12 v bulb resistance.?

Thread Starter

quadhed

Joined Jan 13, 2016
23
Hi. I'm using a 12 v bulb that has a max rating of 16v and 40 mA. My question is: The resistance of the bulb is 41 Ohms. Wouldn't the 12v/40 ohms exceed the max rating? Something must be wrong with the specs? Right?
 

Reloadron

Joined Jan 15, 2015
6,204
You are measuring the cold resistance. The resistance increases substantially as the bulb heats up.
This. The resistance between a cold and hot filament can be 10X or greater. Measure the actual bulb voltage and bulb current and now calculate the actual bulb resistance. With bulb in circuit and illuminated. :)

Ron
 

Thread Starter

quadhed

Joined Jan 13, 2016
23
This. The resistance between a cold and hot filament can be 10X or greater. Measure the actual bulb voltage and bulb current and now calculate the actual bulb resistance. With bulb in circuit and illuminated. :)

Ron
I see. What you say makes a lot of sense! Thank you for your knowledge. I'll do that.
 

Audioguru again

Joined Oct 21, 2019
3,833
An incandescent light bulb usually burns out only at the moment it is turned on when the current, rise in temperature and physical expansion of the filament are massive.
 

dendad

Joined Feb 20, 2016
3,939
This charactoristic make incandescent lamps a very good test for power supplies. They draw max current on startup from cold so give the power supply a bit of a fright.
 

Tonyr1084

Joined Sep 24, 2015
6,452
12V at 40mA is about 500mW (half a watt). Assuming this is operating parameters, 12V divided by 40mA is 300Ω. That seems a bit high, but I suppose it's possible. We ARE talking about a low wattage bulb. [edit] Strike the following comment: That is - IF - we are talking about an incandescent bulb. 40mA is closer to the operating current of an LED. Most common (simple) LED's generally operate at a max current of 30mA, and may typically be operated at 20mA by design. SuperBright LED's may withstand higher currents. So I want to ask "Are we talking about an incandescent lamp or an LED?" [end edit]
 
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Tonyr1084

Joined Sep 24, 2015
6,452
Re-examining the original post, I see you have measured the filament resistance. LED's won't have a resistance that I know of. At least I've never thought to test one for resistance. Hey! Good opportunity to learn something. I'm going to grab an LED and measure its resistance right now.

[edit] Nope. No resistance in either polarity. [end edit]
 

Ian0

Joined Aug 7, 2020
3,754
Filament resistance is proportional to absolute temperature. If room temperature is 298K (it's 293K in Britain, because we like to wear our jumpers) and filament lamps tend to run at about 2700K, you can see why the cold current is 9 times the running current. Halogen lamps run hotter (3000K to 3300K) so have greater inrush. You can also see why outdoor lamps fail on frosty nights!
 

ElectricSpidey

Joined Dec 2, 2017
1,962
Re-examining the original post, I see you have measured the filament resistance. LED's won't have a resistance that I know of. At least I've never thought to test one for resistance. Hey! Good opportunity to learn something. I'm going to grab an LED and measure its resistance right now.

[edit] Nope. No resistance in either polarity. [end edit]
Of course LEDs have resistance...you just can't measure it with an ohm meter.

Take voltage and current readings then do the math for resistance.

Some meters have a high enough voltage to do a diode check, but they mostly read voltage not ohms.
 

Audioguru again

Joined Oct 21, 2019
3,833
The tiny dim incandescent light bulb was used and called "a grain of wheat bulb" 60 years ago.

A 2V red LED at 10mA would be calculated to be a resistance of 2V/10mA= 200 ohms.
 

AnalogKid

Joined Aug 1, 2013
9,480
Semantics: A diode is described as having an *equivalent* resistance. A "normal" resistor's value is independent of the current through it or the voltage across it. With a theoretically perfect diode, the voltage across it is completely independent of the current through it; from milliamps to amps, the voltage across it is a constant. For any particular current, you can calculate the equivalent resistance *at that current* with Ohm's Law.

This can have all kinds of fun implications for circuits. For example, you can use a diode as a variable resistor in an analog circuit by varying a relatively large DC current through it with a relatively small AC signal superimposed. I have schematics from the 1960's for both a bass guitar amp and a TV station intercom, two wildly different applications, and both use use a variable-DC-biased diode as an electronically-variable resistor in a gain-limiting circuit.

ak
 

DickCappels

Joined Aug 21, 2008
7,955
As per @ElectricSpidey in post #11:

This is the IV curve from an LED from a Christmas light string. I suspect there is some intentional resistance built-into the LED, but the resistance is apparent in the linear part of the curve. About 12 ohms.

1633825452434.png

1633825602085.png
 

dcbingaman

Joined Jun 30, 2021
495
You are measuring the cold resistance. The resistance increases substantially as the bulb heats up.
I have heard, if you use a current limiting circuit (or a constant current source) to feed an incandescent bulb, the inrush current is eliminated and you can extend bulb life. The inrush current is very hard on the bulb filament.
 

AnalogKid

Joined Aug 1, 2013
9,480
Very true. It is much easier for a DC circuit than an AC circuit. For an AC circuit, an alternate approach is to start with a standard TRIAC dimmer circuit, and modulate the gate drive to ramp up the RMS voltage to the bulb. Not exactly constant current, but still very life-extending.

ak
 

DickCappels

Joined Aug 21, 2008
7,955
This can give you an idea of how great the inrush current it compared to the operating current. Most incandescent lamps run from 3300° K (warm) to about 6300° K (called "daylight").
1634288108788.png
 
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