BOOST CONVERTER

In boost regulator, the average output voltage V_o is greater than the input supply voltage V_s , hence the name ‘boost’.

Boost converter circuit diagram is shown below.

Assumptions :-

(1) The value of capacitance be large enough that the output voltage remains almost constant.

(2) Lossless circuit.

(3) The load current remains constant at I_o .

BOOST CONVERTER WORKING

The boost converter operation can be divided into two modes.

Mode-1 :- ( 0 ≤ t ≤ T_ON )

During this mode, chopper remains ON. With chopper ON, the diode is reverse biased by the initial voltage of the capacitor and it acts like an open circuit.

Therefore, the buck converter circuit diagram during this mode looks like

Current flows in loops 1 and 2.

INDUCTOR CURRENT AND VOLTAGE :-

CAPACITOR CURRENT :-

In loop-2, the capacitor discharges through the load.

As, i_c = - I_o and I_o= constant , therefore i_c is constant with negative sign.

CAPACITOR VOLTAGE :-

Capacitor voltage variation is explained by the basic capacitor equation

Since capacitor current i_c is a constant, therefore its integral will be linear and hence the capacitor voltage decreases (negative i_c) linearly.

BOOST CONVERTER WAVEFORMS

Mode-2 :- (T_ON ≤ t ≤ T )

During this mode, chopper is OFF. But since i_L cannot become zero immediately therefore, it takes path through the diode. Diode acts as a closed switch.

The circuit during this mode looks like

CAPACITOR CURRENT :-

i_c = i_L – I_o from the circuit.

Since i_L decreases linearly from I_max to I_min and I_o is constant. Therefore, i_c decreases linearly from ( I_max- I_o)to ( I_min - I_o ) in T_off time.

ENERGY TRANSFER :-

Since i_L is decreasing, the inductor voltage reverses polarity i.e becomes negative. So in this mode, energy is supplied by the source and inductor to load and capacitor.

By KVL in the above circuit,

V_s – V_o = 0

V_L = - ( V_o - V_s )

AVERAGE INDUCTOR VOLTAGE OVER FULL SWITCHING CYCLE