In general, d.c. chopper consists of power semiconductor devices (SCR, BJT, power MOSFET, IGBT, GTO, MCT, etc., which works as a switch), input d.c. power supply, elements $(\mathrm{R}, \mathrm{L}, \mathrm{C}), \text { etc}$ and output load.The average output voltage across the load is controlled by varying on-period and off-period (or duty cycle) of the switch.

A commutation circuitry is required for SCR based chopper circuit. Therefore, in general, gate-commutation devices based choppers have replaced the SCR- based choppers. However, for high voltage and high-current applications, SCR based choppers are used. The variations in on- and off periods of the switch provides an output voltage with an adjustable average value. The power-diode $\left(\mathrm{D}_{\mathrm{F}}\right)$ operates in freewheeling mode to provide a path to load-current when switch (S) is OFF. The smoothing inductor filters out the ripples in the load current Switch S is kept conducting for period $T_{\text { on }}$ and is blocked for period $T_{\text { off }}$ .The chopped load voltage waveform is shown in Fig.2 .

During the period $T_{\text { on }},$ when the chopper is on, the supply terminals are connected to the load, terminals. During the interval $T_{\text { off }},$ when the chopper is off, load current flows through the freewheeling diode $D_{F} .$ As a result, load terminals are short circuited by $D_{F}$ and load voltage is therefore, zero during $T_{\text { off }}$ In this manner, a chopped d.c. voltage is produced at the load terminals. From Fig. 2, the average load-voltage $E_{0}$ is given by

$$E_{0}=E_{\mathrm{dc}} \frac{T_{\mathrm{on}}}{T_{\mathrm{on} +\mathrm{off}}}-----(1)$$

where,

$T_{\text { on }}=$ on-time of the chopper,

$T_{\text { off }}=$ off-time of the chopper

$T=T_{\text { on }}+T_{\text { off }}=$ chopping period

If $\alpha=\frac{T_{\text { on }}}{T}$ be the duty cycle, then above equation becomes,

$$E_{0}=E_{\mathrm{dc}} \cdot \frac{T_{\mathrm{on}}}{T}-----(2)$$ $$E_{0}=E_{\mathrm{dc}} \cdot \alpha-----(3)$$

Thus, the load voltage can be controlled by varying the duty cycle of the chopper.

Also, $$\qquad E_{0}=\frac{T_{\text { on }}}{T} E_{\mathrm{dc}}=T_{\text { on }} \cdot f \cdot E_{\mathrm{dc}}-----(4)$$

where $\qquad f=$ chopping frequency

From Eq. 3 it is obvious that the output voltage varies linearly with the duty cycle. It is therefore possible to control the output voltage in the range zero to $E_{\mathrm{dc}}$ .

If the switch S is a transistor, the base-current will control the ON and OFF period of the transistor switch. If the switch is GTO thyristor, a positive gate pulse will turn-it ON and a negative gate pulse will turn it OFF. If the switch is an SCR, a commutation circuit is required to turn it OFF.

The average value of the load current is given by

$$I_{0}=\frac{E_{o}}{R}=\frac{\alpha \cdot E_{\mathrm{dc}}}{R}-----(5)$$

The effective (RMS) value of the output voltage is given by

$$\begin{aligned} E_{o(\mathrm{RMS})} &=\sqrt{\frac{E_{\mathrm{dc}}^{2} \cdot T_{\mathrm{on}}}{T}}=E_{\mathrm{dc}} \cdot \sqrt{\frac{T_{\mathrm{on}}}{T}} \\ &=E_{\mathrm{dc}} \sqrt{\alpha} \end{aligned}$$