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So, i'm finally getting around to finishing up my pond scum light. I've built a voltage divider. I was wanting to run the light at either full power (14.4v) or 1/2 (7.2v). Everything is wired right, switch works and output voltage is good but I can't get the lamp (mr16 20w 12v) to fire at 7.2v. Am I missing something here? My divider is a simple circuit.



R1 and R2 are both 1000 ohm 1/2 watt.
 

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Hi Nocturnus so if I understand correctly you want to dim your lights for some reason?
The reason your voltage divider does not light up your 12 volt light is that you are trying to pull a lot of current through a 1000ohm resistor.
Using Ohms law V=IR and also VxI=W we have 20/12 which is 1.67Amps and to get this through a 1000ohm resitor there will be little or no voltage to light the bulb.
The simplest solution would be to just put a resistor in series to limit the current and since the resistor will just turn the dropped voltage into heat not quite sure I would bother at all as it really is a waste of battery power.
Best yet would be a simple voltage regulator something like a 7805 or 7812 and then it is total variable voltage with a stable current and not too much wasted as heat way more efficient.
 

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For normal operation P=VI .... 20w = 12v x 1.7A
V = IR so 12v= 1.7A x 7.2ohms... your bulb is about 7.2 ohms. Put that into your circuit across R2 and you see most of the voltage is across R1.
Redo your voltage divider with the bulb as R2.... R1 is 7.2 ohms. I dont think a 12v bulb will glow at 7V though.... mine seemed to get very dim around 10. Also the bulb resistance changes with current so it really a case of experimenting with a few small resistors(actually big resistors with small ohm values). Cutting your bulb V down to 12V from 14.4 should see an increase in runtime.
 

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Following znomit's suggestion will give you 7.2V across the bulb [as you desire] , but why are you trying to dim the bulb? I assume your goal is a longer run time?

Well, consider the bulb powered directly by a 14.4V 1Ah battery. The bulb consumes 28.8W (14.4V*2A), so it will last 14.4*1 Wh/28.8W=.5 hour.

Setting up the voltage divider up as znomit suggests will push your 14.4V through equivalent resistance of 14.4ohms, assuming bulb resistance is the same at 7.2v. It won't be, but it's good enough for approximation. Now you're consuming a total of 14.4W, so your run time is now 1 hour, but your bulb power is only 7.2W.
So you double run time, but reduce bulb electrical power by a factor of 3.Light output power will probably be far below 1/3 of nominal output. IF the bulb will even light. IMO a voltage divider is not the best way to increase run time for a halogen setup.

You'd probably get better results with a linear regulator IMHO. Something like an LM317 but with a higher current rating for your application. If it were me I'd breadboard out a few different resistor options. 10w might be a good place to start.
 

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Best solution.... dual lighthead with a 10w bulb you can switch to.
Even more besterer solution, cheap led torch from dealextreme for the times you don't need the 20w halogen.
 

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+1 that didn't occur to me.

Dual bulb you'd be wasting the least amount of power, and you'd have the option of running both lamps at once for technical sections.

:)
 

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Or dual power packs, one 14.4V and one 12V with a switch between them to the same lamp wires.
 

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Nocturnus said:
So, i'm finally getting around to finishing up my pond scum light. I've built a voltage divider. I was wanting to run the light at either full power (14.4v) or 1/2 (7.2v). Everything is wired right, switch works and output voltage is good but I can't get the lamp (mr16 20w 12v) to fire at 7.2v. Am I missing something here? My divider is a simple circuit.



R1 and R2 are both 1000 ohm 1/2 watt.
As others have mentioned it won't work because a voltage divider is only used in circuits where you want a reference voltage, not to power a load but for voltage measurement purposes.

Instead of quoting several people separately I will comment on a few things in one post.

A) You can't directly use Ohm's law to determine the result of adding a resistor because there is a change in the resistance of the bulb based on the filament temperature, by more than the value of the resistor you'd ultimately need to add. This increase in resistance of the filament is basically the only reason the filament doesn't melt.

B) A single resistor (or paralleled resistors to achieve the precise value you might need) would work to dim it but not save much battery power since the light dims so quickly with a minor voltage drop, but there is nothing to gain by using a regulator except a finer granularity of control, electrically speaking it is no less lossey, no more efficient nor better operation than a resistive drop to the same target voltage.

