# What happens when the voltage applied to fully charged capacitor becomes lower than initial voltage?

I have 24 volt capacitor and I charged it fully using 24 volt power supply. What happens if the power supply voltage becomes 20 volt which is connected to the capacitor that is fully charged at 24 volt.

Car's cigarette lighter socket is used as source power supply. I am trying to use capacitors to provide more current at peak times of load. Power supply voltage can be changing from 12.6 volt to 14.4 volt.

• By "power supply" do you mean an ideal voltage source? If not, can your power supply sink current as well as source current? – Elliot Alderson Oct 28 '19 at 18:16
• Details of how the output of the power supply are constructed are required to answer this question. Some supplies do not allow current to flow into the outputs, others do uncontrollably, and still others are able to do it under control so actively bring the capacitor down to match its voltage. – DKNguyen Oct 28 '19 at 18:17
• Car's cigarette lighter socket is used as source power supply. I am trying to use capacitors to provide more current at peak times of load. Power supply voltage can be changing from 12.6 volt to 14.4 volt. – James Oct 28 '19 at 18:36
• What kind of load are we talking about here? How many Amps? How long do the capacitors need to "help supply" the 14v? – rdtsc Oct 28 '19 at 21:48

If the supply voltage is changed quickly enough, the the capacitor starts sourcing voltage, the current flows backwards into the supply.

Bypass capacitors are used to regulate voltage, but mostly for short term voltage drops from cables or trace inductance. The capacitor can supply voltages to the load in the event the voltage drops.

• Car's cigarette lighter socket is used as source power supply. I am trying to use capacitors to provide more current at peak times of load. Power supply voltage can be changing from 12.6 volt to 14.4 volt. – James Oct 28 '19 at 18:37
• So if the supply voltage is changed slowly there is no change to the capacitor voltage? – Elliot Alderson Oct 28 '19 at 18:37
• @ElliotAlderson Yeah, if the leakage current exceeded the current going back to the supply, not likely but theoretical. – Voltage Spike Oct 28 '19 at 18:38
• Bypass capacitor can handle the volt changes in DC? I just need constant voltage at load and excess voltage in capacitor should be prevented from reaching Car's power supply circuit. – James Oct 28 '19 at 18:59
• Yeah, but not for very long unless you have a lot of capacitance and that presents it's own problem, you have to know what the load is and the resistance/inductance in the line to make a calculation – Voltage Spike Oct 28 '19 at 19:01

If the source voltage (the car battery) becomes lower than the capacitor's voltage then the capacitor will try to charge the capacitor. Current will flow from the capacitor to the battery until their voltages are once again equal.

It's important to note that the magnitude of the current, and therefore the time taken to equalize the voltages, depends on the resistance of the wiring between the battery and the capacitor and the value of the capacitance.

• What will happen to the voltage across the load which needs constant voltage. I need a constant voltage at the load. And supply voltage will be coming from the alternator when car engine is running. – James Oct 28 '19 at 18:55
• If you truly need a constant voltage then you will need to use a dc-dc converter that accepts the range of input voltages you get from the battery and provides a constant output voltage. You can't get a truly constant voltage by adding a finite amount of capacitance. – Elliot Alderson Oct 28 '19 at 18:58
• Elliot, yes i too thought of it. And can I prevent the voltage from capacitor going back to power supply just by adding a diode? Is it advisable to add two power sources at different voltages like this and give it to a dc-dc converter? Thanks for valuable inputs. – James Oct 28 '19 at 19:03
• You can prevent current from flowing from the capacitor back to the battery using a diode, but if you use an appropriate dc-dc converter you won't need the capacitor at all. – Elliot Alderson Oct 28 '19 at 19:06
• Basically using the capacitors to store current and provide it along with power supply when more power is needed by load than the power supply momentarily. – James Oct 28 '19 at 19:10

As they are in parallel, ideally the capacitor would follow exactly the voltage that is applied to it. If the battery voltage changes immediately, the voltage drop between the capacitor and the battery will generate a current with the value: $$I=\frac{V}{R}$$ (being V the voltage drop and R the resistance of the cable that connects the capacitor to the battery).

But, in the real world, the formula for the current on a voltage change on a capacitor over time is actually $$i(t)=\frac{du(t)}{dt}\times C$$ (being u(t) the voltage drop over time and C the capacitance you are using). In this formula, you can see that the current will be higher if the derivative of the voltage drop is higher. What this tells us is that the current will be proportional to the voltage drop change rate (quicker change means higher current and viceversa).

This current will discharge the capacitor and decrease its value over time since the capacitor voltage will be decreasing as well. When the capacitor voltage is the same as the battery, the current will be 0 so it will stop discharging.