LED bulb resistor??

[winchman]

I'm using a green LED night light replacement bulb in my shop door monitoring system. It's made to operate on 120vac and draw 0.25 watt, but I'm using a 12vdc wall wart. It comes with a lifetime warranty. The bulb looks like this, but it's green:



Here are the guts:



It worked fine for several months, but then it quit. The ohmmeter shows it's open circuit. I thought it would be OK to operate it at 12vdc, but apparently not. It didn't appear to be as bright on 12vdc as it was on 120vac.

What size resistor do I need in series with the bulb to make it last longer?

[ed_h]

You really need the LED specs to say for sure, but a 1K resistor should give you around 12 mA peak current, which should be fairly safe. You could probably go higher on the current, but too high and life will be shortened.

Ed

[MotorradMike]

Generally, you can not measure an LED or any diode, with an ohmmeter.
You need a "Diode" range which provides enough voltage to cause current flow in the diode.

5th range on this meter.

[Mike Nash]

The bulb looks like this, but it's green:

There is no way the "looks like" applies to the guts you have (outside of base and globe maybe.)

If it's brown and creamy and lying in the backyard it must be dark chocolate, because it sure ain't sweet.

[6PTsocket]

The guts are are four diodes forming a full wave rectifier to change AC to DC and a couple of resistors to limit current. All of that will be replaced with a single resistor. A white LED typically has voltage drop of around 3.5 volts. You say it is 1/4 watt LED. Watts =volts x amps so the current is around 71 ma. You want to run on 12 volts DC. That means that the bulb drops 3.5 volts and the resistor drops the other 8.5 volts. Resistor= volts÷current. 8.5÷.071 = 120 ohms. There is 0.6 watts dissipated by that resistor so you are going to need a 1 watt resistor. For long life and to be on the safe side I would try a 150 or 180 ohm resistor. It should still be plenty bright. Another thing to consider is those wall worts can output more than they are rated for under light load and might be putting out more like 15 volts. Check it. If it is high, go to 200 or 220 ohms. You have a multimeter. Measure the current and stay under 71 ma.Adjust the resistor value as needed.

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[RichR]

The guts are are four diodes forming a full wave rectifier to change AC to DC and a couple of resistors to limit current. All of that will be replaced with a single resistor. A white LED typically has voltage drop of around 3.5 volts. You say it is 1/4 watt LED. Watts =volts x amps so the current is around 71 ma. You want to run on 12 volts DC. That means that the bulb drops 3.5 volts and the resistor drops the other 8.5 volts. Resistor= volts÷current. 8.5÷.071 = 120 ohms. There is 0.6 watts dissipated by that resistor so you are going to need a 1 watt resistor. For long life and to be on the safe side I would try a 150 or 180 ohm resistor. It should still be plenty bright. Another thing to consider is those wall worts can output more than they are rated for under light load and might be putting out more like 15 volts. Check it. If it is high, go to 200 or 220 ohms. You have a multimeter. Measure the current and stay under 71 ma.Adjust the resistor value as needed.

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No, he said:

It's made to operate on 120vac and draw 0.25 watt

[winchman]

I left the bulb intact (with all the guts inside the globe) when I was using it on 12vdc, so the diodes and two resistors were in the circuit. They're folded together inside the aluminum base.

The bulbs come in clear or green, but they look identical because both have clear globes. The package in the photo says "Clear" at the lower right corner. The package my bulbs came in says "Green".

[6PTsocket]

No, he said:

A white LED has around a 3 5 volt drop. If you run it on 4 volts or 1000, the remainder of the voltage must be dissipated elsewhere in the circuit. As far as the actual bulb is concerned, there is no such thing as 120 volt AC LED. When you buy a bulb with a base, the actual LED is powered by the voltage reducing circuitry in the base.

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[6PTsocket]

The guts are are four diodes forming a full wave rectifier to change AC to DC and a couple of resistors to limit current. All of that will be replaced with a single resistor. A white LED typically has voltage drop of around 3.5 volts. You say it is 1/4 watt LED. Watts =volts x amps so the current is around 71 ma. You want to run on 12 volts DC. That means that the bulb drops 3.5 volts and the resistor drops the other 8.5 volts. Resistor= volts÷current. 8.5÷.071 = 120 ohms. There is 0.6 watts dissipated by that resistor so you are going to need a 1 watt resistor. For long life and to be on the safe side I would try a 150 or 180 ohm resistor. It should still be plenty bright. Another thing to consider is those wall worts can output more than they are rated for under light load and might be putting out more like 15 volts. Check it. If it is high, go to 200 or 220 ohms. You have a multimeter. Measure the current and stay under 71 ma.Adjust the resistor value as needed.

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I just realized that you said your LED is green If it is just green colored plastic over a white LED than everything iI said was accurate. If it is actutually a green led you have to drop 10 volts in the resistor.

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[6PTsocket]

Generally, you can not measure an LED or any diode, with an ohmmeter.
You need a "Diode" range which provides enough voltage to cause current flow in the diode.

