P&P Review Notes
Frank Merat
368-4572 (work)
216-368-6039 (work FAX)
216-291-0602 (home & FAX)
e-mail: flm@po.cwru.edu
http://vorlon.cwru.edu/~flm/eit/home.html
NOTE: Please inform me of your opinion of the relative emphasis of the topics in
this review material by any of the above FAX or e-mail number addresses or
simply write me your comments and send them by US mail to
Frank Merat
4398 Groveland Road
University Heights, Ohio 44118
Your input will be used to revise next year's exam review course.
Kirchoff’s Law’s
3. An ammeter is being designed to measure currents over the five ranges indicated in the
accompanying illustration. The indicating meter is a 1.00 milliampere movement with an
internal resistance of 50 ohms. The total resistance ( R1 + R2 + R3 + R4 + R5 ) is to be 1000
ohms. Specify the resistances Ra and R1 through R5 .
1.5mA
R1
15mA
R2
Ra
150mA
R3
1.5A
15A
R4
50W
R5
For 1.5mA
1.5mA
1.0mA
Ra
1000W
50W
For 15mA
1mA( R1 + 450W + 50W) = (15 - 1)mA(1000 - R1 )
1.0mA
450W
R1
15mA
50W
1000W-R1
There is 0.5 milliampere through the 1kW resistor.
Since the voltages must be equal:
1000(0.5mA) = 1mA( Ra + 50W)
or
Ra = 450W
R1 + 500W = 14000 - 14 R1
15 R1 = 14000 - 500 = 13500
R1 = 900
PE Review
Copyright F. Merat
Page 1
For 150mA
1mA( R2 + 900W + 450W + 50W) = (150 - 1)mA(1000 - (900W + R2 ))
R2 + 1400W = 149(100 - R2 )
1.0mA
900W
450W
150 R2 = 14900 - 1400 = 13500
R2 = 90
R2
150mA
50W
1000W-(900W+R2)
For 1.5A
1mA( R3 + 900W + 90W + 450W + 50W) = (1500 - 1)mA(1000 - (900W + 90W + R3 ))
1.0mA
900W
450W
R3 + 1490W = 1499(10 - R3 )
90W
1500 R3 = 13500
R3
R3 = 9W
1.5A
50W
1000W-(900W+90W+R3)
For 15A
1mA( R4 + 9W + 90W + 900W + 450W + 50W) = (15000 - 1)mA(1000 - (900W + 90W + 9W + R4 ))
1.0mA
R4 + 1499W = 14999(1 - R4 )
900W
15000 R4 = 13500
90W
R4 = 0.9W
9W
R4
15A
50W
1000W-(900W+90W+9W+R4)
And finally R5 = 1000W - (900W + 90W + 9W + 0.9W) = 0.1W
PE Review
Copyright F. Merat
Page 2
1. A voltmeter is being designed to measure voltages in the full-scale ranges of 3, 10, 30
and 100 volts DC. The meter movement to be used has an internal resistance of 50 ohms
and a full-scale current of 1 mA. Using a four-pole, single-throw switch, design the
voltmeter.
The meter circuit is easily designed using the equivalent circuit of the meter.
@100 volts
Rext
Rcoil=50W
100V
1mA FS
V 100volts
=
= 10 5 W
I
1mA
Rext = 10 5 W - 50 = 99950W
R=
@30V R =
V 30volts
=
= 30000W and Rext = 30000 - 50 = 29950W
I
1mA
@10V R =
V 10volts
=
= 10000W and Rext = 10000 - 50 = 9950W
I
1mA
@3V R =
V 3volts
=
= 3000W and Rext = 3000 - 50 = 2950W
I
1mA
You can design several different types of meter circuits using this data.
99950W
29950W
9950W
2950W
100V
2950W
7000W
20000W
70000W
30V
10V
3V
3V
10V
1mA FS
30V
1mA FS
PE Review
Copyright F. Merat
100V
Page 3
Thevenin’s Theorem, Norton’s Theorem
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.
10
8
terminal voltage
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.
6
4
2
0
0
2
4
6
8
terminal current (amperes)
10
12
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
1/6-W
Evaluating the expression,
V - 10 9 - 10
=
I-0
6-0
+
10 V
V - 10
1
=I
6
-
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
PE Review
1/6-W
Copyright F. Merat
Page 4
Power and Energy Calculations
10. The measurement system shown below is used for a balanced load of 1 kW with a
lagging power factor of 0.8. Determine the wattmeter readings.
See Section 3-16 for an explanation of the two-wattmeter method of three phase power
measurement.
Ia
+
Pa
+
Van
-
-
Ib
+Vbn+
Pb Vbc-
neutral
Vcn
+
Ic
The neutral is an artificial point used to make the two-wattmeter analysis easier.
Since it is not specified assume an ABC phase sequence. For this problem
PF = cosq = 0.8 lagging so that q = +36.87∞ .
Vc
Va
Vb
The voltages and currents are then specified by
Ia = I– - q = I– - 36.87∞
Van = V–0∞
Vbn = V– - 120∞
Ib = I– - q - 120∞ = I– - 156.87∞
PE Review
Copyright F. Merat
Page 5
Ic = I– - q + 120∞ = I– + 83.13∞
Vcn = V– + 120∞
Recall the power triangle and write the expressions for the voltages and currents as seen
by the wattmeters.
imaginary
S
q
P
Q
real
For wattmeter A:
Vac = Van - Vcn = V–0∞ - V– + 120∞ = V (1.732– - 30∞)
Ia = I– - 36.87∞
Sac = Vac Ia* = V (1.732– - 30∞)( I– - 36.87∞) = VI (1.732– - 30∞)(1– + 36.87∞) = VI (1.732–6.87∞)
*
Or, in rectangular form
Sac = VI (1.72 + j 0.207)
For wattmeter B:
Vbc = Vbn - Vcn = V– - 120∞ - V– + 120∞ = V (1.732– - 90∞)
Ib = I– - 156.87∞
*
Sbc = Vbc Ib* = V (1.732– - 90∞)( I– - 156.87∞) = VI (1.732– - 90∞)(1– + 156.87∞) = VI (1.732–66.87∞)
Or, in rectangular form
Sbc = VI (0.68 + j1.59)
Vline - line
3
In this problem Van , Vbn , and Vcn are phase voltages. Vac and Vbc are line-line voltages.
