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Old 05-18-2010, 10:35 PM   #1
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Re: 12V DC to 9V DC?

Go down to your local auto parts store and ask for a Standard RU37 (or RU37T in Tru Tech...cheaper). It's a ballast resistor for Dodge ignitions and should cost you less than $10 while having the capability of supporting the current draw of multiple fixtures. Feed 12 volts in one side and run parallel branches (individual wires to each fixture) off the other side. Done deal.

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Old 05-19-2010, 02:40 PM   #2
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Re: 12V DC to 9V DC?

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Originally Posted by DarrenDriven
Now we're talking! I called my nearest store and they have two in stock at their warehouse, one of them will be waiting for me at the store by noon. SWEET!

OK, so I also bought some LED puck lights that are intended to run on three AAA batteries each. So anyone have a cheap and easy solution to convert 12V to 4.5V? I can buy a 4.5V cell phone car charger on eBay for about $5.00 and hack it up, but if someone had an idea for a local product in that same price range then I would rather do that since I hate to wait for stuff to ship.
Just wire three puck lights in series for a total voltage of 13.5, which is really what your fully charged house batteries put out anyway.
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Old 05-19-2010, 04:50 PM   #3
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Re: 12V DC to 9V DC?

I am watching this from the edge of my seat. I might(more like will) learn something. I always need to learn more 'bout 'lectricity.
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Old 05-19-2010, 07:50 PM   #4
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Re: 12V DC to 9V DC?

1. LEDs need to be current regulated, not voltage regulated. If the fixtures have internal current regulation based on the 9 volts or 4.5 volts, then a regulated lower voltage derived from the 12 volts (actually 13-15 volts) will work. A resistor is a cheap way of regulating current when the voltage is stable, but it must be calculated for the supply voltage and the draw of the load. If the load varies, like an audio amp, or if the voltage varies, like a battery being charged and discharged, then a simple resistor creates problems when used to create a voltage drop.

If the fixtures are cheap, and the designers expect to protect the LEDs by the internal resistance of the batteries, then providing a constant 9 or 4.5 volts may provide too much power for them to survive.

2. Incandescent bulbs of the same wattage can be daisy-chained in series to run on a higher voltage, but this may cook the puck lights unless they have internal current regulation. Depending on the design, if the voltage drops to 12.5 when running on battery, they could go out instead of getting dim like a bulb. LEDs are diodes, diodes are electrical one-way valves, and they require a certain amount of pressure (voltage) to hold them "open" (on). This is what you experienced on the "right" side of the bus.

In a series circuit , the available voltage is divided by the inverse of the resistances. If you used mis-matched incandescent bulbs, the dimmer ones would get over-voltage, and the brightest ones would get under-voltage. Your resistor and 5-watt bulb are mis-matched. The resistance of the RU37 is so small compared to the resistance of the 5-watt bulb that the bulb uses most of the power, not the resistor. You would probably see the voltage drop down if you tested with a halogen headlight instead of a 5-watt bulb.

If I had two 12-watt, six-volt bulbs, I could series-wire them and use the pair on 12 volts. Each bulb draws 12w/6v or two amps, because the resistance of each one is 6v/2a or 3 ohms. In series, they add to 6 ohms at 12 volts, draw 2 amps through the pair, and each gets 12v/2a or six volts. Each bulb is still 12 watts at 6 volts, and the battery supplies 12v x 2a or 24 watts of power.

If I had a 6-watt bulb and a 36-watt bulb, and series-wired them together, it would be different. The 6 watt bulb normally draws 6w/6v, or 1 amp. Its resistance is 6v/1a, or 6 ohms. The 36-watt bulb draws 36w/6v, or 6 amps. The resistance is 6v/6a, or 1 ohm. In series, they add to 7 ohms at 12 volts, and draw 12v/7a or roughly 1.7 amps. Ignoring the fact that resistance in incandescent bulbs changes as they heat up, the 6-watt bulb with the higher resistance gets 6/7 of the shared voltage, or about 10.3 volts. The 6-watt bulb shines at 17.7 watts for as long as it lasts. The 36-watt bulb with the lower resistance gets 1/7 of the shared voltage, or about 1.7 volts. It glows at 2.9 watts until the 6-watt bulb burns out, and then goes out because the circuit is open.

Getting back to your problem, providing a stable 9.0-volt regulated step-down source, and then from that supply powering either pucks in series pairs (going on and off together) and/or the Harbor Freight LEDs will probably work for you, if the fixtures have either an internal regulator, or a resistor calculated to never give full brightness.
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Old 05-19-2010, 07:53 PM   #5
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Re: 12V DC to 9V DC?

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Originally Posted by DarrenDriven
OK, so if I can find one source of 4.5V then I can wire all of these puck lights in parallel.

It seems that iPod car chargers are 5V. They are all over eBay for VERY cheap. Do you think that 5V will harm something built for 4.5V?
If the pucks have active current regulation, no. If they use series resistors, the chargers might cook the LEDs, depending on how "close to the redline" the LEDs are run on battery.
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Old 05-19-2010, 08:39 PM   #6
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Re: 12V DC to 9V DC?

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You just melted my brain. Hehe
Me too.
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Old 05-19-2010, 10:28 PM   #7
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Re: 12V DC to 9V DC?

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Originally Posted by DarrenDriven
I hooked up all three pucks in parallel to three AAA batteries and each light I turned on caused the other light(s) to dim. So, I guess what I need to figure out is how many watts were being drawn and then find a 4.5V DC converter that supplies around the same amperage, right?
I tell you what - connect ONE puck to three AAA batteries through your meter, and tell me how many milliamps the puck draws, (on bright if it has multiple levels) and I can give you a simpler answer.
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Old 05-19-2010, 11:34 PM   #8
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Re: 12V DC to 9V DC?

