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In the figure shown below, 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. What is the PEAK-TO-PEAK source current ripple in amperes for the chopper?

My calculations are :

\$\triangle I=\frac{TV_o(1-D)D}{L}\$

\$T=\frac{1}{25*10^{4}}\$

D=0.4

\$V_o=\frac{V_s}{1-D}\$

\$\triangle I=(\frac{1}{25*10^{4}})(\frac{12(0.4)}{L})\$

\$\triangle I=(\frac{1}{25*10^{4}})(\frac{12(0.4)}{100*10^{-6}})\$

\$\triangle I=(\frac{1}{25*10^{4}})(\frac{4.8}{100*10^{-6}})\$

\$\triangle I=\frac{48}{25*10}\$

Are these calculations correct? enter image description here

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  • \$\begingroup\$ It would help if you added units to the variables. \$\endgroup\$
    – jippie
    Apr 2, 2013 at 20:48

1 Answer 1

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If the converter is in continuous mode,

\$ V_L = L \dfrac{\Delta i}{\Delta t}\$

\$ \Delta t = \dfrac{0.4}{250kHz} = 1.6 \mu s\$

\$ \Delta i = \dfrac{V \Delta t}{L} = \dfrac{12V \cdot 1.6 \mu s}{100 \mu H} = 192mA \$

If you simulate this circuit in LTspice (or some other Spice simulator), you get very similar results. I got 196mA, but I didn't use an ideal MOSFET.

enter image description here

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  • \$\begingroup\$ @ Madmanguruman : Thanks. I could locate the error in my calculations. \$\endgroup\$ Apr 3, 2013 at 9:45

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