Project 3

Solution: 

A step-down DC/DC converter is a device capable of converting an unregulated input voltage into an regulated output voltage lower than the input voltage.A buck converter steps down the applied DC input voltage level directly. 

Given:

• Vin = 48V
• Vout = 12V
• Output Current = 10A

**Inductor current ripple(dI/I0) is less than 40% of the average inductor current and Output voltage ripple of the capacitor is less than 5%**

To find:

• Duty ratio, (Resistor as Load), L and C

So,we are going to use MOSFET as switch to construct the buck converter because MOSFET has high switching frequency and it use for low power rating, the MOSFET switching frequency is between 20khz - 2Mhz.

Calculations: Taking the Frequency of Mosfet as 40khz:

For finding the Load resistance 

Vout=Rl(load Resistance) * Il(Load current)

Rl=Vout/Il 

Rl=12/10=1.2ohm.

Duty Ration =Vout/Vin

                  =>12/48=0.25

Duty Ration%= 0.25*100 =>25%

For finding the  inductance Value:

L=[Vout * (Vin - Vout)]/[dIl * fs* Vin]               

dIL= Inductor Ripple current is should be in between 20% to 40% of the output current:

dIl= (0.2 to 0.4) * Il

dIl= 0.2 * 10A

    =>2A

so, L=[Vout * (Vin - Vout)]/[dIl * fs* Vin] 

     L=[12 * (48 - 12)]/[2 * 40000* 12] =0.0001125=112.5 micro Henry

For calculating the Capacitance:

C=(dIl)/(8 *fs*dVout)                                                 dVout=Equivalent series resistance(ESR) - dIL

C=2/(8*40000*2.4)                                                    dVout=1.2ohm - 2A

C=1/384000=>2.6micro Farad                                  dVout=2.4V