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Summary EAD410 Inductor design practice problem 1 with answers April 2019

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EAD410 Inductor design practice problem 1 with answers April 2019

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EAD 410 inductor design practice problem 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 230±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 15% of the nominal load current while the peak-peak output ripple
voltage is less than 2% of the nominal output voltage. Rated load power is 2000
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 10 m
and the semiconductor devices have rise- and fall-times of 200 ns with a tolerance
of 10%.

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  0.912
choose
 max  0.91

 for a given input voltage and a fixed turns ratio, peak secondary voltage is
fixed/constant

vˆ pri , min  281.424V
vˆsec, min  41.82V
n  6.7294




1

, vˆ pri , max  344.85V
vˆsec, max  51.248V
V s  264V  0.7426
Vo  36V

0.7426    0.91


(ii) Capacitance and inductance of output filter components to meet
specifications

L Vs  216V  2.055H
Vo  36V

L Vs  264V  5.878H
Vo  36V

C Vs  216V  7.23F
Vo  36V

C Vs  264V  7.23F
Vo  36V

Choose
Lo  5.878H
Co  7.23F

(iii) For realising the output filter inductor, determine the type of core
material, core flux density, to be used for this application. An important
consideration is size. Give reasons for your choices and tabulate any
relevant data used in decision making,

INDUCTOR DESIGN
The design inputs are:
IL,pk-pk=8.33 A Powdered Iron material Amorphous Iron
IL,pk=59.727 A Bc,pk=1.0 T Bc,pk=1.5 T
IL,rms55.612 A μr=75 μr=3000
Idc= 55.56 A Psp=550 W/kg Psp=35 W/kg (251mW/cc)
(3949mW/cc)
f=100 kHz
L= 5.88 μH
Ta=500C ; Ts=1000C


o A DC inductor is to be designed. Hence, it is not necessary to employ
ferrite cores as flux ripple component is fairly small.
o Consider amorphous iron or powdered iron core material.

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