My name is Barry Stansfield, I began my electronics journey just six weeks ago!

I desperately need your help! I am new to electronics and I have been trying to build a simple LED –capacitor circuit on a breadboard. It is proving mission impossible, the capacitors won’t work! I’ve tried everything, I have used over two dozen different capacitors and almost every combination of resistor and transistor that I can think of. I thought that it may have been the breadboard so I tried setting up the circuit on three different breadboards and the result was the same. The capacitor simply isn’t charging, it’s supposed to stay on when the switch is turned off thus allowing the LED to remain lit for a couple of extra seconds, but when I turn the switch off the LED goes off immediately! Surely every capacitor I come across is not bad ? Without solving this problem I cannot move on to more complex circuits, please, please help, I am at my wits end what am I doing wrong ?

Attached is a picture of the circuit, please don't leave me standing for too long.

Barry

 

Try connecting the circuit like this and let us know if it still does same thing.

 

Try connecting the circuit like this and let us know if it still does same thing.

Ok thanks ill give it a try, fingers crossed

 

NOTE on your picture in post#1, the led doesn't have a series resistor, so it will burn out.

 

What's the output voltage of your power supply?

The likely problem is that the output voltage of your supply is going to 0V when you shut it off, thus rapidly discharging the capacitor.

Instead of switching off the supply, try physically removing the one of the wires from the breadboard to shut it off.

The diode that bushrat has in his circuit prevents the supply from discharging the cap.

With a 1 kΩ resistor, you would need a 1000 uF capacitor to get a time constant of about 1 second.

You definitely want a resistor in series with the LED to limit its current because you have enough base current to significantly overdrive it.

 

I have been using different power supplies 3V, 6 V and 9V it makes no difference but I will try, what is this "protection diode" that bushrat speaks of ?I have never heard of that.

Barry

 

It's not really being used as a protection diode. I explained it's purpose above. In the circuit as he shows, it actually serves no purpose. Later this afternoon I'll try to post a slight variant on his circuit that shows the role that it is intended to play.

 

It works you were right, it works! When the wire marked with the red circle is pulled out the LED stays on for about ten seconds. What I don’t understand is why ? I thought that the whole point of a capacitor was to maintain power even after the power source had been shut off, if I can just turn the battery off and then it fails then what’s the point ? Have I misunderstood how a capacitor works ?

Also you mentioned a series resistor for the LED, is the 1K resistor on its own inadequate, and also where would I apply the series resistor ? Please forgive me for being so pedantic but I have Aspergers syndrome and so in order for me to process this info I must understand every aspect.


Thank You

Barry

 

Like this ? Please be aware that I haven't plugged in the negative, this is just so that you can see the circuit a little better. Barry

 

When you turn off the power supply, how can the LED stay on? There is nothing to provide power to it!

When you remove the wire, the power supply is still there to provide current to the LED. The rest of the circuit determines how long the LED will draw power from the supply.

 

But doesn't the capacitor store voltage and current ? doesn't it keep some reserve power and then release that power once the battery is turned off ?

Barry

 

But doesn't the capacitor store voltage and current ? doesn't it keep some reserve power and then release that power once the battery is turned off ?

Barry

Yes, but not nearly enough to feed that LED for long.

Without a current limiting resistor, it's hard to say how much current that LED is pulling (you REALLY need to install a current limiting resistor in series with the LED). Let's say that it is only drawing 100 mA (and it is probably drawing a lot more given a base current into the transistor on the order of around ten milliamps). How long can a 1000 uF capacitor (and you have never told us the values of the components you are using, forcing us to guess)? About a tenth of a second -- and that's to discharge it completely. It will fall below the voltage necessary to keep the LED forward biased in a fraction of that. And as current is pulled off to feed the LED, that charge is no longer there to slowly turn the transistor off, so it turns off much faster as well.

 

But doesn't the capacitor store voltage and current ? doesn't it keep some reserve power and then release that power once the battery is turned off ?

You don't have your circuit wired as shown in the diagram. The experiment removes the connection between the cap and battery, but leaves the battery connected to the LED. This allows the LED to stay lit as long as there's enough voltage on the cap to turn the transistor on.

The way you have it wired, the capacitor provides power to the LED and that discharges it faster than if it was only supplying base drive to the transistor.

The experiment is faulty. As pointed out, it doesn't include a resistor between the LED cathode and the transistor collector to limit current in the LED. A 9V "transistor" battery can't put out a lot of current, but it's still a bad design practice to omit current limiting. When it's done, it's for cost savings and the manufacturer would rather sell you another one than have the design be robust.

EDIT: Just noticed that you're doing a different experiment...

 

For the original circuit, when you turn off the power supply, there is no power source for LED to operate no matter the capacitor was charged or not.
What you should do instead is to remove the wire connected between one pole of capacitor and positive rail of power supply.

 

For the original circuit, when you turn off the power supply, there is no power source for LED to operate no matter the capacitor was charged or not.
What you should do instead is to remove the wire connected between one pole of capacitor and positive rail of power supply.

If turning off the power supply doesn't actively drive the top rail to ground, but instead makes the power supply output look like an open, then the capacitor can provide power to the LED (through the very lead that you are saying to remove). But the LED draws many, many times the current that the base resistor does and so the capacitor discharges so rapidly that it appears that it turns off immediately.

 

If turning off the power supply doesn't actively drive the top rail to ground, but instead makes the power supply output look like an open, then the capacitor can provide power to the LED (through the very lead that you are saying to remove). But the LED draws many, many times the current that the base resistor does and so the capacitor discharges so rapidly that it appears that it turns off immediately.

I see but that method may not be a good one to observe the charging and discharging of capacitor as the cap should have drive both transistor and LED.

 

I see but that method may not be a good one to observe the charging and discharging of capacitor as the cap should have drive both transistor and LED.

I agree and I and others have previously pointed out several shortcomings with this circuit. If the cap is intended to be a timing capacitor, then it should not also be used to power anything else. If it is going to be used for both, then the current draw for the other purpose should be at least roughly regulated. Adding a current limiting resistor in series with the LED would help that, as well as limiting the current to reasonable levels.

Almost makes you wonder if the person that wrote that experiment every actually took the time to perform it and,if so, to actually think about the results.

 

Like this ? Please be aware that I haven't plugged in the negative, this is just so that you can see the circuit a little better.

A couple comments about your bread boarding technique.

If you're going to use different colored wires, be consistent in their usage. For example, use the dark/black/brown wires for ground and the red/orange for power.

When bending/flexing component leads, you need to make sure to not apply stress where the lead exits the component. Hopefully your guide mentioned that.

If you rearrange the components, you don't need any abnormally bent leads.