Then
P = 3 3Vphase Iline cosq = 3Vphase Iline cosq
Using the numbers for this problem
1000 watts = 3Vphase Iline (0.8)
or VI=416.67 watts
For a balanced load P = 3Vline - line Iline cosq and Vphase =
As a check on our calculations
Pa = Re{Sac } = 1.72VI = 1.72( 416.67) = 716.67watts
Pb = Re{Sbc } = 0.68VI = 0.68( 416.67) = 283.33
These powers add up to exactly 1000 watts so the answer looks good.
PE Review
Copyright F. Merat
Page 6
Transient Analysis
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.
iC
R1
i2
VS
+
+
R2
C
-
VC
-
The boundary conditions for the capacitor C are
1. no DC current flow
2. v(0 + ) = v(0 - )
R2
VS since there is no DC current through C.
By inspection, vC (0 - ) =
R1 + R2
(a) When the switch is open
R2
V
vC (0 ) vC (0 ) R1 + R2 S
VS
+
i2 (0 ) =
=
=
=
R2
R2
R2
R1 + R2
+
iC
i2
Then, iC (0 + ) = -i2 (0 + ) =
+
R2
VC
C
-
-VS
R1 + R2
-
(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
+
PE Review
Copyright F. Merat
Page 7
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
iL(0+)
+
VR
+
R2
-
VR = -iL (0 + ) R2 = -
VS R2
R1
di VL VR
VS R2
=
=
=dt L
L
LR1
VL
-
(c)
VL = -
VS R2
R1
PE Review
Copyright F. Merat
Page 8
10. The switch of the circuit shown has been open for a long time and is closed at t=0.
VC1
50W
+
-
0.01f
VC1
+
-
10V
+ 0.01f
1000W
100W
-
Determine the currents flowing in the two capacitances at the instant the switch closes.
Give the magnitude of the capacitances and directions.
The initial circuit when the switch is open for a long time is very simple.
VC1
+
-
50W
0.01f
10V
+
-
100W
The initial conditions are VC1 (0 - ) = 0 and VC2 (0 - ) = +10V
When switching capacitors look like ideal voltages sources which retain their initial
voltages, i.e., the circuit at t=0+ looks like
10V
50W
+
-
0.01f
+
10V
+
-
0.01f
1000W
-
100W
0V
PE Review
Copyright F. Merat
Page 9
which can be reduced to
50W
i1
i2
10V
+
-
iC2
10V
+
-
iC1
1000W
100W
By inspection (and being careful to follow the passive sign convention)
10V 1
i1 =
= Amp
50W 5
10V
1
i2 =
Amp
=
100W 10
Then,
1 1
= 0.3 Amp
iC1 = +
5 10
1
iC2 = -i2 = - Amp
10
PE Review
Copyright F. Merat
Page 10
Fourier Analysis
3. A train of rectangular pulses is applied to an ideal low-pass filter circuit. The pulse
height is 5 volts, and the duration of each pulse is 2ms. The repetition period is 10ms.
The low-pass filter has a cut-off frequency of 500 Hz. What percentage of the signal
power is available at the output of the filter?
The ideal filter
1
f
fo=500Hz
+5V
-5
-1
+1
+5
9 10
t(msec)
T
As drawn this pulse train is an even function. That was my option since I prefer even
function series.
f (t ) =
•
1
2pn ˆ •
2pn ˆ
ao + Â an cosÊ
t + Â bn sinÊ
t
Ë T ¯ n =1
Ë T ¯
2
n =1
The last term of this expression for f (t ) can be ignored. All bn = 0 since f (t ) is an even
function and sine is odd.
v( t ) =
ao •
2pn ˆ
+ Â an cosÊ
t
Ë T ¯
2 n =1
ao = 2 favg = 2
1
[(5V )(2m sec)] = 2V
10 m sec
t +T
2 1
2pn ˆ
an =
f (t ) cosÊ
t dt
Ú
Ë
T t1
T ¯
Do the integral from t=-0.005 to t=+0.005 seconds. For that case, the integral reduces to
PE Review
Copyright F. Merat
Page 11
0.001
2
2pn ˆ
4 ¥ 5 Ê 0.01ˆ Ê 2p ˆ
an = 2
5 cosÊ
t dt =
nt
sin
Ú
Ë
¯
0.01 0
0.01
0.01 Ë 2pn ¯ Ë 0.01 ¯ 0
0.001
wn =
2pn 2p
=
n
T
0.01
fn =
n
n
=
= 100 n Hz
T 0.01
=
10 Ê p ˆ
sin n
pn Ë 5 ¯
The ideal filter will pass ao , a1 , a2 , a3 , a4 and a5
Power is rms voltage squared.
2
2
1 ÈÊ ao ˆ
Vrms
a12 a22 a32 a42 a52 ˘
Pout =
= Í
+
+
+
+
+ ˙
2
2
2
2
2˚
R
R ÎË 2 ¯
The first term is different since the rms value of the dc term is the dc term
other terms the rms voltage is given by Vrms =
a0 = 2
ao
. For all
2
Vpeak
. Computing the terms we get
2
10 Ê 3p ˆ
sin
= 1.01
3p Ë 5 ¯
10 Ê 4p ˆ
a4 =
sin
= 0.47
4p Ë 5 ¯
10 Ê 5p ˆ
=0
a5 =
sin
5p Ë 5 ¯
a3 =
10 Ê p ˆ
sin
= 1.87
Ë 5¯
p
10 Ê 2p ˆ
a2 =
sin
= 1.51
2p Ë 5 ¯
a1 =
The output power is then
2
V2
1È 2
(1.87)2 (1.51)2 (1.01)2 (0.47)2 ˘ 4.51
Pout = rms = ÍÊ ˆ +
+
+
+
+ 0˙ =
R
R ÎË 2 ¯
2
2
2
2
R
˚
Using the definition of rms to determine the power for the input pulse waveform
2
+0.001
˘
1È 1
1 1
5
V2
25
25(0.002) =
Pin = rms = Í
dt
˙ =
Ú
R
R ÍÎ T -0.001 ˙˚
R 0.01
R
The percentage power passed by the filter is then
4.51
4.51
Pout
= R =
@ 90%
5
5
Pin
R
PE Review
Copyright F. Merat
Page 12
Transfer Functions
7. For the circuit shown, the current source is iS (t ) = 0.01 cos 500t
10W
A
+
+
iS(t)
30W
-
0.001F
Vc
0.25H
-
Find the transfer function G(s) =
VC (s)
IS ( s )
Using KCL at node A (+ out), and that the voltage at A is VC :
VC
10 + 0.25s
IS
VC
30
0.001sVC
VC
VC
+ 0.001sVC +
=0
30
10 + 0.25s
1
1
ˆ
IS = VC Ê + 0.001s +
Ë 30
10 + 0.25s ¯
Then
1
V (s)
VC
G( s ) = C
=
=
1
IS (s) V Ê 1 + 0.001s +
ˆ 10 + 0.25s + 0.03s(10 + 0.25s) + 30
C
Ë 30
10 + 0.25s ¯
30(10 + 0.25s)
- IS +
G( s ) =
30(10 + 0.25s)
300 + 7.5s
=
40 + 0.25s + 0.03s(10 + 0.25s) 40 + 0.55s + 0.0075s 2
7.5(s + 40)
0.55
40 ˆ
0.0075Ê s 2 +
s+
Ë
0.0075
0.0075 ¯
s + 40
G(s) = 1000 2
s + 73.33s + 5333
G( s ) =
PE Review
Copyright F. Merat
Page 13
5. Given the circuit shown where the amplifiers are ideal unity-gain voltage amplifiers
(infinite input resistance and zero output resistance). Amplifier A3 is a summing
amplifier, so its output is the sum of its two inputs. Determine the transfer function of the
resulting filter, identify it and its principal parameters.
