Exam (elaborations) EAD 410 (EAD410)

EAD 410 Magnetic design practice problem 2 2019


A power processor is required to feed an application at a voltage of 36 VDC.
Galvanic isolation is required and it is decided to employ a full-bridge DC-DC
converter with transformer isolation to realise the interface. The switching
frequency is 100 kHz and a surface temperature of 1000C is to be assumed. Input
DC voltage is obtained by rectifying the available AC supply of 240±10% V, 50
Hz. The peak-peak ripple in the input DC voltage should not exceed 5% of the
peak DC voltage.

The secondary winding is centre-tapped and an LC filter is connected at the output
terminal of the high frequency rectifier. It keeps the peak-peak output current
ripple less than 10% of the nominal load current while the peak-peak output ripple
voltage is less than 5% of the nominal output voltage. Rated load power is 1000
W.

The high-frequency rectifier diodes have forward voltage drop of 1.5 V. The full-
bridge high-frequency inverter used to convert the input DC voltage into a high-
frequency AC waveform employs semiconductor devices with a voltage drop of 2
V when conducting. The output filter inductor has a series resistance of 30 m
and the semiconductor devices have rise- and fall-times of 150 ns with a tolerance
of 12%.

Data sheets for various core materials should be obtained and consulted where
necessary.

Stating any assumptions made, determine:

i. Practical range of waveform duty ratio

 max  0.9328
choose
 max  0.932

vˆ pri ,min  293.833V
I L,ave,min  27.778 A
vˆsec,min  41.13V
n  7.144




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, vˆ pri ,max  360.02V
vˆsec,max  50.395V
 Vs  264V  0.7607
Vo  48V
0.7607    0.932

ii. Size of input low-frequency rectifier filter capacitor to meet specifications

isw,nom,min  I pri ,nom,min  3.888 A
C Pin Vs 216V  1217.3W
Vo 36V

I dc Vs 216V  4.087 A
Vo 36V

Cin Vs 216V  2.682mF
Vo 36V


Pin Vs  264V  1246.482W
Vo 36V

I dc Vs  264V  3.424 A
Vo 36V

Cin Vs  264V  1.849mF
Vo  48V


iii. Capacitance and inductance of output filter components to meet
specifications

L Vs  216V  4.691H
Vo 36V

L Vs  264V  16.515H
Vo 36V

C Vs  216V  0.965F
Vo 36V

C Vs  264V  0.965F
Vo 36V
Choose
Lo  16.515H
Co  0.965F

iv. Practical values of voltage and current ratings for high-frequency rectifier
and inverter semiconductor devices




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