badass951

I have been reading over some previous threads reegarding LED alternatives and DIYs, however I can not seem to find an answer to my question.

I am very new to using LEDs to replace factory bulbs, although a few years ago a friend of mine had an extra LED replacement bulb for an autometer gauge so I replaced the one in my boost gauge with no problems since. I recently noticed it plugs in to the same size socket as the ones in the instrument cluster. I plugged one in real quick and it worked.

http://www.autometer.com/cat_accesso...st.aspx?pid=27

Anyway, I understand than an led will draw a small fraction of the current that a factory cluster bulb will, however why is there a need for a resistor in this case? Just from experience I know that my boost gauge led has had a 12v source with no resistor and no problems over the course of a few years. Also, since the cluster has power going through the dimmer switch it already has a higher resistance than my boost gauge source.

Are your calculations for a resistor based on the amperage that each factory cluster bulb draws? Like I said I don't know much about leds so I am sure I am way off, but I was under the impression you used resistors when a component would draw more current and damage the circuit.

Thanks in advance!

RKD in OKC

I thought the resistors used with LED bub replacements were so the LED would see less current so it would dim with the factory dimmer.

badass951

Or maybe so the circuit would draw more current - to be equal to a factory bulb so that it would work with the dimmer? Therefore if dimmer function were not a concern no resistor is necessary?

RKD in OKC

Googled the resistor thing. The current for the incandesent bulbs could be enough to overheat and burn out the LED's, hence the resistor.

VWaddict

Well, not quite... that's an oversimplification, probably because the answer is rather involved.

It's nothing to do directly with current, since ANY load on a voltage only draws the current it 'needs'. A 120V household electrical socket is rated at 15Amps, but not everything which you plug into it draws ANYTHING LIKE that much current.

Here's a full explanation:

LEDs have a fixed 'forward-bias' working voltage, referred to as "Vfwd"; (Volts Forward).

There is NO such thing as a 12-volt LED, LED forward voltages vary by colour and construction, but for example; a Gallium-Arsenide red-coloured LED usually has a forward-bias voltage of about 2 Volts. (Numbers in this example are being rounded up or down to the nearest whole number for simplicity with the math.)

Supplying 12 volts and connecting directly to an LED or any other 'avalanche'-type device (i.e. a device which 'attempts' to reduce the voltage across its terminals by reducing its resistance) would result in the LED reducing its resistance until the point where too much current will flow, destroying the LED, which typically can ONLY work up to a maximum continuous current of 0.02A (20 milliamperes). Compare that with a 5-watt incandescent bulb, which would consume about 400milliamperes (400mA).

Ohms law states that V (Electrical potential, expressed in volts) = I (electrical current, expressed in Amperes) multiplied by R (electrical resistance, expressed in Ohms, [symbol Ω]).

In any series chain, a known resistance and an UNKNOWN resistance (the LED will vary its resistance MASSIVELY trying to maintain 2.0 Volts across itself in our Gallium Arsenide [GaAs] red LED example).

So if we feed 12.0 Volts into a 1,000Ω resistor in series with a 2.0 Volt LED, the LED will adjust its resistance to maintain 2.0 Volts across it. The fixed resistance will therefore have the remaining 10 volts distributed across it.

Ohms law says that 1,000Ω with 10 volts across it will pass 10mA. Therefore the LED will have adjusted its resistance to be 200Ω. (if 10 Volts = 1,000Ω at 10mA, 2 Volts must = 200Ω, since the current is the same everywhere in a serial chain.

Reduce the resistance to 500Ω, and twice the current will flow. (half the resistance = double the current) reduce the resistance by ten (100Ω) and ten times the current will flow... 100mA is already enough to blow up just about ALL typical LEDs. -Reduce the current infinitely (i.e. reduce it to zero) and the forward current increases infinitely. -Now you're trying to
'lose' ten volts across zero ohms, and Ohms law states that current = voltage divided by resistance... if he resistance is zero, we're dividing by zero, and any number divided by zero equals infinity.

Since this math can get a little more than most people want to concern themselves with when "replacing a light bulb", most "plug-in" LED solutions are MORE than an LED; they also include a series resistor.... oh, and a rectifier also, because otherwise, plugging an LED in backwards would destroy the junction, if the supply voltage is greater than the LED's reverse breakdown voltage (expressed as PIV or Peak-Inverse-Voltage).

More than you probably wanted to know, but that's the definitive answer.

Keith

billthe3

Since the autometer gauge is built to run with the incandencent bulb, I belt the led bulb that you have that is built into the socket has a resistor built into it already.


Like these.

badass951

Thanks!

I guess I assumed they did make 12v leds since I have seen them at radio shack and they appeared to have no resistors built in (they were the exact same style as the ones you used to update your instrument cluster - excellent writeup by the way). However, I was discussing plug-in style leds from autometer and was curious why no one has gone the simple route of using these leds? Prior to your post, from my understanding of ohm's law I assumed that if I had a 12v source and a load that wanted to draw say 20mA, I would just assume that the led would have a resistance of 600 ohms. I guess what I failed to realize was the 12v led from autometer was indeed different than the leds you were using for the fact that the ones from autometer are already modified to work in a 12v socket and the ones you were using were not.

Anyway, thanks this still does answer my question. I was mainly curious if there was something I might have been missing, as I could not figure any need for a resistor with a 12v led (which I know now is not a 12v led but instead an led already modified to work in a 12v circuit).