall effect
Transformers
Page
1
2,3
4
5
6
7
12
13
14
15
Concentrates
Source/load relationships
Matrix analysis of circuits
Transformers
Reactances
Wheatstone bridge
Oscillator
25
27
31
33
35
38
Transient oscillator
40
Chapter 2
Part 1 - Warmups
1. Using the same diameter nichrome wire as in example 2.1, determine the length needed
to obtain a resistance of 1 ohm at 100∞ C. Determine the resistance at 20∞ C.
The characteristics of #30 AWG wire are:
diameter (mils) diameter
in2
(circular mils)
10.03
100.5
7.894¥10-5
W/1000ft @20∞ C
pounds/1000 ft.
103.2
0.3042
for nichrome wire @20∞ C
r20=1.08¥10-6W-m
a20=17¥10-3/∞ C
r100= r20[1+ a20(T-20)] = 1.08¥10-6[1+ 17¥10-3 (100-20)]
r100= 1.08¥10-6[1+ 1.36] = 2.55¥10-6W-m
2
mm ˆ
2
p Ê10.03 ¥ 10 -3 in ¥ 25.4
2
2
p (0.255 ¥ 10 -3 m)
Ë
¯
dˆ
pd
in
Ê
A=p
=
=
=
= 5.10 ¥ 10 -8 m 2
Ë 2¯
4
4
4
R=r
l
A
1W = (2.55 ¥ 10 -6 W - m)
l = (1W)
R20
l
5.10 ¥ 10 -8 m 2
5.10 ¥ 10 -8 m 2
= 0.02 m
2.55 ¥ 10 -6 W - m
(1.08 ¥ 10
(l = 0.02 m) =
-6
W - m)(0.02 m)
5.10 ¥ 10 -8 m 2
= 0.42W
Note: Circular mils is nothing more than the diameter (expressed in thousandths of an
inch) squared.
-1-
2. For the circuit shown in example 2.2, assume the switch has been closed for a long time
and is opened at t=0. Determine (a) the current through the capacitor at the instant that the
switch is opened, (b) the time derivative of the capacitor voltage at the instant the switch is
opened, and (c) the voltage that will exist across the capacitor after a long time.
R1
iC
i2
+
+
VS
R2
C
-
VC
-
The boundary conditions for the capacitor C are
no DC current flow
v( 0 + ) = v( 0 - )
R2
By inspection, vC (0 - ) =
VS since there is no DC current through C.
R1 + R2
(a) When the switch is open
iC
i2 (0 + ) =
i2
VC
C
R2
) = v (0 )
-
C
R2
R2
VS
VS
R1 + R2
=
=
R2
R1 + R2
Then, iC (0 + ) = -i2 (0 + ) =
+
R2
vC (0
+
-
(b) Using the derivative relationship i = C
dv
for C
dt
iC (0 + )
dv
-VS
+
t
=
0
=
=
(
)
dt
C
( R1 + R2 )C
(c) vC (t = •) = 0
-2-
-VS
R1 + R2
3. For the circuit shown in example 2.3, assume that the switch has been closed for a long
time (the final conditions of example 2.3 apply), and the switch is opened at t=0.
Determine at the instant the switch opens (a) the current in the inductance, (b) the time rate
of change of current in the inductance, and (c) the voltage across the inductance.
R1
iL
i2
+
+
VS
R2
L
-
VL
-
The boundary conditions for an inductor are iL (0 + ) = iL (0 - )
(a) iL (0 - ) =
VS
since the inductor is a DC short
R1
Therefore, iL (0 + ) =
VS
R1
(b) For an inductor, v = L
di
dt
VS R2
R1
di VL VR
VR
=
=
=- S 2
dt L
L
LR1
iL(0+ )
+
+
VR
VR = -iL (0 + ) R2 = -
VL
R2
-
-
(c)
VL = -
VS R2
R1
-3-
Example 2.4
A transformer consists of a primary winding with 500 turns and two secondary windings
of 125 turns and 30 turns. The 125 turn winding has 60 ohms connected to its terminals,
and the 30 turn secondary winding has 3 ohms connected to its terminals. If the primary
winding is connected to a 120 volt 60 Hz source, determine the current rating for each
winding.
N 2=125
60W
N 3=36
3W
+
120 V
N1=500
-
4. Repeat Example 2.4 with the 120 volt source connected to winding 2 and the 60 ohm
load attached to winding 1.
i2
500
N
V1 = 1 V2 =
(120) = 480volts
125
N2
+
i1
V3 =
i1 =
N3
36
V2 =
(120) = 34.56volts
N2
125
N2=125
120V
V2
-
+
60W
V1 N1 =500
i3
-
480V
= 8amps
60W
+
N3=36
V3
-
34.56V
= 11.52 amps
i3 =
3W
To find i2 we use conservation of power
v2i2 = v1i1 + v3i3
120i2 = ( 480)(8) + (34.56)(11.52)
120i2 = 3840 + 398.13
i2 = 35.32 amps
-4-
3W
Example 2.5
Obtain an expression for the characteristic
curve shown over the range from 8 to 12
amps. Obtain linear circuit models from the
derived expressions.
10
terminal voltage
8
6
4
2
0
0
2
4
6
8
10
12
terminal current (amperes)
5. Obtain a linear model of voltage versus current for the characteristic shown in example
2.5 over the current range of 0 to 6 amps. Obtain the current source and voltage source
equivalent circuits.
Use the two-point method to fit a straight line to the data
V - V0 V1 - V0
=
I - I0
I1 - I0
Read two points from the graph
point 0: 10 volts, 0 amps
point 1: 9 volts, 6 amps
Evaluating the expression,
V - 10 9 - 10
=
I-0
6-0
V - 10
1
=I
6
1/6-W
+
10 V
-
6V - 60 = - I
6V + I = 60
- I + 60
1
= - I + 10
6
6
NOTE: Thevenin is usually easier to
visualize
V=
I = -6V + 60
60 A
1/6-W
-5-
6. A 5 volt, 25 amp d.c. source has voltage regulation of 1%. Obtain an equivalent circuit
for this source.
The equivalent circuit is
R
+
VNL
VR =
VNoLoad - VFullLoad
¥ 100%
VFullLoad
VNoLoad - 5V
¥ 100%
5V
Solving for VNoLoad
5% = VNoLoad ¥ 100% - 500%
5 + 500 = VNoLoad (100)
505
VNoLoad =
= 5.05
100
1% =
R
+
5.05 V
R=
5.05 - 5 0.05
=
= 0.002W
25 A
25
-6-
7. All silicon diodes have a reverse breakdown voltage essentially the same as the
avalanche voltage of a zener diode. This results in three distinct regions of operation: one
for the normal forward region, one for the avalanche region, and an intermediate region
where the device behaves essentially as an open circuit. Obtain the linear models in these
three regions for a 1 amp, 2 watt silicon diode with a reverse breakdown voltage of 5.5
volts and a maximum reverse voltage of 5.6 volts.
