low-loss voltage regulator

Look at some of the maxim and linear technology parts. mike

Switching regulator is the only way. National, Maxim, LTC.

What's the total LED current?

John

But

Two things, I hope you're current limiting the LED with a resistor (unless they are already regulated.) Two, do what the other guy said and look for a more efficient regulator. If you do that, you can have really low dropout. So you may be able to loose two or three of your batteries to save weight. Since you have them all in series now, you'd have the same battery life with 3/4s the weight. 5 cells even at the lowest you should take them (.95V) would still put out 4.75 volts, which is easy to design around.

What kind of LED's need a precise 3.6 Volt supply? Usually the I/V relationship of LED's is such that it makes more sense to drive them with a current source than a voltage source.

Anyway, if you want to design a step-down regulator with an input range of 5-10 Volts and an output of 3.6, you can probably find a good solution in a TI or National or Linear Technologies chip. The only major external components will be filter inductors and filter capacitors.

What you are looking for is a step-down regulator or buck regulator. The first one I found at National's website, which may or may not be suitable for your application is the LM2734Y.

Good luck!

--Mac

I read in sci.electronics.design that snipped-for-privacy@sbcglobal.net wrote (in ) about 'low- loss voltage regulator', on Mon, 7 Feb 2005:

You should NOT run LEDs from a constant-voltage supply. The current is an exponential function of voltage and temperature. You need a constant- current supply.

Apart from fairly small differences in quiescent current, ANY linear voltage regulator will dissipate the same power, and consequently produce the same heat. The only way to reduce the heat waste is to use a switching regulator - google for "simple switcher". What current are you running at?

Paul Burke

Do you have all the LEDs in parallel?

Connect them in groups of two in series and with a small dropping resistor in series with each pair.

Mark

Mark

You cannot get around P = I * E in a linear regulator. So, at 10V in, the regulator must drop 6.4V; multiply by the current to the load (guess 20mA per LED times 40 = 800mA) --> about 5 watts dissipated ===> need goodly heatsink. Solution: a switching buck regulator.

Well - in the current design just loosing some of the batteries would improve the efficiency with close to no effect to the battery life... 3.6V + LDO drop (say 200mV) should not take more than maximum 5 cells!

Besides that I second using a switcher dedicated for LED-driving. Meaning: Outputting constant current so the power isn't wasted in series resistors...

Cheers, Anders

If he uses the right part, he should be able to run a buck regulator in current limiting mode. Also, and I haven't worked this out, he should be able to do something by using a very small output capacitor.

Tam

You'd be better off redesigning the thing and using a switching technique to drive the LEDs, like this:

Leon

No, the LED's are not regulated. I'm not current limiting them at this point. My (simplistic) thinking is that any resistor is going to cause I^2R power loss. Also, you mention "really low dropout". Pardon my ignorance, but what is "dropout?" Thanks, Bruce

Thanks for the note, Mac. I looked at the LM2734Y you mention. It says it is "configured to convert 5Vinput to 1.8V output at 1 Amp". Since the spec sheet on the LED's says their optimum voltage is 3.6, it sounds as if the LM2734Y will come in shy of the mark. Bruce

HI, Paul, I'm running at about 1.2 A when the battery pack (7 cells) is fully charged. i.e. when the voltage regulator is delivering its nominal 3.6 V. Bruce

I third the switcher solution. Do you have a vendor/product number to recommend? Thanks, Bruce

Where did you read that?

The LM2734Y output Voltage can be set by a resistor divider to any Voltage down to 0.8 V. Of course, since it is a step-down regulator, the input Voltage must be higher than the output by some amount (not sure how much).

Here is a URL for the datasheet:

Good luck with your project. ;-)

--Mac

All linear regulators require that the input Voltage be somewhat higher than the output Voltage. Usually there is a minimum difference below which the regulator is not guaranteed to work.

For example, 2 Volts might be a typical value for older linear regulators. So if you want to regulate down to 3.6, you need to make sure you keep the input above 5.6. And the 2 Volts is called the dropout Voltage, because the regulator will (or might) drop out of regulation if the difference ever gets that low.

Some newer regulators might allow you to operate with the input Voltage down to Vout + 0.8 Volts or so. This would then be called a Low Drop Out Voltage regulator, or LDO.

--Mac

But

onto

low