5th range on this meter.

The diode check function was made for rectifiers and bipolar transistor junctions with a forward drop of around 0.7 volts. The test voltage may not be high enough for an LED with around a 2 volt forward drop for most colors and as high as 5 volts for a white one. That test was never intended for LEDs.

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[winchman]

My cheap ohmmeter doesn't have a diode setting.

I put the resistor on the left in the picture (red, red, orange = 33K ohm, right?) into the monitor circuit along with a new bulb just like the other one. That ought to add some life to the bulb.

It's not nearly as bright now, but I'll probably be able to see it at night from 250 feet away.

[MotorradMike]

The diode check function was made for rectifiers and bipolar transistor junctions with a forward drop of around 0.7 volts. The test voltage may not be high enough for an LED with around a 2 volt forward drop for most colors and as high as 5 volts for a white one. That test was never intended for LEDs.

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The meter puts out over 2.5V on that range with enough current to dimly light almost every LED I've ever tested with it. Fluke may have designed it just for the junctions you suggest but it works as-is for most LEDs too.

[Mcgyver]

How to did you input the 12 VDC - where was the connection,

so until the electronic guys show up and do this properly......the photo of the resistors diodes is 'as it was', not your creation, right? thats a 22k resistor...so ignoring the diode drops because they are comparatively so small, at 120 AC = 170 peak, the peak current through the resistor is about 8ma. I don't get why it even worked with 12V, as 12VDC into the rectifier unless the hook up somehow bypassed the 22k resistor would be like 0.5 ma. To get the same current at 12V with a 2V drop....1.2k resistor should work.

[6PTsocket]

My cheap ohmmeter doesn't have a diode setting.

I put the resistor on the left in the picture (red, red, orange = 33K ohm, right?) into the monitor circuit along with a new bulb just like the other one. That ought to add some life to the bulb.

It's not nearly as bright now, but I'll probably be able to see it at night from 250 feet away.

That resistor is far too big. The LED drops around 3.5 volts and the resistor drops the rest. Originally it was dropping 116.5 volts (120volts-3.5volts.) . Now it only it needs to drop 8.5 volts. (12volts -3.5volts.). What is the value of the other resistor? I can't quite tell in the picture. You are running that LED at 3 ma when it is probably good for at least 50ma if it is a 1/4 watt LED. No wonder the light output is low. There are many tutorials on line for calculating the value of an LED resistor. The math is really simple. It is a lot better than hit or miss. That method yields dim bulbs and burned out ones.

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[Paul Alciatore]

I don't agree with some of the numbers being paraded around here. If it is made to operate on 120 VAC at the 0.25W power level, then it must have been originally made to draw only 0.0021 Amps. (2 mA). That sounds low, but it is what the numbers say.

For a reality check, I see two resistors in the photo. They are in series with the rectifier circuit so their values will add together. One appears to have color code bands of red(2), brown(1), and orange(3) which gives a value of 21,000 Ohms. The other one appears to be a 10 Ohm value which adds little to give a total resistance value of 21,010 Ohms. Now, subtracting a couple of Volts (I am making a reasonable guess) for the LED's Voltage drop: 120 - 2 = 118 Volts for the resistor to drop. 118 V/21,010 Ohms = 0.0056 Amps (5.6 mA).

These two calculations may differ by a factor of about 2.5 times, but they are both a lot lower than the numbers that others are assuming here. It would seen that the original design of this circuit was to draw between 2 and 5.6 mA, not the 12 or 71 mA that others here are assuming. My conclusion is that the LED in these bulbs is a high efficiency one.

Now, you say that you powered the original bulb with 12 VDC from a wall wart apparently without any alterations to the bulb. OK, it may be 12 V or it could be higher if the wall wart does not include a regulator. These devices, when unregulated, often are designed with some extra Voltage so, when they are supplying current to a load and their output Voltage goes down, they drop down to their rated/stated output Voltage. And while running it at that 12 Volts or perhaps a few Volts more, it burned out. At 15 Volts, the device should have been drawing no more than 0.8 mA or so. If it was designed to run at 2 to 6 mA, that current/power level should not have burned it out.

I think it did not burn out from too much current. There was probably some other cause for the failure. Electronic devices often just fail and some fail long before the average life span for that particular part. If you were happy with the light level while running it at 12 VDC, I would just put a new bulb in and see what happens. It will very likely last a lot longer than that first one.

I'm using a green LED night light replacement bulb in my shop door monitoring system. It's made to operate on 120vac and draw 0.25 watt, but I'm using a 12vdc wall wart. It comes with a lifetime warranty. The bulb looks like this, but it's green:



Here are the guts:



It worked fine for several months, but then it quit. The ohmmeter shows it's open circuit. I thought it would be OK to operate it at 12vdc, but apparently not. It didn't appear to be as bright on 12vdc as it was on 120vac.

What size resistor do I need in series with the bulb to make it last longer?