# Newbie question

Discussion in 'General Electronics Discussion' started by 8bit, Oct 30, 2013.

1. ### 8bit

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Oct 30, 2013
I have just been dabbling with simple componets for the first time and was trying an idea out that doesn't seem to work. Can someboby tell me why? I have a 9v battery connected to a capacitor, resistor and a LED, all in series, and I assumed that the capacitor would charge then discharge lighting the LED. It doesn't.

2. ### davennModerator

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Sep 5, 2009
hi 8bit
welcome to the forums

well DC doesn't flow through a capacitor, so the result is that the capacitor
is basically a "hole" gap in the circuit.

There's no need for a capacitor anyway, the LED will light directly from the battery via the current limiting resistor

cheers
Dave

3. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
Those components on their own are insufficient to make an oscillator.

Depending on how you wire it up, you might get a small (or long) flash as you apply power then nothing. Other possibilities include that the LED turns on, and stays alight briefly after the power is removed.

To explain what is happening (and not happening) in your circuit we would need to see a schematic and/or a photo of what you have constructed.

4. ### 8bit

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Oct 30, 2013

I was just playing with it all, trying to learn as I go. I assumed the capacitor would charge then discharged lighting the LED as it did so. I guess I was wrong.

Last edited: Oct 30, 2013
5. ### 8bit

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Oct 30, 2013
I'm really struggling with this which is embarassing as this is real basic stuff but I just don't understand why I can't get this to work. Can't you charge a capacitor and have it discharge through something?? I feel my electronic journey is over before its begun.

How can you check to see if a CAP is not frazzled?

6. ### Harald KappModeratorModerator

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Nov 17, 2011
A circuit diagram would be helpful. I Assume your circuit looks something like this:

When you close the last connection, the battery will charge the capacitor. As the symbol of the capacitor indicates, it is an open circuit (basically two plates separated by a gap). This charging will allow an initial current for a very short time (approx. 3*R*C). This time will be too short for you to notice the LED flashing.
Once the capacitor is charged, the voltage across the capacitor is the same as the battery voltage and no more current will flow.
In effect you see no LED lighting.

The capacitor has no reason to discharge in this circuit. Therefore you will not see an alternating charge-discharge action.

If you want to nake the LED flash, look up a simple two-transistor astable multivibrator. Here is a complete tutorial explaining the function of such a circuit.

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7. ### BobK

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Jan 5, 2010
Think about it. What would cause the capacitor to change from charging to discharging? Since there is no change in the circuit, there is nothing to trigger that change. You need at least a switch to initiate that change.

If the switch could be controlled electronically, i.e. charge until a certain voltage is reached, then discharge until another lower voltage is reached, you would have what is called a relaxation oscillator, and the LED would flash.

Bob

8. ### 8bit

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Oct 30, 2013
Maybe I'm doing something wrong but either the LED is very dimly lit or it lights straight away. I guess I'll have to keep playing around.

I assumed once I disconnect the battaery the capacitor would discharge and light the LED.(?)

Last edited: Oct 31, 2013
9. ### 8bit

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Oct 30, 2013
After the LED has gone out and the CP has fully charged I then disconnected the battery expecting the CP to then discharge and light the LED briefly but it didnt. Should this not happen?

10. ### davennModerator

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Sep 5, 2009
as we said to you earlier
not if its in the circuit configuration you described and what Harald drew a circuit for

did you actually read and understand the responses by Harald, Bob and myself ?
I suspect not, going by your previous 2 posts

Dave

11. ### 8bit

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Oct 30, 2013
Sorry, I've always been a bit of a slow learner. :s

My understanding so far is this. The current will flow until bascially the charge in the CP equals the charge of the battery. That said, I was wondering if you removed the battery wouldn't the CP discharge and light the LED briefly?

12. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,501
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Jan 21, 2010
1) to light the LED current can only go in one direction. In your circuit, that is in the direction of charging the capacitor

2) If you disconnect the battery there is no circuit, so no current can flow.

3) if you replace the battery with a short, the current will not flow because the diode (LED) stops it.

HOWEVER, if you take the capacitor out of the circuit and turn it around, current will flow again as it is first discharged, then charged in the opposite direction.

13. ### 8bit

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Oct 30, 2013
Thats great, I think I've got it now...finally!

14. ### 8bit

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Oct 30, 2013
Last edited: Nov 8, 2013
15. ### BobK

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Jan 5, 2010
It has circuitry that switches from charging to discharging after a set voltage is reached.

Bob

97
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Oct 30, 2013

Thanks Bob