1A
I
-5.5 V
-5.5 V
0.6 V
0.6V
2V
V
-5.6 V
Linear model
Theoretical
1.4 W
Powerrated = vrated ¥ irated
vrated =
Powerrated 2 watts
=
= 2Volts
irated
1Amp
0.6 V
V - 2 0.6 - 2
=
= 1.4
I -1
0 -1
V - 2 = 1.4 I - 1.4
V = 1.4 I + 0.6
In region I:
+
-
In region II: nothing, diode is an open circuit
In region III: find the rated current using
breakdown voltage
Vmax Imax = 2 watts
(5.6Volts) Imax = 2 watts
Imax = 0.36 Amps
0.28 W
5.5 V
V - ( -5.5) -5.6 - ( -5.5)
=
I-0
-0.36 - 0
V + 5.5 = 0.28 I
V = 0.28 I - 5.5
+
-7-
8. A three-terminal device is described by the following z-parameter equations.
VIN = 250iIN + 5iOUT
VOUT = -100iIN + 25iOUT
Obtain an equivalent circuit for this device.
Re-writing in matrix form. This is the z-parameter model.
È VIN ˘ È 250 5 ˘ È iIN ˘
ÍV ˙ = Í-100 25˙ Íi ˙
˚ Î OUT ˚
Î OUT ˚ Î
IIN
Z11
Z 22
+
V IN
+
Z 12VOUT
Z 21VIN
IIN
V OUT
-
250W
25W
+
V IN
IOUT
IOUT
+
5IOUT
-100IIN
V OUT
-
or draw reversed as
25W
IOUT
+100IIN
V OUT
+
-8-
9. Common base transistor configurations are often described in terms of common base yparameters: yib, yrb, yfb and yob. (The b’s in the subscripts indicate that the parameters were
obtained from a common base configuration with the emitter as the input terminal and the
collector as the output terminal.) A common circuit model for the transistor used in this
configuration is shown. Obtain the y-parameters for this common base circuit.
aie
re
e
ie
rc
c
+
ic
+
veb
rb
-
vcb
b
This is algebraically a very complex problem. Finding two-port parameters is usually
algebraically complex.
The y-parameters are defined by
È I1 ˘ È y11 y12 ˘ È V1 ˘
ÍI ˙ = Íy
˙Í ˙
Î 2 ˚ Î 21 y22 ˚ ÎV2 ˚
to find y11 and y21 short V2 :
I1 = y11V
I2 = y21V1
to find y12 and y22 short V1 :
I1 = y12V2
I2 = y22V2
Let’s start by shorting V2 to get:
V1-I1re
re
I1
+
V1
rb
rc
aI1
I2
Using KCL at the common node:
V -Ir V -Ir
I1 = 1 1 e + 1 1 e + aI1
rb
rc
Ê 1 1ˆ
Ê 1 1ˆ
I1 (1 - a ) = V1 Á + ˜ - I1re Á + ˜
Ë rb rc ¯
Ë rb rc ¯
Êr +r ˆ
Êr +r ˆ
I1 (1 - a ) = V1 Á b c ˜ - I1re Á b c ˜
Ë rb rc ¯
Ë rb rc ¯
-9-
I1 (1 - a )(rb rc ) = V1 (rb + rc ) - I1re (rb + rc )
I1 (1 - a )(rb rc ) + I1re (rb + rc ) = V1 (rb + rc )
V1 (rb + rc )
I
y11 = 1 =
V1 V1 (1 - a )(rb rc ) + re (rb + rc )
[
]
y11 =
(rb + rc )
(1 - a )(rb rc ) + re (rb + rc )
y21 =
I2
V - I r + aI1
I
=- 1 1e
= -1 + 1 (re - a )
V1
V1
V1
(rb + rc )
(r - a )
(1 - a )(rb rc ) + re (rb + rc ) e
(a - 1)(rb rc ) - re (rb + rc ) + (rb + rc )(re - a )
y21 =
(1 - a )(rb rc ) + re (rb + rc )
y21 = -1 + y11 (re - a ) = -1 +
arb rc - rb rc - re rb - re rc + rb re + rc re - arc - arb
(1 - a )(rb rc ) + re (rb + rc )
arb rc - rb rc - arc - arb
y21 =
(1 - a )(rb rc ) + re (rb + rc )
y21 =
y21 = -
(1 - a )rb rc + a (rb + rc )
(1 - a )(rb rc ) + re (rb + rc )
Using definitions
I1 = y12V2
I
y12 = 1
V2
I
y22 = 2
(3)
V2
Now short V1 :
V2
I2
rc
aI1
I2
re
V’
rb
V'
V - V'
= 2
- aI1
(1)
re || rb
rc
where the first expression is the lower
resistances, and the second term is the upper
loop
V'
I1 = (2)
re
I2 =
Substituting (2) into (1) gives:
Ê 1 aˆ
V - V'
V' V2
I2 = 2
+a
=
- V' Á - ˜
rc
re
rc
Ë rc re ¯
Using (3)
-10-
I2 =
Ê 1 aˆ
V2
- I2 (re || rb )Á - ˜
rc
Ë rc re ¯
Ê
r r r - arc ˆ V2
I2 Á 1 + e b e
˜=
re + rb rc re ¯ rc
Ë
Ê (r + r )r + r r - arb rc ˆ V2
I2 Á e b c e b
˜= r
(re + rb )rc
Ë
¯
c
I2
re + rb
re + rb
y22 =
=
=
V2 (re + rb )rc + re rb - arb rc (1 - a )rb rc + re (rb + rc )
Using (2),
V'
I
re
y12 = 1 =
V2
V2
Using (1)
Ê 1 aˆ
V
I2 = 2 - V ' Á - ˜
rc
Ë rc re ¯
-
Ê r - arc ˆ
V2
= V' Á e
˜
rc
Ë rc re ¯
V'
Solving for - :
re
V' Ê
V ˆ Ê rc ˆ
- = Á I2 - 2 ˜ Á
˜
re Ë
rc ¯ Ë re - arc ¯
and substituting into our expression for y12
Ê
V2 ˆ Ê rc ˆ
I2 rc
V2
Á I2 - ˜ Á
˜
rc ¯ Ë re - arc ¯ re - arc re - arc
Ë
y12 =
=
V2
V2
1
1
I
rc
=
y12 = 2
[ y22 rc - 1]
V2 re - arc re - arc (re - arc )
Substituting for y22
È
˘
1
(re + rb )rc
y12 =
- 1˙
Í
(re - arc ) Î (1 - a )rb rc + re (rb + rc ) ˚
(r + r )r - (1 - a )rb rc - re (rb + rc )
y12 = e b c
(re - arc ) (1 - a )rb rc + re (rb + rc )
r r + r r - r r + arb rc - re rb - re rc
y12 = e c b c b c
(re - arc ) (1 - a )rb rc + re (rb + rc )
I2 -
y12 =
[
]
[
]
[
]
rb [arc - re ]
- rb
=
(re - arc ) (1 - a )rb rc + re (rb + rc ) (1 - a )rb rc + re (rb + rc )
-11-
10. For a photodiode with characteristics shown in figure 2.10 and used in the circuit for
example 2.11 with a 5 volt source and a 10,000 ohm load, determine the load voltage as a
function of illumination level. Over what range of illumination levels will this result be
valid? Assume the curves extend to -5 volts on the third quadrant of the characteristic
curves.
