What resistor to reduce 18v to 14v?

[Anonymous]

I am running in Lionel TMCC with 18v to track power. I'm using SC2s to turn block power on and off. I want to install indicator lights in my control panel so I know that a block is powered. I have 18v bulbs (T-1 18v 0.026A) and would like to reduce voltage to the bulbs to 12-14v to reduce light intensity. I'm at a loss as to what type of resistor to use or if this is even a good idea. Thanks for any help and guidance.

[BR60103]

Your post has been on awhile with no help, so I'll stick my neck out. You need to get a multimeter out and measure the resistance of the bulbs. You will need a resistor proportionate to the voltage drop you want and the resistance of the bulb. e.g. if your bulb is 1000 ohms and you want to drop from 18 to 12 volts, you need a 500 ohm resistor. (1000 + 500 divided by 1500) means that 1000/1500 of the 18V gooes to the bulb and 500/1500 of 18V goes to the resistor.
The wattage of the resistor should be enough to handle the bulb; check the temperature of the bulb as the resistor may heat up (that power has to go somewhere).
I'm not sure if the bulb resistance changes when it lights up.
Caould you justify putting two bulbs in series? that would cut you down to 9V each.
--David

[Anonymous]

The short ans. according to Ohms law is a 1400 ohm resister rated at .01 amps

[lionelsoni]

Unfortunately, the lamp resistance can change by as much as a factor of 10 between cold and lit, so it is useless to try to measure it with an ohmmeter. You need to measure the current that it draws when operated at its normal voltage. You can do this with an ammeter and a power supply at that normal voltage. AC ammeters are not very common. If you have an ammeter, it is most likely a DC ammeter. So you need to make the measurement with a DC voltage, not a train transformer. You could use an automobile battery, which is about 12 volts.
Anyway, when you get the current measurement, divide the voltage difference by it. For example, if your measurement was 100 milliamperes, 18 minus 12 volts is a 6-volt drop, so divide 6 volts by 100 milliamperes to get 60 ohms. Then pick the closest standard resistor value, perhaps 56 ohms. To find the power rating needed, square the current and multiply by the resistance; so: (100 mA)*(100 mA)*(56 ohm) = .56 watt. Then double or triple this to give yourself a safety margin. I would use a 56 ohm, 1watt resistor for this example.

Bob Nelson

[lionelsoni]

I just noticed that you already know the current rating of the lamp, 26 mA. From this, we can make a pretty good guess at the value of the resistor. Incandescent lamp current varies about as the .55 power of voltage. So, at 14 volts, your 18-volt lamp should be drawing about 87 percent of 26 mA, or about 23 mA. Divide 4 volts by this and get 177 ohms; so use 180 ohms. The power is about 95 milliwatts; so a 1/2-watt resistor is plenty big.
If you want to go down to 12 volts, use 270 ohms, 1/2 watt.

Bob Nelson

[webenda]

I see the answers are homing in on the correct value. 180 ohms is a good choice. The following spreadsheet example calculates the value for a ballast resistor, taking into account that the lamp's resistance is lower at lower voltage.

Copy and past into a blank spreadsheet cell A1. It should populated the rest of the cells automatically. Input your transformer voltage in cell B2. Input voltage and current ratings for the bulb you are using in cells B3 and B4. Input voltage you wi***o operate lamp at in cell B5. Copy formulas from D6 through D13 into cells B6 through B13 and D15 through D16 into B15 through B16. (Get rid of the quote mark to change it from text to formula.)The correct value for a ballast resistor is calculated in cells B12 and B13. Enter the closest common value available in cell B14.

It is not necessary to use the exact value calculated in cell B12. In the example, the closest value found at Radio Shack is 150 ohms. In the case of wattage rating, more is better. Use a half-watt resistor to dissipate 0.1 watts. It will run cooler than a quarter watt resistor.

