I bought some L.E.D.s on fleebay with resiters. The green leds came with a resister dropping voltage to 3v. BUT I have some that are free flowing. I have hooked up three or four resisters now and I am getting readings of 11.75v on the other end.

Do they burn out? I thought when a resister goes nothing gets through. Are these cheap or am I missing something. I get a 465ohm reading with them.

I hooked one up and it was 3.15v removed my burned out led replaced it with another one, checked again and I am getting 11.74v again. Whats up with these??

 

You may be applying too much DC voltage or unfiltered DC voltage or AC.

The resistors will limit the current and supply a proper voltage for the led ONLY for a given applied voltage. AC will destroy leds if greater than the maximum reverse voltage allowed by the led.

+?-------------resistor---------------led--------------gnd

 

You may be applying too much DC voltage or unfiltered DC voltage or AC.

The resistors will limit the current and supply a proper voltage for the led ONLY for a given applied voltage. AC will destroy leds if greater than the maximum reverse voltage allowed by the led.

+?-------------resistor---------------led--------------gnd

This is correct. You need to measure the series current flowing in the circuit to determine if it is at or below the maximum allowable current for the specific LED being used. If it's too high then you either have to lower the applied DC voltage or increase the resistance of the series resistor.

Lefty

 

But before you start connecting things up again and wind up with more burned-out parts...

What is/are the rating(s) of the LEDs?
What I'm asking for here is the typical Vf (forward voltage) @ current. It will look something like:
typ Vf=2.3v@20mA
There may also be maximum and minimum specifications, but right now I'd just like to know the typical.

Next, how are you trying to power the LEDs? An automotive electrical system? Your PC power supply? What is the supply voltage?

Note that LEDs should not be powered by AC, as the reverse voltage will damage them.

 

Sorry, should have specified. these where bought specifically for 12vdc auto. They are on a motorcycle.
none running voltage is 11.75vdc with the resister on it is still 11.75vdc.

 

voltage for the green 5mm led is 3-3.6v. Don't have the amps rating but they were listed as 12vdc automotive ready lights with resisters.

 

These are all in parallel with 12vdc going to each led. They are being used as idiot lights

 

OK. When your motorcycle is running, the generator will be charging the electrical system at about 13.8 volts. You should check it to make sure.

You need to connect the resistor in series with the LED, one resistor for each LED. The shorter LED lead is the cathode, which needs to be connected to the more negative side.
I will assume for the moment that the LEDs are rated for Vf=3v@20mA.
Rlimit >= (Vsupply - Vf(LED))/DesiredCurrent
Rlimit >= (13.8v - 3v) / 20mA
Rlimit >= 10.8v/0.02A
Rlimit >= 540 Ohms. The closest standard E24 value is 560 Ohms. Chart: http://www.logwell.com/tech/components/resistor_values.html

Let's see what your current might be with a 465 Ohm resistor when your engine is running:
Current(LED) = (Vsupply - Vf(LED)) / Rlimit
Current(LED) = (13.8v - 3v) / 465
Current(LED) = 10.8 / 465
Current(LED) = 23.2mA
That isn't terribly high, but 560 Ohms would be more safe.

 

Any chance the polarity is backwards? If there's no current flowing through the LED, then there'll be no voltage drop across the resistor.

 

OK, Sorry for the ignorance. I am a 50 years old movie stuntman and still love to learn.

Series is pos to neg to pos to neg? doesn't that drop the voltage to each led by the number in series? 6 2vdc leds with no resisters = 12v

 

Any chance the polarity is backwards? If there's no current flowing through the LED, then there'll be no voltage drop across the resistor.

OK I may be amiss here. I was measuring the voltage drop across the resister with no led present. just wired in the resister and measured the voltage from the back side of the resister and the neg lead.

 

What I am not getting here is the 12v reading on the back side of the resister. ( no led in place, probes on the back side of the resister and neg wire.)

 

OK, Sorry for the ignorance. I am a 50 years old movie stuntman and still love to learn.

That's why we're all here

Series is pos to neg to pos to neg?

It's like connecting together several extension cords to make one long one.
Resistors and coils/inductors don't care about polarity. Semiconductors (like diodes, LEDs, transistors, MOSFETs) will not work properly and/or will get burned up if you get the polarity wrong.

doesn't that drop the voltage to each led by the number in series?

Yes, it does.

6 2vdc leds with no resistors = 12v

OK, there you run into a problem. While it seems to make perfect sense to connect six 2V LEDs in series to drop 12v, it doesn't work that way in the "real world". Each LED is unique. While a batch of LEDs may have a "typical Vf @ current", that Vf may vary as much as 10%. 75% to 85% may fall quite close to the typical, but there will be some that are off by a fair margin.

