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As we have already discussed, when a filter capacitor is connected to the output of a rectifier circuit, that capacitor charges to the peak rectified voltage during any half cycle that the rectifier is producing an output voltage. The capacitor then provides energy to the load until the rectifier produces anothe peak to charge the capacitor.
But what happens if we rearrange the circuit components a bit? Can we get the charge placed on the capacitor during one half-cycle to be added to the rectifier output during the other half-cycle? How far might we extend such an arrangement, and what would be the cost in both output ripple and available output current?
There are multiple possibilities, and we will examine the main ones on this page.
Consider the dual-output half-wave rectifier circuit shown to the left. This is the same circuit we looked at on the page on rectifiers. If we simply add two filter capacitors to the two outputs, we'll have two output voltages: one negative, and one positive. Each output will have a significant amount of ripple as soon as load current is drawn from it, but the peak output voltage will be equal to the peak voltage of the whole transformer secondary winding.
Now, suppose we make one change, as shown in the circuit diagram to the right. This change is simply to use the negative output as our ground reference, and take the positive output voltage as our only output from the power supply. Since each capacitor charges to the peak secondary voltage, the output voltage from this circuit will be the sum of the two capacitor voltages, or twice the peak voltage of the secondary winding.
This circuit, then, operates in such a way as to produce an output voltage that is twice the transformer secondary voltage. Therefore, it is known as a voltage doubler. More accurately, it is a full-wave voltage doubler, because it uses both half-cycles of the incoming ac wave.
Of course, doubling the output voltage comes at a price. Each capacitor is charged individually from its rectifier, but they appear in series to the output. Therefore, the available output current is only half the current that would be available from a half-wave rectifier by itself. This makes intuitive sense as well: you cannot get more energy out of the circuit than you put into it. Therefore, if you double the output voltage, you must cut the current in half to maintain the same power level.
The other factor is the ripple voltage. Each capacitor is recharged while the other is discharging, so there is some cancellation of the ripple voltages. Nevertheless, the output ripple of this circuit is significant, and will normally require either additional filtering or regulation to be usable by most electronic circuits.
If we rearrange the diode and capacitor in the negative half of the voltage double circuit above, we get the circuit shown to the right. This time, one end of the secondary winding is grounded, so that is our reference point. The ungrounded end will be driven alternately negative and positive with respect to ground.
This circuit operates in a manner that is not quite as straight-forward as the original voltage doubler we examined. To understand the operation of this circuit clearly, we need to take a detailed look at it during successive half-cycles of the ac input from the transformer. We will initially assume ideal components and that C1 = C2.
With real-world components, of course, there is a small voltage drop across each diode when it is forward biased. Also, any load on this circuit will draw current from C2 at all times, thus discharging this capacitor to some extent. However, on each positive half-cycle, C1 will recharge C2 from the voltage it had at the start of the half-cycle halfway up to +2vp.
Note that the output current capacity of this circuit is still only half the current capacity of a normal rectifier circuit. Any attempt to draw additional current from the voltage doubler will simply cause C2 to discharge faster, thus reducing the output voltage. It is never possible to get more power out of the voltage doubler than goes into it.
We can speed up the charging and recharging of C2 if we make C1 larger than C2. For example, if C1 = 100µf and C2 = 10µf, C1 can transfer much more charge to C2 on each positive half-cycle, and the voltage on C2 will increase much faster than the voltage on C1 will decrease. Of course, this also means that the output current capacity is even more limited, since C2 will discharge rapidly as well as charging rapidly.
Voltage doublers are very useful in situations where the load current is relatively light, and the required voltage is higher than is available from a standard transformer. For example, many standard transformers have output voltages of 6.3 vac, 12.6 vac, and 25.2 vac. This dates back to the days of vacuum tubes, before transistors and integrated circuits were developed. Vacuum tubes require a heated filament, much like the filament in an incandescent light bulb, and most such filaments were standardized to operate at either 6.3 vac or 12.6 vac. Hence, many transformers were designed to provide power to these filaments, and the voltage ratings for transformer secondary windings have remained standard even though most applications for vacuum tubes have been turned over to semiconductor devices.
Therefore, if you find that you have an application that requires a 45 volt dc power supply, you might well start with a 25.2 volt transformer and use a voltage doubler to obtain the necessary voltage.
One very useful feature of the modified voltage doubler circuit above is that it is expandable. In the circuit to the right we have re-drawn the diodes and capacitors, and added a third section. The result here is a voltage tripler — the output voltage is triple the transformer secondary voltage.
The first two sections, consisting of C1-D1 and C2-D2, still operate as a voltage doubler exactly as described above. With the addition of C3-D3, however, we see an additional effect. On the negative half-cycles when D1 is forward biased, D3 is also forward biased by the voltage on C2. Therefore C2 and C3 are effectively connected in parallel by D1 and D3, while D2 remains reverse biased and therefore does not conduct. As a consequence of this parallel connection, C2 shares its charge with C3 and both capacitors get charged towards a voltage of +2vp.
On the positive half-cycles, D1 and D3 are off, and D2 is on. This allows C2 to recharge from C1, but it also connects C1 and C3 in series, thus increasing the output voltage even more than the voltage doubler circuit.
Of course, the output voltage from this circuit will fluctuate on alternate half-cycles so that a considerable amount of filtering will be required to produce a smooth dc output. Once all capacitors are fully charged under no-load conditions, this output will vary from +2vp on negative half-cycles to +4vp on positive half-cycles. The average output voltage will be +3vp, so this circuit is a voltage tripler.
The story doesn't end here, either. If we add another C-D section with C4 in series with C2 and D4 echoing the placement of D2, we will have a voltage quadrupler. Furthermore, since C2's left end is connected directly to ground, the fluctuations applied by the transformer winding to C1 and C3 do not apply. Without a load, C4 will charge to almost +4vp and remain there. Of course, with a load, C4 will discharge relatively rapidly. After all, when we quadruple the output voltage, the available output current must be quartered to avoid the attempt to draw more power from the output than is provided to the input.
With ideal components, there is no theoretical limit to how far this circuit can be extended. In the real world, however, there are always practical limits. Between diode voltage drops and charge lost in imperfect capacitors, any attempt to extend this circuit beyond about 10 sections will fail to provide any useful voltage increase. Also, the presence of any load will drastically reduce the output voltage of a high-order voltage multiplier.
Nevertheless, there are practical uses for such circuits. Consider a typical electronic flashgun for photography. It is powered by a battery composed of two to four AA or AAA-sized cells, at 1.5 volts each. When you turn it on, you hear a high-pitched whine for several seconds, and then the circuit is ready to fire the actual flash tube.
What you hear is an oscillator circuit, which generates an ac output when powered from a dc source. The whine is a harmless side effect. The ac output is applied to a high-order voltage multiplier which builds up enough voltage to operate the flash tube. When the flash is triggered, the capacitors all discharge through it, thus providing one momentary burst of light from the flash tube. Then the voltage multiplier must recharge for several seconds before the circuit is ready to fire the flash tube again.
Any application that calls for a brief application of high voltage at infrequent intervals is a good candidate for a high-order voltage multiplier circuit.
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