3mH
50W
V1
A1
+
1µf
Vin
A3
1µf
+
Vout
50W
A2
V2
-
3mH
1
s10 -6
1
=
-6
-3
-6 2
1
+ 50 + s3 ¥ 10 -3 1 + 50 ¥ 10 s + 3 ¥ 10 ¥ 10 s
-6
s10
V1
1
=
-5
Vin 1 + 5 ¥ 10 s + 3 ¥ 10 -9 s 2
3 ¥ 10 -9 s 2
V2
s3 ¥ 10 -3
=
=
1
-3
1 + 5 ¥ 10 -5 s + 3 ¥ 10 -9 s 2
Vin
+
50
+
3
¥
10
s
-6
s10
Vo V1 V2
1 + 3 ¥ 10 -9 s 2
=
+
=
Vin Vin Vin 1 + 5 ¥ 10 -5 s + 3 ¥ 10 -9 s 2
Next, we need to put it into a standard form which we can then look up in a table of
filters
s2
1+ 2
wo
It is that of a band-reject filter
s
s2
1+
+ 2
Qw o w o
Identify the equivalences,
1
= 3 ¥ 10 -9 gives w o2 = 3.33 ¥ 108 , or w o = 18257 rad sec (about 2905 Hz)
2
wo
1
1
1
= 5 ¥ 10 -5 gives Q =
=
= 1.09
-5
-5
5 ¥ 10 w o 5 ¥ 10 (18257)
Qw o
V1
=
Vin
PE Review
Copyright F. Merat
Page 14
2. Plot the gain and phase response of the system given by the transfer function:
Hs =
s + 6500 s + 8500
s + 350 s + 100,000
SOLUTION:
Putting the expression into standard form
1 + s
1 + s
6500 8500
6500
8500
Hs =
s
s
350 100,000 1 +
1 +
100,000
350
The gain is given by
20log H jw
+ 20 log 1 + j w
+ 20 log 1 + j w
6500
8500
w
w
– 20 log 1 + j
– 20 log 1 + j
100,000
350
= 20 log 1.57
The first term is +4 db. The other terms are evaluated graphically.
The phase angle is given by
w
w
+ Tan– 1 w
– Tan– 1 w – Tan– 1
100,000
6500
8500
350
The first term evaluates to zero degrees. The other terms are evaluated graphically.
–H jw = –1.57 + Tan– 1
PE Review
Copyright F. Merat
Page 15
Gain (magnitude) response:
+20 db
+10 db
0 db
-10 db
-20 db
1
100
10,000
106
108
w
PE Review
Copyright F. Merat
Page 16
Phase response:
1
100
10,000
106
108
1
100
10,000
106
108
+90∞
+45∞
0∞
-45∞
-90∞
w
PE Review
Copyright F. Merat
Page 17
Complex Impedance
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 corresponding circuit for this problem is then
50W
+
Z1
200mH
VS
Z2
1µF
100W
At DC the inductor is a short, the capacitor is an open.
50W
+
20V
VT1 = 20Volts
RT1 = 50W
PE Review
Copyright F. Merat
Page 18
At w=2236 we compute the equivalent impedances to get
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∞)
PE Review
Copyright F. Merat
Page 19
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
PE Review
Copyright F. Merat
Page 20
Multiplying out
j 50 X2 - X1 X2 = 5000 + j100 X1 + j100 X2
Equating the real and imaginary parts
(1)
X1 X2 = -5000
(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 = -
X 2 is inductive so X2 = wL2 = 100
100
= 0.05 H
L2 =
2000
PE Review
Copyright F. Merat
Page 21
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
(1)
Case 3:
360Vrms
R+jX
120V
-j60W
Ê
ˆ
- j 60
VOUT = Á
˜ 120
Ë - j 60 + R + jX ¯
(- j 60)(120) = 360
R + j ( X - 60)
5.184 ¥ 10 7
2 = 129600
R 2 + ( X - 60)
R 2 + ( X - 60) = 400 (2)
2
PE Review
Copyright F. Merat
Page 22
Case 4:
51.43Vrms
R+jX
120V
Ê
ˆ
+ j 60
VOUT = Á
˜ 120
Ë + j 60 + R + jX ¯
(+ j 60)(120) = 51.4
+j60W
R + j ( X + 60)
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
PE Review
Copyright F. Merat
Page 23
6. A parallel combination of a resistance (10W), capacitance (88.5µf), and inductance
(66.3 mH) has 60Hz, 230 volt (RMS) applied. Obtain the:
(a) reactances of C and L.
(b) admittance of each circuit element.
(c) phasor diagram for the currents, using the applied voltage as the reference.
(d) admittance diagram for the circuits, including the total admittance.
(e) input current as a sine function, taking the applied voltage as a reference. (Is the
circuit inductive or capacitive?)
(f) power factor.
(g) power triangle.