ILIGHT
0 fc
100
200
300
400
500
4
1
+
1
If (µA)
0
RL
-
I0
kILIGHT
(a)
-8
-4
-3
-2
-1
0
(b)
(a) Biasing circuit for photodiode of figure
2.10
(b) equivalent circuit
-12
-5
RL
2
2
-4
IL
1
Vf (volts)
Fig. 2.10 Typical photodiode characteristics
i - 0.7 10.3 - 0.7
=
= 0.192 where the so-called “dark” current is estimated
IL - 0
500 - 0
to be about 0.7µA for 0 fc (footcandles). The other point comes from the estimated line for
the 500 fc illumination level.
From example,
i = 0.7 + 0.0192 I L (i is in microamperes)
(1)
Considering the load electrical circuit using KVL
5 = Vf + i(10 kW)
(2)
Substituting (1) into (2) gives
5 = Vf + (0.7 + 0.0192 I L ) ¥ 10 -6 (10 kW)
5 - Vf = 0.007 + 0.000192 I L
But 5 - Vf = VLOAD is simply the voltage across the load resistor.
The equation VLOAD = 0.007 + 0.000192 I L is valid when the illumination I L >0 and breaks
down when the diode becomes forward biased, i.e., the voltage across the load resistor
becomes greater than the voltage across the photodiode. At this point the voltage across the
diode drops to zero and the entire 5 volts appears across the load resistor.
At this breakdown point,
VLOAD = 0.007 + 0.000192 I L = 5Volts
and the corresponding light level is
5 - 0.007
IL =
ª 26, 000 footcandles.
0.000192
-12-
Example 2.12
The characteristics shown in figure 2.11 have voltage divisions of 0.2 volts and current
divisions of 1 amp. Obtain an equivalent circuit for a diode which accounts for the
temperature dependence with T1=20∞ C and T2=80∞ C.
10
If
Vs/RL
T1
T2
If (amps)
8
+
+
6
-
VS
4
2
-
Vs
Vf
RL
0
0.6
1.2
1.8
2.4
3
Vf (volts)
Figure 2.11
Temperature characteristics
for a diode
11. Repeat example 2.12 for T1=0∞ C and T2=100∞C.
Using T1=0∞ C and T2=100∞ C.
From the graph for T1 we can estimate the two points on graph to be (0.95V,0Amps) and
(1.4Volts,1.5Amps). Using the two point method
Vf - 0.95 1.4 - 0.95
=
if - 0
1.5 - 0
Vf - 0.95 = 0.3i f
Vf = 0.3i f + 0.95
There are many ways to find the second curve for T2=100∞ C. The slope remains the same
but the y-intercept changes.
Note that for T2=100∞ C at i f = 0 , Vf = 0.6 . The T2 curve is then
Vf = 0.3i f + 0.60
Assume that the y-intercept is a linear function of temperature and again use the two point
method. Using the points (0.95,0) and (0.60,100) for the curve we get an equation for the
y-intercept of
Co - 0.95 0.60 - 0.95
=
= -0.0035
T -0
100 - 0
Co = -0.0035T + 0.95
The overall equivalent circuit is
Vf i f , T = 0.3i f + Co = 0.3i f - 0.0035T + 0.95
(
)
-13-
Example 2.13
Copper has an electron density of 1028 electrons per cubic meter. The charge on each
electron is 1.6¥10-19 coulombs. For a copper conductor with width 5 mm, height 5 mm,
and length 10 cm, determine the Hall voltage for a current of 1 amp when the magnetic field
intensity is 1 amp/m, B=µoH and µo=4p¥10-7.
12. Use the dimensions given in example 2.13 but assume the conductor is an n-type
semiconductor with 1019 electrons per cubic meter. The current is 0.1 amp. Obtain an
equivalent circuit for a transducer using the Hall voltage to measure magnetic field
intensity. Assume that the voltage pickups are capacitative so there is no connection of the
output to the current circuit.
l=10cm
y
h=5mm
x
w=5mm
v
z
H=1A/m
r
r r r
The fundamental equation for the Hall effect is F = q E + v ¥ B = 0
Since therer is only a y-component of the E field we can rewrite this as
r
Ey + v ¥ B = 0
(1)
(
(
)
)
y
The current I x in the x-direction is given by I x = nqvx A where
n=number of carriers per square meter
q is the charge on the carriers
vx is the x-component of the carrier velocity and
A=wh is the conductor cross sectional area.
Solving,
I
I
vx = x = x
nqA nqwh
(2)
Rewriting (1) and substituting (2) into it
Ey - vx Bz = 0
Vh Ê I x ˆ
-Á
˜ m o Hz = 0
h Ë nqwh ¯
Solving for Vh
Ê
ˆ
0.1Amp
Vh = (0.005m)Á 19
4p ¥ 10 -7 ) Hz = 1.57 ¥ 10 -5 Hz
(
˜
-19
Ë (10 )(1.6 ¥ 10 )(0.005m)(0.005m) ¯
-14-
Example 2.14
A simplified small signal amplifier circuit is shown. The voltage source Vin is a variable
signal source and can be considered to be an ideal voltage source. The transistor has
parameters hie=500, hre=0, hfe=50, and hoe=0.0001. Determine the Thevenin equivalent
circuit (as seen from terminals a and b) in terms of the unknown turns N1 and N2 on the
right-hand transformer.
