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How do I reduce voltage for using low voltage devices?

(Example using 3 volt device. -Editor)

As you are using a fixed supply ( constant 12v DC)we can now work this out using resistors. As the light is designed to work from 3 volts (2 batteries) we need to 'lose' 9 volts. I will assume the light only draws 50mA from the battery.
So- 12volts - 3 Volts = 9 Volts.
9 volts / 50mA = 180 ohms.
Also, we need to know how big the resistor needs to be to get rid of the heat this is going to generate. As heat equals power, we can use P=VI,
P = 9 x .05 gives us .45 watts, so a half watt resistor will work but get very hot, I would recommend a 1 watt resistor and plenty of space around it at least.

The 180 ohm, 1 watt resistor will at least give a good starting point. If the light is a little dull, go down to a 168 ohm unit. If too bright, go up to 220 ohm. As the resisters cost about 5 cents each, getting a few around 180 ohms will not be exspensive, just make sure they are rated at 1 watt or more.

Ron Powell
November 11, 1998

One other suggestion, if the flashing light is meant to be stationary...ie at a railroad crossing.
You can buy very inexpensive AC to DC adapters that will do the job. I picked one up two days ago at a discount store here, it plugs into the wall, is UL approved, has a slide switch to adjust the output voltage from 3 VDC to 12 VDC in 1.5 volt steps, handles up to 500ma (there are also larger units available) and the whole thing only costs $ 7.99...That's Canadian dollars, so it shouldn't be more than $5.00 US

If on the other hand you want your flashing lights on a moving train, different story. If it is in the engine, you can generate a fixed three volts with the standard diode circuit in series with the motor; if it's in a car, you basically replace the motor with a 12 volt automotive light plus the diodes.

Knut Schartmann
November 10, 1998

Using resistors to drop voltage is only practical where the input voltage and the load current stays the same. In the case of model trains and light globes , neither of these conditions are met.

If your maximum voltage from the power source is 12 volts and your light uses 100mA of current then-
12 volts - 3 volts (this is what the light normally gets from the battery) then ohms law says we need a 90ohm resistor
12v - 3 v = 9v
9v/.1A= 90 ohms.

BUT , you only need 90ohms when 12 volts is applied, at 6 volts you only need 30 ohms and at 3 volts you do not need any! (assuming you want the light to start as soon as possible after power is applied )

Better ways would be to :
Feed the circuit/lamp from either batteries (the easiest way) The advantage to this is that the light will still work even with the train stopped.

OR

A small bridge rectifier ( a 500mA type would do) feeding a LM7805 5 volt regulator. The circuit/lamp would then be fed via 3 x diodes or a 27 ohm resistor (I prefer the diode system) This has the disadvantage that around 7 volts would be required to make it work correctly.

Sorry if this is confusing- this 'green steam' stuff is nearly as hard to control as the white stuff!

Ron Powell
November 10, 1998