The circuit is
230V,60Hz
10W
88.5µf
66.3mH
The angular frequency is w = 2p (60) = 376.99 rad sec
(a)
X L = jwL = j (377)(66.3 ¥ 10 -3 ) = j 25
1
1
XC =
= -j
= - j 30
jwC
(377)(88.5 ¥ 10 -6 )
(b)
1
BL =
= - j 0.04 mhos
j 25
1
BC =
= + j 0.0334 mhos
- j 30
1
1
G= =
= 0.1mhos
R 10
(c)
230–0∞
iR =
= 23–0∞amps
10
230–0∞
iC =
= 7.67– + 90∞amps
- j 30
230–0∞
iL =
= 9.2– - 90∞amps
j 25
PE Review
Copyright F. Merat
Page 24
iC=7.67
iR=23
iL=9.2
(d)
BC=0.0334
G=0.1
BL=0.04
Yt = 0.1 + j 0.0334 - j 0.04 = 0.1 - j 0.0066 = 0.1002– - 3.776∞
(e)
i = YV = (0.1002– - 3.776∞)(230–0∞) = 23.00 - j1.518 = 23.05– - 3.776∞
The circuit is inductive since the angle is negative. Recall the phasor diagram
iC
V
iL
(f)
PF = cos( -3.776∞) = 0.9978
(g)
p = vi* = (230–0∞)(23.05– - 3.776∞)* = 5301.5– + 3.776∞ = 5289.99 + j 349.14
The power triangle will look like this
5301.5VA
349.14VAr
5289.99 watts
PE Review
Copyright F. Merat
Page 25
Laplace Transforms
2. For the circuit shown, find (a) the system transfer function, (b) the response to a unit
step at the input, and (c) the response to a unit pulse with a duration of millisecond.
1000W
input
10µf
100W
output
Compute the parallel impedance of the 10µf capacitor and the 100W resistance and then
use a voltage divider relationship.
Compute the parallel impedance
1
100
100
-5
s(10 ¥ 10 -6 )
100
Z parallel =
= 10 -s3 =
1
1 + 10 s 1 + 10 -3 s
100
+
s(10 ¥ 10 -6 )
10 -5 s
(a) system transfer function
100
100
-3
-3
100
Vout
1 + 10 s
=
= 1 + 10 s
=
1000 + s + 100 s + 1100
Vin 1000 + 100
-3
1 + 10 -3 s
1 + 10 s
1
s
100 1 A
B
Vout =
= +
s + 1100 s s s + 1100
(b) response to a unit step
As + A1100 + Bs = 100
Equating the real parts A1100 = 100 or A =
1
11
Equating the imaginary parts As + Bs = 0 or B = - A = -
1 Ê
1 ˆ
1 1
1
1 ˆ
Vout = 11 - Á 11 ˜ = Ê ˆ - Ê
˜
Á
s Á s + 1100 ˜ 11 Ë s ¯ 11 Ë s + 1100 ¯
¯
Ë
PE Review
Copyright F. Merat
1
11
Page 26
Inverse Laplace transforming
Vout (t ) =
1
1
u(t ) - e -1100 t u(t )
11
11
(c) response to a unit pulse with duration of 1msec.
Vin (t ) = u(t ) - u(t - 10 -3 )
Laplace transforming this gives Vin (s) =
1 e -0.001s
s
s
The output is then given by
Vout (s) =
1 Ê1
1 ˆ 1 -0.001s Ê 1
1 ˆ
- e
Ë
¯
Ë
11 s s + 1100
11
s s + 1100 ¯
Vout (t ) =
1
1 -1100 ( t -10 -3 )
1 - e -1100 t )u(t ) - e
u(t - 10 -3 )
(
11
11
PE Review
Copyright F. Merat
Page 27
6. A Butterworth filter has its transfer function given below. Determine its corner
frequency, and its response to a 1 volt step input.
G( s ) =
100, 000
s + 26.13s + 341.4 s 2 + 2613s + 10, 000
4
3
From a table of Butterworth filter transfer functions [our book no longer has one], for n=4
2
2
Ê s ˆ
Ê s ˆ
s
s
q1 = 22.5∞ , Á ˜ + 0.765
+ 1 and q 2 = 67.5∞ , Á ˜ + 1.848
+1
wo
wo
Ë wo ¯
Ë wo ¯
We need to identify w o so divide by 10,000 to get 1 in denominator.
100, 000
10, 000
2
Ê s ˆ + 26.13 Ê s ˆ + 341.4 Ê s ˆ + 2613 Ê s ˆ + 1
Ë 10 ¯
10 Ë 10 ¯
100 Ë 10 ¯
1000 Ë 10 ¯
By inspection w o = 10
G( s ) =
4
3
From the table of filter transfer functions, the poles are at ±22.5˚ and ±67.5˚
Putting this into standard form for a transfer function [again from a table of transfer
functions] we have
10
2
ÈÊ s ˆ
Ê s ˆ + 1˘ ÈÊ s ˆ + 1.848Ê s ˆ + 1˘
.
+
0
765
ÍË ¯
Ë 10 ¯ ˙˚ ÍÎË 10 ¯
Ë 10 ¯ ˙˚
Î 10
The step response for this function is truly NASTY to compute. Do by partial fraction
expansion.
G( s ) =
2
B' s + C'
D' s + E'
1 A'
R( s) = G(s) =
+
+
2
2
s
s
s
s s Ê sˆ
+ 0.765Ê ˆ + 1 Ê ˆ + 1.848Ê ˆ + 1
Ë 10 ¯
Ë 10 ¯
Ë 10 ¯
Ë 10 ¯
Put into more standard form for Laplace analysis.
A
Bs + C
Ds + E
R( s) = + 2
+ 2
s s + 7.65s + 100 s + 18.48s + 100
This transforms to an expression of the form
r(t ) = Au(t ) + B' ' e - at cos bt + C' ' e - at sin bt + D' ' e - ct cos dt + E' ' e - ct sin dt
where I did not bother to compute the coefficients.
