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SULIT

UNIVERSITI MALAYSIA PERLIS

Peperiksaan Pertengahan Semester

Sidang Akademik 2016/2017

April 2017

PLT207 – Power Electronics

[Elektronik Kuasa]

Masa : 1 jam 30 minit

(ANSWER)

Please make sure that this question paper has ELEVEN (11) printed pages including this

front page and appendices before you start the examination. [Sila pastikan kertas soalan ini mengandungi SEBELAS (11) muka surat yang bercetak termasuk muka

hadapan dan lampiran-lampiran sebelum anda memulakan peperiksaan ini.]

This question paper has TWO (2) questions. Answer ALL questions. Please write your

answer in the answer paper. [Kertas soalan ini mengandungi DUA (2) soalan. Jawab SEMUA soalan. Sila tulis jawapan anda dalam

kertas jawapan.]

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Question 1 [Soalan 1]

(a) Briefly explain the Effective Values voltage or Root-Mean_Square (rms) voltage

value with suitable equation [Terangkan secara ringkas tentang Voltan Nilai Berkesan atau voltan punca-min kuasa dua (pmkd)

dengan persamaan yang sesuai]

(3 Marks / Markah)

- The effective value of the voltage or current is also known as the root Mean Square (rms)

value. The effective value of the periodic voltage waveform is based on the average power

delivered to a resistor

- The RMS value is the effective value of a varying voltage or current. It is the equivalent

steady DC (constant) value which gives the same effect. For example, a lamp connected to

a 6V RMS AC supply will shine with the same brightness when connected to a steady 6V

DC supply.

- The effective value is the square root of the mean of the square of the voltage hence the

term root mean square

(b) The voltage and current for a device (using the passive sign convention) are

periodic functions with T=30 ms described by [Voltan dan arus untuk satu peranti (menggunakan tanda lazim yang pasif) adalah fungsi berkala

dengan T=30 ms digambarkan oleh]

0

12)(

Vtv

A

A

A

ti

4

5

8

)(

mstms

mst

3015

150

mstms

mstms

mst

3010

106

60

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Determine: [Tentukan:]

(i) the instantaneous power [kuasa serta merta]

(2 Marks / Markah)

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(ii) the average power [kuasa purata]

(2 Marks / Markah)

(iii) the energy absorbed by the device in each period [tenaga diserap oleh peranti ini di dalam setiap tempoh]

(2 Marks / Markah)

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(iv) the rms values of the voltages and current. [nilai pmkd voltan dan arus]

(2 Marks / Markah)

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(c) A half-wave rectifier has a source of 240 V rms at frequency 60Hz and RL load

with R=15 Ω and L=80 mH . Given β =4.35 rad. Determine : [Satu penerus setengah gelombang mempunyai sumber 120Vrms pada 60Hz dan beban RL dengan

R = 15 Ω dan L = 80mH. Diberikan iaitu β = 4.35 rad. Tentukan:]

(i) an expression for the load current [suatu ungkapan untuk arus beban]

(3 Marks / Markah)

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(ii) the average current [arus purata]

(2 Marks / Markah)

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(iii) the power absorbed by the resistor [kuasa yang diserap oleh perintang]

(2 Marks / Markah)

(iv) power factor [faktor kuasa]

(2 Marks / Markah)

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Question 2 [Soalan 2]

(a) Explain the basic operation of single phase full-wave bridge rectifier. Draw the

circuit diagram and sketch the voltages and current waveforms. [Terangkan operasi asas penerus titi gelombang penuh satu fasa. Lukiskan gambarajah litar dan

lakarkan bentuk gelombang voltan dan arus]

(4.Marks / Markah)

- The objective of a full-wave rectifier is to produce a voltage or current that is purely dc or

has some specified dc component.

- Full-wave rectifiers have some fundamental advantages. The average current in the ac

source is zero in the full-wave rectifier, thus avoiding problems associated with nonzero

average source currents, particularly in transformers.

- The output of the full-wave rectifier has inherently less ripple than the half-wave rectifier.

- For the bridge rectifier, Diodes D1 and D2 conduct together, and D3 and D4 conduct

together. Kirchhoff’s voltage law around the loop containing the source, D1, and D3

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(b) A single phase bridge rectifier has an RL load with R = 15 Ω and L = 30 mH. The

ac source is 120 V rms and 60 Hz. Evaluate : [Penerus titi satu fasa mempunyai beban RL dengan R = 12 Ω dan L = 20 mH. Sumber au 120 V

rms dan 60 Hz. Nilaikan:]

(i) the average load current. [purata arus beban.]

