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Powering my 40 watt light bulb


rbewley

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...power your 24V bulb with 18V battery? ...buy bigger battery, or get a bulb for lower voltage...low voltage ultra-bright bulbs

 

HAVE:

18V battery

18V light with housing (for front of bike)

 

NEED:

low amp/low V rear light with housing (that'll talk to the 18V battery)

 

12V bulbs fry out on 18V battery. ??What sort of bulb(with housing) do you suggest?? Add some sort of DC voltage regulator?

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Simplest way out would be to use your 12V bulbs with the 18V battery.

If you caonnect them as it is they will fry out no doubt about that, connect in between the battery and the bulb a potentiometer, adjust its value till you get a 12V output voltage across your bulb. You can get small potentiometers very easily, and then adjust them by tightening a nut. If you do this, you would have in essence attached a regulator in your circuit, if you increase resistance of potentiometer even further, then voltage will drop and bulb will glow less, if you reduce resistance, glow will increase but beyond a point (12v) your bulb life will go down rapidly. Use a multimeter to observe voltage while you set resistance and you should be done.

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to work out what potentiometer to get, use the equation V = IR, re-arrange that to

R = V/I

 

R = resistance [ohms]

V = voltage [volts]

I = current [amps]

 

the only thing is that you will need to know the current for that equation, which you will have to measure.

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to work out what potentiometer to get, use the equation V = IR, re-arrange that to

R = V/I

 

R = resistance [ohms]

V = voltage [volts]

I = current [amps]

 

the only thing is that you will need to know the current for that equation, which you will have to measure

 

You do all those calculations only if you want to use a resistance in the circuit, with a potentiometer, you just set up a voltage measuring device across your bulb and adjust your potentiometer till the voltage comes to 18V

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can you use what, the potentiometer? yes, a potentiometer is a type of variable resistor, you can attach it like you would attach a normal resistor, before the component, in this case LEDs in series.

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When you get your potentiometer, be careful in the way you attach it. It would typically have 3 leads, 1 is common, 1 is adjustable the third is set to maximum resistance. Be sure not to use the lead which is already set to maximum, otherwise you'll defeat the purpose of the potentiometer.

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Easiest way to do this is just use an ohmmeter when the pot is set to about mid-level. Often the poles aren't marked.

 

Can't agree with you more.

I had to use a potentiometer in a circuit, took me some time to procure its data sheets from the net and figure out the leads and their purposes.

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yes, thats the whole point of a resistor [a pot = potentiometer = a kind of variable resistor] resistors reduce the "juice" so to say...

 

yt, how did you come to your 5.4 watts, he said "and say .3amps" i think he was estimating of making it up :)

V=IR

theres: 12V "and say .3amps" for that we would need 40ohms resistance. thats not enough, i dont think to reduce 18V --> 12V, so i think he is making the .3amps figure up ;)

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yt' date=' how did you come to your 5.4 watts, he said "and say .3amps" i think he was estimating of making it up :)

V=IR

[/quote']

that`s fine for Ohms law, I`m using the POWER EQUASION! P=VI

 

and in this case 18 X .3 = 5.4 ;p

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no no, misunderstanding !

 

i was proving that the .3amps figure was innacurate and unrealistic

 

i always remembered power equation as PIV or P=IV, it just seem natural PIV like a pivot or sumin!

 

:D

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Battery is 18V' date='

The LED array only needs 12V and say .3amps

[/quote']

he wasn`t saying it was a literal value, it`s Hypothetical, as an example for the question he asked. it doesn`t have to be realistic :)

 

"and say" could just as easily be replaced with "and maybe".

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yes sure, but it was still an innacurate figure which additionally was unneeded which resulted in your 5.4Watts figure being 'theoretically' correct, but as the figures you used in the equation werent real nor realistic the 5.4 figure was also not real or realistic ;)

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why not?

he didn`t say HOW MANY leds were in the array, he just said "Array" it`s perfectly realistic!

 

I have an led array that draws 150ma at 240 volts here, what would happen if I wired 10 of them up in parallel?

 

240 volts at 1.5 amps!!! it SOUNDS alot, but it would be "unrealistic"!

 

I see no problem at all with 12v @ .3 amps at all? it can be done in 8 leds!

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yeah, spose, good point!

 

you win, moving on!

 

i was looking at buying very bright LEDs, is there any difference between ultra-bright, high-brightness and super-bright LEDs? which is brightess?

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you must think im really dumb, it dont say the candle-power, nor anything which i can make sense of, but i have now found hyper-bright! only cost 79p [uk money]

 

Light output typical at

IF=20mA 3.5cd or 1.5cd

 

i'd get the brighter one, but what is 'cd' a measurement of, i've never heard of it before [is that not something i should say!?]

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ROFLOL :))

 

you must think im really dumb' date=' it dont say the candle-power, nor anything which i can make sense of,

 

Light output typical at

IF=20mA 3.5cd or 1.5cd

 

i'd get the brighter one, but what is 'cd' a measurement of, i've never heard of it before [is that not something i should say!?']

cd means CanDle :)))

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ROFLOL indeed, really ROFLOL or more collapse in disbelief!

 

so 3.5 candle power isnt that bright is it? or is it? because you can get torches of a few thousand/million CD power, or will this seem bright, as it is from a small LED?

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3.5 cd will be more than the 1.5 obviosly. and also VERY BRIGHT! aviod dirrect eye contact with the beam!

 

the torches are different again, most are Halogen based, and consume LOTS of power. nothing at all to do with LEDs though, don`t confuse the 2

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