Challenge

Project 1

AAmit Kumar9 Mar 2024

Buck converter:

It is part of dc-to-dc converter use for step down the voltage. A Buck converter steps down a DC voltage from the input to the output. The circuit operation depends on the conduction state of the MOSFET: On-state: The current through the inductor increases and the diode blocks.

The buck converter is so named because the inductor always “bucks” or acts against the input voltage. The output voltage of an ideal buck converter is equal to the product of the switching duty cycle and the supply voltage.

CCM:

Buck Converter Continuous conduction mode (CCM) was analysed in continuous conduction mode (CCM), i.e., the inductor current was always greater than 0.

Given data for design the buck converter:

Vout = 30v

Vin = 40-50v (we select 50v)

Power (P) = 500w

Switching frequency assume = 50khz

Now calculate:

Duty cycle (D):

D = Vout/Vin

= 30/50

=0.6

Resistance value (R):

R = Vout^2/P

= 30x30/500

= 1.8ohm

Inductor current:

I = P/Vout

= 500/30

= 16.67amp

Inductor current ripple is < 10%

∆IL = 0.1 X 16.67

= 1.67 amp

Inductor value (L):

L = (Vout / fs*∆IL) *(1-D)

= (30/50x1000x1.67)*(1-0.6)

=0.1437mH

Capacitor value (Cout):

Cout= (1-D) / 8*L*fs^2*(∆V0/V)

=(1-0.6) / 8*0.1437mH*(50k)^2*(0.01)

=0.4 / 28740

= 13.9179µF

Capacitor such that the output voltage ripple is < 10%.

Now we are find out the parameter:

Parameter	Value
Output voltage Vout	30v
Input voltage Vin	50v
Output current I	16.67Amp
Duty cycle D	60%
Resistance R	1.8Ω
Inductor L	0.1437mH
Capacitor C	13.9179µf
Ripple current ∆IL	1.67amp
Power P	500w
Switching frequency (Assume) fs	50Khz
inductor current ripple	<10%
capacitor output voltage ripple	<10%

Parameters setting:

Model:

Output:

Given data for design the buck converter:

Vout = 300v

Vin = 400-500v (we select 500v)

Power (P) = 5kw