(2 Marks / Markah)

(ii) the power absorbed by the load. [kuasa yang diserap oleh beban.]

(2 Marks / Markah)

(iii) the power factor. [factor kuasa.]

(2 Marks / Markah)

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(c) Figure 2 shows a single phase ac voltage controller with a resistive load. Explain

the principle of operation for the circuit [Rajah 2 menunjukkan satu pengawal voltan au satu fasa dengan satu beban rintangan. Terangkan

prinsip operasi untuk litar tersebut:]

(4 Marks / Markah)

- The SCRs cannot conduct simultaneously.

- The load voltage is the same as the source voltage when either SCR is on.

- The load voltage is zero when both SCRs are off.

- The switch voltage vsw is zero when either SCR is on and is equal to the source voltage

when neither is on.

- The average current in the source and load is zero if the SCRs are on for equal time

intervals. The average current in each SCR is not zero because of unidirectional SCR

current.

- The rms current in each SCR is 1/√2 times the rms load current if the SCRs are on for equal

time intervals. (Refer to Chap. 2.)

Figure 2 [Rajah 2]

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(d) The single –phase ac voltage controller of Figure 2 has a 240-V rms source and a

load resistance of 35 Ω. Determine :

[Pengawal voltan satu fasa dalam Rajah 2 mempunyai punca 240 V-pmk dan beban rintangan 35 Ω

.Tentukan :]

(i) the delay angle required to deliver 800 W to the load. [sudut lengah yang diperlukan untuk menghasilkan 800W kepada beban.]

(2 Marks / Markah)

BONUS (graph not given)

(ii) the rms current in each SCR. [arus pmk bagi setiap SCR.]

(2 Marks / Markah)

(iii) the power factor. [faktor kuasa.]

(2 Marks / Markah)

-ooOoo-

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APPENDIX A [LAMPIRAN A]

COMMONLY USED FUNCTIONS

2 2

sin( ) sin cos cos sin

cos( ) cos cos sin sin

sin 2 2sin cos

cos 2 1 2sin 2cos 1

sin sin 2sin cos2 2

sin sin 2cos sin2 2

cos cos 2cos cos2 2

cos cos 2sin sin2 2

1sin sin

A B A B A B

A B A B A B

A A A

A A A

A B A BA B

A B A BA B

A B A BA B

A B B AA B

A B

2

2

cos( ) cos( )2

1cos cos cos( ) cos( )

2

1sin cos sin( ) sin( )

2

cossin

sin 2sin

2 4

sin( ) sin( )sin sin

2( ) 2( )

sincos

sin 2cos

2 4

co

A B A B

A B A B A B

A B A B A B

nxnxdx

n

x nxnxdx

n

m n x m n xmx nxdx for m n

m n m n

nxnxdx

n

x nxnxdx

n

2

sin( ) sin( )s cos

2( ) 2( )

sinsin cos

2

cos( ) cos( )sin cos

2( ) 2( )

m n x m n xmx nxdx for m n

m n m n

nxnx nxdx

n

m n x m n xmx nxdx for m n

m n m n

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APPENDIX B [LAMPIRAN B]

MATHEMATICAL FORMULAE

Power Computations

Fourier Series 0 n 0 n 0 0 n 0 n

n 1 n 1

f (t) a a cos(n t) b sin(n t) a C cos n t

RMS of function f(t)

0

0

t T

2rms

t

1F f (t)dt

T

2

2 2 nrms n,rms 0

n 0 n 1

CF F a

2

Instantaneous Power p(t) = v(t)i(t)

Average/real power:

0 0

0 0

t T t T

t t

1 1P p(t)dt v(t)i(t)dt

T T

If v(t)=Vdc dc avgP V I

If i(t)=Idc avg dcP V I

If v(t) and i(t) are

sinusoidal waveforms rms rmsP V I cos

, pf cos

If v(t) and i(t) are

nonsinusoidal periodic

waveforms

n,max n,max

0 0 n n

n 1

V IP V I cos

2

Power in a resistor 22rms

rms

VP I R

R

Apparent power rms rmsS V I Power factor

rms rms

P Ppf

S V I

Uncontrolled Half-Wave Rectifier

R load

Average output voltage m

o avg m

0

V1V V V sin( t)d( t)

2

RMS output voltage

2 m

rms m

0

V1V V sin( t) d( t)

2 2

RL load

Instantaneous output

voltage m

o

V sin t for 0 tv t

0 for t 2

Average output voltage

β

mo avg m

0

V1V =V = V sin ωt dωt = 1 cosβ

2π 2π

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Instantaneous output

current

t /m

22 1

Vsin t sin e for 0 t

i t Z

0 for t 2

L Lwhere Z R L , tan , and

R R

RL-Source load

Instantaneous output

current

t /dcm

/ 1dc dcm

m

VVsin t Ae for t

i t Z R

0 otherwise

V VVwhere A sin e , sin

Z R V

L-Source load

Instantaneous output

current m dc

1V cos cos t V t for t

i t L

0 otherwise

RL load with freewheeling diode

Instantaneous output

voltage m

o

V sin t for 0 tv t

0 for t 2

m m mo 0 02

n 2,4,6...

