Inductor calculation.txt

CALCULATIONS :-

given values :

Vin = 21.5 V         5% ripple
Pin = 45 W
Iin = 2.88 A         20% ripple
Vout = 96 V
Efficiency = 95%
Fs (switching frequency) = 20 Hz

Time period = 1/Fs = 1/20000 = 0.00005 s

Output power = 95/100 * 45 
             = 42.75 W
             
Duty Ratio 
D for a three level boost conveter 
Vout/Vin = 3/(1-D)
96/21.5 = 3/1-D
D = 0.328125

Iin = 2.88 A
ripple is 20%
ripple current = 0.2*2.88 A
               = 0.576 A 
Peak current = 2.88 + 0.576 
             = 3.456 A

calcualtion of boost coverter inductance value

L = ( Vout * Ts )/ (16 * ∆i)
  = ( 96*0.00005)/16*0.576
  = 0.5275 mH

L = 520.75 uH

Vout = 96v
ripple 5 %
ripple voltage = 0.05*96
               = 4.8v

THREE LEVEL BOOST COVERTER HAS 2 CAPACITORS IN PARALLEL 
HENCE THE VOLTAGE DROP ACROSS EACH CAPACITOR IS 2.4V

CALCULATION OF CAPACITOR 

C = ∆i * D /( Fs * ∆Vout )
  = 0.576 *0.328125 / 20000 * 2.4
  = 3.9375 uF
  
Voltage across each capacitor is Vout/2  = 48 V

Inductor Design 

GIVEN 
Bm = 0.2T
Kw = 0.35 
J = 3 * 10 ^6 A/m^2
Kc = 1

AREA PRODUCT
AcAw = 2E / (Bm*Kw*J*Kc)

WHERE 
E = 1/2 * L * I^2
  = 0.5 * 0.50275 mH * ( 3.456) ^2
  = 0.0031502
  
AcAw = 2*0.0031502/(0.35* 0.2 * 3*10^6 * 1)
     = 29618.1174 mm^4