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Tutorial 7 Solution

The document provides solutions to 4 questions: 1) Analyzes the output voltage of a diode circuit under positive and negative pulses. 2) Calculates the current gain, supply voltage, and base resistor of a common emitter amplifier. 3) Models the magnetic circuit of a core with and without airgaps and calculates the inductance, current, and energy. 4) Shows how to relate the hysteresis loss of a core between two operating conditions defined by different voltages, frequencies, and flux densities.

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Mihir Kumar Mech
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25 views3 pages

Tutorial 7 Solution

The document provides solutions to 4 questions: 1) Analyzes the output voltage of a diode circuit under positive and negative pulses. 2) Calculates the current gain, supply voltage, and base resistor of a common emitter amplifier. 3) Models the magnetic circuit of a core with and without airgaps and calculates the inductance, current, and energy. 4) Shows how to relate the hysteresis loss of a core between two operating conditions defined by different voltages, frequencies, and flux densities.

Uploaded by

Mihir Kumar Mech
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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Solutions for Tutorial-7

Q 1. Solution:
(a) For positive pulse of vi :
The diode would be ON and vo = 2V + 0.7 V = 1.3 V
The capacitor charges to 10 V + 2 V 0.7 V = 11.3 V
For negative pulse of vi :
The diode would be OFF and vo = 10 V 11.3 V = 21.3 V

(b) In the positive swing of applied voltage, the diode gets short-circuited being forward
biased. Thus the charging time constant would be almost zero.
In the negative swing of applied voltage, the diode gets open-circuited being reverse
bias. Thus the discharging time constant is = RC = (56 k)(0.1F) = 5.6 ms
In positive cycle, the very low time constant ensures that the capacitor gets charged
quickly to the applied voltage and also keeps the diode ON. As the diode is assumed to
have only fixed voltage drop with no resistance, the output voltage remains constant.
For the negative cycle, note that the half period of the applied signal happens to much
smaller than the effective time constant .

T=

1
T
= 1 ms,
= 0.5 ms << = 5.6 ms
f
2

So the capacitor does not get enough time to discharge appreciably before the input
signal switches its polarity, thus the output voltage remains unchanged.

Q2. Solution:
(a) I E =

I
VE
2.1 V
3.09 mA
=
= 3.09 mA , assuming I C I E , = C =
= 154.5
RE 0.68 k
IB
20 A

(b) VCC = VR + VCE + VE = (2.7 k )(3.09 mA ) + 7.3 V + 2.1 V = 17.74 V


C

(c) R B =

VR

IB

VCC VBE VE 17.74 V 0.7 V 2.1 V 14.94 V


=
=
= 747 k
IB
20 A
20 A

Q3. Solution:

c
c

a
NI

(a)
Figure 1(a)

+
Magnetic equivalent circuits for Q1

NI

eq

The equivalent magnetic circuit can be drawn as shown in the figures.


a)

(b)

The reluctances are computed as follows:


2

For the center core:


For the right/left path:

5 10
H-1 = 0.332 105
7
4
2400 4 10 5 10
15 102
H-1 = 0.995 105
a = b =
2400 4 107 5 10 4
c =

H-1
H-1

eq = c + (a // b) = 0.8295 105 H-1


Therefore, Inductance
Since Bmax is 2T,

L = N2/eq = 4002/(0.995105) = 1.93 H


max = 2510-4 = NImax/eq
Imax = 2510-40.8295 105/400 =0.207 A
Wf = (1/2) Imax2L = 0.50.20721.93 = 0.041 J

b)With the airgaps,

g =

0.5 10 2
= 79.577 105 H 1
4 107 5 104

0.5

0.

For the center core:


c =

-1
5 102
= 0.332 105 H
7
4
2400 4 10 5 10

Figure 1(b)

For the left/right path: a = b =

0.5 102
14.5 102
H-1 = 80.54 105 H-1
+
7
4
4 10 5 10
2400 4 107 5 10 4

eq = c + (a // b) = 40.602 105 H-1


L = N2/eq = 4002/(40.602 105) = 0.0394 H
Maximum B occurs in the central limb,
max=BmaxA = 2510-4 = NImax/eq
Imax = 2510-440.602 105/400 =10.15 A
Wf = (1/2) Imax2L = 0.510.1520.0394 = 2.03 J
Flux in the air gap = (1/2) max= (1/2)2510-4 = 510-4Wb
Energy stored in the two air gaps
Wg= 2(1/2)gmax2g = [(510-4]279.577105 = 1.989 J

Q4. Solution:
Let the two operating conditions be specified by

V1,f1, Bm1andV2, f2, Bm2

Then,
V1 = Kf1Bm1,

and

f2Bm2/f1Bm1 = V2/V1

For hysteresis loss, Ph1 = KhBm11.6f1,

V2 = Kf2Bm2,
Bm2 = (V2 /V1) (f1 / f2) Bm1 = (220/110) (60/50) 0.8 = 1.92 T
Ph2 = KhBm21.6f2

Ph2 = (Bm2/Bm1)1.6 (f2/f1) Ph1 = (1.92/0.8)1.6 (50/60) 200 = 676.4 W

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