These designs are according my ideas.There may be errors in design
I have done simulation only.Hardware will be done march 2016
Datas given in this IEEE paper
POWER =400W
VIN=24V
IIN=16A
VOUT=400V ,
IOUT =1A
Lm =48μH, Lk =0.25μH
the turns ratio of coupled inductor n = Ns/Np =4
OPERATING FREQUENCY OF THE SWITCH (MOSFET HERE) =50KHz
Gain formula of converter = ( (1+n) / (1−D) )+n
VOLTAGE GAIN CALCULATION
Gain = VOUT / VIN
VIN = 24V
VOUT = 400V
Gain = 400V / 24V =16.6
DUTY CYCLE CALCULATION
Voltage gain of converter = ( (1+n) / (1−d) )+n
16.6 = ( (1+4) / (1−d) )+ 4
16.6 = ( 5 / (1−d) )+ 4
12.6 S= 5 / (1−d)
(1−d) = 5 / 12.6
(1−d) = 0.396
D = 1− 0.396
D = 0.604
Coupling coefficient calculation
We have,
The coupling coefficient of coupled inductor, k = Lm / (Lm + Lk)
Lm =48μH, Lk =0.25μH
K = 48μH / (48μH + 0.25μH) = 0.995
COUPLED INDUCTOR DESIGN
Here core used is etd-59.From the datasheet of ETD-59 core
AL=1.5uH
We have,
M=k √LP * LS = 48μH
In paper winding ratio is given about 1:4
Let N be No. of turns in primary and 4N be No. of turns in secondary
LP=N2AL
LS=16N2AL
M = k√16N2AL * N2AL
M = k √16N4 * AL2
M= k * 4 N2 * AL
48μH =.995 * 4 N2 *1.5uH
48μH =5.97 uH * N2
N2 = 48μH / 5.97 uH
N2 = 8
N=2.83 =3
No. of turns in primary = 3
No. of turns in secondary =4N =4 * 3
No. of turns in secondary = 12
LP= NP2 * AL
LP= 9* 1.5 uH
LP= 13.5 uH
LS= NS2 * AL
LS= 122 * 1.5uH
LS= 144 * 1.5 uH
LS= 216 uH
OUTPUT CAPACITOR VALUE
FOR A CAPACITOR VOLTAGE CURRENT BASIC RELATION IS
I = C * dV / dt
dV is output ripple voltage. Assume that output ripple voltage is about 0.02% of output voltage
dV = 0.02% * 400V
dV = 0.08 V
C = I * dt / dV
We have dt = duty ratio/frequency
C = I * D / (F * dV )
C = 1A* 0.604/ (50000 Hz* 0.08 V)
C = 151 uF =180uF (Standard value)