Quote:
Originally Posted by DarrenDriven
So I picked up the RU37 and got it home. I hooked up one side to my DC power supply and the other side to my multimeter. I'm still showing 13.8V. I put a small 5w light bulb in line and got the same results. Is there something I am missing? I don't want to hook it up to my LED lights because I am afraid it will just blow it again.

***edit* I should also mention you have your meter in series in that picture for checking voltage! Bad juju! I'm surpised you didn't let the smoke out of the meter. Hook the circuit up so you go power supply + to ballast resistor to light bulb and back to power supply negative. Then take your meter and touch one wire to each of the blades on the resistor. This puts you in parallel which is a much happier way to check voltage drops.


You're missing a ground! In order for there to be current there must be a drop in voltage and vice versa. The resistor isn't going to provide a load on the system until there is a complete path for power like back to the ground on the power supply.

I threw up a quick drawing in Visio to show you what I mean. Think of the first picture as your setup with the power supply. Without a ground your switch is essentially open so when you stick a meter at the point shown you get 12 volts. Hook up that ground and you get 0 volts.



But you are supposed to have about 9 volts, right? You won't see anything like that until it is hooked up in the working circuit going through the fixtures. 6 AA batteries at a nominal voltage of 1.5 gives you 9 volts and you have a realistic range of 1.6-1.4 volts or so between brand new batteries and some pretty tired units giving you an actual range of 8.4-9.6 volts...and to add even MORE fuel to the fire, AA batteries under load really are only capable of providing about 1.1-1.3 under load giving you 6.6-7.8 volts. A ballast resistor usually gives you about 8 volts or so so I think you should be fine. There are some finer points to LED's, but I *think* that should do it for you. At least it was a cheap effort!
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Old 05-20-2010, 08:36 AM   #9
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Re: 12V DC to 9V DC?

1. Get your 5.0 volt cell charger, and wire the pucks in parallel. Put a 5-ohm resistor in each puck between the +5.0 volt supply and the +4.5 volt battery terminal. Connect the minuses directly to the 5.0 volt ground. You are good to go!

Each puck/resistor will still draw 100 mA, so the number of pucks you can power depends on the charger you get. I would stick to half capacity for long-term use, i.e: 4 pucks on an 800 mA supply, 5 pucks on a 1000 mA (or 1.0 A) supply.

Explanation: you have to "throw away" 0.5 volts of the 5.0 volts to provide the 4.5 volts the pucks like. The 5-ohm resistor is selected to do this at 100 mA. It will turn 1/20 watt into heat, and the pucks will use 0.45 watts. With a resistor, each puck will consume 1/2 watt. (The cell phone charger will probably burn up another 0.7 watt per puck or slightly more of the 12 volts, as the draw for each one must be at least 100 mA.)

The math: Since you need to lose 0.5 volts (E), and provide up to 0.1 Amps (I), and since resistance (R) is E/I, then 0.5/0.1 = 5. To be sure the resistor can handle the power (P), and P=E x I, I checked 0.5 x 0.1 and got 0.05 watt. If this were a headlight bulb, you would need a heavier resistor like the RU37 to handle the heat dissipation. A small 1/2 watt or even 1/4 watt resistor will work fine on the pucks.


2. I re-checked the photo after TE_03 had his concerns. If you had no ground, I don't think you would have a reading. But it looks like you were still measuring the power supply output, not the resistor output. In your drawing, remove the ground from the middle. As you have it drawn, the RU37 would get the full 12 volts of the supply, and the would LED get zero volts from ground to ground.

These are the measurements you could make, with the ground at 9v? removed:
Measure from +12 on the left to the ground on the right to find the supply voltage.
Measure from 9v? in the middle to the ground on the right to find the voltage used by the LED.
Measure from +12 on the left to 9v? in the middle to find out how much voltage the RU37 is "throwing away" as heat.
Disconnect one point at either +12 volts, ground, or 9v? and put the meter into the circuit break in series to measure the current (but you knew that.)

The amount of volts the resistor "throws away" depends on the current draw of the load. That's why a corroded battery cable with one-half ohm of resistance won't sweat powering a 2-amp lamp, but might catch fire trying to supply 220 amps to a 2400-watt inverter. In a design like the pucks, the resistance must be matched to the current draw and voltage dropped.
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Old 05-20-2010, 04:34 PM   #10
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Re: 12V DC to 9V DC?

Quote:
Originally Posted by Redbear
2. I re-checked the photo after TE_03 had his concerns. If you had no ground, I don't think you would have a reading. But it looks like you were still measuring the power supply output, not the resistor output. In your drawing, remove the ground from the middle. As you have it drawn, the RU37 would get the full 12 volts of the supply, and the would LED get zero volts from ground to ground.
That's what I meant by no ground. The resistor is never actually being grounded so there is no current flow...which means he's measuring supply voltage.


Quote:
The amount of volts the resistor "throws away" depends on the current draw of the load. That's why a corroded battery cable with one-half ohm of resistance won't sweat powering a 2-amp lamp, but might catch fire trying to supply 220 amps to a 2400-watt inverter. In a design like the pucks, the resistance must be matched to the current draw and voltage dropped.
This is very true. It is my sincere hope that the ballast resistor is the correct size for the job as I *think* his total draw should be about the same as the factory ignition system on those trucks. One of the unique features of those ballast resistors is that there resistance is somewhat temperature related meaning they kind of self regulate their resistance based on how hot they are which is directly proportionate to the current flow in the system.
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