1kW
+
a
+
V in
-
+
b
Ideal 1:1
Ideal N1:N2
13. For the circuit of example 2.14, obtain the Norton equivalent circuit.
The transistor model is
iB
500W
For the transformer, if the +v terminal is at
the dot then the current is into the dot.
iS
+
+
0
50I B
-
V1
10kW
V1
-
-
+
See p. 2-6 of your reference book
The resulting overall circuit is then
iC
1000W
VIN
IB
iSC
+
500W
50IB
10kW
-
+
+
N1 :N2
-
Note the results of the dot convention. The current flowing into the dot results in current
flowing out of the top of the transformer secondary. The voltage source’s polarity is
reversed by the transformer as shown in the first figure. The output is a short circuit
because we are determining the short circuit current.
This gives
- vin
- vin
=
1000 + 500 1500
Continuing through the transistor,
iB = +
-15-
50
vin
1500
By the dot conventions for the output transformer, the output current is going into the dot.
N
N
The current magnitude is transformed as 1 and the voltage magnitude as 2 (Power is
N2
N1
conserved).
iC = -50iB = +
Since we are looking for a Norton equivalent circuit, the output is short circuited and the
load voltage is zero. Consequently, we are only looking at the short-circuit current. In this
context the transformer also appears as an ac short and we can ignore the 10kW output
impedance of the transistor.
N1
N
50
1 N1
=vin 1 = vin
N2
1500 N2
30 N2
This gives us the short circuit current in the direction shown. To get the Norton equivalent
circuit we need to also compute the Thevenin resistance. To get this resistance we short
vin , NOT the transistor current source 50iB and look at the resulting circuit.
iSC = -iC
i1
1000W
i2
+
500W
50IB
+
10kW
Z
-
N1 :N2
-
This leaves only the 10kW resistor which gets transformed by the transformer as
2
Ê N2 ˆ
RN = Á ˜ 10 kW
Ë N1 ¯
The final Norton equivalent circuit is then
iSC
Z
or
1 Ê N1 ˆ
Á ˜ VIN
30 Ë N2 ¯
ÊÁ N 2ˆ˜ 2
10k
Ë N 1¯
-16-
Example 2.14
A simplified small signal amplifier circuit is shown. The voltage source Vin is a variable
signal source and can be considered to be an ideal voltage source. The transistor has
parameters hie=500, hre=0, hfe=50, and hoe=0.0001. Determine the Thevenin equivalent
circuit (as seen from terminals a and b) in terms of the unknown turns N1 and N2 on the
right-hand transformer.
1kW
+
a
+
V in
-
+
b
Ideal 1:1
Ideal N1:N2
14. Consider the circuit of example 2.14 with an 8 ohm speaker attached to terminals a-b.
ÊN ˆ
Show that the necessary turns ratio Á 1 ˜ to provide maximum power transfer is 35.36.
Ë N2 ¯
Using the Norton equivalent circuit from the previous problem
1 Ê N1 ˆ
Á ˜ VIN
30 Ë N2 ¯
ÊN ˆ
Á 2 ˜ 10k
Ë N1 ¯
2
2
ÊN ˆ
The maximum power transfer occurs when RN = Á 2 ˜ 10 kW = 8W
Ë N1 ¯
ÊN ˆ
8W
Solving for the turns ratio, Á 2 ˜ =
= 0.0283
10 kW
Ë N1 ¯
Ê N1 ˆ
1
= 35.36
Á ˜=
Ë N2 ¯ 0.0283
-17-
8W
Example 2.16
The circuit shown has Z1=1µf and Z2=200mH. The voltage source is represented by the
partial Fourier series
vS = 20 cos1000t - 5 cos 4000t
Determine the voltage across the 100 ohm resistor.
50W
Z1
+
VS
100W
Z2
15. The elements Z1 and Z2 are interchanged in example 2.16, and the voltage vS takes on
the value of
vS = 20 + 5 cos 2236t
Find the voltage across the 100 ohm resistor.
The resulting circuit is then
50W
+
Z1
200mH
VS
Z2
At DC the inductor is a short, the capacitor is an open.
50W
+
20V
VT1 = 20Volts
RT1 = 50W
At w=2236 we compute the equivalent impedances to get
-18-
1µF
100W
50W
+
Z1
j447.2W
5V
Z2
-j447.2W
Z1 = jwL = j (2236)(0.2 H ) = j 447.2W
1
1
Z2 =
= - j 447.23W
=
jwC j (2236)(1 ¥ 10 -6 )
The Thevenin voltage for this AC circuit is found using a voltage divider
- j 447.2
VT 2 =
(5) = - j 44.7
50 + j 447.2 - j 447.2
The Thevenin resistance is that of the two impedances in parallel
(50 + j 447.2)(- j 447.2) = 3999. - j 447.2
ZT 2 = (50 + j 447.2) || ( - j 447.2) =
50 + j 447.2 + ( - j 447.2)
To find the output voltage we use superposition
@DC
50W
Using a voltage divider we get
100
VOUT1 =
(20) = 13.33Volts
+
100 + 50
20V
100W
@w=2236 rad/sec
(4000-j447.2)W
+
-j44.7V
100W
Using a voltage divider we get
100
VOUT 2 =
(- j 44.7)
100 + 4000 - j 447.2
VOUT 2 = 0.12 - j1.08 = 1.08– - 83.78∞
The measured voltage across the 100W resistor is then
V100 W = VDC + VAC = VTH1 + VTH 2 = 13.33 + 1.08 cos(2236t - 83.78∞)
-19-
16. The circuit of example 2.17 has rb =200 ohms, b=50, rc =2500 ohms, and re =10 ohms.
V
Find the voltage gain OUT when a 1000 ohm load is placed across the VOUT terminals.
VIN
Note that the output current I2 is
V
- OUT
1000
for this load.
+
I1
I2
bi 1
rb
V1
+
rc
V2
-
-
Vin
V out
+
+
V1
V2
Re
-
-
Figure 2.16. Combination of 2-port networks with common currents.
