# What is the PEAK-TO-PEAK source current ripple in amperes for the chopper?

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? • It would help if you added units to the variables. Apr 2, 2013 at 20:48

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