Also, as LEDs get hotter, their Vf actually decreases. This can cause a "Chernobyl effect", or a runaway thermal condition resulting in an LED being destroyed due to overcurrent.

LEDs need to have either an active current regulator (a circuit), or a passive current limiter (a simple resistor).

For a motorcycle, the electrical system might vary from a low of 11.5v (when the battery is nearly dead) to over 14v when the engine is running. This makes a big difference to an LED circuit.

If you're using simple resistors, you have to plan for the higher voltage and deal with the LEDs getting dimmer at the lower voltage range.

Current regulators aren't difficult to make, but if you are making several of them it will certainly cost more than just a resistor.

 

What I am not getting here is the 12v reading on the back side of the resister. ( no led in place, probes on the back side of the resister and neg wire.)

OK, what's going on there is that the resistor has no load on it. Your meter has a very high resistance. When you are measuring the "back side" of the resistor to ground (negative), your meter is completing the circuit.

A typical digital multimeter may have 10 megohms or more internal resistance. That's a huge resistance compared to the 465 Ohm resistor you measured earlier. So, when you complete the circuit with your digital multimeter, most of the voltage in the circuit is dropped across your multimeter.

If you are using an older analog-type meter that has a needle (d'Arsonval movement) the internal resistance is usually 10,000 Ohms per volt. You were saying earlier that you were measuring around 11.54v with your meter, so that would be 115,400 Ohms resistance to ground compared to the 465 Ohm resistor. No wonder you were reading so much voltage, right?

Once you connect an LED in series with the resistor to ground, you will be able to measure the actual Vf of the LED, by putting your negative meter lead on ground, and the positive meter lead on the junction between the LED and the resistor.

By the way, if your battery is only reading 11.54v, you really need to get it charged up right away. Lead-acid batteries start to sulphate when their voltage falls to 12.5v; they have a short life after that. Use a low amperage charger; 1-2A.

 

That is all very interesting. My batt is 12.7v at the pos terminal. My meter is a digital. The wiring harness on the other hand has a considerable v drop. At the instruments where I am measuring the voltage after the resister is 11.75v.

so if I am understanding this I have to measure the voltage with the led in place. That makes sense because when I did the measurement of 3.15v the led was connected.

So another question. since the charging system will run voltage up to 13.8v can I use some kind of regulating resister or diode to limit the voltage to 12v to the leds?

 

That's why we're all here
OK, there you run into a problem. While it seems to make perfect sense to connect six 2V LEDs in series to drop 12v, it doesn't work that way in the "real world". Each LED is unique. While a batch of LEDs may have a "typical Vf @ current", that Vf may vary as much as 10%. 75% to 85% may fall quite close to the typical, but there will be some that are off by a fair margin.

Also, as LEDs get hotter, their Vf actually decreases. This can cause a "Chernobyl effect", or a runaway thermal condition resulting in an LED being destroyed due to overcurrent.

LEDs need to have either an active current regulator (a circuit), or a passive current limiter (a simple resistor).

For a motorcycle, the electrical system might vary from a low of 11.5v (when the battery is nearly dead) to over 14v when the engine is running. This makes a big difference to an LED circuit.

If you're using simple resistors, you have to plan for the higher voltage and deal with the LEDs getting dimmer at the lower voltage range.

Current regulators aren't difficult to make, but if you are making several of them it will certainly cost more than just a resistor.

dimming leds at lower voltage is fine. Do I understand you right in that: I have to measure each led to make sure of the voltage? and, resister every led even when they are daisy chained together in series?

 

That IS interesting. Perhaps your headlight was on, causing a voltage drop?

Try measuring from the hot side of the resistor to ground, and then the not-connected side of the resistor to ground. You should measure the same voltage.

Then try starting the motorcycle, and measure the voltage when running the engine at around 1,000 RPM.

That will give a baseline as to what to plan for.
Right now:
Current(LED) = (Vsupply - Vled)/Rlimit
Current(LED) = (11.75 - 3.15) / 465 Ohms
Current(LED) = 8.6 / 465 = 18.5mA (rounded off)

That is a safe current for most standard LEDs. However, we don't currently know what your voltage is when the motor is running.
If it's 14v, then: (14 - 3.15)/465 = 23.3mA; perhaps a bit high, but not ridiculously so.

 

That IS interesting. Perhaps your headlight was on, causing a voltage drop?

Try measuring from the hot side of the resistor to ground, and then the not-connected side of the resistor to ground. You should measure the same voltage.

