BOOST CONVERTER DESIGN

PROBLEM

DESIGN AN BOOST CONVERTER WITH FOLLOWING REQUIREMENTS

• V_IN=4V
• I_IN=6A
• V_OUT=16V ,
• I_OUT =1A

ANSWER

ASSUMPTIONS

FOR DESIGN OF THE CIRCUIT WE NEED FOLLOWING ADDITIONAL PARAMETERS AND WE WILL ASSUME THESE VALUES CONSIDERING THE FACT THAT OUTPUT WILL NOT IDEAL BUT CLOSE TO THE IDEAL STATE

• OUTPUT RIPPLE CURRENT = 30% OF OUTPUT CURRENT = 0.3A
• OPERATING FREQUENCY OF THE SWITCH (MOSFET HERE) =30KHz
• OUTPUT RIPPLE VOLTAGE = 0.5% OF OUTPUT VOLTAGE

DUTY CYCLE CALCULATION

DUTY CYCLE = (1- V_IN / V_OUT)

=(1- 4 / 16)

=.75

INDUCTOR VALUE CALCULATION

WHEN THE MOSFET IS ON, EQUIVALENT CIRCUIT IS GIVEN BELOW

IN ABOVE STATE, VOLTAGE ACROSS THE INDUCTOR WILL BE EQUAL TO THE VOLTAGE OF THE BATTERY V_IN = V_L

FOR AN INDUCTOR, VOLTAGE‑CURRENT BASIC RELATION IS

V_L = L · dI / dt

THEN, L = V_L·dt / dI

L = V_IN·dt / dI

HERE dt = DUTY CYCLE / FREQUENCY AND dI IS THE RIPPLE CURRENT

L = V_IN·DUTY CYCLE / (FREQUENCY·dI)

L = 4·0.75 / (30KHz·0.3)

L = 333 µH

CAPACITOR VALUE CALCULATION

FOR A CAPACITOR, VOLTAGE‑CURRENT BASIC RELATION IS

I = C·dV / dt

I·dt = C·dV = Q (CHARGE)

IN THE FIGURE BELOW, THE FIRST WAVEFORM SHOWS THE CURRENT THROUGH THE CAPACITOR. THE RIPPLE IS SMOOTHED BY THE CAPACITOR, WHICH ABSORBS EXCESS CURRENT AND DISCHARGES WHEN THE CURRENT FALLS BELOW THE AVERAGE, MAINTAINING A CONSTANT DC VOLTAGE. THE SHADED AREA IN THE SECOND PART SHOWS THE CHARGE STORED IN THE CAPACITOR.

INDUCTOR AND CAPACITOR CURRENTS: THE SHADED AREA FORMS A TRIANGLE, DIVIDED INTO TWO RIGHT TRIANGLES.

THEN, CHARGE Q = AREA OF TRIANGLE = (1/2)·(T_ON/2)·(ΔI/2) + (1/2)·(T_OFF/2)·(ΔI/2)

UNDER WORK…….