The document describes a dc battery charging circuit using a half-wave rectifier connected to a 230V 50Hz source through an 8 ohm resistor. It asks to (a) find the average charging current, (b) find the power supplied to the battery and dissipated in the resistor, (c) calculate the supply power factor, (d) find the charging time if battery capacity is 1000Wh, and (e) find the rectifier efficiency and peak inverse voltage of the diode.

1. Example1
A dc battery of constant emf E is being charged through a resistor using half-
wave diode rectifier. For source voltage of 230 V, 50 Hz and for R =8Ω, E
=150 V,
(a) Find the value of average charging current,
(b) Find the power supplied to battery and that dissipated in the resistor,
(c) Calculate the supply pf,
(d) Find the charging time in case battery capacity is 1000 Wh and
(e) Find rectifier efficiency and PIV of the diode.
Dr. Oday A. Ahmed

2. the value of average charging current
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3. the power supplied to battery
the power dissipated in the resistor
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4. Calculate the supply pf,
Find the charging time in case battery capacity is 1000 Wh
(Power delivered to battery) ×(charging time in hours)= Battery capacity in Wh
Hence, charging time is
Dr. Oday A. Ahmed

5. rectifier efficiency
PIV of the diode
𝑃𝑜𝑤𝑒𝑟 𝑑𝑒𝑙𝑖𝑣𝑒𝑟𝑒𝑑 𝑡𝑜 𝑏𝑎𝑡𝑡𝑒𝑟𝑦
𝑇𝑜𝑡𝑎𝑙 𝑖𝑛𝑝𝑢𝑡 𝑝𝑜𝑤𝑒𝑟
What is the PIV of the single-phase diode Rectifier Supply R-Load
Dr. Oday A. Ahmed

6. Dr. Oday A. Ahmed
A single-phase 230V: 1kW heater is connected across I-phase 230V, 50Hz
supply through an SCR. For firing angle delays of 45° and 90°, calculate the
power absorbed in the heater element.
230V, 1kW Heater
Example2

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Solution
Heater resistance:
For α=450, RMS output voltage:
𝑉
𝑜𝑟 =
2.230
2 𝜋
𝜋 −
𝜋
4
+
1
2
𝑠𝑖𝑛900 = 155.071 V Power
absorbed by heater element for α=450

8. Dr. Oday A. Ahmed
Power absorbed by heater element for α=450
𝑉
𝑜𝑟
2
𝑅
=
155.071
230
2
× 1000 = 454.57 𝑤𝑎𝑡𝑡
For α=900, RMS output voltage:
𝑉
𝑜𝑟 =
2. 230
2 𝜋
𝜋 −
𝜋
2
+
1
2
𝑠𝑖𝑛1800 = 115 V
Power absorbed by heater element for α=900
𝑉
𝑜𝑟
2
𝑅
=
115
230
2
× 1000 = 250 𝑤𝑎𝑡𝑡

9. Dr. Oday A. Ahmed
In a single-phase full-wave diode bridge rectifier, the diodes have a reverse
recovery time of 40 µs. For an AC input voltage of 230 V, determine the
effect of reverse recovery time on the average output voltage for a supply
frequency of (a) 50 Hz and (b) 2.5 kHz.
Solution
D 1 and D2 will not be off at ωt =π but will continue to conduct until t=( π/ω )+trr
Example3

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With zero reverse recovery time, average output voltage,

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For f= 50 Hz and trr = 40 µs, the reduction in the average output
voltage,
Percentage reduction in average outpace voltage

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(a) For f= 2.5k Hz, the reduction in the average output voltage,
Percentage reduction in average output voltage

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Example 4
A separately-excited DC motor, shown below, drives a rated load torque of 85N.M at
1200rpm. The field circuit resistance is 200Ω and the armature resistance is 0.2Ω. The
field winding connected to single-phase 400V source fed via single-phase full controlled
AC-DC converter with zero degree firing angle. The armature circuit is also fed through
another full converter from the same single phase 400V. With magnetic saturation
neglected, the motor constant is 0.8Volte-sec/Amp-rad.
For ripple free armature and field currents, determine:
a. Rated armature current
b. Firing angle delay of armature converter at rated load.
c. Speed regulation at full load.

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Some Points before solving
this Example
Variable-Speed DC Motor Drive
Methods

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basic DC motor types

16. Under steady-state operation, time
derivatives is zero. Assuming the motor is
saturated
For field circuit,
Vf = If Rf
Eg = Kv ω if
The armature circuit
Va = Ia Ra + Eg
Va = Ia Ra + Kv ω If
Kv is the motor voltage constant
(in V/A-rad/s)
The motor back emf, which is also known as speed voltage, is expressed :

17. Steady-state Torque and Speed
The motor speed can be easily derived :
If Ra is a small value (which is usual), or when the motor is slightly
loaded, i.e, Ia is small
That is if the field current is kept constant, the motor speed depends only on
the supply voltage.
The developed torque is :
The required power is :
Td = Kt If Ia = B ω + TL
Pd = Td ω

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It is a two quadrant drive

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either converter 1 operates to supply a positive armature voltage Vo, or converter 2
operates
to supply negative armature voltage –Vo. Converter 1 provides operation in first and
fourth quadrants, and converter 2 operation in second and third quadrants. It is four
quadrant drive and provides four modes of operation: forward motoring, forward
braking (regeneration), reverse motoring, and reverse breaking (regeneration).

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Return to Example
For ripple free armature

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a. Rated armature current
Electromagnetic torque Te
K has the unit Volte-sec/Amp-rad and equal to 0.8
rated armature current is

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a. Firing angle delay of armature converter at rated load.
For separately-excited DC motor:

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a. Speed regulation at full load.
At the same firing angle of 57.630 motor emf at no load,
no-load speed is equal to
the speed regulation is:

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