The circuit is then
rb=200W
bi1=50i 1
rc =2500W
1000W
Re=10W
We will convert both the upper and lower circuits to z-parameters and then combine them.
for z-parameters
È V1 ˘ È z11 z12 ˘ È I1 ˘
ÍV ˙ = Í z
˙Í ˙
Î 2 ˚ Î 21 z22 ˚ Î I2 ˚
For an open output I2 = 0 and
V1 = z11 I1
-20-
V2 = z21 I1
I1
I2
V1
rb=200W
bi 1=50i 1
rc =2500W V2
By inspection
V
z11 = 1 = 200W
I1
-(50 I1 )(2500)
V
z21 = 2 =
= -1.25 ¥ 10 5
I1
I1
if I1 = 0
V1 = z12 I2
V2 = z22 I2
and
V
z12 = 1 = 0
I2
V2
z22 =
= 2500W
I2
For the 10W resistor
I1
I2
V1
Re=10W
È10 10 ˘
Z=Í
˙
Î10 10 ˚
The total admittance matrix is then
200
0 ˘ È10 10 ˘ È 210
10 ˘
È
=
ZT = Í
+
5
˙
˙ Í
˙ Í
Î-1.25 ¥ 10 2500 ˚ Î10 10 ˚ Î-124990 2510 ˚
To find the voltage gain we write the network equations resulting from ZT
V1 = 210 I1 + 10 I2
(1)
(2)
V2 = -124990 I1 + 2510 I2
For the output,
V2 = - I2 RL
Using (2) and (3)
Ê V ˆ
V2 = -124990 I1 + 2510Á - 2 ˜
Ë RL ¯
-21-
V2
2510
V2 = -124990 I1
1000
1 + 2.51
I1 = V2 = -2.81 ¥ 10 -5 V2
124990
Substituting this result into (1) and using (3)
-V2 ˆ
V1 = 210( -2.81 ¥ 10 -5 V2 ) + 10Ê
Ë 1000 ¯
V2 +
V1 = -5.90 ¥ 10 -3 V2 - 0.01V2 = -1.59 ¥ 10 -2 V2
V2
= -62.9
V1
-22-
17. Using y-parameters, obtain the total circuit y-parameters for the circuits indicated by the
dashed lines. Hint: first find the y-parameters of the two indicated two-port networks, then
combine them to obtain the total network y-parameters.
Rf
+
+
Vin
Vout
+
-
rb
V1
g mV1
rc
-
The solution of this problem is similar to that of problem 16 except that y-parameters will
be used.
È I1 ˘ È y11
ÍI ˙ = Íy
Î 2 ˚ Î 21
y12 ˘ È V1 ˘
y22 ˙˚ ÍÎV2 ˙˚
For V1 = 0 :
I1 = y12V2
I2 = y22V2
For V2 = 0 :
I1 = y11V1
I2 = y21V1
I2
Rf
I1
V2
V1
I1
Rf
I2
y12 =
1
I1 - I2
=
=V2
V2
Rf
y22 =
1
I2
=
V2 R f
È 1
Í R
Therefore, YR = Í 1f
ÍÍÎ R f
1
I1
=
V1 R f
1
I
I
y21 = 2 = - 1 = V1
V1
Rf
y11 =
1
Rf
1
Rf
-
˘
˙
˙
˙
˙˚
-23-
For V2 = 0
For V1 = 0
I2
I1
V1
rb
gm V1
rc
I1
V2=0
V 1=0
I1 1
=
V1 rb
I
g V
y21 = 2 = m 1 = + gm
V1
V1
y11 =
And, therefore, Ytransistor
Ytotal
Ytotal
I2
rb
I1
=0
V2
I
1
y22 = 2 =
V2 rc
y12 =
È1
Ír
=Í b
Ígm
ÍÎ
0 ˘˙
1˙
˙
rc ˙˚
È 1 0˘ È 1
Ír
˙ Í R
= Ytransistor + YR = Í b 1 ˙ + Í 1f
Ígm
˙ Írc ˙˚ ÍÎ R f
ÍÎ
1 ˘
È 1 +1
ÍR
r
Rf ˙
˙
= Í f 1b
1 1˙
Ígm +
R f R f rc ˙˚
ÍÎ
1
Rf
1
Rf
-
˘
˙
˙
˙
˙˚
-24-
gmV1
rc
V2
CONCENTRATES
Example 2.16
The circuit shown has Z1=1µf and Z2=200mH. The voltage source is represented by the
partial Fourier series
vS = 20 cos1000t - 5 cos 4000t
Determine the voltage across the 100 ohm resistor.
50W
Z1
+
VS
100W
Z2
1. In the circuit shown in example 2.16, the voltage source is operating at a frequency of
2000 rad/sec. Specify the impedances Z1 and Z2 so that maximum power is transferred to
the 100 ohm load resistance. (Hint: Z1 and Z2 must be reactive or else they will absorb
some of the power.) In specifying Z1 and Z2, give their values in microfarads and/or
millihenries. There are two possible sets of answers.
When faced with a source-load problem always Thevenize the source circuit. From the
above circuit
jX2
VOC =
VS
jX1 + jX2 + 50
where we have assumed that Z1 and Z2 are purely reactive, i.e., Z1 = jX and Z2 = jX2
Similarly,
(50 + jX1 ) jX2
ZT = (50 + jX1 ) || jX2 =
50 + jX1 + jX2
The resulting circuit is then
ZT
+
VOC
100W
For maximum power transfer we require that ZT = (100)* = 100W or
(50 + jX1 ) jX2 = 100
50 + jX1 + jX2
Multiplying out
-25-
j 50 X2 - X1 X2 = 5000 + j100 X1 + j100 X2
Equating the real and imaginary parts
X1 X2 = -5000
(1)
(2)
50 X2 = 100 X1 + 100 X2
Solving (2) for X2 and substituting into (1)
-50 X2 = 100 X1
100
X2 =
X1 = -2 X1
-50
X1 ( -2 X1 ) = -5000
X12 = 2500
Therefore,
X1 = ±50W
The corresponding solutions for X2 are found by substituting this result into (1):
X1 X2 = -5000
5000
5000
== - / + 50W
X2 = ±50
X1
Case 1: X1 = +50W, X2 = -100W
X1 is inductive so X1 = wL1 = 50
50
= 0.025 H
L1 =
2000
1
= -100W
wC2
Solving for the capacitance gives
1
C2 =
= 5mf
(2000)(100)
X2 is capacitive so X2 = -
Case 2: X1 = -50W , X2 = +100W
1
= -50W
wC1
Solving for the capacitance gives
1
C1 =
= 10mf
(2000)(50)
X1 is capacitive so X1 = -
X2 is inductive so X2 = wL2 = 100
100
= 0.05 H
L2 =
2000
-26-
2. For the circuit shown, obtain the node voltage equations in matrix form. Solve these
V
equations for V1 and V2 . (The actual output is V1 - V2 .) Obtain the voltage gain OUT for this
VS
emitter -coupled amplifier.
50W
V4
I1
500W
V3
50I1
1000W
1000W
50I1
+
V1
VS
50W
1000W
V2
500W
I2
1000W
1000W
This is a complex problem which requires that we know how to construct an admittance
matrix representation for the circuit.
