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I am planning a 220V AC to 9V DC rectifier circuit.

In order to provide a stable DC output, I've decided to put a capacitor. But how do I find the rating of the capacitor I should use?

In my circuit, the current flowing will be a maximum of 5mA, and the power will not be greater than 50mW.

Keeping these in mind, can anyone suggest what capacitor I should use?

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  • \$\begingroup\$ Show your circuit diagram. What are your voltage ripple requirements? \$\endgroup\$
    – winny
    Feb 2, 2018 at 21:07
  • \$\begingroup\$ Oops, I couldn't understand the meaning of ripple requirements. Diagram added. \$\endgroup\$ Feb 2, 2018 at 21:12
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    \$\begingroup\$ Please note that your 9 Vac secondary will produce about 13 Vdc due to top wave rectification. \$\endgroup\$
    – winny
    Feb 2, 2018 at 21:30
  • \$\begingroup\$ @winny in that case, I'll manage it by putting a resistor in the path before the capacitor, so that the potential drop across the resistor is around 4V. \$\endgroup\$ Feb 2, 2018 at 21:32
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    \$\begingroup\$ That would require a constant-ish current load, which I assume you don’t have. What you need is to have a transformer with fewer secondary turns to produce a lower voltage or a regulator following your capacitor to drop the 13 V down to 9 V. An LM7809 or similar would do nicely. \$\endgroup\$
    – winny
    Feb 2, 2018 at 21:35

1 Answer 1

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The choice of the capacitor value needs to fulfil a number of requirements. In the first case the value must be chosen so that its time constant is very much longer than the time interval between the successive peaks of the rectified waveform:

\$ R_{load} C >> \frac{1}{f}\$

where: \$R_{load}\$ = the overall resistance of the load for the supply

\$C\$ = value of capacitor in Farads

\$f\$ = the ripple frequency - this will be twice the line frequency a full wave rectifier is used.

As there will always be some ripple on the output of a rectifier using a smoothing capacitor circuit, it is necessary to be able to estimate the approximate value. Over-specifying a capacitor too much will add extra cost, size and weight - under-specifying it will lead to poor performance.

For cases where the ripple is small compared to the supply voltage - which is almost always the case - it is possible to calculate the ripple from a knowledge of the circuit conditions:

Full wave rectifier:

\$V_{ripple} = \frac{I_{load}}{2fC}\$

Half wave rectifier:

\$V_{ripple} = \frac{I_{load}}{fC}\$

These equations provide more than sufficient accuracy. Although the capacitor discharge for a purely resistive load is exponential, the inaccuracy introduced by the linear approximation is very small for low values of ripple.

It is also worth remembering that the input to a voltage regulator is not a purely resistive load but a constant current load. Finally, the tolerances of electrolytic capacitors used for rectifier smoothing circuits are large - ±20% at the very best, and this will mask any inaccuracies introduced by the assumptions in the equations.

More: http://www.radio-electronics.com/info/circuits/diode-rectifier/rectifier-filtering-smoothing-capacitor-circuits.php

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    \$\begingroup\$ This does not answer the OP's question. Also, there is no load resistor in the diagram that the OP has provided. Answering questions here requires your own words instead of copying and pasting what a website has said. Next time, read the entire question. \$\endgroup\$
    – user103380
    Feb 2, 2018 at 21:56
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    \$\begingroup\$ @KingDuken, I disagree, the answer is perfectly fine, it encourages OP to think first, and to learn something about rectifiers. The OP did provide Rload, indirectly, by mentioning load current. Next time please re-acquaint yourself with Ohm's law before downvoting and lecturing on answer styles. \$\endgroup\$ Feb 2, 2018 at 23:33
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    \$\begingroup\$ @KingDuken We've been given the required voltage, and worst-case current. So calculating the worst-case Rload is trivial. \$\endgroup\$
    – Simon B
    Feb 3, 2018 at 0:12

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