I have a 24v to 12v converter which feeds 20 LEDs in series. The LEDs are rated at 12v. I have to replace a faulty LED. The one i have is a 24v LED. with I=V/R does this mean the LED will draw more current than the 12v LED.

 

Could you point to the LED specs? It is kinda hard to answer without the information.

Your LEDs are not simple parts, they are hybrids, which means they have some circuitry already incorporated. I doubt you could replace the 12VDC model for the 24VDC model, but more information is needed.

Welcome to AAC, by the way!

 

I have a 24v to 12v converter which feeds 20 LEDs in series. The LEDs are rated at 12v.[/I]

12V isn't enough to power qty 20 of any LED that are connected in series, most require a bare minimum of 1.7V each which means 20 of them in series would require 34V.

Obviously we're missing some key inforation here.

 

12V isn't enough to power qty 20 of any LED that are connected in series, most require a bare minimum of 1.7V each which means 20 of them in series would require 34V.

Obviously we're missing some key inforation here.

He wanted to say in parallel.

 

In that case I doubt replacing a defective 12V unit with one that requires 24V will work satisfactorily, it probaby won't light up much if at all.

 

Hi

Yes i meant in parallel. I put the LED into the circuit and it worked. It was a high intensity LED with a 28Vopr rating. The input voltage to the circuit is 12v. Due to this the LED was dimmer. But that suits me fine and isn't a problem as it is now the same brightness as the other LEDs in the circuit which are rated at 12Vopr. The problem i have is that even when i switch the LED switch off, they all go out apart from the new LED and stays on very slightly. This is no good as i used it as an indicator behind a lens and it looks slightly on. I have heard it could be leakage current which is directly proportional to the intensity of light or forward voltage but I don't understand why the others are fine. what is forward voltage?

 

Vf (forward voltage) is the voltage that is measured across a diode (or light-emitting diode, generically a P-N junction) when a current is flowing through the P-N junction in the forward-biased direction (ie: the anode is more positive than the cathode).

The Vf will vary as a function of the current. It is nearly a logarithmic function.

Here is an example plot and table of Vf vs current for a 1N4148 switching diode:

LEDs have a much higher Vf than a standard silicon diode.

However, once a light-emitting diode starts conducting, even a minuscule amount of current can produce detectable light. Your converter/power supply probably has one or more capacitors in it's output filter that provides this current. It can take a LONG time to discharge a cap via an LED.

One easy way that you might bleed the output cap(s) down quickly is to use a small 12v light bulb (like an automotive instrument panel bulb) in parallel with the other LEDs. Incandescent lamps have a non-linear resistance; when they are operating at normal voltage, the filament has a relatively high resistance. When the filament cools, it has a relatively low resistance. This is just the kind of response that you need to discharge the output cap(s) quickly.

 

Ok thanks for the good advice.
I don't understand how the other LEDs in the circuit don't do the same ( and slightly stay on ) like the new LED?

Regards
Alan

 

The Vf of LEDs can vary considerably, even in the same batch. I have seen variations up to 10% in a batch of 200.

You've added another LED that is not even the same type as the original 12v LEDs. It apparently has a lower Vf than the other LEDs, even though it is rated for a higher input voltage.

 

Thanks

Ok so my new LED has a lower Forward Voltage than the rest.

(1) And is this the reason it is staying slightly on ?
(2) Is the "Forward Voltage" the same thing as the "Voltage drop" across an LED.
(3) So even when the supply is switched of to it, there is still a voltage across it ?

 

Thanks

Ok so my new LED has a lower Forward Voltage than the rest.

Apparently so, as it is the only one remaining lit when you turn the power off.

(1) And is this the reason it is staying slightly on ?

Yes; it is the path of least resistance.

(2) Is the "Forward Voltage" the same thing as the "Voltage drop" across an LED.

Yes.

(3) So even when the supply is switched off to it, there is still a voltage across it ?

There is a charge stored in your supply output capacitor(s).
As the cap(s) discharge through the LED, the LED current decreases, which slows the rate of discharge.

I was fiddling with an LED circuit awhile back; powered it using a small supply that had an output cap. I was surprised that the LED remained dimly lit for over 2 hours after I had turned the supply off!

If you don't want to use a light bulb, you could simply use a fixed 2.4k Ohm resistor across the output of the supply. The initial discharge rate will be about 5mA, which won't load your supply very much, but will help to turn the LED off more quickly.