Step-down chopper variable frequency, 5 ms ON time

In a step-down chopper the average output is $V_o = kV_s$ with $k = T_{on}/T$. Variable-frequency (frequency modulation) control keeps either $T_{on}$ or $T_{off}$ constant and changes the period $T$ to change $k$.

Operation in variable-frequency mode

  +---CH---+--------+
  |        |        |
 Vs      FD^      Load  vo
  |        |        |
  +--------+--------+

 constant Ton, larger T -> lower k
 vo |__      __      __
    |  |____|  |____|  |____
 constant Ton, smaller T -> higher k
 vo |__  __  __  __  __
    |  ||  ||  ||  ||  |

• The switch is turned ON for a fixed time $T_{on}$; the next turn-on is delayed (or advanced) to set $T$.
• Increasing $T$ (lower frequency) lowers $k$ and $V_o$; decreasing $T$ raises them.
• Drawbacks: the chopping frequency varies over a wide range, so input/output filter design is difficult; low frequency at small $k$ increases current ripple and may make load current discontinuous; possible interference with signalling and telephone lines. Hence constant-frequency (PWM) control is usually preferred.

Numerical

Given: $T_{on} = 5$ ms (constant), $k = 0.4$.

$$\begin{aligned} T &= \frac{T_{on}}{k} = \frac{5}{0.4} = 12.5\ \text{ms} \quad (f = 80\ \text{Hz}) \\ T_{off} &= T - T_{on} = 12.5 - 5 = 7.5\ \text{ms} \end{aligned}$$

Average output voltage:

$$V_o = kV_s = 0.4\,V_s$$

The supply voltage is not stated; taking $V_s = 200$ V (the value used in the same textbook circuit):

$$V_o = 0.4\times200 = 80\ \text{V}$$

Answer: $T_{off} = 7.5$ ms; $V_o = 0.4V_s$ ($= 80$ V for $V_s = 200$ V).