Voltage dropping via resistor

I need to drop the voltage for a meter light, from +30dc to +12dc, I would prefer to do this simply by using a resistor. Can somebody tell me how to calculate it or maybe even what size I should use?

Ohm's Law. Find out how much current the light draws. That's I. You're dropping 18 volts (30 - 12) across the resistor. That's V.

R = V / I

To figure out how much power the resistor has to dissipate,

P = V x I

In both equations, V is in volts, I is in amperes, R is in ohms and P is in watts. Oversize the resistor by at least a factor of two: if it will be dropping 0.8 watts, use at least a 1.6 watt resistor -- in practice, 2 watt.

Peace,
Paul

i know this is ohms law 101, but still, for the life of me, i have the most rediculously hard time figuring out how exactly i figure out how much current a light bulb draws.

current meter in series with a dc source of the correct voltage for the bulb?

i guess i've never had the correct voltage for the bulb *and* needed a dropping resistor at the same time.

and i suppose this is where a handy bench psu comes in handy.

Do you know the voltage and wattage for the bulb?

I = P (watts) / V

So if it's a 12 volt bulb, 1 watt, then I = 1 / 12 = 0.0833 (amps).

Ohm's law is the easiest and most useful math you could learn for DIY stuff.
http://www.sengpielaudio.com/calculator-ohm.htm

[quote author="enthalpystudios"]i know this is ohms law 101, but still, for the life of me, i have the most rediculously hard time figuring out how exactly i figure out how much current a light bulb draws.

current meter in series with a dc source of the correct voltage for the bulb?

i guess i've never had the correct voltage for the bulb *and* needed a dropping resistor at the same time.

and i suppose this is where a handy bench psu comes in handy.[/quote]

In general lamps are intended to be driven from a voltage source and will typically be rated for a working voltage and nominal power at that voltage. Their cold and hot impedance varies quite a bit so it isn't trivial to determine empirically.

I would approach this assuming that if we size the resistor to deliver the current it needs to make rated power at it's nominal voltage, it will behave and stabilize near there. By all means calculate, then breadboard this up and make adjustments from there.

If you only know the working voltage but not target current draw, the simplest approach is to just use trial and error. Start with a larger resistance than you'll need and slowly come down or parallel with others until you get to your desired working voltage. Brightness and lamp life is not linear with current or voltage so for longer life run it lower than 12V if that gives you acceptable light. Also if the 30V is unregulated anticipate high line and low line effects.

When you don't get what you expected that's called experience.

JR

The other folks have suggested good ways to figure out the resistor needed. Here's another: find the current draw by finding out what model number the bulb is, and looking it up in a catalog. Most meter-illumination bulbs are pretty standard deveices, and will show up in standard catalogs.

Peace,
Paul

John, that is good information for the thread. I have designed motor/lighting controllers for commercial robotics and what you say certainly corresponds to my findings. Some of the more interesting findings with PWM drive is the ability to pulse with a higher voltage but with shorter durations to achieve a higher perceived brightness. If you get the ratio correct then you can keep the filament heated during the low periods without the pumping effect.

lots of coolness.

:thumb:

if you didn't know the current and it wasn't for a lightbulb and you needed to drop the DC voltage would the best way be a voltage divider?

-richie

[quote author="capnspoony"]if you didn't know the current and it wasn't for a lightbulb and you needed to drop the DC voltage would the best way be a voltage divider?[/quote]

A voltage divider is only really good if the load resistance is much higher than the resistances in the divider - otherwise the load becomes a significant part of the divider circuit.

[quote author="Scodiddly"][quote author="capnspoony"]if you didn't know the current and it wasn't for a lightbulb and you needed to drop the DC voltage would the best way be a voltage divider?[/quote]

A voltage divider is only really good if the load resistance is much higher than the resistances in the divider - otherwise the load becomes a significant part of the divider circuit.[/quote]
and if you want there is a simple way of using a buffer after a resistor divider.
just use a bjt emittor follower count the 0.7v b-e drop at the calculation of your resistor divider and there you have it
you also can add a small cap at the base.
i recently buile a very simple voltage 300v supply with tip51 ( or tip50? can also be)

you only have to see that your supply has no ripple etc...that its clean
just divide with resistor and hopla
if you have a ripple you can use a zener/resistor combination and the ripple is filterd as long as it doesnt fall under the required output voltage.

just for info. it's really basis and simple but most of the time it does the job

now in this case just calculate the current needed.
if you want to do it with a bjt then you also have to know that you dissipate some heat and in this case it would be waste

[quote author="Svart"] Some of the more interesting findings with PWM drive is the ability to pulse with a higher voltage but with shorter durations to achieve a higher perceived brightness. If you get the ratio correct then you can keep the filament heated during the low periods without the pumping effect.

lots of coolness.

:thumb:[/quote]

Perceived brightness of bulbs is proportional to temperature of the filament, temperature of the filament depends on power, so for the same perceived brightness you need to dissipate the same power, regardless is it pulsed or constant.


Also, with each heating/cooling cycle mechanical tensions occure in the filament that shorten it's life...

