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I connected a 12V led strip to the circuit of the cigarette lighter of my car.

Unfortunately, when the car is turned on, the power is cut off for a few seconds (3-4s) and the led strip goes out.

I thought about using a capacitor to keep the LEDs on with a circuit like this:

enter image description here

The current charges the capacitor, and when the power is cut off the capacitor keeps the led strip on for a few seconds. I measured the current across the led strip and it absorbs 70mA.

Can you give me some suggestions on what values to use for the capacitor and resistor (if useful?) or is it impossible?

The meaning of the diode would be to prevent that the current goes to the electronics of the car, since I don't know what's upstream.

Sorry but I know little about electronics.

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    \$\begingroup\$ Depends how much of a dip in brightness you can accept after 4 seconds. The capacitor value will probably be impractically large. \$\endgroup\$
    – Finbarr
    Commented May 3, 2022 at 16:16
  • \$\begingroup\$ Is the power entirely cut, or is there just a voltage sag? If there's a sag, then you can address this with a SMPS so long as you're willing to modify the strip circuit. \$\endgroup\$
    – Reinderien
    Commented May 4, 2022 at 0:22

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It's probably not worth the trouble to protect the huge inrush current from blowing a fuse then make it large enough in Farads using a Ultracap to sustain some Amps for low sag in voltage. C = Ic Δt/Δv. With some charge current limiter and an expensive large Ultracap it is possible, but not practical.

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  • \$\begingroup\$ Thanks. It’s definitely not worth it \$\endgroup\$
    – antoniom96
    Commented May 3, 2022 at 17:46
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    \$\begingroup\$ It might take a small SLA battery and float charger and switch logic \$\endgroup\$ Commented May 3, 2022 at 17:58
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70mA times 5 seconds is 350mC (milli-Coulombs). Let's assume your LED strips can drop from the normal 13.2V down to 11.2V (2 volt drop) and that's still an okay brightness for you. Then the capacitance needed is 350mC divided by 2V = 175mF (milli-Farads).

A Google search tells me that supercapacitors are available in this range, but not for 12V rating. You may install several in series to increase the voltage rating, noting that this decreases the capacitance proportionally. E.g. 3x 1F 5.5V supercapacitors in series will give you 1/3F (333mF) with a maximum voltage of 16.5V.

Note it will be a supercapacitor or ultracapacitor; boring old normal capacitors don't store this much energy (they stop around 1mF).

The amount of capacitance needed is of course lower if you can tolerate a lower voltage drop.

There's no reason to add a resistor between the capacitor and the LEDs. The LED strip has its own resistors where needed.

There is a reason to add a resistor between the capacitor and the car battery (either side of the diode, doesn't matter which) - to limit the "in-rush current" when the car is turned on and the battery starts charging the empty capacitor very quickly. Probably only a few ohms are needed.

Is it worth it? Up to you I suppose. Tony Stewart EE75 doesn't think so, but it's your car.


This answer does not address the "load dump" concern. I am not at all familiar with this subject area, but I have heard there are requirements for car systems to withstand "load dump" voltage spikes up to 100V for 0.6 seconds. Conceivably you could design the resistor and capacitor to absorb the excess energy, but my quick estimation didn't give good numbers, so you might need a protection circuit to block it instead.


Hey, I think they sell capacitors marketed for car audio amplifiers, for basically this exact circuit. They put them in parallel with the battery so the amplifier can take more power than the battery can put out, for short amounts of time. The diode and resistor probably aren't included.

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    \$\begingroup\$ You can buy a 180mF "normal" capacitor with a 25V rating. It's just that it's nearly the size of a Coke can and about a hundred times the price. \$\endgroup\$
    – Finbarr
    Commented May 3, 2022 at 17:23

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