The standard (“formal”) admittance matrix formulation is:
˘ È V1 ˘
È Â I1 ˘ È Â Y11 Â Y12
˙ ÍV ˙
Í
˙ Í
˙Í 2 ˙
ÍÂ I2 ˙ = ÍÂ Y21 Â Y22
˙Í ˙
Í
˙ Í
˙Í ˙
Í
˙ Í
˚Î ˚
Î
˚ Î
where
 Ii =sum of current sources feeding into node i
(into=positive, out=negative)
 Yii =sum of all admittances connected to node i
 Yij = -sum of all admittances connected between nodes i and j
Vi =node voltages (including dependent sources)
The first step in solving the problem is to Nortonize the voltage source, converting it into a
current source
50W
+
VS
-
IS=VS/50
fi
Now, convert all resistances into admittances (I will use the unit of millimhos for
convenience) and label the nodes.
-27-
50W
node 4
2mmhos
V3
node 3
I1
50I1
IS
1mmho
node 1
40mmhos*
1mmho
1mmho
50I1
node 2
2mmhos
1mmho
I2
node 0
1mmho
Ground
Note that node 0, the reference node, must always be present.
*The 40millimhos is the combination of the two 50W resistors in parallel.
Constructing the admittance matrix representation
0 ˘ È V1 ˘
-1
È -50i1 ˘ È1 + 1 0
Í +50i ˙ Í 0 1 + 1
0 ˙ ÍV2 ˙
-1
2
Í
˙=Í
˙Í ˙
-1 2 + 1 + 2 + 1 + 1
-2 ˙ ÍV3 ˙
Í50i1 - 50i2 ˙ Í -1
Í
˙ Í
Í ˙
IS
0
40 + 2 ˙˚ ÎV4 ˚
-2
Î
˚ Î 0
Consider node 1:
A current source of 50i1 leaves node 1, therefore the current is -50i1
The sum of all admittances connected to node 1 is 1+1 millimhos
There are no admittances directly connected between nodes 2 and 1 or 4 and 1, therefore
Y12 = Y14 = 0
There is 1 millimho connected between nodes 1 and 3 so Y13 = -1 millimho since all offdiagonal terms are negative.
Actually each row of the matrix represents a KCL equation for that node and this is the way
I prefer to generate the matrix. At node 1, assuming that currents into the node are
positive,
-50i1 - 1(V1 - 0) - 1(V1 - V3 ) = 0
which can be re-written as
-50i1 = (1 + 1)V1 - V3 = 0
which is exactly the first row of the matrix equation.
The rest of the problem is simply solving the matrix:
È -50i1 ˘ È 2 0 -1 0 ˘ È V1 ˘
Í +50i ˙ Í 0 2 -1 0 ˙ ÍV ˙
2
Í
˙=Í
˙Í 2 ˙
Í50i1 - 50i2 ˙ Í-1 -1 7 -2 ˙ ÍV3 ˙
Í
˙ Í
˙Í ˙
IS
Î
˚ Î 0 0 -2 42 ˚ ÎV4 ˚
Expressing i1 and i2 in terms of independent variables V3 and V4
V - V3
i1 = 4
= 2(V4 - V3 )
500
V
i2 = 3 = 2V3
500
and converting the source term
-28-
VS
= 20VS
50
Then,
-100V4 + 100V3
È
˘ È 2 0 -1 0 ˘ È V1 ˘
Í
˙ Í 0 2 -1 0 ˙ ÍV ˙
100V3
Í
˙=Í
˙Í 2 ˙
Í50(2V4 - 2V3 ) - 50(2V3 )˙ Í-1 -1 7 -2 ˙ ÍV3 ˙
Í
˙ Í
˙Í ˙
20VS
Î
˚ Î 0 0 -2 42 ˚ ÎV4 ˚
È-100V4 + 100V3 ˘ È 2 0 -1 0 ˘ È V1 ˘
Í
˙ Í 0 2 -1 0 ˙ ÍV ˙
100V3
Í
˙=Í
˙Í 2 ˙
Í 100V4 - 200V3 ˙ Í-1 -1 7 -2 ˙ ÍV3 ˙
Í
˙ Í
˙Í ˙
20VS
Î
˚ Î 0 0 -2 42 ˚ ÎV4 ˚
Rearranging the equations,
È 0 ˘ È 2 0 -101 +100 ˘ È V1 ˘
Í 0 ˙ Í 0 2 -101
0 ˙ ÍV2 ˙
˙=Í
Í
˙Í ˙
Í 0 ˙ Í-1 -1 207 -102 ˙ ÍV3 ˙
Í20V ˙ Í
Í ˙
-2
42 ˙˚ ÎV4 ˚
Î0 0
S˚
Î
This equation can now be solved for the variables we are interested in ( V1 and V2 ) using
any method you want.
I highly recommend using a calculator which can solve systems of equations since this is
where I usually make math errors. However, I will use expansion by minors to solve for
V1 and V2 since the matrices are only 4¥4.
È 0 -101 100 ˘
-20VS Í 2 -101
0 ˙
Í
˙
ÍÎ-1 207 -102 ˙˚
V1 =
0 ˘ È0 -101 100 ˘
È 2 -101
Í
2 -1 207 -102 ˙ - 1Í2 -101 0 ˙
Í
˙ Í
˙
ÍÎ 0
-2
42 ˚˙ ÎÍ0 -2
42 ˙˚
The coefficients of all other minors are zero and not shown. Recall that solving for V1
requires that the column vector of sources be substituted for column 1. Note also that the
signs of the minor coefficients alternate.
Solving:
V1 =
-20VS [(2)(207)(100) - (100)( -101)( -1) - (2)( -101)( -102)]
2[(2)(207)( 42) - (2)( -2)( -102) - ( -1)( -101)( 42)] - 1[(2)( -2)(100) - (2)( 42)( -101)]
V1 =
-20VS [ 41400 - 10100 - 20604]
-20VS [10696]
=
2[17388 - 408 - 4242] - 1[ -400 + 8484] 2[12738] - 1[8084]
-29-
V1 =
-213920
VS = -12.29VS
17392
Similarly,
È 2 -101 100 ˘
-20VS Í 0 -101
0 ˙
Í
˙
-20VS [(2)( -101)( -102) - ( -1)( -101)(100)]
ÍÎ-1 207 -102 ˙˚
=
V2 =
0 ˘ È0 -101 100 ˘
17392
È 2 -101
Í
˙
Í
˙
2 -1 207 -102 - 1 2 -101 0
Í
˙ Í
˙
ÍÎ 0
-2
42 ˙˚ ÍÎ0 -2
42 ˙˚
where we saved a little effort by recognizing that the denominator is the same as before.
