With a diode, we can build a rectifier to produce a d.c. voltage that is nearly equal to the peak value
of input a.c. voltage. We can also use diodes and capacitors to build a circuit that will provide a d.c output that is multiple of the peak input a.c. voltage. Such a circuit is called a voltage multiplier. For example, a voltage doubler will provide a d.c. output that is twice the peak input a.c. voltage, a voltage tripler will provide a d.c. output that is three times the peak input a.c. voltage and so on. While voltage multipliers provide d.c. output that is much greater than the peak input a.c. voltage, there is no power amplification and law of conservation of energy holds good. When a voltage multiplier increases the peak input voltage by a factor n, the peak input current is decreased by approximately the same factor. Thus the actual power output from a voltage multiplier will never be greater than the input power. In fact, there are losses in the circuit (e.g. in diodes, capacitors etc.) so that the output power will actually be less than the input power.
A half-wave voltage doubler consists of two diodes and two capacitors connected in a manner as
shown in Fig. 6.48. It will be shown that if the peak input a.c. voltage is VS(pk), the d.c. output voltage will be 2 VS (pk) provided the diodes are ideal (this assumption is fairly reasonable). The basic idea in a voltage multiplier is to charge each capacitor to the peak input a.c. voltage and to arrange the capacitors so that their stored voltages will add.
Circuit action. We now discuss the working of a half-wave voltage doubler.
(i) During the negative half-cycle of a.c. input voltage [See Fig. 6.49 (i)], diode D1 is forward biased and diode D2 is reverse biased [See Fig. 6.49 (i)]. Therefore, diode D1 can be represented by a short and diode D2 as an open. The equivalent circuit then becomes as shown in Fig. 6.49 (ii).
As you can see [See Fig.6.49 (ii)], C1 will charge until voltage across it becomes equal to peak value of source voltage [VS (pk)]. At the same time, C2 will be in the process of discharging through the load RL (The source of this charge on C2 will be explained in a moment). Note that in all figures electron flow is shown.
(ii) When the polarity of the input a.c. voltage reverses (i.e. during positive half-cycle), the circuit conditions become as shown in Fig. 6.50 (i). Now D1 is reverse biased and D2 is forward biased and the equivalent circuit becomes as shown in Fig. 6.50 (ii).
Referring to Fig. 6.50 (ii), it is easy to see that C1 (charged to VS (pk) )and the source voltage (VS) now act as series-aiding voltage sources. Thus C2 will be charged to the sum of the series peak voltages i.e. 2 VS(pk)
(iii) When VS returns to its original polarity (i.e. negative half-cycle), D2 is again turned off (i.e. reverse biased). With D2 turned off, the only discharge path for C2 is through the load resistance RL. The time constant (= RL C2) of this circuit is so adjusted that C2 has little time to lose any of its charge before the input polarity reverses again. During the positive half-cycle, D2 is turned on and C2 recharges until voltage across it is again equal to 2 VS (pk).
∴ D.C. output voltage, Vdc = 2 VS (pk)
Since C2 barely discharges between input cycles, the output waveform of the half-wave voltage doubler closely resembles that of a filtered half-wave rectifier. Fig. 6.51 shows the input and output
waveforms for a half-wave voltage doubler.
The voltage multipliers have the disadvantage of poor voltage regulation. This means that d.c. output voltage drops considerably as the load current increases. Large filter capacitors are needed to help maintain the output voltage.
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Reference: Principles Of Electronics By V K Mehta And Rohit Mehta
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