C) As others have mentioned the bulb is optimized for a certain voltage and it's brightness will decrease much more than the decrease in current, to the point where it's hardly worthwhile to save battery life, but to reduce light so you don't blind someone yet not be in total darkness would make more sense.

D) Although the regulator or resistor are very lossey, neither will just turn the dropped volts into heat. Either way, you have less total current flowing from the battery. Given the reactivity of the filament to the current as it heats up, the resistor might actually be more efficient but in a trivial, not practically significant way.

E) TO keep the bulb bright enough to be usable you wouldn't want to drop more than about 2V, so if the light were drawing 2A and if we ignored the filament change (which we can't, but we don't have the data to determine this except finding that out experimentally), you'd then need 4W of power handling from the resistor so adding a bit of margin at least a 7W resistor.

F) I don't know your light housing, if it could take a dual bulb. The other option would be a bulb with dual elements, like you find in some automobile headlights. If you really really want to stick with halogen this might be the best way, although typical automobile headlights draw a lot more current, so I'm wondering about bulbs a motorcycle or scooter might use (I don't know, just a random speculation).

G) LED is definitely the final solution to saving battery power, the best one perhaps. There are however other options. A step down switching regulator plus a DPDT switch would allow you to more efficiently convert the higher voltage of the battery pack to a lower one without so much loss as through a regulator or resistor.

TI (Texas Instruments) among others have made little drop in modules but I don't remember if those have the specs you need like 2A current, often they were meant as drop in replacements for an LM317 or LM780(n). They aren't really anything more than several small surface mount parts on a PCB with pin legs and a laser trimmed resistor to get really close to the target voltage, otherwise their design isn't so unique - closer than a precise voltage than you really need to get which means using some switching controller and that part's datasheet example circuit should be good enough for the purpose if you had to build something from scratch to meet your exact needs. It still won't go very far towards saving battery power.

I guess I don't really have a main point other than I'd prefer an LED light if it were built from the ground up, but how much loss you can accept to save time and money with an existing light I can't decide for you. If you don't need very much light you might even be able to put 4 led in series mounted in the existing light in series with a small value (for minimal power loss) inline resistor to limit current a little) and a dual throw switch to choose between the original bulb or those.

If you stuck with 1/2W LED or less you wouldn't need to bother with heatsinking so their inclusion in an existing light could be rather compact, basically only holes in the reflector and after aiming them, epoxy the back of them so they stay aimed correctly in same direction as the bulb center does.

One last thought would be that I wonder if anyone now makes LED automobile headlight bulbs that are drop-ins for the halogens, for legacy electrical systems on cars not originally equipped with LED headlights, meaning they are designed to run off the automobile electrical system which is already expected to peak at 13.8V or higher, via LED dies in series or a built-in regulator board such that you need do nothing to your battery pack, that it would accept the (was it 14.4V?) existing pack and run from that, meaning you might be able to reuse much if not all of your existing setup just needing to modify if the bulb size or base needs a different socket.
 

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With halogen (or other incandescent) bulbs, the power consumption goes up or down with every percent of voltage change (from the nominal design voltage of the bulb). But while the lumens move up or down exponentially (to the power of 3.6) the consumption is pretty damn linear at 0.55 of the voltage change. So 10% more voltage (or less) results in 5.5% more (or less) consumption. The lumens however change by either a 40.8% increase at 110% of nominal voltage to a 31.6% decrease at 90% of nominal voltage.

In other words, while over-volting sees big gains in efficiency of light output vs power consumption, under volting does not. In such situations, a seperate lower wattage bulb is more practical. To put this in perspective... a 12V-20W bulb normally draws about 1.66Ah, but at 13.2V would now be drawing 1.75Ah but giving the equivalent light of a 12V-28.16W bulb (if such a bulb existed). In the other direction, running 10.8V to the same bulb only reduces the current draw to 1.57Ah but the light drops to the equivalent of a 12V-13.69W bulb (again if such a thing existed). A 12V-14W bulb would only draw 1.17Ah in comparison.
 
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