Resistor Value for 14 volt GOW Incandescent Light bulbs

Try equal number of bulbs in series, like 2 parallel bulbs in series with 2 parallel bulbs. See how that looks even tho it seems like not enough voltage.

Otherwise you'll have to find out how much current each bulb draws at 16 v

Another approach to lowering AC track voltage are pairs of diodes.  Each pair drops the voltage by about 0.7V.  There have been many threads describing this technique.  More assembly than a single resistor but provides flexibility in finding the right voltage/brightness.  The key being they are dirt cheap...2 cents a piece in this case:

Bridge rectifiers are easier to use for this task, one bridge equals four diodes.  Let's see, $4.10 including shipping, they come from a US seller so you get them quick.  Since each bridge is four diodes, you have less wiring and it's still dirt cheap to solve the problem.

Search eBay item: 300940929957

I have a pack of 12-14 volt GOW bulbs that draw 75 ma. If you wanted to use resistors, here are what the values would need to be, based on dropping 6 V across the resistor in each case and based on 75 ma per bulb:

1 Bulb: 82 ohm, 1/2 watt

2 Bulbs: 39 ohm, 1 watt

4 Bulbs: 22 ohm, 2 watt

6 Bulbs: 15 ohm, 3 watt

The resistor values are chosen based on standard commonly available values. Note the higher wattage resistors will likely need to be wire wound. Note also that various smaller resistor values can be connected in series to provide the values needed.

If you have any way of measuring the actual load of your GOW bulb @ design voltage, it would be better to use that value as it may be different. I am guessing that not all GOW bulbs are created equally.

Rod

I have GOW bulbs that draw 20ma and 40ma, so they certainly aren't all created equal.  I also have some that seem to be about 9V rating, they're way too bright on 12V, but fairly dim on 6V.  I think those draw 50ma.

Of course, for the stated issue, I think one diode may be all you need.  There's nothing to say you need to run them at 100% rated voltage, I frequently use 12V bulbs with a diode on track power.  That's one of my favorite methods of lighting number boards, the bulbs aren't super bright (and hot), and I get plenty of defused light for the number boards.

Rod - thanks for the values.

John - a question - don't know much about diodes

You said one diode would be enough but another post states they drop only 0.7 volts. Are you saying that rectifying the voltage to DC is the major advantage here? I am assuming the 1N4004.

My apologies to all for asking a lot of questions - I want to standardize for all product I will be building. I figure I can attach a diode/resistor to the end of a power lead and let the customer decide whether or not to cut it off depending on power source.

Thanks

Joe

The difference is with a single diode you're only using half the AC waveform, so it basically chops the voltage in half.  If you use the back-to-back diodes, or are running on DC, they only drop about .6 to .7 volts for each diode.

It's cutting the power in half...

Half wave rectifying the sine wave results in half the power going to the resistor (bulb). So since power is voltage squared, Vrms (half wave) = Vrms (full wave)/ root 2. Or say 18 Vrms divided by 1.414 = 12.73 Vrms.

So as far as powering your bulb with a half sine wave, it's like using 12.73 Vrms instead of 18 Vrms

cjack posted:

Half wave rectifying the sine wave results in half the power going to the resistor (bulb).

Well, I think it's "none of the above"...not half the voltage, not half the power.  If the load is a plain resistor, I agree it would be half the power.  But for an incandescent filament bulb, there is a material drop in its resistance when you run it at half-wave rectification...say a 25% decrease in resistance varies with bulb.  The filament is running much cooler so its resistance decreases and I suggest you get significantly more than half the power.

I think the OP should just try a diode and see if it's useful as a 2-cent solution.  But if not, then to implement the idea of allowing the end-user to simply clip-off one component, I think he needs to stock a set of different resistor values.  Though I hasten to point out that if in clipping off that one component means applying full 18V track voltage to the 14V GOW bulbs, I'd look into taking out some life-insurance policies!

Excellent - I will standardize on the diode and use 14 volt GOW's.

Thanks to everyone!

Joe

...dang non linear bulbs...kind of why I said resistor to get the right numbers across. Bulbs are ballast resistors, constant current for a lot of their range.

But these are 14 V bulbs ?, so they should be fairly happy with 12.73 Vrms...happier than with 18 Vrms, eh?

Well, not as non-linear as a semiconductor junction (LED)!    If Joe wants a single component selector scheme, I think it needs to be a resistor using the calculation method you provided.

I find that 12-14V bulbs do very nicely on 18V track power with a diode.  They're decidedly not as bright as powering them with 12VDC.  Remember, there is that diode drop as well.  Here's two identical bulbs, the one on the right has 18VAC through a diode and the one on the left is powered with 12VDC.  Both values measured using a Fluke true-RMS bench meter.  The DC powered one is noticeably brighter.

Longer lasting too maybe.

I'm sure they'll last longer if you don't run them 100%.