I’m working with a car battery that has a voltage drop when cranking the engine. As a result of this, the stereo restarts even though it was running already.

To resolve the issue of stereo rebooting due to low voltage current, I bought some capacitors that will sit in between the stereo and the battery to provide power to stereo until the battery’s voltage comes back up. I’ve decided to get high Farad (10-20F) capacitors, but I can only get any practical sized capacitors with high F that only have low voltage ratings (2.7v). So, I am in the process of connecting these in series to get the voltage to at least 12v. I am aware of the loss of potential F as a result of this, so my question relates predominantly to safety and failure -

  1. Let’s say I connect six 2.7v capacitors in series resulting in 16.2v. Am I correct in assuming that the 12v coming in from the battery is safe enough for the capacitors to get charged without fail? Of course the negative end of the series will be grounded.
  2. Is it safe for the capacitors to be connected to a power source (battery) 24/7?
  3. If the stereo is not in use, would the capacitors use up battery’s power?

Thanks in advance for help!

  • 2
    \$\begingroup\$ Not going to work. No amount of capacitance is going to supply enough current during cranking the starter, to prevent a voltage drop. Getting a larger capacity battery may help. FWIW, I've never seen a car stereo that didn't "restart" when starting the engine. \$\endgroup\$
    – Norm
    Commented Aug 20, 2018 at 21:42
  • \$\begingroup\$ Most modern cars cut the power to the stereo, headlights, wipers, fans, etc., when the starter is turned on. This is by design to make maximum power available for cranking and spark generation. \$\endgroup\$
    – Transistor
    Commented Aug 20, 2018 at 21:48
  • \$\begingroup\$ Thanks for the replies, I’m aiming to keep a steady supply of current to the stereo using capacitors for a couple of seconds until the battery’s voltage comes back up, rather than to rectify the voltage drop in the battery power itself. \$\endgroup\$
    – James
    Commented Aug 20, 2018 at 21:57
  • \$\begingroup\$ Perhaps, you need a battery with more CCA capacity to raise the voltage. The rating is done for a drop of 5V from 12.5 to 7.5V at rated CCA. \$\endgroup\$ Commented Aug 20, 2018 at 23:08
  • 1
    \$\begingroup\$ Could be crazy but you might be better off: 1. Working out the voltage drop (say it drops to 8V - you might know this but you didn't say it). 2. Work out the needed current (work out the current the stereo needs during normal operation) 3. Simply place a boost power module at the stereo (with the input being the battery, output being the stereo) which can handle that low of a Vin, the required current output at 12V, and gives a constant 12V Vout. \$\endgroup\$
    – DSWG
    Commented Aug 21, 2018 at 0:54

3 Answers 3


My friend built a very similar concept for his boat and it worked fine. So what you are attempting may be possible.

The first thing to note is that you only want to power your stereo. You don't want your capacitor to try and power the starter. You can isolate your capacitor and stereo from the main 12V via a diode. That way your stereo and capacitor can keep their voltage high when the input voltage from the battery dips. If you don't want to use a diode its your choice, Maxwell makes some nice automotive starting capacitors (300F range) but they are several thousand dollars. Your radio by itself probably needs just a few Farads.

The next thing to do is find out how much capacitance you actually need. It may be that the amount of capacitance is way too big to be practical so don't even bother. On the other hand it could be that you don't need several farads and you may be able to buy a single 1F cap rated to 25V (so no need to worry about your stacking problem).

To determine capacitance you need to know three things.

  1. How much wattage the stereo is drawing.

    a) Measure the battery voltage (V_batt) powering the unit.

    b) Insert an amp meter in series with the 12V power leads on the stereo.

    c) Measure the current draw with the unit on.

    d) Power = Current x V_batt

  2. What voltage does the stereo actually reset at?

    a) Find some way to gradually reduce the input voltage into the stereo and see at which point it resets (V_reset).

3) How long you crank for (t_crank)

After you have V_batt and Current and V_reset you can calculate capacitance. I am going to make the assumption that the stereo draws the most wattage at 12V and lower wattage at lower voltages. In the worst case you then have...

C = 2 * t_crank * Power /(V_batt^2 - V_reset^2)

So for example if the battery is 12V and the radio resets at 8V and the radio draws 50W and you crank for 3s then the capacitance becomes...

C = 2 * 3s * 50W / (12V ^2 - 8V ^2) = 3.75F.

You would want to pick the capacitor to be larger so that you have some margin (say 5F in this case).

Note that you may need to make a small correction to the formula above to account for the diode drop (depending on which one you use). Also you may want to use two diodes and a resistor to limit inrush when you first connect your capacitor.

If it turns out that you need several farads and you need to stack them, then it can be safely done. What you need to do is make sure the voltage across them remains balanced. This can be done by putting a Zener diode in parallel with each capacitor. The zener diode should have a breakdown voltage less than your capacitor rating. Also the sum of the zener voltages should be greater than your battery voltage.

The capacitors and zeners shouldn't draw any significant power when the car is off.


If you put a relay between the power source and the Capacitor to disconnect the battery while starting you could use it. The capacitor in series would be about 1.5F to 3.0F depending on the values you use. Leaving the capacitor hooked up all the time will not hurt anything butenter image description here disconnect for long term storage.

  • 1
    \$\begingroup\$ Probably better to use the built in schematic editor \$\endgroup\$
    – BeB00
    Commented Aug 20, 2018 at 23:56

You need a couple of experiments to determine what is happening. Normally the starter switch will cause the power to the stereo to be interrupted during starting, as "Transistor" suggests. That means you don't have a voltage drop, you have a switch opening up. That is different from "voltage sag" due to the heavy load from the starter. If the switch opens up, in some ways that is easier for you than a voltage drop, because you will not need a diode to prevent your capacitor from being loaded by the starter. A large capacitor may work, but you need to measure the load current you are trying to supply to the stereo in order to size the capacitor correctly. The capacitor voltage will change like this: dv = (i*dt)/C, where dv is the voltage change, dt is your time (say, 2 seconds), i is the current into the stereo, and C is your capacitor. If you have 10F capacitors, but have six in series, C=10F/6 = 1.66F. With a 1 amp stereo load, you will drop 1.2V in your two seconds. This will probably be OK for your stereo. But if your giant stereo is a 10A load, you will drop 12V and you don't have a chance. You would be better off with a gel-cell auxiliary battery.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.