Then try starting the motorcycle, and measure the voltage when running the engine at around 1,000 RPM.

That will give a baseline as to what to plan for.
Right now:
Current(LED) = (Vsupply - Vled)/Rlimit
Current(LED) = (11.75 - 3.15) / 465 Ohms
Current(LED) = 8.6 / 465 = 18.5mA (rounded off)

That is a safe current for most standard LEDs. However, we don't currently know what your voltage is when the motor is running.
If it's 14v, then: (14 - 3.15)/465 = 23.3mA; perhaps a bit high, but not ridiculously so.

Headlight not connected at time of testing. The runinig bike is 13.8v@2500rpm.

I have 2 leds@3v and 2@2v . the one that is constantly on burned at about 5 minutes.

I tested the connection and got 2v on the led side. The led needs 3v to opperate. (this was the burned one) I connected a new one and got 3.15v with a new resister.

I tested the original led that had stopped working. before the resistor 11.75v to ground side of led. after the resistor 11.75v to ground side of led. so I am guessing since the led was cooked and no load was on the resistor I was measuring 11.75.

When the new working led went on I had a 3.15v load.

I did try a new led connecting to the old led leads but it didn't work either. Thats when I found out there was only 2v coming through.

So I should plan for 13.8v? I am getting confused now. what is the minimun V a 3v led will work with?

I have 5 LEDs I am working with. If I connect them in series (2v, 2v, 3v, 3v) with a 12-13.8v supply, what do I need for a resistor?

 

dimming leds at lower voltage is fine. Do I understand you right in that: I have to measure each led to make sure of the voltage? and, resister every led even when they are daisy chained together in series?

If you are "daisy chaining" the LEDs in series, that's a very different ballgame.

With LEDs that have a Vf of 3.1v @ 20mA, about the most I'd recommend to use in series on a motorcycle is three.
Rlimit >= (Vsupply - VfLED(total)) / DesiredCurrent
Rlimit >= (14 - (3 x 3.1)) / 20mA
Rlimit >= (14 - 9.3) / 0.02A
Rlimit >= 4.7 / 0.02 = 235 Ohms.
The closest standard value is 240 Ohms. Chart: http://www.logwell.com/tech/components/resistor_values.html

Something I haven't mentioned before is the resistor wattage rating.

Let's look at the previous case, where you have a single LED and a single resistor.
If your voltage supply is 14v, your LED drops 3.1v, then the resistor drops the remaining 10.9v across itself. It also has 23.3mA flowing through it.
Power in Watts = Voltage x Current, so 10.9v x 23.3mA = 0.254 Watts.
We double this result for reliability, to 0.508 Watts. Since it's so close, you could use a resistor rated for 1/2W, but generally you want to go to the next higher rating.

With the current example using 3 LEDs in series with a 240 Ohm resistor dropping 4.7v, you have 4.7v x 20mA = 0.094 Watts; doubled it's 0.188 Watts. You can use a 1/4 Watt resistor in this case.

But what happens to the current if the electrical system drops to 12v?
The LEDs are still going to be dropping about 3.1v each, for a total of 9.3v.
12v - 9.3v = 2.7v; 2.7v / 240 Ohms = 11.3mA. The LEDs will be about half as bright as when the engine is running.

The problem isn't nearly as severe if you have just one LED and one resistor. Let's look at the 465 Ohm resistor with one LED in series for a moment.
With 14v, we've already calculated the current will be 23.3mA.
With 12v, then current = (12v - 3.1v)/465 = 8.9/465 = 19.13mA. You'll hardly notice the difference.

Similar things happen if your voltage goes too high - except then you start popping LEDs. Run the numbers the same way I've been showing you - what would happen if your regulator caused the charging voltage to go to 14.6v, and you had three LEDs in series with a 240 Ohm resistor?

 

OK, so make two strings.
One 2v, and one 3v in each string.
Rlimit = (13.8v - (2v+3v)) / 20mA
Rlimit = 8.8/0.02 = 440 Ohms.
Since it's so close, we'll try 430 Ohms.
LEDcurrent = 8.8/430 = 20.46mA. That should be pretty comfy.

What if the generator was charging hard at 14.5v?
14.5v -5v= 9.5v; 9.5/430 = 22.1mA - still not horriffic.
Check the wattage: 9.5 x 22.1mA = 0.21; x2=.42, need a 1/2W resistor.
Check the current with a discharged battery:
11.7 - 5 = 6.7v; 6.7/430 = 15.6mA. You'll notice that it's dimmer, but it'll still be fairly bright.