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I have a DC measuring digital voltmeter. The power for the voltmeter is coming out of a charging rectifier. The frequency out of the charging rectifier is anywhere between 20 and 60Hz. Operating voltage of the voltmeter is about 12V.

The voltmeter consumes max 30mA. What capacity is required for the decoupling capacitor to supply the voltmeter to continue working during power dropouts?

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  • \$\begingroup\$ As a minimum, you need to specify the voltage requirement for your voltmeter as well as how long do you want it to work after a power outage. This will allow the calculation of how much energy must be stored in the capacitor. Combining the energy requirement with the voltage would then allow the calculation of the needed capacitance. \$\endgroup\$ – Barry Aug 21 at 19:21
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    \$\begingroup\$ How is the 'about 12V' derived from the 'charging rectifier', and what is it charging? \$\endgroup\$ – Bruce Abbott Aug 22 at 7:31
  • \$\begingroup\$ @bruceabbott About 12v is because if the generaror which delivers the ac is running not fast enough, voltage will be below 12v. The charging rectifier tops voltage at 13.8V for charging a lead acid battery. That's also the reason why frequeny can vary. \$\endgroup\$ – JohnDoe Aug 22 at 16:06
  • \$\begingroup\$ Are you trying to power the voltmeter from the charger output? Will a 12V battery be connected at the same time? \$\endgroup\$ – Bruce Abbott Aug 22 at 23:04
  • \$\begingroup\$ @bruceabbott I see where you are aiming at. When a battery to charge is connected, the volt meter reading is fine as there are no power spikes. Unfortunately most of the time no car battery will be attached that's why I want to get rid of the flickering / malfunction of the volt meter reading. \$\endgroup\$ – JohnDoe Aug 23 at 4:10
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If the volt meter uses a constant 30mA at 12V, then it could be modeled as a resistive load. The load would look like 12V/0.03A= 400Ω

You could then find the RC time constant: a 100uF capacitor and the 400Ω load would take 0.04 seconds to drop to 63% of 12V or roughly 4V.

The problem with using a capacitor to smooth out the voltage is it will immediately drop, if the voltmeter requires a constant voltage (within a certain tolerance like ±0.3V) then a very large capacitor would be required. This leads to large inrush currents and problems with the rectifier.

A better way would be to use a voltage rail higher than 12V and voltage regulator with a capacitor to regulate to 12V. Keep in mind that with realistic values for capacitors it will be difficult (not impossible) to design a circuit to maintain voltage for more than ~1 second at 30mA. Batteries are also a great way to provide energy.

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  • \$\begingroup\$ "... would take 0.04 seconds to drop to 63% of 12 V or roughly 8 V." Upside down. It would drop by 63% of 12 V to roughly 4 V. \$\endgroup\$ – Transistor Aug 21 at 21:43
  • \$\begingroup\$ @Transistor Thanks, I'd say I have a case of the Monday's but it's Wednesday \$\endgroup\$ – Voltage Spike Aug 21 at 21:45
  • \$\begingroup\$ @voltagespike The volt meter is actually very forgiving concerning operation voltage anywhere between 9 to 30V so the capacitor specification for this purpose doesn't need to be terribly sharp. \$\endgroup\$ – JohnDoe Aug 22 at 16:14
  • \$\begingroup\$ @JohnDoe That's a good thing, then you could probably go with just a cap \$\endgroup\$ – Voltage Spike Aug 22 at 16:25

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