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Half-Wave Rectifiers

  Figure 81 shows a half-wave rectifier circuit. The voltage source v_S is an AC source

$$v_S(t) = V_m \sin (\omega t)$$

of frequency $\omega$ radians per second, and V_m is the maximum or peak voltage. Note that $\omega = 2\pi f$, where f is the frequency in Hz. Generally, the source v_S is the secondary winding of a transformer.

  

Figure 81: Half-wave rectifier.

The diode is the component which does the rectification, since it permits current flow in one direction only. The resistor R_L represents the resistance of the load drawing the power.

Let's analyse this circuit assuming the diode is ideal. When v_S > 0, the diode is forward biased, and so switched on; therefore v_out = v_S. But when v_S < 0, the diode is reverse biased, i.e. switched off, and hence v_out = 0 V. This is illustrated in Figure 82.

  

Figure 82: Half-wave rectifier waveforms.

Exercise. Sketch the diode voltage v_D.

The load voltage waveform v_out is always positive, and so has a non-zero DC component, the average value V_AVG which we calculate as follows:

$$\begin{array}{rl} V_{AVG} & = \displaystyle{\frac{1}{T}} \int... ...a t) dt \\ \\ & = \displaystyle{\frac{V_m}{\pi}} . \end{array}$$ (77)

If we use the practical diode model to take into account the diode voltage drop, then we need to reduce V_m by 0.7 V when forward biased:

V_m(out) = V_m - 0.7 V

and the average voltage becomes

$$V_{AVG} = \frac{V_{m(out)}}{\pi} .$$ (78)

This means that the average voltage is reduced by the forward bias voltage drop across the diode.

The peak inverse voltage (PIV)  is defined as the peak voltage across the diode when reverse biased: (note that with zero current the voltage drop across R_L is zero)

PIV = V_m

The diode must be capable of withstanding this voltage.

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