# help with simple circuit

Discussion in 'General Electronics Chat' started by kibbles18, Jun 28, 2010.

1. ### kibbles18 Thread Starter New Member

Jun 27, 2010
5
0
i had a question that i posted earlier so i did an experiment. i bought a .01
50WVDC max capacitator from radioshack. i also bought a green LED of 2.1VDC. i hooked up the LED to a 9v and it turend on bright then diminsihed. i could repeat this w/the same battery. so then i hooked the capacitator up to the 9v battery and waited ~10 seconds. i hooked up the LED to the capacitator and nothing happened. can someone tell me where i did wrong?

2. ### Markd77 Senior Member

Sep 7, 2009
2,803
595
It is probably a single flash that is too fast to see. Have you got resistors in there? If not your LED wont last long, in fact I'd be surprised if it's still working.

3. ### Wendy Moderator

Mar 24, 2008
20,993
2,731
Use a resistor (always use a resistor). You may have blown the green LED from the 9V battery if you didn't.

LEDs, 555s, Flashers, and Light Chasers

A large capacitor will act much like a battery, start at 100µF, and move up.

4. ### hgmjr Moderator

Jan 28, 2005
9,030
215
Most likely you damaged/destroyed the LED. LEDs when powered by voltages greater than their rated voltage will require the use of a current limiting resistor to drop the difference between the LEDs rated voltage and the voltage available from the power source.

hgmjr

5. ### kibbles18 Thread Starter New Member

Jun 27, 2010
5
0
ok well i obviosuly need to get a resistor to drop the volts down. but could it have anything to do with the capacitor? how long does it take to charge one with a 9v to power a LED? i dont think the 9v was able to supply full voltage, it was one i scrapped from an old controller :/

6. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
784
I did an experiment.

A 100uF capacitor charged to 9V gives a series connected 1.2kΩ & LED about 0.5 second glow time. A 2200uF gives about 10 second glow time.

Clearly you won't see a anything with a .01uF capacitor.

7. ### Markd77 Senior Member

Sep 7, 2009
2,803
595
It will charge to the battery voltage almost instantly, the only limit is the internal resistance of the battery. It isn't going to get higher than the battery voltage.

8. ### Mike33 AAC Fanatic!

Feb 4, 2005
349
25
You can work it out, just grab a resistor and your favorite capacitor. Big caps (oh, 10uF, 100uF...) work well. The TIME CONSTANT is the time in seconds it takes for the capacitor to reach 63% of its total capacity, and is equal to the resistance in ohms times capacitance in farads. OR, resistance in MEGOHMS X Capacitance in MICROFARADS, which is simpler.

So, for a 100k resistor in series w/a 10uF cap, you get .1 megohms x 10uF = 1second. The cap is 63% charged in 1 second. Times 5, to get 99% or so charged, = 5 seconds. It takes 5 time constants for a cap to be considered "charged" or "discharged".
It works the same whether you are charging or discharging the cap, by the way....

9. ### Wendy Moderator

Mar 24, 2008
20,993
2,731
There is a wrong assumption there. It doesn't drop the voltage down, it controls the current. LEDs are current controled devices, once you hit the break over voltage it doesn't change too much.

I strongly recommend you read the 1st 2 chapters of the link I gave you.

10. ### kibbles18 Thread Starter New Member

Jun 27, 2010
5
0
i will print them out to read them in the car cuz i am going somewhere for the day