V2 =
20VS [20604 - 10100]
= +12.079VS
17392
AV =
V1 - V2 -12.29VS - 12.079VS
=
= -24.37
VS
VS
Expansion by minors:
Check sign:
-30-
3. For the circuit shown, obtain the loop current equations. Solve for the current in the 200
ohm resistance connected to the secondary of the ideal transformer.
100W
+
200W
I1
2V
200W
50I1
1000W
200W
ideal 4:1
As usual we will Thevenize the dependent current source to make the loop analysis easier.
1000W
50I1
-
1000W
fi
50,000I1
+
We will also reflect the load resistance through the transformer to make analysis simpler.
3200W
200W
where ( 4) (200) = (16)(200) = 3200W
2
ideal 4:1
Now, redrawing the circuit for analysis
100W
1000W
+
-
2V
200W
-
200W
50,000I1
I1
3200W
+
Writing the loop equations
-2 + 100i1 + 200i1 = 0
+50000i1 + 1000i2 + 200i2 + 3200i2 = 0
0 ˘ È i1 ˘
È2 ˘ È 300
Í0 ˙ = Í50000 4400 ˙ Íi ˙
˚Î 2 ˚
Î ˚ Î
-31-
I2
We are interested in the transformer primary current i2 . Solving for i2 :
È 300 2 ˘
Í50000 0 ˙
˚ = 0 - (50000)(2) = -75.76 mA
i2 = Î
0 ˘ (300)( 4400) - 0
È 300
Í50000 4400 ˙
Î
˚
The transformer secondary current is then
-75.76mA¥4=-303mA
with the current as shown below
200W
303mA
-32-
4. A 60 Hz source has its voltage measured under various loads, with the results shown.
Case Load
Voltmeter reading
1
open circuit
120 volts rms
2
60W resistance
72 volts rms
3
60W pure capacitance
360 volts rms
4
60W pure inductance
51.43 volts rms
Determine the voltage for a load of 120 ohms pure capacitance.
We will assume that the internal impedance of the source is complex and in series with the
source.
Case 1: under no load the open circuit voltage is 120 volts rms
Case 2:
72Vrms
R+jX
(60)(120)
= 72
(60 + R) + jX
60W
120V
Ê
ˆ
60
VOUT = Á
˜ 120
Ë 60 + R + jX ¯
5.184 ¥ 10 7
= 5184
(60 + R)2 + X 2
(60 + R)2 + X 2 = 10000
Case 3:
360Vrms
R+jX
120V
Ê
ˆ
- j 60
VOUT = Á
˜ 120
Ë - j 60 + R + jX ¯
(- j 60)(120)
-j60W
R + j ( X - 60)
= 360
5.184 ¥ 10 7
2 = 129600
R 2 + ( X - 60)
R 2 + ( X - 60) = 400 (2)
2
Case 4:
-33-
(1)
51.43Vrms
R+jX
120V
Ê
ˆ
+ j 60
VOUT = Á
˜ 120
Ë + j 60 + R + jX ¯
(+ j 60)(120)
+j60W
R + j ( X + 60)
= 51.4
5.184 ¥ 10 7
2 = 2645.045
R 2 + ( X + 60)
R 2 + ( X + 60) = 19598.911 (3)
2
Solving equations (2) and (3)
R2 + X 2 - 120 X + 3600 = 400
R2 + X 2 + 120 X + 3600 = 19598.911
R2 + X 2 - 120 X = -3200
R2 + X 2 + 120 X = 15998.911
Subtracting
-240 X = -19198.911
X=79.995W
Substituting this result back into equation (2) gives
2
R2 + (79.995 - 60) = 400
2
R2 + (19.995) = 400
R2 = 0.200
R = 0.447W
The output voltage for a capacitive -j120W load can then be calculated using these results.
Ê
ˆ
- j120
- j14400
(- j120)120
VOUT = Á
=
= 359.91 - j 4.021
˜ 120 =
Ë R + jX - j120 ¯
0.447 + j 79.995 - j120 0.447 - j 40.005
VOUT = 359.93– - 0.64∞
VOUT ª 360 volts rms
-34-
5. A bridge network is used to measure the temperature of a chemical bath.
R1
R2
+
24 V
m
1kW
1kW
R1 and R2 are identical thermistors with negative temperature coefficients of resistance of 4%/∞ C. At 25∞ C they are both 1500W. The meter (m) has an internal resistance of 800W.
Resistance R2 is held at 25∞ C.
(a) Determine the meter current when R1 is at 20∞ C.
(b) Determine the meter current when R1 is at 30∞ C.
The first step is to calculate the different values of R1.
R1=1500W @ 25∞ C
temperature coefficient -4%/∞ C
@20∞ C
R1 = 1500(1 - 0.04(T - 25∞C ))
R1 = 1500(1 - 0.04(20 - 25)) = 1500(1 + 0.04(5))
R1 = 1500(1 + 0.2) = 1800W
@30∞ C
R1 = 1500(1 - 0.04(T - 25∞C ))
R1 = 1500(1 - 0.04(30 - 25)) = 1500(1 - 0.04(5))
R1 = 1500(1 - 0.2) = 1200W
-35-
R1
+
24 V
R2=1500W
800W
V1
V2
1kW
1kW
We will write the nodal equations as shown below:
24 - V1 V1 - V2
V
- 1 =0
1800
800
1000
24 - V2 V1 - V2
V
+
- 2 =0
1500
800
1000
Re-writing
24 - V1 - 2.25V1 + 2.25V2 - 1.8V1 = 0
