LED circuit issue

[mxracer700]

Hi All,

Thanks in advance to anyone who can shed some light here.

Building a simple circuit to add LED headlights a child's battery ride on car.

Each headlight has 3 white 10mm water clear LED wired in series. (20,000mcd,4.0v, 20ma) each for a combined draw of 12v. There is parallel wiring between the left and right headlight assemblies and then it's wired from the left one back to the battery and switch. Using an online LED calculator I added the suggested 1 ohm resistor to the neg lead off the left assembly LED's only as a test. It lasted seconds and blew all 3 LED. Yet the right headlight assembly which has been connected directly to power (no resistor) for a while has not failed yet. After the first set on the left blew and because the other set with no resistor did not, I replaced the 3 dead LED and almost immediately they blew again while the right is still running fine. To clarify the power comes to the left side assembly first and then parallels over the to the right and ends there.

1. It is perhaps my false understanding that if a string of series LED matches the voltage (3 x 4v =12v) a resistor is not required as long as the power source is constant. T or F?

2. If that is False then why did the setup blow with the suggested resistor in place and why not the other side with no resistor?

3. Does it matter if the resistor is added to the pos or neg side of LED's both single or series?

Any info on this would be huge help, scrambling to get this project ready for Santa to deliver.

Cheers, MXR.

 

[MikeMl]

...
1. It is perhaps my false understanding that if a string of series LED matches the voltage (3 x 4v =12v) a resistor is not required as long as the power source is constant. T or F?

2. If that is False then why did the setup blow with the suggested resistor in place and why not the other side with no resistor?

3. Does it matter if the resistor is added to the pos or neg side of LED's both single or series?

Any info on this would be huge help, scrambling to get this project ready for Santa to deliver.

Cheers, MXR.

1. is FALSE FALSE FALSE. A slight increase in the LED temperature will cause the current to increase exponentially. Can you say thermal runaway?

2. The slightest deviation away from the source being >12.0000V will cause the current to increase exponentially. Can you say thermal runaway?

3. No.

Running LEDs without current regulation is like dancing on the head of a pin, or trying to balance a knife on its tip.

 

[mxracer700]

Okay thanks Mike, 100% I need resistors. Struggling to understand why one set keeps running with no resistor and the other does not. Also why with the correct (seemingly) resistor the first set blew at all?

 

[MikeMl]

Okay thanks Mike, 100% I need resistors. Struggling to understand why one set keeps running with no resistor and the other does not. Also why with the correct (seemingly) resistor the first set blew at all?

The seemingly correct resistor was so small so as to be useless for regulating current. For these LEDs, the current limiting resistor should have dropped no less than ~3V @20mA, so R= E/I = 3/0.02 = 150Ω.

That means you cannot run three 4V LEDs in series from 12V, and still have something left over for the resistor's voltage drop. You should reduce the string to only 2 LEDs in series per string, run the two strings in parallel, and use two strings, each with its own resistor.

I have never seen a LED with a forward drop of 4V. Can you post a link to its data sheet?

As to the second part of your question, I already answered it: Running LEDs without current regulation is like dancing on the head of a pin, or trying to balance a knife on its tip.

 

[Mosaic]

I recently did an LED driven Arc Reactor Replica for child wearability.

The ballast resistors for the LEDs proved to be a prob as the Vf of the LEDs are close to the typical Vcc of the supply, about 0.4 V typical. The Vcc could be as much as 1V above Vf which makes keeping the LEDs bright & fed by a decent current as the battery discharges a bit questionable with just a ballast resistor approach.
Consider:
A 0.4V across a ballast resistor = 20mA @ 20 Ohms. But at 1V , 20 ohms => 50mA! Overdrive.

I used a simple current mirror with discrete PNPs for current mirror sourcing and NPNs to ground for uC driven PWM switching of the LEDs. This permitted the elimination of the ballast resistors and much better brightness across the range of battery voltages.

 

[mxracer700]

The seemingly correct resistor was so small so as to be useless for regulating current. For these LEDs, the current limiting resistor should have dropped no less than ~3V @20mA, so R= E/I = 3/0.02 = 150Ω.

That means you cannot run three 4V LEDs in series from 12V, and still have something left over for the resistor's voltage drop. You should reduce the string to only 2 LEDs in series per string, run the two strings in parallel, and use two strings, each with its own resistor.

I have never seen a LED with a forward drop of 4V. Can you post a link to its data sheet?

As to the second part of your question, I already answered it: Running LEDs without current regulation is like dancing on the head of a pin, or trying to balance a knife on its tip.

Thanks Mike that now makes sense to me. I will do as you say and I'm sure it will be fine. These LED are sold by a local chain of electronic supply stores called Sayal. The brand is Knight Lites and I can't find a spec sheet on them as even their own website is down. Individually packaged the specs I outlined are marked on every package. I questioned it too but was told "it's in their computer" as that so it must be correct. Take that for what it is worth. I googled the PN and not much came up. KSW-1691-1P

 

[mxracer700]

I recently did an LED driven Arc Reactor Replica for child wearability.

The ballast resistors for the LEDs proved to be a prob as the Vf of the LEDs are close to the typical Vcc of the supply, about 0.4 V typical. The Vcc could be as much as 1V above Vf which makes keeping the LEDs bright & fed by a decent current as the battery discharges a bit questionable with just a ballast resistor approach.
Consider:
A 0.4V across a ballast resistor = 20mA @ 20 Ohms. But at 1V , 20 ohms => 50mA! Overdrive.

