Question

Determine the values of inductor and filter capacitor for a buck converter with following specifications: Input= 48V, output = 24V, Load curernt = 6A, switching frequency = 20kHz, ripple current = 30%

Answer 1

Design procedure of Inductor and capacitor in a buck converter vary depending on the data. More and more the specifications better will be the design. So for the data given above we can have a simple design calculation as follows:

Assumption made : Buck converter operates in continuous current mode.

is the equation for voltage across the inductor during ON time of buck converter. We can use this as the governing equation for inductor design. We will rearrange the above equation as below :

During switch-ON:

---------------- (1)

Assuming the resistance of the inductor coil as zero, and . Where D is the duty ratio. And T is the switching period. is the current ripple.

Further for a buck converter. During Switch ON period, . Back substituting all these in equation (1) we get,

.----------------------(2)

Where is the switching frequency. Also = 30 percent of the output current, which is 1.8 A.

Substituting the given values in equation (2), we get,

L = 0.3333 mH. This is not an hard wired value. This can be adjusted according to the availability.

Similarly from the principle of conservation of charge the design equation for capacitor is given by :

-------------(3)

where is the ripple in output voltage. Since the ripple in output voltage is not specified we will assume it as 10 percent of output voltage = 2.4 V. Substituting these values in equation (3) we get,

C = 4.6875 F.

Again this is not a hard wired value. And can be changed slightly depending on the availability in the market.