24 - V2 + 1.875V1 - 1.875V2 - 1.5V2 = 0
In matrix form,
-2.25
˘ È V1 ˘
È24 ˘ È1 + 2.25 + 1.8
=
Í24 ˙ Í -1.875
1.5 + 1.875 + 1˙˚ ÍÎV2 ˙˚
Î ˚ Î
È24 ˘ È 5.05 -2.25˘ È V1 ˘
Í24 ˙ = Í-1.875 4.375˙ ÍV ˙
Î ˚ Î
˚Î 2 ˚
Solving,
È24 -2.25˘
Í24 4.375˙
105 - ( -54)
159
159
˚ =
V1 = Î
=
=
= +8.895
È 5.05 -2.25˘ (5.05)( 4.375) - ( -1.875)( -2.25) 22.094 - 4.219 17.875
Í-1.875 4.375˙
Î
˚
È 5.05 24 ˘
Í-1.875 24 ˙
166.2
(5.05)(24) - (-1.875)(24)
˚ =
V2 = Î
=
= +9.2979
È 5.05 -2.25˘ (5.05)( 4.375) - ( -1.875)( -2.25) 17.875
Í-1.875 4.375˙
Î
˚
-36-
Im ( @ 20∞C ) =
V1 - V2 8.8951 - 9.2979
=
= -0.504 mA
800W
800
At 30∞ C the node equations are:
24 - V1 V1 - V2
V
+ 1 =0
@node 1
R1
800
1000
24 - V2 V1 - V2
V
@node 2
+
- 2 =0
R2
800
1000
24 - V1 V1 - V2
V
- 1 =0
1200
800
1000
24 - V2 V1 - V2
V
+
- 2 =0
1500
800
1000
In matrix form,
È- 24 ˘ È- 1 - 1 - 1
Í 1200 ˙ Í 1200 800 1000
Í 24 ˙ = Í
1
Í˙ Í
800
Î 1500 ˚ Î
1
˘
˙ È V1 ˘
800
1
1
1 ˙ ÍÎV2 ˙˚
˙
1000 800 1500 ˚
Multiplying the first row by 1200 and the second row by 1500 we get:
1.5
˘ È V1 ˘
È-24 ˘ È-1 - 1.5 - 1.2
=
Í-24 ˙ Í 1.875
-1.5 - 1.875 - 1˙˚ ÍÎV2 ˙˚
Î
˚ Î
1.5 ˘ È V1 ˘
È-24 ˘ È -3.7
Í-24 ˙ = Í1.875 -4.375˙ ÍV ˙
Î
˚ Î
˚Î 2 ˚
1.5 ˘
È-24
Í-24 -4.375˙ 105 + 36
˚ =
= 10.5421
V1 = Î
1.5 ˘ 13.375
È -3.7
Í1.875 -4.375˙
Î
˚
È -3.7 -24 ˘
Í1.875 -24 ˙
˚ = 88.8 + 45 = 10.0037
V2 = Î
1.5 ˘
13.375
È -3.7
Í1.875 -4.375˙
Î
˚
Im ( @ 20∞C ) =
V1 - V2 10.5421 - 10.0037
=
= +0.67mA
800W
800
-37-
6. An oscillator circuit is shown below. Determine the minimum value of K such that
eo=E. Also, find the frequency at which eo=E.
10-9F
5000W
+
+
1000W
E
KE
5000W
0.001H
400W
10-9F
eo
As usual we Thevenize the source so that we can combine resistances.
5000W
+
KE
5000W
5000KE
fi
-
Redrawing the circuit and writing the loop equations
10000W
10-9F
+
+
+
E
1000W
10-9F
10 -3H
5000KE
i1
i2
400W
i3
-
-5000 KE + 10000i1 + (i1 - i2 ) jw10 -3 = 0
jw10 -3 (i2 - i1 ) + (i2 - i3 )
1
=0
jw10 -9
1
1
i - i2 ) +
i3 + 400i3 = 0
-9 ( 3
jw10
jw10 -9
In matrix form,
È
-3
Í
È+5000 KE ˘ Í10000 + jw10
Í
˙ = Í - jw10 -3
0
Í
˙ Í
ÍÎ
˙˚ Í
0
0
Í
Î
- jw10
-3
1
+ jw10 +
jw10 -9
1
jw10 -9
-3
-38-
0
1
jw10 -9
1
1
+
-9
jw10
jw10 -9
˘
˙
˙ È i1 ˘
˙ Íi2 ˙
˙Í ˙
˙ ÍÎi3 ˙˚
+ 400 ˙
˚
eo
Since we are only interested in i3:
˘
È
-3
Í10000 + jw10 -3
- jw10
+5000 KE ˙
˙
Í
1
-3
˙
Í - jw10 -3
+ jw10 +
0
jw10 -9
˙
Í
˙
Í
1
0
0
˙
Í
-9
jw10
˚
Î
=
i3
˘
È
-3
˙
Í10000 + jw10 -3
- jw10
0
˙
Í
1
1
˙
Í - jw10 -3
+ jw10 -3 +
jw10 -9
jw10 -9
˙
Í
˙
Í
1
1
1
+
+ 400 ˙
0
Í
-9
-9
-9
jw10
jw10
jw10
˚
Î
i3 =
Ê
1 ˆ
+5000 KE( - jw10 -3 )Á ˜
Ë jw10 -9 ¯
Ï 4
ˆ
Ê
ˆ Ê
j2
j2
1 ˆÊ
1 ˆÊ
1 ˆ
-3 Ê
-3
-3 ¸
+ 400˜ - ( - jw10 -3 )( - jw10 -3 )Á +
+ 400˜ - Á Ì(10 + jw10 )Á jw10 +
˜Á+
˜Á˜ (10000 + jw10 )˝
Ë
¯
Ë jw10 -9
¯ Ë jw10 -9 ¯ Ë jw10 -9 ¯
jw10 -9 ¯ Ë jw10 -9
˛
Ó
i3 =
i3 =
Ï
2 ¥ 10 22
10
2
10
¥
Ì
w2
Ó
+5000 KE ¥ 10 6
2 j ¥ 1015
4 j ¥ 1015
10 22 j1015 ¸
+ jw 4000 + 4 ¥ 108 + 2 +
˝
w
w
w
w ˛
+5 ¥ 10 6 KE
ÏÊ
Ê
10 22 ˆ
5 ¥ 1015 ˆ ¸
10
j
2
.
04
¥
10
+
4000
w
ÌÁ
˜
Á
˜˝
w2 ¯ Ë
w ¯˛
ÓË
For eo @ E for oscillation we require i3 to be real which requires 4000w 4000w 2 = 5 ¥ 1015
Solving for w gives w 2 =
Using this value,
5 ¥ 1015
= 0 or
w
5 ¥ 1015
= 1.25 ¥ 1012 or w = 1.12 ¥ 10 6 rad / sec
4000
+5 ¥ 10 6 KE
(400) = E
10 22
10
2.04 ¥ 10 1.25 ¥ 1012
Solving for K gives
+5 ¥ 10 6 K ( 400)
1.24 ¥ 1010
2480
or
K
=
=+
= +6.2
=
1
6
10
9
+5 ¥ 10 ( 400)
400
2.04 ¥ 10 - 8 ¥ 10
-39-
The following problem is similar to a problem I encountered on my PE exam (morning
session)
Sketch the steady-state output of the following circuit. Be sure to indicate values of
voltages and time. Assume the op-amp operates on ±15 power supplies.
R4
R3
+
Vo
-
IDEAL
R2
C1
The component values are C1=0.1µf, R2=5kW, R3=10kW and R4=10kW.
-40-
-41-