# modifying existing LED cirtuit

• posted

Hello all,

i have a basic question to ask, i tried to find it on the web but with no luck. i am trying to use a LED array taken from a lamp sold in a discount store for 2\$. It is quite simple, 5 whites LEDs (in parallel), a push button switch and a 10 Ohms (brown, black,black, gold) resistor. It is powered by 3 AAA batteries. Since that makes 4,5 volts, i taught naively that i could hook it up directly on the 5 volts from a PC power supply. Well you guessed it, the resistor became real hot real fast!

Now to my question, can i simply replace the resistor with on that is the proper strength to limit correctly the current to the LEDs?

I've include a rude diagram of the actual array.

Sorry for the long link i can't find the proper way to do it :(

Robbob

• posted

Yes, you must increase the resistance. Normally one would use a constant current source but the resistor is suppose to be acting like one. It is a cheap way that doesn't usually work whenyou alter the parameters.

Easiest way would be to measure the current using the batteries then find a resistance using the 5v that gives the same current.

So if it was normally drawing 0.25A with the 10Ohm resistor at 4.5V then you might need 30Ohms or so when you go to 5V to get approximately the same current.

Basically LED's are diodes and when they are forward biased their internal resistance drops significantly(almost acting as a short). It seems giving the extra 0.5V causes the LED's to act as a short(so you have 0.5A through the resistor).

Also they could be using the internal resistance of the batteries to help limit the current which the computer supply doesn't have(or which is much much smaller).

e.g., you might need a 50Ohm resistor in series to get the same current through the led's.

• posted

Assuming, for a moment, that the AAA batteries weren't being limited by their own internal R (which should be under 1/2 an Ohm) and that the R in question is supposed to run at about

1/8th watt and no more at 4.5V, then I'd figure this:

P = I^2 * R, therefore I = sqrt(P/R) or about 100mA.

This is consistent with the 5 LEDs taking 20mA for their drive. So it seems reasonable from this perspective.

This 100mA via R=10 suggests the R drops about 1V. In short, the white LEDs operate on about 3.5V, or so.

A change in V from 4.5 to 5.0 would be dV = 5 - 4.5 = 0.5. dV/R = dI, so this suggests about another 50mA into the 5 LEDs, or about 10mA into each assuming equal distribution (which won't be the case, but may be okay for rough guessing right now.)

The new estimate of 150mA, now at a 1.5V across R instead of the earlier 1.0V, would be about 1/4 watt instead of 1/8 watt. About doubling the power that needs to be dissipated. I suspect, but didn't look up just now, that the thermal resistance of an 1/8watt to 1/4watt resistor must be in the area of 200C/W. 1/4W at 200C/W is about 50C. If ambient is about 25C, that's 75C... which is getting hot. Beforehand, it would have been perhaps 50C, which feels less concerning when you touch it.

Assuming you still want the same brightness, and let's call that 100mA or so, then you want to drop 1.5V (I'm assuming that my earlier 3.5V figure was about right for the LEDs) and that means 1.5V/0.1A or about 15 ohms. Which is a standard value. And the dissipation should still be about in the same area, say 150mW. So getting one that is about the same size as the current one should be okay.

Jon

• posted

I think that's good.

It probably wouldn't be that much current. That would imply a drop of 2.5V across the R, leaving 2V for the white LEDs. Which no white LED works at, so far as I know. That's a red LED voltage.

A change of .5V across 10 ohms would be a change of 50mA, not

500mA.

I checked. AAA alkalines are spec'd to about .150-.300 ohms, when fresh. This is significantly less than the 10 ohms. But it is correct to keep it in mind, just in case. In this case, I don't think it is that important, though.

As I wrote elsewhere, I'd go with 15 ohms (maybe 18) just to start. But your point about measuring the current at 4.5V holds, too. And I'd go that way, if an ammeter is available, and work out the R from there (after also measuring the 5V supply voltage.)

Jon

• posted

You should replace it with 5 resistors, one for each led. Thats how it should have been done in the first place. First guess, 8 times the original resistor.

• posted

Yes,

Yes.

Figuring 20mA per LED and about 1.5V drop for each, would be

1.5/0.02 = 75 ohms. So yup-ish, yet again.

