Step-down chopper variable frequency, 3 ms ON time

In variable-frequency (frequency modulation) control, $T_{on}$ is kept constant and the duty cycle is changed by varying the chopping period $T$ (hence $T_{off}$).

Given: $V_{dc} = 200$ V, $T_{on} = 3$ ms (constant), $k = 0.3$, resistive load.

OFF time

$$\begin{aligned} k &= \frac{T_{on}}{T} \;\Rightarrow\; T = \frac{T_{on}}{k} = \frac{3}{0.3} = 10\ \text{ms} \\ T_{off} &= T - T_{on} = 10 - 3 = 7\ \text{ms} \end{aligned}$$

Chopping frequency $f = 1/T = 100$ Hz.

Average output voltage

$$V_o = kV_{dc} = \frac{T_{on}}{T_{on}+T_{off}}V_{dc} = 0.3\times200 = 60\ \text{V}$$

 vo  200|___         ___
        |   |       |   |
      0 |___|_______|___|_______ t
         3ms  7ms
        |<--- T = 10 ms --->|

Answer: $T_{off} = 7$ ms; average output voltage $V_o = 60$ V.

Note: variable-frequency control is simple, but a wide frequency range makes filter design hard and may cause interference; for low $k$ the OFF time becomes very long, so the load current may become discontinuous.