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Old 10-24-07, 05:40 PM   #1
ElJamoquio
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Voltage Regulator

I'm powering a 6V, 10W light that I already have with some 7.2V (nominal... actual 8.4 or so) battery packs I already have.

So... what are the super-efficient (hopefully super-small, too) voltage regulators I should use?

...and by 'nearly loss-less', do the manufacturers mean loss-less on just an amperage basis, or loss-less on a power basis?

Thanks!
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Old 10-25-07, 07:33 AM   #2
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Power. A nearly lossless regulator will draw as little power for itself as possible. This could mean that it has a very low voltage dropout (ie. the voltage it takes for itself), the current it draws, or a bit of both.
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Old 10-25-07, 09:03 AM   #3
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Power. A nearly lossless regulator will draw as little power for itself as possible. This could mean that it has a very low voltage dropout (ie. the voltage it takes for itself), the current it draws, or a bit of both.
Right. Actually it matters little whether it's referring to voltage, amperage or power(watts) since power(watts) is the product of voltage times amperage. If loss is measured in any of the three it's all relative.
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Old 10-25-07, 09:39 AM   #4
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Originally Posted by ElJamoquio View Post
I'm powering a 6V, 10W light that I already have with some 7.2V (nominal... actual 8.4 or so) battery packs I already have.

So... what are the super-efficient (hopefully super-small, too) voltage regulators I should use?

...and by 'nearly loss-less', do the manufacturers mean loss-less on just an amperage basis, or loss-less on a power basis?

Thanks!
In general losses and efficiency are computed on a power basis although a creative marketing type might do otherwise.

For maximum efficiency (90% or better) you need to get a switching type regulator. A disadvantage of this type is the production of electrical interference but this generally won't be a problem unless you listen to AM radio while cycling.

Thee simplest type of regulator is a linear one. Its efficiency is generally lower than the switching type. If you are regulating 8.4 down to 6 the efficiency will be about (6/8.4) X 100 = 71.4%. The good thing is that no interference is generated.
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Old 10-25-07, 02:28 PM   #5
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Originally Posted by ElJamoquio View Post
I'm powering a 6V, 10W light that I already have with some 7.2V (nominal... actual 8.4 or so) battery packs I already have. So... what are the super-efficient (hopefully super-small, too) voltage regulators I should use?
If thats a mr11 lamp, you don't need a regulator to power it with the battery. Many incandescent bulbs will tolerate a wide range of drive voltage. This is a fairly common strategy for getting lots of light out of a small lamp/reflector package, used for years in bike lighting systems.

Using 7.2v battery to directly power a 6v lamp means you'll get much brighter light and much shorter lamp life, although it may still burn for 100s of hours overvolted 20%. This is usually considered a good compromise since lamps are fairly inexpensive.

http://nordicgroup.us/s78/
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Old 10-25-07, 02:37 PM   #6
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If thats a mr11 lamp, you don't need a regulator to power it with the battery. Many incandescent bulbs will tolerate a wide range of drive voltage. This is a fairly common strategy for getting lots of light out of a small lamp/reflector package, used for years in bike lighting systems.

Using 7.2v battery to directly power a 6v lamp means you'll get much brighter light and much shorter lamp life, although it may still burn for 100s of hours overvolted 20%. This is usually considered a good compromise since lamps are fairly inexpensive.

http://nordicgroup.us/s78/

What about LEDs? Sorry, I know almost nothing about electronics, but all these DIY headlamp questions are making me curious.

If I do the math for my battery pack correctly, do I need a driver?
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Old 10-25-07, 03:08 PM   #7
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What about LEDs? Sorry, I know almost nothing about electronics, but all these DIY headlamp questions are making me curious.

If I do the math for my battery pack correctly, do I need a driver?
LEDs are current-driven devices, so you have to make sure the current is limited instead of the voltage. The easiest way to do this is by using a current-limiting resistor, whose value is based on the battery voltage, the LED current, and the LED's forward voltage drop at that current. This method gets more efficient as the battery voltage approaches the forward voltage of the LED.

An LED driver is definitely recommended for any high-power LED, like a Luxeon or CREE. The driver can accept a very wide range of input voltages, as long as that voltage is at least 1-2V above the forward voltage of the LED. However, the driver maintains a 90%+ efficiency no matter what the input voltage is, unlike a current limiting resistor.

Another advantage of the driver is that as the battery voltage drops, the LED will maintain its level of brightness. Using a current limiting resistor, the LED would slowly dim as the battery voltage drops.
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Old 10-25-07, 03:31 PM   #8
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What about LEDs? Sorry, I know almost nothing about electronics, but all these DIY headlamp questions are making me curious.

If I do the math for my battery pack correctly, do I need a driver?
Are you really running an LED or several LED's that draw 10 watts? If so you really do need a driver! LED drivers can be purchased very reasonably here.
http://www.dealextreme.com/products.dx/category.917
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Old 10-25-07, 03:57 PM   #9
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If the light you are powering is a halogen (incandessant... Spelling) then the more volts the better. A 6 volt will do good up to 9v i would say. But at 9v your really pushing the filament. With leds its always good to buy one of those led specific regulators.
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Old 10-25-07, 04:33 PM   #10
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Are you really running an LED or several LED's that draw 10 watts? If so you really do need a driver! LED drivers can be purchased very reasonably here.
http://www.dealextreme.com/products.dx/category.917
Not running anything at all yet, just trying to get info. I see these DIY instructions that just say buy this and solder - that's fine, I can do that - but I want to know why.
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Old 10-25-07, 08:42 PM   #11
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Originally Posted by seeker333 View Post
If thats a mr11 lamp, you don't need a regulator to power it with the battery. Many incandescent bulbs will tolerate a wide range of drive voltage. This is a fairly common strategy for getting lots of light out of a small lamp/reflector package, used for years in bike lighting systems.

Using 7.2v battery to directly power a 6v lamp means you'll get much brighter light and much shorter lamp life, although it may still burn for 100s of hours overvolted 20%. This is usually considered a good compromise since lamps are fairly inexpensive.

http://nordicgroup.us/s78/
Yup, I'm powering it at the higher voltage now, but I wanted to get more run time (hopefully possible with a high-efficiency regulator) and protect the bulb (actually that's a secondary consideration, but still).
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Old 10-25-07, 08:43 PM   #12
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If the light you are powering is a halogen (incandessant... Spelling) then the more volts the better.
Yes, I'm running a halogen. Why do you consider more voltage better?
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Old 10-25-07, 08:43 PM   #13
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...And thanks to all...
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Old 10-26-07, 06:14 PM   #14
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Because it will put out alot more: 20% more voltage gives you almost double the output. It also makes the halogen setup more efficient, since you get more light with a bit more voltage.
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Old 10-27-07, 11:26 AM   #15
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Buy a purpose built IC. The problem with a 5V regulator is that as the battery discharges it might not be able to power the regulator. The ICs used for flashlights can raise the voltage (drop the A.)
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