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I have a capacitor charging from a supply voltage by a resistor and diode (resistor to limit current and diode to prevent energy flowing backwards when supply voltage drops).

The voltage drop on the diode is 0.18V. The cap rating and the supply voltage are the same.

I want to eliminate the voltage drop to gain the extra energy storage, without having to resort to something like a boost converter. Is there a simple way?

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Pick a diode with an even lower forward voltage, like LSM115J, which typically drops 80 mV at 10 mA. Your initial current will probably be much larger, but the low voltage drop only becomes important at the end of charging, when the current is much lower. Eventually the voltage drop will still decrease further as the current gets lower, so the ultimate voltage difference may be even lower.

If the voltage is 12 V, then an 80 mV drop will cost you 1.3 % in the energy you can store.

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If you were absolutely desperate for every last bit of stored energy, then a MOSFET that was turned on during charging and off when charged would allow close to zero drop.

If you have a long period to top up the capacitor then a high value Resistor from cap to supply will top up the cap to close to supply in time. It will of course also discharge when supply drops - you need to decide on the tradeoffs between charge rate and leakage rates.

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Better yet, use a FET as an "active diode". In fact, a number of manufacturers have products called "ideal diodes", which are small ICs with a FET and a control circuit integrated, such as these products from Linear Technology. You can also build such a circuit from discrete parts if you prefer.

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  • \$\begingroup\$ Won't the FET's body diode conduct when reverse polarized, when it's supposed to prevent discharging the capacitor? \$\endgroup\$ – Federico Russo Aug 30 '12 at 14:41
  • \$\begingroup\$ Yes, the trick is to hook the MOSFET up "backwards", reversing the usual drain and source connections. It'll work just fine this way in this sort of application. \$\endgroup\$ – Dave Tweed Aug 30 '12 at 14:51
  • \$\begingroup\$ But MOSFETs are not symmetrical, and datasheets won't tell you their behavior when source and drain are reversed. \$\endgroup\$ – Federico Russo Aug 30 '12 at 16:47
  • \$\begingroup\$ What can I tell you? It's commonly done, and it works well. You just need to make sure you don't exceed Vgs in this configuration. Look at all the examples in the product sheet I linked to above; they're all hooked up this way. \$\endgroup\$ – Dave Tweed Aug 30 '12 at 18:03

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