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7 November, 2013Power electronics1

Single phase half-wave uncontrolled rectifier with R-Load

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Single phase half-wave uncontrolled rectifier with R-L Load

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Power electronics3

Single phase half-wave uncontrolled rectifier R-L Loadwith freewheeling diode

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Single phase half-wave uncontrolled rectifier R-L Loadwith freewheeling diode


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Single phase half-wave uncontrolled rectifier with DC- Load


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Single phase half-wave uncontrolled rectifier with L and DC- Load

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Single phase full-wave uncontrolled rectifier with R- Load


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Single phase half-wave controlled rectifier with R-Load

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Single phase half-wave controlled rectifier with R-L-Load

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Single phase full-wave controlled rectifier with R-Load

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Single phase Full-wave controlled rectifier with R-L-Load

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1

Q2-

Single phase half-wave controlled rectifier is connected to 220 V, 50Hz supply to feed 5

resistor in series with 10mHinductor if the firing angle = 30o.

(a) Find the DC and rms value of output voltage.

(b) Draw the waveforms of load, resistor, inductor voltages and load current.

Solution:

a)

() () () ( )

() [( ) ( )

]

() [( ) ( ) ]

() [( ) ( ) ]

()


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2

( ) ()

() ()

b)


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Problem

In the rectifier shown in the following figure assume Vs= 220 V, 50 Hz, L =10 mHand Ed=

170 V. calculate and plot the current and the diode

voltage drop along with supply voltage, Vs.

Solution:


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1

Example

A diode circuit shown in Fig. below with R=10 , L=20mH, and VS=220 2 sin314t.

a)Determine the expression for the current though the load in the period 0 t 2and

determine the conduction angle .

b)If we connect freewheeling diode through the load as shown in Fig. below Determine

the expression for the current though the load in the period of 0 t 3.


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2

Solution:(a) For the period of 0 t , the expression of the load current can be obtained as

following


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3


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(b) In case of free-wheeling diode as shown in Fig.2, we have to divide the operation of this

circuit into three parts. The first one when 0 t (D1 ON, D2 OFF), the second case

when t 2(D1 OFF and D2 ON) and the last one when2 t 3(D1 ON,

D2 OFF).

In the first part (0 t ) the expression for the load current can be obtained as In case (a).

Then:

the current at t = is starting value for the current in the next part. Then

In the second part t 2 , the expression for the load current can be obtained as

following:


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5

The current att = 2 is starting value for the current in the next part. Then A

i(2)= 0.095103A.

In the last part (2 t 3) the expression for the load current can be obtained:


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1

PE-Problem 2013-2014

Battery voltage E = 12Vand its capacity is 100Wh.The average charging current should be

Idc=5A. The primary voltage is Vp = 120V, 60Hz, and the transformer turn ratio is n = 2:1.

Calculate:

a.

Conduction angle of the diode.

b. The current-limiting resistance,R.

c.

The power ratingPR ofR.

d. The charging time h0 in hours.

e.

The rectifier efficiency .

f.

The PIV of the diode.

()


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2


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Q1-

Single phase half-wave diode rectifier is connected to 220 V, N1/N2= 5 supply to feed 10pure

resistor.

Draw load voltage and current and diode voltage drop waveforms along with supply voltage.

Then, calculate:

(a) The rectification efficiency

(b) Ripple factor of load voltage

(c) Peak Inverse Voltage (PIV) of the diode.

Solution:

VP= 220 V N1/N2= 5 R= 10


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Q2-

single-phase diode bridge rectifier has a purely resistive load of R=10ohms and, VS =44V.

Draw load voltage and current and diode voltage drop waveforms along with supply voltage

Then, Determine

(a) The efficiency,

(b) Form factor,

(c) Ripple factor,

(d) The peak inverse voltage (PIV)

(e) Input power factor.

Solution:

VS= 44 V R= 10


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Alazhar University Gaza

Faculty of Engineering & I nformation

Technology

Fi eld of Study: ME, CCE

Course Code: I TCE 4310

Course Name:Power Electroni cs

f irst Semester 2012/2013

05/12/2012, 60minutes

Mid term

Question Paper

(Total Points: 100)

St. Name: ________________________ St. Reg.#: ______________

/: _______________________ /: ______________

Answer the following Questions.

Question -1Single-phase diode bridge rectifier has a purely resistive load of R = 10 ohmsand, VS = 44 V.

Draw the source voltage and current, load voltage and current, Diode voltage and current.