It is Ok to dropp excess of voltage by resistor, but you should remember that power dissipation on it will be sufficient (say, if a 12V bulb dissipates 1W, there will be about 2W on the resistor).

Good side is, less inrush current on the bulb when switched on will prolonge its life.

Also, with each heating/cooling cycle mechanical tensions occure in the filament that shorten it's life

Click to expand...

True but you are talking a few %, nothing signifigant. You will get more percentages in manufacturing error that limit the lifespan of a filament than you will create by overdriving the filament, within reason of course.

Also, filaments are not instantaneously heated and cooled, there is a clear ramping period which can be augmented by the frequency and duty cycle. Find the rampdown period of the filament and bring your frequency to the point where the falling edge-rising edge period is roughly equal to the total rampdown period. adjust the duty to bring the ramp's lowest point back up. Add a dash of flywheel diodes and debug to taste.

What you are doing is pulsing current only when needed to keep the filament hot. It's a balancing act but it does work and can conserve power. It's really just a trick and if the pulse frequency is too low you can see the strobe effect in the filament.

Perceived brightness of bulbs is proportional to temperature of the filament, temperature of the filament depends on power, so for the same perceived brightness you need to dissipate the same power, regardless is it pulsed or constant

Click to expand...

Only if you are comparing the same type of filaments to each other. If you compare a halogen to a non halogen incandescent then all bets are off. Halogen gas concentrations, percentages of each gas, impurities in the filaments, impurities in the gas can have a measurable effect on the brightness while still consuming(wasting) the same amount of power.

What happens when you compare apples to oranges? How about a case study:

You mention perceived brightness, lumens (measured in candela), which has nothing to do with the actual power output but are generally used to quantify each other. I used 5w LEDs which have 0% IR output, 75% blue and around 24% yellow (1%other colors). The perceived brightness(with a lux meter) was measured to be nearly identical to that of a 25w halogen bulb while consuming 1/4th the current @1/2 the voltage. The LED was able to illuminate an object 20ft away with a 6deg collimator. On the other hand, the 25w halogen put out a whopping 65% IR, 20% reds, 10% yellows and the rest was a peaky mess all over the spectrum. That means that 16.25 watts of power was wasted as IR.

But that is only half the story. The human eye is MUCH more sensitive to blues than reds. This also gives humans a false sense of brightness, where a 25w halogen causes mild discomfort when pointed into the eye while a 5w LED of the same measured brightness causes actual pain in the eye of the engineer who wasn't quick enough to leave the room before being chosen as a guinea pig.

Guinea pigs were not harmed during the testing process. Rogue engineers are another matter..

:green:

LEDs are not bulbs. Bulbs to be brighter have to be heated to higher temperature, it means more power dissipation, regardless of form of powering current.

If you read carefully I never said that you can cheat Mr. Ohm. You gain a higher perceived brightness because the voltage pulse is higher but the current needed when the pulse goes high also pulses, this creates little current spikes which account for the current that seems to be "lost" during the filament cool down periods. It's all there and Mr. Ohm is happy but our goal is to gain a higher perceived brightness without signifigantly shortening the life of a filament. Constant voltage application at a higher level than the bulb is spec'd at will dramatically reduce the lifespan of the bulb for a small amount of increased brightness. With PWM, our voltage pulses are much higher than the specified bulb voltage but the average voltage (duty % vs. voltage) level is equal to that of the max the bulb can handle. As long as we keep the filament above a certain temperature our current draw is actually fairly low as well.

Again, we apply a pulse to "heat" the filament (actually the far infrared heat is a lower order byproduct of our goal to vibrate the filament fast enough to generate photons which themselves do not signifigantly heat an object) and we give the filament a low period during which it is not drawing current but is still vibrating and giving off light and heat. Before it reaches a level that we determine is too dim, we apply voltage/current to "heat" it up again. If we let it "cool" down too much we start to see the filament strobe and it's rather annoying. The fix is to move the pulse frequency up to a much higher one so that we can lower the duty cycle and still not see a stobing effect yet retain our perceived brightness. maybe that was a better explaination?

It's just an offshoot of SMPS design really.

[quote author="Svart"]

It's just an offshoot of SMPS design really.[/quote]

Armchair-theoretical one.

I'm currently working on some PWM drive circuitry for ultra-bright LEDs, and I have heard about this higher peak voltage for increased brightness stuff, but things seem to end up a wash when we use the luxmeter in the lab... As far as the percieved effect, we havent done too much work.

As for bulbs I cant say much...

-Mike

PM and I'll give you some of my findings. I used luxeon V lambs.

so is it a G1176 to a sifam meter? just curious ran across this problem two years ago....

Svart;

I have some development for your theory.

PWM creates a potential barrier in bulb's filament, it's height depends on a difference between current and no current. Electrons recombinating generate photons of frequency depending on height of the barrier. However, in real life photons have low frequency so they are invisible. Why? Because of a parasite thermal current!!!
Now, if to put a straight long wire into a transparent flexible tube and keep temperature in it close to -273C we can use such light sources to eluminate highways, streets, squares... Great perspectives!

So you are saying that if we supercool the filaments they should become superconducting and emit light?