Capacitor question -- LED not staying on?
Here is my setup--
12V adaptor red wire ---> LED positive & 2200 uf 25V capacitor positive.
All three grounded together. (Also tried separate ground).
When I unplug the 12V adaptor, LED doesn't stay on.
Why isn't the capacitor keeping the LED on? Do I need a resistor or doide?
I don't really know technical terms pls dumb it down for me.
- PhilomelLv 72 months agoFavorite Answer
If you have the cap, LED and adapter in parallel then the adapter is supplying current to the cap and the LED. They are charging up to the forward voltage of the LED which is around 1.5V if it id a standard LED. The capacitor is also only charged up to ~1.5V. when you remove the adapter power, the voltage collapses to zero almost instantly and the LED extinguishes.
If you want the Led to stay on you will need much more than 2200uF.
Put several caps in parallel.
Put a resistor of ~ 1K between the caps and the LED.
- 2 months ago
The adaptor may have a voltage regulating bleed resistor to prevent the output voltage from rising too high and this will be in parallel with your circuit. Linear regulated adaptors are usually based around the LM317 variable voltage regulator. This regulator is set by resistors across the output, the regulators adjust terminal and the ground rail. Again these resistors will be in parallel with your circuit. Switched mode adaptors again will have a resistive divider across the output but it will be of higher impedance value therefore it will take that bit longer to discharge.
- derframLv 72 months ago
It's unclear to me exactly what you are trying to do. If you are trying to use the charger to charge the cap and then allow the cap to provide short term power to the LED, then you at least need to add a resistor to your circuit.
Add a 10K resistor in series with the LED, and then connect this series network across the capacitor. Charge the capacitor with the 12 volt adapter, then remoce the power source. The capacitor will provide current to the LED for a *short* time. You can make the LED brighter by reducing the value of the resistor, but that will also reduce the time the LED will stay lit.
- 異域秦後人Lv 72 months ago
 Power supply output has very low impedance that drains away the electricity stores in capacitor right away while unplug the power source 120Vac side.Cut off the 12V output side might work !
 LED is a very high power type, therefore 2200uF is not enough to keep it light up. Increase it into 1F and try again.
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- Dale-ELv 72 months ago
The Led drops less than a Volt so, even though your capacitor has a 25 volt rating the LED will shunt most of the circuit voltage and prevent the capacitor from charging up very much at all.If you wire your LED in series with the Capacitor, you need a double throw switch to complete the circuit when it disconnects from the battery. Wire it this way. Battery wired together with both capacitor and one end of switch. Other end of switch wired to other end of battery (your ground). Switch flipper throw wired to one end of LED and other end of LED to capacitor. Notice: Both capacitor and LED need to be poled properly according to battery voltage. Other wise it will not work at all, or possibly the capacitor will heat up and pop. Also maybe consider putting a 10 Watt incandescent bulb between the LED and the capacitor to help preserve the LED.
- VamanLv 72 months ago
I can confuse you more. Use q/c+R dq/dt=0 v is the voltage q is the charge and t is the time. Decay time = R c. So you can guess that if you connect with 0 resistance, the charge will decay fast. Only way is to increase the decay time. Try connect a small resistance of some ohms in series and see whether the led glows for a longer period. Use twiddle type resister where one can tune the resistance. Please post your observation.
- JimLv 72 months ago
Capacitors only have a little actual power, you need a battery to keep it on.