COMMA DELINEATED SPREADSHEET:

What Ballast Resistor for Lamp?,VALUES,UNITS,FORMULA
Voltage transformer (Vt),18.0,Volts,Given
Voltage lamp (Vl),18.0,Volts,Given
Current lamp (Il),0.0260,Amps,Given
Voltage desired (Vd),14.0,Volts,Given
Resistance lamp (Rl=Vl/Il), 692,Ohms,=B3/B4
Ratio (Ratio=Vd/Vl),0.778,,=B5/B3
Resistance cold (Rc=Rl/10),69.2,Ohms,=B6/10
Resistance desired [Rd=Ratio(Rl-Rc)+Rc],554,Ohms,=B7*(B6-B8)-B8
Current desired (Id=Vd/Rd),0.0253,Amps,=B5/B9
Voltage ballast [Vb=(Vt-Vd)],4.00,Volts,=B2-B5
Resistance ballast (Rb=Vb/Id),158,Ohms,=B11/B10
Watts ballast (Wb=VbId),0.101,Watts,=B11*B10
R = 100 120 150 180 220 270 330 390,150,Ohms,Closest common value
Voltage Lamp [VL=RdVt/(R+Rd)],14.2,Volts,B9*B2/(B14+B9)
deviation from Target dT=100(VL-Vd)/Vd,1

Wayne

[lionelsoni]

Wayne,
I'm afraid that lamp resistance doesn't vary in a straight line as a function of lamp voltage, as you assumed in your spreadsheet. If current varies as the .55 power of voltage, then resistance goes as the .45 power. I believe that the ballast resistance r should be

r = (18 - v) / ((v / 18)^.55 * .026)

where v is the desired lamp voltage on a lamp which draws .026 A at 18 V, and the caret denotes exponentiation.

Bob Nelson

[jprampolla]

Hi Folks,
I am not an expert in electronics, so perhaps I am missing something important here, but a simple way I reduce the voltage to a light being powered by AC is to use a diode in series with the bulb to reduce the voltage by approximately half. In DC applications, I use diodes in a string to reduce voltage by about .7 volts per diode. Although not fancy electronics, it does work and has helped me many times. Perhaps someone else could recommend the proper ratings for the diodes; I just oversize them to be on the safe side.
Hope this helps.

Take care, Joe.

[lionelsoni]

Nothing wrong with that, Joe, as long as 9 volts is acceptable to "stangtrain". If more than one lamp is involved, as I think is his case, it would be best to connect half the diodes one way and the other half the other way. Otherwise, the combined DC current from many lamps could saturate the transformer.
I would think that the ubiquitous and cheap 1N4001 would be amply rated in voltage and current for any such toy-train-lamp application.

Bob Nelson

[jprampolla]

Hi Bob,
Thanks for that information about the diode rating and about orienting half of the diodes in opposite directions.
Take care, Joe.

[Anonymous]

instead of trying to determine the resistance of the various bulbs. you can do what I did; that is purchase or you may have a low cost train transformer,and hook lights up to the transformer..this will allow you to vary the voltage,and if it is critical, you can install a small voltmeter in the line to monitor the voltage..

Good luck

[webenda]

Bob,

You are correct.

I measured three lamps at different voltages to make sure my linear method worked. I noticed the deviation from a straight line, but since the lamps varied up to 10% from their rated parameters and I use 10% resistors for ballast, my method was close enough.

Your method, however, is exact (as close as my 0.5% meter can measure.) Better yet, a spreadsheet using the formula you offered is shorter.

Comma Delineated Spreadsheet
Using Bob Nelson's Formula
--------------------------
DESCRIPTION,VALUES,UNITS,FORMULA
Transformer,18.0,Volts,Given
Lamp Voltage,18.0,Volts,Given
Lamp Current,0.0260,Amps,Given
Desired Volt,14.0,Volts,Given
Desired Amps,0.0226,Amps,=(B5/B2)^.55*B4
Ballast Volt,4.00,Volts,=B2-B5
Ballast Ohms,177,Ohms,=B7/B6
Ballast Used,180,Ohms,Closest common value
Ballast Watts,0.091,Watts,=B7*B6

Wayne