In the figure shown, the chopper feeds a resistive load from a battery source. MOSFET Q is switched at 250 KHz, with a duty ratio of 0.4. All elements of the circuit are assumed to be ideal. Average source current in Amps in steady state is
In the figure shown, the chopper feeds a resistive load from a battery source. MOSFET Q is switched at 250 KHz, with a duty ratio of 0.4. All elements of the circuit are assumed to be ideal. Average source current in Amps in steady state is Correct Answer <span class="math-tex">\(\frac{5}{3}\)</span>
Concept:
Volt - sec balance:
VS ∝T + Vs (1 - D) – V0 (1 - D)T = 0
\({{\rm{V}}_0} = \frac{{{V_s}}}{{1 - \; \propto }}\)
Ampere – sec balance:
- Io ∝T + (IL – I0) (1 - ∝) T = 0
- Io ∝ T + IL (1 - ∝) T – I0 (1 - ∝) T = 0
\({{\rm{I}}_L} = \frac{{{I_0}}}{{1 - \propto }}\)
Ripple current:
VL(ON) = Vs
\(\begin{array}{l} L\frac{{\Delta {\rm{I}}}}{{ \propto T}} = {V_s}\\ \Delta {\rm{I}} = \frac{{ \propto {V_s}}}{{FL}}\\ \Rightarrow {I_{Lmax}} = {I_L} + \frac{{\Delta {{\rm{I}}_L}}}{2}\\ = \frac{{{{\rm{I}}_0}}}{{1 - \propto }} + \frac{{ \propto {V_s}}}{{2FL}} \end{array}\)
Calculation:
The circuit is a boost converter
\(\begin{array}{l} {V_o} = \frac{{{V_{dc}}}}{{1 - d}} = \frac{{12}}{{1 - 0.4}} = 20V\\ {I_o} = \frac{{{V_o}}}{R} = 1A\\ {I_s} = \frac{{{I_o}}}{{1 - d}} = \frac{1}{{0.6}} = \frac{5}{3}A \end{array}\)