PE Review
Copyright F. Merat
Page 28
Solid-state Device Characteristics and Ratings
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
Powerrated = vrated ¥ irated
vrated =
1.4 W
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
V - ( -5.5) -5.6 - ( -5.5)
=
I-0
-0.36 - 0
V + 5.5 = 0.28 I
V = 0.28 I - 5.5
PE Review
0.28 W
5.5 V
+
Copyright F. Merat
Page 29
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
(
)
PE Review
Copyright F. Merat
Page 30
Transducers/Sensors
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)
RL
2
2
-4
IL
(b)
-8
-12
-5
-4
-3
-2
-1
0
1
Vf (volts)
(a) Biasing circuit for photodiode of figure
2.10
(b) equivalent circuit
Fig. 2.10 Typical photodiode
characteristics
Fit alinear function to the variation of current with respect to light level gives
i - 0.7 10.3 - 0.7
=
= 0.192 where the so-called “dark” current is estimated to be about
500 - 0
IL - 0
0.7µA for 0 fc (footcandles). The other point comes from the estimated line for the 500
fc illumination level.
(1)
i = 0.7 + 0.0192 I L (i is in microamperes)
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 is given by
VLOAD = 0.007 + 0.000192 I L = 5Volts
and the corresponding light level is given by
5 - 0.007
IL =
ª 26, 000 footcandles.
0.000192
PE Review
Copyright F. Merat
Page 31
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
PE Review
Copyright F. Merat
Page 32
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
˚
PE Review
Copyright F. Merat
Page 33
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
+
- 2 =0
@node 2
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
.
.
3
7
1
5
13.375
È
˘
Í1.875 -4.375˙
Î
˚
Im ( @ 20∞C ) =
PE Review
V1 - V2 10.5421 - 10.0037
=
= +0.67mA
800W
800
Copyright F. Merat
Page 34
Conductivity/Resitivity
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=r 20[1+a 20(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
5.10 ¥ 10 -8 m 2
5.10 ¥ 10 -8 m 2
= 0.02 m
l = (1W)
2.55 ¥ 10 -6 W - m
R20 (l = 0.02 m) =
(1.08 ¥ 10 W - m)(0.02m) = 0.42W
-6
5.10 ¥ 10 -8 m 2
Note: Circular mils is nothing more than the diameter (expressed in thousandths of an
inch) squared.
PE Review
Copyright F. Merat
Page 35
Semiconductors
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
(1)
Ey + v ¥ B = 0
(
(
)
)
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
(2)
nqA nqwh
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) ¯
PE Review
Copyright F. Merat
Page 36
Frequency Response
8. For a series RLC circuit of 10 ohms, 0.5 henrys, and 100µf, determine the resonant
frequency, the quality factor, and the bandwidth.
0.5H
10W
100µf
Z(s)
Z (s) = sL + R +
1
sC
Z (s) =
1
1
s + 10 +
2
2
s(10 ¥ 10 -6 )
Z (s) =
1È
2 ¥ 10 4 ˘ 1 È s 2 + 20 s + 2 ¥ 10 4 ˘
s
+
20
+
=
˙
2 ÍÎ
s ˙˚ 2 ÍÎ
s
˚
Z (s) =
1 2
s + 20 s + 2 ¥ 10 4
2s
[
The standard form is s 2 +
]
w0
s + w 02
Q
w 02 = 2 ¥ 10 4
w 0 = 100 2 = 141.2 rad sec
w 0 100 2
=
= 20
Q
Q
Q=
100 2
= 5 2 @ 7.07
20
Q=
w0
Dw
Dw =
w 0 100 2
=
= 20
Q
5 2
PE Review
Copyright F. Merat
Page 37
9. For a parallel RLC circuit of 10 ohms, 0.5 henrys, and 100µf, determine the resonant
frequency, the quality factor, and the bandwidth.
Y(s)
0.5H
10W
100µf
Use the same approach as for a series RLC circuit but use Y(s) instead.
1
1 1
1
1
=
+ + sC =
+
+ s(10 2 ¥ 10 -6 )
1
Z (s) sL R
sÊ ˆ 10
Ë 2¯
2 1
20 + s + 10 - 10 -3 s 2
Y (s) = +
+ 10 -4 s =
s 10
10 s
Y (s) =
Y (s) =
s 2 + 1000 s + 20, 000
10 4 s
As before the model is s 2 +
w0
s + w 02
Q
w 02 = 20, 000
w 0 = 141.42 rad sec
w0
= 1000
Q
Q=
w0
= 0.141
1000
w0
Dw
w
141.42
Dw = 0 =
= 1000
Q
0.141
Q=
PE Review
Copyright F. Merat
Page 38
Frequency Selective Filters
4. Specify R2 and capacitances C1 and C2 to cause the following circuit to be a
Butterworth low-pass filter with a -3dB (corner) frequency of 150 Hz. The amplifier is to
be considered an ideal unity gain amplifier (infinite input resistance and zero output
resistance).
10kW
10mH
V1
5W
R2
A
+
+
C2
C1
Vin
Vout
-
-
1
sC1
1
V1
=
=
4
1
Vin
+ 10 4 1 + s10 C1
sC1
1
1
Vo
sC2
=
= 2 -2
1
V1
+ ( R2 + 5) + s10 -2 s 10 C2 + ( R2 + 5)sC2 + 1
sC2
Vo
V V
1
1
= 1 o =
¥ 2 -2
4
Vin Vin V1 1 + s10 C1 s 10 C2 + ( R2 + 5)sC2 + 1
The standard form for a Butterworth filter from a table of standard transfer functions (3rd
order) is
Vo
1
1
=
¥
2
Vin 1 + s
Ê s ˆ Ê s ˆ
w o ÁË w ˜¯ + ÁË w ˜¯ + 1
o
o
We want w o = 2pfo = 2p (150) = 300p
Identifying terms
10 -4
4
= 106nf
10 C1 = 300p , or C1 =
300p
10 2
1
=
C
= 113mf
10 -2 C2 =
,
or
2
(300p )2
(300p )2
2
1
1 (300p )
, or R2 =
- 5 = 4.4W
( R2 + 5)C2 =
300p
300p 10 2
PE Review
Copyright F. Merat
Page 39
Power Factor Correction
A load is connected to a voltage of 1320 volts at 60Hz. The load dissipates 100kW with
a 0.867 lagging power factor. Specify the capacitance needed to correct the power factor
to:
(a) 0.895 lagging
(b) 0.95 leading
(Give the voltage and volt-ampere-reactive ratings at 60Hz for the capacitors.)
