Electronic Design

Handy Circuit Options Boost Output Of Capacitive Supplies

Capacitive supplies are very attractive for applications that don't require isolation from the mains supply. Simple, inexpensive, lightweight, and reliable, they run cool and don't generate interference. However, they suffer from a significant disadvantage: If you need an output current in excess of tens of milliamperes, the dropping capacitor tends to become bulky. This is especially true when the output must be referenced to one of the mains terminals, as is often the case with triac-driving circuitry. The only option then is to use the half-wave voltage-doubler topology, which unfortunately halves the available output current.

Another solution is to use a current-multiplying rectifier to boost the output current. By adding some components to the standard half-wave doubler circuit (Fig. 1a), you obtain the current-doubling variant (Fig. 1b).

In Figure 1b, another capacitor (C3) is inserted into the path of the positive half-cycle current. Thus, it receives the same charge as C2, the main output capacitor.

During the negative half-cycle, D3 is blocked while Q1 conducts via R2, connecting the positive side of C3 to that of C2. On the negative side, D1 provides a path from C3 to the ground, which effectively parallels both capacitors. This doubles the charge that's transferred each cycle to the output, also doubling the available output current. D4 resets the charge of C1 during the negative half-cycles. With this improvement, we're now on an equal footing with a full-wave circuit, but with a common terminal between the input and the output.

Further improvements are also possible. Up to this point, the negative half-cycle hasn't actively contributed to the current output. The circuit of Figure 2 changes that situation. During the negative half-cycle, C4 is charged via D5 and D7, and Q2 is blocked because it's negatively biased. When the input becomes positive, Q2 conducts via R4, as does D6, connecting C4 to the input of the previously examined circuit. This effectively doubles its input current. Because it operates "blindly" on any current, whatever its origin, it also gets doubled. The end result is an output current equal to four times that of the initial circuit of Figure 1a.

It's worth saying a word about component selection, especially for those unfamiliar with capacitive supplies. For safety reasons, it's essential that C1 be an X-rated model (that is, suitable for connection across the mains). A 100-nF value was used in this example. The value of R1 isn't critical. It simply limits the inrush current when the supply is plugged in and the ac waveform is at or near its maximum. But it should be able to withstand the full mains peak voltage, which is 325 V in the example.

But it's better to plan for the worst case. If the circuit is plugged in carelessly, the capacitor may be charged to a maximum, then disconnected briefly and connected again to the mains waveform at the opposite maximum. If this happens, the peak voltage seen by the resistor is twice the mains peak voltage, or 650 V in this example. If a 100-Ù resistor is used, as in the example, the peak current will exceed 6 A. Although this peak is of a short duration, it heavily stresses the resistor.

Therefore, you should either select a component specifically designed for pulse operation or use an oversize standard resistor. A 1-W standard film resistor appears to work well with up to 1-µF capacitors. Then, if the resistor is too small and eventually fails, the good news is that it will fail open-circuit. Carbon-composition resistors have almost completely disappeared nowadays, and they should never be used in this application because of their propensity to fail in unpredictable ways, posing a serious fire hazard.

R6 is optional. It's a discharge resistor preventing the continued presence of a high voltage on the mains plug terminals after the equipment is unplugged. Although the energy stored in C1 isn't high enough to be harmful, it can give an unpleasant shock to the user. C2, the filter capacitor, should be dimensioned according to your ripple requirements (not forgetting that the ripple frequency equals the mains frequency). C3 is identical to C2. The value of C4 is half that of C2, but it sees a double voltage. The peak current ratings for Q2 and Q1, respectively, are the output current and twice that value. Their base resistors must be computed accordingly. The diodes aren't critical, but they should not be underdimensioned, due to the aforementioned turn-on peak current. The 1N4001 diodes are perfectly suitable.

This circuit reduces size at the expense of some complexity. The cost is on average similar to that of the standard solution, being rather lower for high currents and vice versa. Although the circuit is fairly reliable, it can't match a plain capacitor in this respect: Each component added also increases the risk of failure. Therefore, if you're looking for the highest possible reliability, the conventional solution of Figure 1a remains preferable.

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