Finding out the Resistor for 1.5 volt Incandescent lamps.

Joe Moltz said on 26Jan06.

The best way is with a multimeter set to read current. Connect the meter in series with the bulb to read the current draw, not parallel like you would to read voltage.

Most bulb packages have the current rating listed on the label. I know Miniatronics does but I have found them to vary somewhat from the package listing. Not much but it can make a difference.

If you are trying to determine a resistor you need to drop voltage for a low voltage bulb, then you will need to know the current rating of the bulb.

If you don’t know the current rating of the lamp, connect it to a single 1.5 volt batter with a multimeter selected to DC mAs connected in SERIES with one lead. If the lamp illuminates then it is a 1.5 volt unit and the meter shows the current value that you need for calculating the resistor. (added by Marcus)

For those that need it, there is a formula to determine the resistor needed to drop the voltage. It is R=E/I where R is the resistor value you are trying to determine, E is the voltage you are trying to dissipate, and I is the current, in AMPS of the bulb. Example, you are trying to use a 1.5 volt bulb on an NCE system that has 13.5 volts on the function output. Therefore you need to dissipate 12 volts. Now, your bulb is rated at 15 milliamps. That is .015 AMPS. 1 milliamp = .001 amp. Plug these figures into the formula and you get R=12/.015. So R=800 ohms. That's the resistance value you need.

You also need to consider the power rating, in WATTS, of the resistor you will need. This formula is P=I*E, where P is the power in watts, I is the current of the bulb and E is the voltage DISSIPATED, not the voltage rating of the bulb. Using that same bulb, I=.015, and E=12 (the voltage being lost). Now the formula looks like this. P=.015*12. Answer, .180 watts. That is more than 1/8 watt but close to 1/4 watt. A 1/4 watt resistor may get a little warm in this case so I would go to a 1/2 watt resistor.

When you work these formulas remember to use the total current for all bulbs that will be connected to a function and on at the same time. In the hood unit diesels, I use two bulbs at each end putting one in each headlight opening. The Miniatronics 1.7mm bulbs fit nicely. So now I am using 30 milliamps (.030 amps)in the formula, which in this case cuts the resistance value approximately in half, and the wattage rating doubles. This would make the wattage rating of the resistor .360 watts. This is getting close to 1/2 watt so keep the 1/2 watt resistor away form plastic parts. Putting the body of the resistor in contact with the lead weight will let the weight act as a heat sink and help some.

Hope this helps.

Joe Moltz

Heavy Metal Thunder Productions

Sykesville, Maryland USA

Chief Engineer JM Northern