1320V, 60Hz
100kW,
PF=0.867 lagging
Determine the initial reactive power before correction.
imaginary
S
q
Q
real
100kW
PF = 0.867lagging = cosq therefore q = 29.89∞
From the power triangle
Q = 100 kW tan q = 100 tan 29.89∞ = 100(0.5747) = + j 57.47kVAR
This reactive power must be corrected as per the problem specification. The desired
power factor is PF = 0.895lagging = cosq' . Therefore, the new angle must be
q' = 26.49∞ . The new reactive power for this angle comes from the new power triangle.
imaginary
S
q’
100kW
Q’
real
Q' = 100 kW tan q' = 100 tan 26.49∞ = 100(0.4984) = + j 49.84 kVAR
The difference in reactive powers must be supplied by the correction capacitor.
PE Review
Copyright F. Merat
Page 40
Q + Qcapacitor = Q'
Qcapacitor = Q' -Q = + j 49.84 - j 57.47 = - j 7.63kVAR
Pcapacitor = VI *
Ê 1320 ˆ
- j 7.63 ¥ 10 3 = (1320)Á
˜*
Ë XC ¯
Solving for the capacitive reactance:
XC * =
(1320)2
- j 7.63 ¥ 10
3
=
1
- j 2p (60)C
Solving for the required capacitance
7630
C=
= 1.16 ¥ 10 5 = 11.6mf
(1320)2 2p (60)
We already know the power rating to be 7.6kVAR and the voltage rating to be 1320
volts.
PE Review
Copyright F. Merat
Page 41
5. An impedance receives a line current which lags the voltage by 30˚. When the voltage
across the impedance is 100 volts (RMS), the impedance dissipates 200 watts. Specify
the reactance of a capacitance to be places in parallel with the impedance which would
make the line current be in phase with the voltage.
The given circuit is
100V RMS,
dissipates 200watts
The current lags the voltage
30˚
current
voltage
V2
V 2 (100)
From the power specification P =
or R =
=
= 50 . This gives G=0.02mhos.
R
200
P
At this point we have I = YV = (G + jB)V = (0.02 + jB)V . Since the phase angle is
known to be 30˚ we can use the relationship between voltage and current to find B.
2
G=0.02
30˚
B
tan 30∞ =
B
or B = G tan 30∞ = 0.02 tan 30∞ = 0.0115. Note that B is actually negative.
G
The resulting circuit is
+jBL
0.02
mhos
-j0.0115
Having the line current being in phase means that the reactance is zero. This requires
jBC + jBL = 0
BC = - BL = -( -0.0115) = 0.0115
XC = -86.6W for the desired power factor correction
PE Review
Copyright F. Merat
Page 42
Transformers
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.
N2=125
60W
N3=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
V3 =
+
i1
36
N3
V2 =
(120) = 34.56volts
125
N2
N2=125
120V
-
+
60W
V1 N1=500
i3
-
480V
= 8amps
i1 =
60W
i3 =
V2
+
N3=36
V3
3W
-
34.56V
= 11.52 amps
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
PE Review
Copyright F. Merat
Page 43
Small signal and large signal
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
+
Vin
-
+
b
Ideal N1:N2
Ideal 1:1
13. For the circuit of example 2.14, obtain the Norton equivalent circuit.
The transistor model is
500W
iB
iS
For the transformer, if the +v terminal is at
the dot then the current is into the dot.
+
0
50IB
10kW
-
+
-
V1
V1
-
+
See p. 28-3 of your reference book
The resulting overall circuit is then
1000W
VIN
IB
500W
50IB
iC
iSC
+
+
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
PE Review
Copyright F. Merat
Page 44
- vin
- vin
=
1000 + 500 1500
Continuing through the transistor,
50
iC = -50iB = +
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).
iB = +
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.
50
1 N1
N1
N
=vin 1 = vin
1500 N2
30 N2
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
1000W
500W
50IB
i1
i2
+
+
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
or
iSC
PE Review
Z
1 Ê N1 ˆ
Á ˜ VIN
30 Ë N2 ¯
Copyright F. Merat
Ê N2 ˆ
Á ˜ 10k
Ë N1 ¯
2
Page 45
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
VS
this emitter -coupled amplifier.
50W
V4
I1
V3
500W
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
-
PE Review
50W
IS=VS/50
fi
Copyright F. Merat
Page 46
Now, convert all resistances into admittances (I will use the unit of millimhos for
V3
2mmhos
node 4
node 3
I1
50I1
IS
1mmho
node 1
40mmhos*
1mmho
1mmho
50I1
node 2
2mmhos
I2
node 0
1mmho
1mmho
Ground
convenience) and label the nodes.
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 0 0 -2 42 ˚ ÎV4 ˚
Î
Expressing i1 and i2 in terms of independent variables V3 and V4
PE Review
Copyright F. Merat
Page 47
V4 - V3
= 2(V4 - V3 )
500
V
i2 = 3 = 2V3
500
and converting the source term
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 ˚
i1 =
È-100V4 + 100V3 ˘ È 2 0 -1 0 ˘ È V1 ˘
˙ Í 0 2 -1 0 ˙ ÍV ˙
Í
100V3
˙=Í
Í
˙Í 2 ˙
Í 100V4 - 200V3 ˙ Í-1 -1 7 -2 ˙ ÍV3 ˙
˙ Í
Í
˙Í ˙
20VS
˚ Î 0 0 -2 42 ˚ ÎV4 ˚
Î
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 ˙
42 ˙˚ ÎV4 ˚
-2
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:
PE Review
Copyright F. Merat
Page 48
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]
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
PE Review
Copyright F. Merat
Page 49
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
+
I2
Writing the loop equations
-2 + 100i1 + 200i1 = 0
+50000i1 + 1000i2 + 200i2 + 3200i2 = 0
PE Review
Copyright F. Merat
Page 50
0 ˘ È i1 ˘
È2 ˘ È 300
=
Í0 ˙ Í50000 4400 ˙ Íi ˙
˚Î 2 ˚
Î ˚ Î
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
300
0
È
˘ (300)( 4400) - 0
Í50000 4400 ˙
Î
˚
The transformer secondary current is then
-75.76mA¥4=-303mA
with the current as shown below
200W
PE Review
303mA
Copyright F. Merat
Page 51
Active networks and filters
1. Consider the OP-AMP circuit shown below.
(a) Determine the system transfer function.
(b) Sketch the gain of the circuit as a function of w. Assume that R1=1000W,
R2=200W, R2=1000W and C=1µF.