V V 2Vv (t) sin t cos n t

2 n 1

R load with output capacitor filter

Instantaneous output

voltage

m

o ( t )/ RCm

V sin t for t , diode onv t

V sin e otherwise, diode off

1 1tan ( RC) tan ( RC) Peak-to-peak ripple of

output voltage o mV V 1 sin

Peak-to-peak ripple of

output voltage with

large RC time constant

mo

VV

fRC

Controlled Half-Wave Rectifier

R load

Instantaneous output

voltage m

o

V sin t for tv t

0 for t 2

Average output voltage

mo

VV 1 cos

2

RMS output voltage

mrms

V sin(2 )V 1

2 2

RL load

Instantaneous output

voltage m

o

V sin t for tv t

0 for t 2

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Instantaneous output

current ( t)/mV

sin t sin e for ti t Z

0 otherwise

Average output voltage m

o

VV cos cos

2

RL-Source load

Minimum triggering

angle 1 dc

min

m

Vsin

V

Instantaneous output

current

t /dcm

/dcm

VVsin t Ae for t

i t Z R

0 otherwise

VVwhere A sin e

Z R

Uncontrolled Full-Wave Rectifier

R load

Instantaneous output

voltage

m

o

m

V sin t for 0 tv t

V sin t for t 2

Average output voltage m

o m

0

2V1V V sin t d t

RMS of load and source

current m

rms

II

2

RL load

Instantaneous output

voltage o o n 0

n 2,4...

mo

mn

v t V V cos n t

2Vwhere V

2V 1 1and V

n 1 n 1

RL-Source load

Average output current o dco

V VI

R

R load with capacitance output filter

Instantaneous output

voltage

m

o ( t )/ RCm

V sin t one diode pair onv t

V sin e diodes off

1 1tan ( RC) tan ( RC) Peak-to-peak voltage

variation o mV V 1 sin

Peak-to-peak voltage

variation with ωRC>>π m

o

VV

2fRC

R load with LC output filter

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Average output current oL R

VI I

R

Average output voltage

for CCM (3ωL>R) m

o

2VV

Controlled Full-Wave Rectifier

R load

Average output

voltage mo m

V1V V sin t d t 1 cos

RMS output current 2

m mrms

sin 2V V1 1I sin t d( t)

R R 2 2 4

RL load

Instantaneous output

current for DCM:

β<π+α

m ( t )/o

2 12

Vi ( t) = sin( t ) sin( ) for t e

Z

L Lwhere Z = + ( L , = , and = ) tanR

R R

Output voltage for

CCM:

1 Ltan

R

o o n 0 n

n 1

v ( t) V V cos n t

mo m

2V1V V sin t d t cos

2 2

n n nV a b

mn

mn

2V cos(n + 1) cos(n 1)a =

n + 1 n 1

2V sin(n + 1) sin(n 1)b =

n + 1 n 1

n = 2, 4, 6 . . .

RL-Source load

Minimum triggering

angle 1 dc

m

Vsin

V

Average output

voltage for CCM m

o

2VV cos

AC Voltage Controller

R Load

Source voltage s mv ( t) V sin( t) Output voltage

m

o

V sin( t) for t and tv ( t)

0 otherwise

RMS load/output

voltage 2

mo,rms m

sin 2V1V V sin t d t 1

22

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RMS load current =

RMS source current o,rms

o,rms

VI

R

Power factor sin(2 )

pf 12

Average SCR current

mSCR,avg

VI 1 cos

2 R

RMS SCR current o,rms

SCR,rms

II

2

RL Load

Output current

t /m

o

22 1

Vsin t sin e for t

Zi t

0 otherwise

Lwhere Z R L , and tan

R

Conduction angle Minimum delay angle min Range of delay angle RMS output current

2o,rms o

1I i ( t)d( t)

RMS SCR current o,rms

SCR,rms

II

2

Average SCR current

SCR,avg o

1I i ( t)d( t)

2