Determine:

(a) The efficiency, (b) Form factor, (c) Ripple factor, (d) The peak inverse voltage, (PIV) of each diode,

and, (e) power factor.

Solution:

()


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Question -2The full-wave controlled bridge rectifierhas an ac input of 120 V at 60 Hzand a 20- load resistor. The

delay angle is 40o. Determine:

a-

the average current in the load

b- the power absorbed by the load

c-

the input power factor

d- Draw the load voltage and current, thyristor

currents and supply current

Solution:

( ) ( )

()

()

() ()()


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Question -3A single-phase half wave converter in the following figure is operated from a 120 V, 60-Hzsupply and

the load resistive load is R =10 . If the average output voltage is 25%of the maximum possible average

output voltage, calculate:

a- the delay angle

b-

the rms and average output currents

c-

the input power factor

Solution:

()

25.0)cos1(5.0

dm

dc

n

V

VV

)cos1(2

m

dc

VV

VVdc

5.13)120cos1(2

7.168

2

)2sin(1

2

m

rms

VV

VVrms

5.372

)240sin(09.2

1

2

7.169


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Alazhar University Gaza

Faculty of Engineering & I nformation

Technology

Fi eld of Study: ME, CCE

Course Code: I TCE 4310

Course Name:Power Electroni cs

f irst Semester 2011/2012

05/01/2012, 60minutes

M id -term

Question Paper

(Total Points: 100)

St. Name: ________________________ St. Reg.#: ______________

/:_______________________ /:______________

Answer the following Questions.

Question -1For the half-wave rectifierof Fig. below, the source is a sinusoid of 220 Vrmsat a frequency of 60 Hz.

The load resistor is 10 . Determine:

(a) The average load current.

(b) The average power absorbed by the load.

(c) The power factor of the circuit

(d) Draw load voltage and current and diode voltage

drop waveforms along with supply voltage

Solution:

The voltage across the resistor is a half-wave rectifier sine wave with peak value

Vm= 220 2= 311 V, the average voltage is Vm/, and average current

a-

Io= Vo/R = Vm/R = 2 (220)/5 = 9.9 A

The rms voltage across the resistor for a half-wave rectified sinusoid is

Vrms= Vm/2 = 2 (220)/2 = 155.56 V

b-

The power absorbed by the resistor is

P = v2rms/R = (155.56)2/10 = 2420 W

The rms current in the resistor is Vm/(2R) = 15.556 A, and the power could also be calculated

from I2rmsR = (15.556)2(10) = 2420 W

c-

The power factor is

PF = P/S = P/Vs,rmsIs,rms= 2420/(220)(15.556) = 0.707


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d-


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Question -2A single-phase half wave converterin the following figure is operated from a 169.7 V, sin 377t, supply.

If the load resistive load is R =10 and the delay angle is = /4, determine:

(a)

The efficiency

(b)The form factor

(c)The ripple factor

(d)The transformer utilization factor

(e)

The peak inverse voltage (PIV)

(f) Draw the waveforms

Solution:R = 10 , = 45o , Vs= 120 V, Vm= 2 * 120 = 169.7 V

a-

Vdc= Vm/2(1+ cos ) = 46.2V

Idc= Vdc/R = 4.62 A

Vrms= Vm/2[1/( - + sin2/2]1/2

= 81.05 V

Irms= Vrms/R = 8.1 A

Pdc= Vdc* Idc= 40.5 * 4.05 = 213.4 W

Pac= Vrms* Irms= 76.1 * 7.6 = 656.5 VA

= Pdc/Pac = 213.4/656.5 = 32.5 %

b- FF = Vrms/Vdc= 81.05/46.2 = 1.754 or 175.4 %

c- RF = FF21 = 1.44 or 144 %

d-

TUF = Pdc/VsIs= 213.4/120*8.1 = 0.22 and 1/TUF = 4.54

e- The PIV = Vm

f- the waveforms


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Question -3Single phase full-wave fully controlled rectifierbridge is connected to 220V, 50 Hzsupply to feed 5

resistor, if the firing angle =30o. Draw the load voltage and current, thyristor currents and supply

current. Then, calculate:

(a) The rectfication efficiency.

(b) Peak Inverse Voltage (PIV) of the thyristor.

Solution:

= (1 + ) = 311

(1 + 30) = 184.8

Idc= Vdc/5 = 36.94 A

= 311.12 + sin(60)2 = 216.8 Irms= Vrms/R = 216.8/5 = 43.35 A

a-= Pdc/Pac = 184.8 * 36.94/216.8 * 43.35 = 72.6 %

b- The PIV = Vm


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