(c) What type of filter is this: low-pass, high-pass, etc.?
R1
+
Vo
-
C
Vs
R2
R3
SOLUTION:
(a) Apply KCl at the inverting input of the op-amp. Assuming that current into the
node is positive we can write
Vs – Vo
V – Vo
V
+ s
– o =0
R1
1
R2
sC
Solving for the transfer function gives
1 + sC V = 1 + 1 + sC V
s
o
R1
R2
R1
Vo
Vs
1 + sC
s + 1
R1
R1C
=
=
1 + 1 + sC
s + 1 + 1 1
R2
R1
R1
R2 C
PE Review
s + 1
R1C
=
1
s +
R1||R2 C
Copyright F. Merat
Page 52
This transfer function has a single pole and a single zero.
(b)
For the given values, the transfer function evaluates to
1
1000 1 ¥ 10– 6
=
1
s +
1000 200
1 ¥ 10– 6
1000 + 200
s +
Vo
Vs
1 + s
s
+
1000
1000
=
= 0.8
s + 1250
s + s
1250
s
1000
The first term is approximately -1.9 dB.
H jw
= 20log 0.8
+ 20log 1 +
(c)
This is a high-pass filter.
– 20log 1 +
s
1250
10 dB
0 dB
-10 dB
-20 dB
1
10
100
1000
104
105
w
PE Review
Copyright F. Merat
Page 53
Two-port theory (see p. 29-16 for definitions)
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
Z22
+
VIN
+
Z12VOUT
Z21VIN
IIN
VOUT
-
250W
25W
+
VIN
IOUT
IOUT
+
5IOUT
-100IIN
-
VOUT
-
or draw reversed as
25W
IOUT
+100IIN
VOUT
+
PE Review
Copyright F. Merat
Page 54
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
bi1
rb
V1
+
rc
V2
-
-
Vin
Vout
V1
+
V2
+
Re
-
-
Figure 2.16. Combination of 2-port networks with common currents.
The circuit is then
rb=200W
bi1=50i1
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 ˚
PE Review
Copyright F. Merat
Page 55
For an open output I2 = 0 and
V1 = z11 I1
V2 = z21 I1
I1
I2
V1
bi1=50i1
rb=200W
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
V
z22 = 2 = 2500W
I2
For the 10W resistor
I1
I2
V1
Re=10W
V2
È10 10 ˘
Z=Í
˙
Î10 10 ˚
The total admittance matrix is then
200
0 ˘ È10 10 ˘ È 210
10 ˘
È
=Í
ZT = Í
+Í
5
˙
˙
˙
Î-1.25 ¥ 10 2500 ˚ Î10 10 ˚ Î-124990 2510 ˚
To find the voltage gain we write the network equations resulting from ZT
(1)
V1 = 210 I1 + 10 I2
(2)
V2 = -124990 I1 + 2510 I2
For the output,
V2 = - I2 RL
PE Review
Copyright F. Merat
Page 56
Using (2) and (3)
Ê V ˆ
V2 = -124990 I1 + 2510Á - 2 ˜
Ë RL ¯
2510
V2 +
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 ¯
V1 = -5.90 ¥ 10 -3 V2 - 0.01V2 = -1.59 ¥ 10 -2 V2
V2
= -62.9
V1
PE Review
Copyright F. Merat
Page 57
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
gmV1
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
V1
V2
I1
Rf
I2
y12 =
1
I1 - I2
=
=V2
V2
Rf
y22 =
1
I2
=
V2 R f
PE Review
1
I1
=
V1 R f
1
I
I
y21 = 2 = - 1 = V1
V1
Rf
y11 =
Copyright F. Merat
Page 58
È 1
Í R
Therefore, YR = Í 1f
ÍÍÎ R f
1˘
Rf ˙
˙
1 ˙
R f ˙˚
-
For V2 = 0
For V1 = 0
I2
I1
V1
rb
gmV1
I1
V2=0
rc
V1=0
I1 1
=
V1 rb
I
g V
y21 = 2 = m 1 = + gm
V1
V1
y11 =
È1
Ír
And, therefore, Ytransistor = Í b
Ígm
ÍÎ
rb
gmV1
rc
V2
I1
=0
V2
1
I
y22 = 2 =
V2 rc
y12 =
0 ˘˙
1˙
˙
rc ˙˚
È 1 0˘ È 1
Ír
˙ Í R
Ytotal = Ytransistor + YR = Í b 1 ˙ + Í 1f
Ígm
˙ Írc ˙˚ ÍÎ R f
ÍÎ
1 ˘
È 1 +1
ÍR
r
Rf ˙
˙
Ytotal = Í f 1b
1 1˙
Ígm +
R f R f rc ˙˚
ÍÎ
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I2
1˘
Rf ˙
˙
1 ˙
R f ˙˚
-
Copyright F. Merat
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Power supplies
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
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Copyright F. Merat
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Oscillators
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
5000KE
i1
10-3H
10-9F
i2
eo
400W
i3
-
-5000 KE + 10000i1 + (i1 - i2 ) jw10 -3 = 0
1
jw10 -3 (i2 - i1 ) + (i2 - i3 )
=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 +
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Copyright F. Merat
- jw10
-
-3
1
jw10 -9
1
jw10 -9
˘
˙
0
˙ È i1 ˘
1
˙ Íi2 ˙
-9
jw10
˙Í ˙
˙ ÍÎi3 ˙˚
1
1
+
+
400
˙
jw10 -9 jw10 -9
˚
Page 61
Since we are only interested in i3:
˘
È
-3
Í10000 + jw10 -3
- jw10
+5000 KE ˙
˙
Í
1
˙
Í - jw10 -3
+ jw10 -3 +
0
jw10 -9
˙
Í
˙
Í
1
0
0
˙
Í
-9
jw10
˚
Î
i3 =
˘
È
-3
˙
Í10000 + jw10 -3
- jw10
0
˙
Í
1
1
-3
˙
Í - jw10 -3
+ jw10 +
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 =
+5000 KE ¥ 10 6
Ï
2 ¥ 10 22 2 j ¥ 1015
4 j ¥ 1015
10 22 j1015 ¸
10
8
w
2
10
4000
4
10
j
¥
+
+
¥
+
+
Ì
˝
w2
w
w
w2
w ˛
Ó
+5 ¥ 10 6 KE
ÏÊ
Ê
10 22 ˆ
5 ¥ 1015 ˆ ¸
10
ÌÁ 2.04 ¥ 10 - 2 ˜ + j Á 4000w ˜˝
w ¯ Ë
w ¯˛
ÓË
5 ¥ 1015
For eo @ E for oscillation we require i3 to be real which requires 4000w = 0 or
w
4000w 2 = 5 ¥ 1015
Solving for w gives w 2 =
Using this value,
+5 ¥ 10 6 KE
(400) = E
10 22
10
2.04 ¥ 10 1.25 ¥ 1012
Solving for K gives
PE Review
5 ¥ 1015
= 1.25 ¥ 1012 or w = 1.12 ¥ 10 6 rad / sec
4000
1.24 ¥ 1010
2480
+5 ¥ 10 6 K ( 400)
K
=
=+
= +6.2
or
=
1
6
10
9
+5 ¥ 10 ( 400)
400
2.04 ¥ 10 - 8 ¥ 10
Copyright F. Merat
Page 62
Transients
Sketch the steady-state output of the following oscillator 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.
