how do I know what size resistor I need

They will work, you may be happier with ones around 470 ohms for a bit brighter lights.

 

thanks!

 

these

Values are explained here.

 

Here is a handy calculator for minimum resistance,maximum brightness ,it assume DC current applied

 

 

 

 

Dale H

If you do buy a bunch of 1000 ohm resistors, you can "make" lower values or higher values using them. For higher values, just add them up and wire in series.  For lower values, use this calculator.

 

Two 1000 in parallel gives you an equivalent 500.

I'm not sure why you'd want to combine resistors when one will do the job...

 

Those 470's are fine for the job.  They'll suffice for pretty much any track voltage with a series diode, the resistor, and the LED.  If you have two LED's to run on AC, then I put them back to back and use the 1K resistor and omit the diode.

 

here is what someone made for me but I do not understand how he did it, he is 80 something and didn't explain it so I could understand.

 

it is violet, black, red, red.  wired to one led.  he said he makes theses all the time and runs them on 18 volt track power.  he gave me one so I can learn but .....

 

the light is bright but could be brighter.

BigdodgeRam;

That color code does not cross to a standard value, I'd go with the Value John gave.

Sometimes the background color throws off the perceived stripe colors.

Thus I always use a meter to verify values.

You'll never go wrong with the meter to check the values, Russel is right.

 

I have a similar question but not concerning LEDs. I want to use the 6v lighting from an old Premier Locosound unit. I need to figure out a resistor value to drop track voltage to 6V.

The resistor value is dependent on the current draw of the lights in question.  You need to know how much current your lighting will be drawing, then we can figure how to drop from 18V to 6V for your current load.  Both the resistor value and wattage need to be considered, especially for incandescent lighting.

 

>>>need to know how much current your lighting will be drawing<<<

 

Thanks but how would I know that?

Assuming a typical 20ma, 3.2 volt LED. Using Ohms law 6-3.2/.02=140 ohms. Use a 150 ohm 1/4 watt resistor.

 

Here is a headlamp circuited from track power

 

www.jcstudiosinc.com/BlogShowThread?id=619

 

Dale H

Dale. this question is NOT about LEDs but 6V incandescent lighting

You need the bulb voltage and amperage,same formula.

 

Dale H

The bulb "amperage" is what I'm trying to figure out.

 

Any rule-of-thumb guess or hunch?

 

They're about the same size and brightness as equivalent LEDs that are used now. Typical modern era diesel dual (small side by side) headlights and taillights.

 

Would they use more or less current than an LED? I dunno and I'm close to being "lost at sea" when it comes to electronics and such.

 

 

I would use LEDs as linked a bit simpler to do. Bulbs use much more power.

 

You at least need to know the bulb voltage. If you know that,insert some diodes in series to reduce voltage. Each diode,in series (in the same direction as current flow) on the DC feed of 6 volts would reduce voltage around .7 volts when current is pulled through them. So if you have a 3 volt bulbs 5 diodes in series would leave about 2.5 volts for a 3 volt bulb. 6 diodes in series would be good for a 1.8 volt bulbs. Dropping voltage with diodes is described here. On a DC output only a string in one direction is needed.

 

www.jcstudiosinc.com/BlogShowThread?id=413

 

Again LEDs would be a lot easier. all you need is a 150 ohm resistor to each LED in proper polarity.

 

 

 

Dale H

 

 

I DO know the bulb voltage - it's 6V. I stated that initially.  I need to step down track voltage of 15-18 down to 6. 

Do you have the means to check the current draw?  Power them from 6V and measure the current.  It's not really that easy to tell you what they draw, I have very similar looking bulbs that draw a wide variance in current.

 

 

OK, so once I figure out their current draw, then how do I come up with a value?

 

If source voltage = '18'

and bulb voltage = '6'

and for now, lets say bulb current draw = 'MA'

 

what would the formula look like?

 

 

 

 

Voltage in volts.

Current in amps.  ( assume 100ma for my sample)

Resistance in ohms.

 

Voltage / current = Resistance

 

(18-6) / .1 = 120 ohms

 

(V * V) / R = W

 

12 * 12 / 120 = 1.2 watts dissipation

 

You'd need at least a 2W resistor, and that would be marginal as it'll get pretty warm.

 

You could use a 7805 or 7806 voltage regulator. Here it is for LEDs but you can drive bulbs IF properly heat sinked and the load is less than 1 amp. There are also higher amperage regulators.  The heat shed by the 7806 regulator from 18 volts would be 12 times the amperage drawn in the circuit. 2, 2 watt, 6 volt bulbs would draw about .66 amps. the regulator would have to shed 8 watts. (.66 amps times 12 volts). This is why we like LEDs.. 2 LEDs .02 amps each. 12 times .04 would be about a half watt shed by the same regulator.