I used a simple current mirror with discrete PNPs for current mirror sourcing and NPNs to ground for uC driven PWM switching of the LEDs. This permitted the elimination of the ballast resistors and much better brightness across the range of battery voltages.

Thanks Mosaic!

You are way beyond my knowledge with that answer but I wanted to let you know I appreciated it none the less!

 

[Diver300]

Using an online LED calculator I added the suggested 1 ohm resistor to the neg lead off the left assembly LED's only as a test. It lasted seconds and blew all 3 LED

I guess that would have been **broken link removed** ?

That calculator takes no account of tolerances. If the supply voltage is an exact multiple of the LED voltage, the resistance is always 1 Ω, for any current.

This one https://www.electro-tech-online.com/tools/led-series-resistance-calculator.php and this one https://mtrak.co.uk/led_calculator.html allow for tolerances and will tell you to have fewer LEDs if needed to give more voltage across the resistor, so better control of the current.

 

[MikeMl]

Most of the White 20mA LEDs I have seen have a Vf of ~3.6V. Three of them would be 10.8V. I'll bet you are using a Sealed Lead Acid battery, which might be 13.8V right off charge, and ~11.5V just before it craters.

Here are two alternative ways of making your headlights.

Note the variation of current through the LEDs as the battery drains.

 

[mxracer700]

Mike, that's an awesome reply and one I can easily understand. Thanks!

1.Did you base your calc on the 3.6v or the 4.0v for the resistor values?
2. That chart really illustrates the batteries inability to deliver current as the voltage drops off. Didn't realize it's as dramatic as it is. I would assume perhaps wrongly if an LED circuit with a higher resistor inline will not be affected by this until the voltage and current drop to the point of matching the fwd drop of the LEDs? Meaning the greater the spread between the voltage range of the power source vs the voltage requirement of the LED's the better off you are in terms of seeing an effect at the LED. However this would come at the expense of battery life as the higher the resistor value the more "wasted" current?
3. At this point I can rewire it anyway that is better. For safety (it will be a lot of work to get at this after the car is built) given the info you have provided I was leaning toward doing them a pair at a time and either doing three pair in series and then connecting them to power parallel or adding two more LED and make it four pair. Each pair would have a resistor of 220 ohm attached to it.

I hear you on the claimed fwd voltage being odd but I can't refute it or verify so for now I am trusting it.

 

[mxracer700]

I guess that would have been **broken link removed** ?

That calculator takes no account of tolerances. If the supply voltage is an exact multiple of the LED voltage, the resistance is always 1 Ω, for any current.

This one https://www.electro-tech-online.com/tools/led-series-resistance-calculator.php and this one https://mtrak.co.uk/led_calculator.html allow for tolerances and will tell you to have fewer LEDs if needed to give more voltage across the resistor, so better control of the current.

Hi Diver,

Thanks for explaining that! Yes the calc (don't know which one but did a few to be sure) did not suggest 3 was not good. As I played with them entering different values to learn it was only when I exceeded the supply voltage with LED voltage did it kick it back and say use less LED's. It's also one thing to suggest using less LED's on a circuit but for a newb like me knowing WHY is just as important. Each of these calculators could easily have a pop up message that says: this design has too little resistance to control the current safely, remove one LED and recalculate". Simple.
You guys have been awesome at the quick and clear education!! Thanks so much.

 

[Mosaic]

Thanks Mosaic!

You are way beyond my knowledge with that answer but I wanted to let you know I appreciated it none the less!

Here's the info on using current mirrors which don't require a ballast resistor and can deliver current over a wider voltage drop range than with a ballast resistor

 

[audioguru]

All LEDs are semiconductors, unlike incandescent light bulbs that are simple pieces of hot wire of a certain length (a simple resistor). The current in a light bulb increases a little if the voltage increases a little. But the current in an LED increases A LOT if the voltage increases a little and you do not even know its actual voltage. A 4V light bulb is designed to operate at 4.0V. But a semiconductor LED has a range of voltage because they cannot make them exactly the same like light bulbs.

Post the datasheet for the LED and look at its range of forward voltage it is probably from 3V to 4V. If it is 3V and you have three in series and use a 1 ohm resistor and the "12V" battery is fully charged at 13.8V then the current is 4.8A so of course the LEDs will instantly burn out.

You used a stupid online calculator that does not know anything about LEDs. It assumes that you tested each LED 's voltage and picked only ones that are exactly 3.0V, and that the "12V" battery is exactly 12.00V. It does not know that an LED's voltage decreases as it warms up causing its current to increase which makes it hotter causing its current to increase which makes it hotter causing its current to increase which makes it hotter ....

 

[mxracer700]

Thanks guys for all the replies and the knowledge that came with them! When I got into trouble I jumped online and picked this forum because it seemed like smart and generous people inhabited it. I was right. My project is done and working great. Merry Christmas and if you don't go in for that sort of thing, all the best for the coming year!

MXR.

 

[Flyback]

Glad your done, but just in case.....use a current regulator diode....
https://www.onsemi.com/pub/Collateral/NSI45020-D.PDF
eg fig 8 with 2 leds.