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Minatronics 1.5V Incandescent Bulbs

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TwinStar:
I'm going to use the Minatronics 1.2 mm 1.5V 30mA bulbs for my E unit headlights. I'm needing assistance in ensuring that I don't blow these things up. I'm assuming a resistor of some sort and while I'd like to just ask for the answer I would prefer to learn how to calculate the value of the resistor that I need for these.

Thanks.

Alan:
Those are incandescent lamps. No resistor needed. An incandescent lamp is a resistor. One that glows brightly!

TwinStar:
Alan:

How do I step the voltage down then? I'm assuming the RP LM-2S light output is track value of approx 14V. Won't that damage the 1.5V bulb?

Alan:
My apologies. I thought you had the supply voltage available already since you bought 1.5v lamps and were asking if a current limiting resistor is needed like with an LED.

You need what's called a resistor network i.e. two resistors connected in series. Since a bulb is a resistor it acts as of of the two resistors in the series. There is a voltage drop across each resistor proportional to its percentage of the overall circuit resistance.

We know the bulb passes 30mA of current at 1.5V. Using Ohm's Law we can determine its on resistance R = V / I which is 1.5 / 0.030 = 50 ohms. [EDIT] Oops, had the decimal point in the wrong place. Fixed now.

Now that we know the bulb has 50 ohms of resistance we can plug that value into a resistor divider formula to determine the value of the second resistor when the supply voltage is 14.8 volts.

Bulb voltage = ( Supply voltage x bulb resistance ) / ( Unknown Resistor + Bulb resistance )
1.5 = ( 14.8 x 50 ) / ( R + 50 )
1.5 = 740 / ( R + 50 )
1.5 * ( R + 50 ) = 740
75 + ( R * 1.5) = 740
( R * 1.5 ) = 665
R = ( 665 / 1.5 )
R = 443
The series resistor needs to be 443 ohms to create a 1.5 volt drop across a 30mA bulb at 14.8 volts. The closest standard value is 470 ohms.

Now that we know the resistor value we calculate the total power so we can choose a resistor of sufficient wattage. Once again using Ohm's Law we calculate the current flowing through the circuit I = V / R.
(We already know this because the bulb spec told us it is 30mA but for sake of making the math complete...)

Current = Voltage / ( Bulb resistance + Series resistor )
I = 14.8 / ( 50 + 470 )
I = 14.8 / 520
I = 0.028

Power (watts) = Voltage * Current
W = V * I
W = 14.8 * 0.028
W = 0.41
The series resistor needs to have a power handling capacity of 0.41 watts. The closest standard value is 0.5 watts. Typically power ratings are halved for cool operation so the resistor really should be 1 watt.

Final answer: On a 14.8 volt system a 470 ohm 1 watt resistor will cause the bulb to glow at full 1.5 volt brightness.

There will be a quiz.  ;D

TwinStar:
Alan:

Thank you! And of course I only have 1/4 and 1/2 watt resistors in stock. Time for Amazon.

The bulbs were selected solely for their size of 1.2 mm. I had tried to use LED's and fiber rod but the size was too big. Another E unit era modeler had successfully used these bulbs so it was a case of size over anything else. These bulbs fit perfectly in the dual light housings that I've installed.

These units have only been worked on for over 5 years. Maybe now I can finish them.

Thanks again.

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