Jon

• posted

A 15-ohm resistor will give you about the same brightness as with the original 10 ohms on a 4.5V supply. A 1/4-watt resistor will still get fairly hot but will not burn out. A 1/2-watt type will run cooler. You could also use two 33-ohm 1/4-watt resistors.

• posted

It wasn't a change but a maximum current assuming 0 led voltage drop. I don't know the LED voltage drop nor the configuration(parallel or series) so I didn't take it into consideration.

I just made up random numbers because it depends on too many unknown factors. If they set the current depending on the knee voltage of the LED's then the internal resistance could be significantly reduced resulting in a significant increase in current.

For example, Lets assumine a standard diode(1N4001) instead of LED's. The knee voltage at room temp is about 0.63V. The bulk resistance is relatively large here and we can estimate the bulk resistance to be about 10Ohms.

This means if we can supply 0.63V we would get about about 0.063A.

Suppose we add an additional 10Ohms series resistance and some voltage V to give the same current above,

V - 0.63 - 0.063*10 = 0 ==> V = 2*0.63 = 1.26V

NOW!!! suppose we have the same circuit but now alter V. Suppose V is set at

4.8V, we can now guess that the diode's bulk resistance is significantly less and current will be

I = (4.8 - 0.8)/10 = 0.4A

In this case the current as increased almost 4x. (The diode's voltage will increase to about 0.8V)

For a supply of 2V a similar effect with the current being approximately

0.12A.

I'm not saying they did this in their design of the LED's but they may have chose a point close to the knee and when the extra voltage was added it was enough to significantly decrease the LED's bulk resistance.

It depends on too many factors to guess at which is why I suggested using a meter and gave him pretty save values to start with incase he doesn't have a meter. My initial guess of 30Ohms may hav been too high as even with a short

5/30
• posted

IIRC White LED's have a very gradual knee voltage and hence more less likely to have issues with current sharing. They design the LED's this way to reduce the cost of using extra resistors. Of course current sharing is still an issue but depending on the design it may be relatively safe to use.

• posted

It has to be included to make any kind of reasoned suggestion, though. And white LEDs are based upon blue ones, which operate in the roughly-3.5V-4V range, depending somewhat on operating current and design.

Well, I used two different bits of knowledge to come at it from two different directions, both of which yielded about the same result -- which I suspect strengthens what I guessed at. It's slightly better than 'random numbers' in this case, I think.

Jon

• posted

There are others here who know better than I do, but so far as I'm aware single-LED whites (as opposed to RGB style) use a blue LED and some phosphor to get the job done. I'm not aware that there is an outright design goal in making the blue LEDs as you suggest above. Can you quantify your meaning? For example, provide a specific white LED example and explain what specific part of an accurate model of it is as you say, above? A gradual knee voltage might imply a modeled series R value that is higher than would be expected, for example. I am curious.

Jon

• posted

and I that taught that it was a simple matter of changing one resistor for an other!

so if i want to keep the current configuration (it is on a nice little pcb with a switch) i need to change the one resistor with a single

15ohms 1/4 watt?
• posted

You made an estimation while mine was not an estimation but a demonstration. Your estimation may be "correct" but if it happens to be too low then you can ruin his LED's. I'm not going to take that chance since it is not mine to ruin. My numbers wern't to give him any values to use which is why I made them higher than nominal just incase he did use them. The idea is that if he uses them and they produce an intensity that is to low he could say "I used

10Ohms and 30Ohms so it must be somewhere inbetween". He then might try 20 ohms or so and realize it probably would work. If the resistor was cool he might decide to go 15ohms or stick with 10 and "risk it". In any case I would bear no responsibility for his actions since I didn't tell him what to use.

After all, if the heat of the resistor was the only concern then there are simple ways to fix that. If it was something I was doing I probably would have doubled the resistance simply to reduce power draw since the light intensity would probably be sufficient.

• posted

As someone else wrote, I might go for a 1/2watt, not 1/4. That will give you a little margin. You could instead parallel two 33 ohm, 1/4 watt resistors; or perhaps 3 of the

47 ohm or 56 ohm variety at 1/4 or 1/8 watt and get by.

Chances are, you have a full 1/4 watt to dissipate there. So you don't want to have only 1/4 watt capability in the part, as that would not give you any margin and the part will be pretty hot when it runs. Better to keep the temp down a bit.

Assuming you really do only want to replace that one resistor, of course.

Jon

• posted

I'll try just that thanks you all

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