Step-up chopper Take a general step-up chopper with
voltage source V_s which is in series with the inductor L,
diode and the load with average voltage V_{ave}. The chopper switch would be in parallel with the series diode and load. Whenever the chopper switch is on, the output is shorted. Using
Kirchhoff's voltage law (KVL) in determining
inductor voltage, L\frac{di}{dt}=V_s and taking the average current within the turn-off time, \frac{\Delta i}{T_{ON}}=\frac{V_s}{L} where T_{ON} is the time were a load voltage is present and \Delta i the change current with respect to T_{ON}. Whenever the chopper switch is off and using KVL in determining inductor voltage with respect to average current within the turn-on time, \begin{align} L\frac{di}{dt}&=V_{ave}-V_s \\ \frac{\Delta i}{T_{OFF}}&=\frac{V_{ave}-V_s}{L}. \\ \end{align} where T_{OFF} is the time were a load voltage is zero. Equating both average current and taking the duty cycle \alpha=\frac{T_{ON}}{T_{ON}+T_{OFF}}, {{equation box 1|indent=::|equation=V_{ave}=\frac{V_s}{1-\alpha}}} where V_{ave} is the average output voltage.
Step-down chopper Taking a general step-down chopper with voltage source V_s which is in series with the chopper switch, inductor, and the load with voltage V_o. The diode would be in parallel with the series inductor and load. The same way by equating the average inductor current during the turn-on and turn-off time, we can get the average voltage by {{equation box 1|indent=::|equation=V_{ave}=\alpha V_s}} where V_{ave} is the average output
voltage, \alpha is the
duty cycle and V_s is the source voltage.
Step-up / step-down chopper Taking a general buck-boost chopper which works as stepup and down chopper, let the voltage source V_s be in series with the chopper switch, reverse biased diode, and the load with voltage V_o. The inductor would be in parallel with the series diode and load. The same way by equating the average inductor current during the turn-on and turn-off time, we can get the average voltage by {{equation box 1|indent=::|equation=V_{ave}=\frac{\alpha V_s}{1-\alpha V_s}}} where V_{ave} is the average output
voltage, \alpha is the
duty cycle and V_s is the source voltage. ==See also==