This problem can only be solved by making some assumptions about the capacitor
(initially uncharged) and the output (assume it starts at 15 volts).
R3
10 kW
Vo =
(15) = +7.5 . This will be
R3 + R4
10 kW + 10 kW
the threshold at which the output switches state, i.e., if V-<+7.5 then Vo=+15 and, if V>7.5 volts, then Vo will switch to –15 volts.
Assuming that Vo = +15 then V+ =
Since the capacitor C1 is initially uncharged the voltage V- varies according to
-
t
R 2C 1
t
-
t
5 ¥10 3 )( 0.1¥10 -6 )
V- = Voe
= 15e (
= 15e 0.5 m sec
This means that V- will initially be at 0 volts but will rise towards +15 with exponential
time constant 0.5msec. I can sketch this and see that the switching will occur at
approximately t=0.5msec. I can also do this mathematically and find out that
-
t
-
t
15e 0.5 m sec = 7.5 , or e 0.5 m sec = 0.5 .
t
= -0.69315 , or t=34.66 msec so
0.5 m sec
my graphical drawing of exponentials was not really all that good.
Taking the logarithm and solving for t gives -
PE Review
Copyright F. Merat
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Vo (volts)
+15
Note: dotted lines show V+7.5
time (msec)
0
0.5
1.0
1.5
2.0
2.5
-7.5
-15
Note that the period changes as the oscillator approaches its steady state values.
PE Review
Copyright F. Merat
Page 64
Characteristic equations
1. A system is described by the differential equation
d 2i
di
(1)
20 sin 4t = 2 + 4 + 4i
dt
dt
di(0)
The initial conditions are i(0) = 0 and
= 4 . Determine i(t ) for t>0.
dt
To find the steady state solution convert to phasors
20 sin 4t = 20 cos( 4t - 90∞) Æ - j 20e jwt
Then (1) becomes
- j 20 = -w 2i + 4 jw + 4i
Since w = 4
- j 20
- j 20
- j 20
i=
=
=
= 1–141.3∞
2
-w + 4 jw + 4 -16 + j16 + 4 -12 + j16
isteady - state = cos( 4t + 141.3∞)
To find the transient solution set the forcing function to zero.
d 2i
di
+ 4 + 4i = 0
2
dt
dt
s2 + 4s + 4 = 0
-4 ± 16 - 4( 4)
s=
= -2
2
Since this is a second order equation we MUST have two linearly independent solutions.
we get the second solution by multiplying by t
Ae -2 t + Bte -2 t
i(t ) = cos( 4t + 143.1∞) + Ae -2 t + Bte -2 t
i(0) = 0 = cos(143.1∞) + A
A = - cos(143.1∞) = 0.8
di
= -2 Ae -2 t + Be -2 t - 2 Bte -2 t - 4 sin( 4t + 143.1∞)
dt
di
= 4 = -2(0.8) + B - 4 sin(143.1∞)
dt t = 0
B = 4 sin(143.1∞) + 1.6 + 4 = 8
The final solution is then
i(t ) = cos( 4t + 143.1∞) + 0.8e -2 t + 8te -2 t
PE Review
Copyright F. Merat
Page 65
An Abbreviated List of Laplace Transform Pairs
f(t) (t>0-)
d(t)
u(t)
TYPE
(impulse)
t
(ramp)
e-at
sin wt
(exponential)
(sine)
cos wt
(cosine)
te-at
(damped ramp)
e –atsin wt
(damped sine)
e –atcos wt
(damped cosine)
F(s)
1
1
s
1
s2
1
s+a
w
s2 + w2
s
s2 + w2
1
s+a2
w
s + a 2 + w2
s+a
s + a 2 + w2
(step)
An Abbreviated List of Operational Transforms
f(t)
Kf(t)
f1 t + f2 t – f3 t + ...
df t
dt
d 2f t
dt 2
d nf t
dt n
t
f x dx
0
f(t-a)u(t-a), a>0
e –atf t
f(at), a>0
tf(t)
t nf t
ft
t
PE Review
F(s)
KF(s)
F1 s + F2 s – F3 s + ...
sF s – f 0 –
s 2F s – sf 0 – –
df 0 –
dt
s nF s – s n–1f 0 – – s n–2
df 0 –
d 2f 0 –
d n–1f 0 –
– s n–3
–
...
–
dt
dt 2
dt n–1
Fs
s
e –asF s
F(s+a)
1f s
a a
dF s
–
ds
d nF s
–1 n
ds n
•
F u du
s
Copyright F. Merat
Page 66
Summary of initial conditions for Laplace transforms:
Inductor:
IL(s)
IL(s)
+
+
VL(s)
sL
iL(0)
sL
s
-
VL(s)
+
LiL(0)
I L ( s) =
i (0)
1
VL ( s) + L
s
Ls
VL ( s) = LsI L ( s) - LiL (0)
Capacitor:
IC(s)
+
VC(s)
-
IC(s)
+
1
sC
1
CvC(0)
sC
VC(s)
+
-
vC(0)
s
IC ( s) = CsVC ( s) - CvC (0)
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VC ( s) =
Copyright F. Merat
v (0)
1
IC ( s) + C
s
Cs
Page 67
Two-port Networks
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PE Review
Copyright F. Merat
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