 

Dale H 

 

http://www.jcstudiosinc.com/BlogShowThread?id=408

The heat dissipated by the regulator would be the same as for the resistor.  OTOH, you wouldn't have to do any calculations, just wire it up.

 

Thanks everyone!

These are easy to use for LEDs

True Chuck, but he's trying to power incandescent lights from AC.

 

Oh. I read "resistor to run LED lights in my train cars?"

The CL2, bridge or diode would work. Wonder if he would need a capacitor...might not see the pulsing at 60 Hz.

Oh...the thread got hijacked.

Ok...incandescent bulbs are ballast resistors. That is to say that the current doesn't change linearly with voltage. A 6 volt bulb will draw near the same current at 3 volts as it does at 6 volts. The best predictable way to run 6 volt bulbs from from 15-18 vac is to use a bridge and a voltage regulator. Alternatively, you could use a variable rheostat to find a brightness you desire and measure the current at that brightness...around 6 volts...then use a power resistor to replace the rheostat. Messy. And power wasting. Any chance of running three 6 volt bulbs in series?

Yes, the thread got hijacked. 

 

Series is a good idea Chuck, didn't think of that.   Solves the issue of power dissipation of the resistor.

 

Yes. For some years with motorcycle lighting folks were using resistors to dump the required current for fault detecting by on board computers when going to LEDs. I was always thinking, why not light a bulb with it in addition?

Reviving an old thread here...

 

I have a stack of 3v grain of wheat bulbs (2.3 ohms resistance, 1.3 amps draw) that I want to repurpose for headlights, drawing directly from track power (0-18 VAC).  Since I want to wire them individually and KISS without ICs, rectifiers, LEDs or any other such modern marvels, can someone point me to an appropriate Radio Shack resistor that'll do the job? 

 

Thanks!

 

Mitch

1.3 amps!!!

 

To drop 15 volts at 1.3 amps you'd be dissipating 20 watts!

 

What kind of GOW bulb draws 1.3 amps?

Thats as much as the entire engine draws, with the smoke unit on lol

My question would have to be, other than the actual specifications of the bulbs, why not use LED's?  A simple resistor and diode, and you have great looking headlights that far exceed incandescent bulbs for directional light, and they run very cool.  Incandescent bulbs don't even cross my mind when I'm fitting new lighting to locomotives, tenders, or rolling stock like passenger cars or cabooses.

 

Well, both my digital and analog multimeters tell me that I have 2.3 ohms of resistance on the bulb.  The digital multimeter reads 2.3 when on the 200 ohm scale, and the analog scale says .23 at the 10x setting.  Am I reading them wrong?   

 

http://www.ohmslawcalculator.c...s_law_calculator.php

 

This tells me that 3 volts and 2.3 ohms of resistance gives me 1.2 amps of current and 3.6 watts of power.  Again, am I misreading things? 

 

Fair enough.  What does a 3v grain of wheat bulb usually draw, then?  :-)

 

Basically, I'm cheap.  I have the bulbs, I want to use them and I don't want to go buying LEDs.    Say!  An LED resistor would probably do the job, too!  What size do they use? 

 

Thanks, all!

 

Mitch

Remember, an incandescent bulb when it's cold measures MUCH less than when it's operating.  Measuring that amount of resistance when cold means nothing.  A You're talking about dissipating about 4 watts in that bulb, that would be almost what you put into a smoke unit!  The front end would be pretty toasty!

 

Typically, a GOW bulb draws in the range of 50-150ma.  I have them as low as 1.5V 20ma.

 

Okey-doke.  Found a good formula for calculating dropping resistor value here:

 

http://www.wiringfordcc.com/gorhlite.htm

 

So!  18 volts - 3 volts is 15 volts.  Assuming that the GOW I have uses 50 ma (or 0.05 amps),  that means a 300 ohm resistor should do the job! 

 

A quick look around Radio Shack shows this:

 

http://www.radioshack.com/prod...sp?productId=2062319

 

So!  Off to the Shack I wander.  I'll keep everyone posted on the results.  :-)

 

Mitch

 

Remember, assuming that 50ma is correct, the resistor is still going to dissipate 3/4 of a watt.  Given the "rule of thumb" of 2x the power rating, you'll need a 2 watt resistor for that task.  If you're wrong on the current, you'll need even more.  A 3v 50ma GOW bulb will not supply much light, so your headlights will be pretty wimpy.

 

I really think you're being penny wise and pound foolish trying to use these bulbs with a giant resistor.  For about 20 cents you can have a warm white LED, a 1N4003 diode, and a 330 ohm 1/4w resistor.  Not only will you not have any heat issues with the power dissipation of the resistor, but you'll have a vastly superior headlight in the bargain.

 

This'll be my last word on the topic.