Yet another LED low voltage driver Questions (Was LED_regulator2.png

Hokay, some measurements of the circuit posted here a few days ago.

The circuit is LED_regulator2.png.

Input V = 1.50VDC.

Input I = 16.3 mA.

Assumed LED V = 3.5V (This is the approx forward voltage of the LED. I tried measuring the V but it measures a few tenths of a volta with a DMM, so I know that most of the current thru the LED is during the flyback pulse across it.

Measured LED current = 4.4 mA. (This is across a 1 ohm resistor placed between the junction of the LED and 4.6 uf cap and the rest of the circuit.

LED power = 3.5V * .0044 mA = 15.4 mW.

Efficiency = 14 mW / 24.5 mW = 62%

and so, despite the use of the LED itself as the rectifier, the efficiency of the circuit is not very good at all. This is the efficiency that I got from many of the circuits I built using the circuit on the left of the pic at

formatting link

For an inductor for this circuit I used a 100 uH inductor that I bought from Mouser. Then I used a 100 uH toroidal inductor I made from a RFI suppression sleeve that I took off a keyboard or mouse cord. I've found that the inductor should have low DC resistance.

I put a 100 uH toroidal inductor across the 560 uH, and the LED current increased by less than a mA, but the total current increased to 22 mA. The increase in brightness was barely perceptible, but at reduced efficiency. So this is apparently not the way to go.

I'll have more experiment results later.

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Watson A.Name - 'Watt Sun'
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Some of the losses are in the regulator. I can't see any way to get around it and keep the .5V to 5V operating range.

To eliminate the regulator, remove everything left of the LED. Connect the negative of the LED to (-). Connect the dangling emitter to (+). It will work more efficiently but for a voltage range of only a few tenths of a volt. Too much will fry it. If that range happens to be very close to what you've got as your power supply, a simple series resistor could be more efficient than the regulator. It would be interesting to try this with FETs too.

current limit resistor

-----R------+-----+ (+) | | | |( | |( | |( 560µH | |( | |( 330pF | | +-||---)-----+-----+ | | | | | E | | PNP +-----B | \ / LED | C | --- | | C | +--R---+----B NPN | | 100K | E | | C | | NPN +-----B | | E | | | | |

------------+-----+-----+ (-)

Reply to
Kevin McMurtrie

In article , snipped-for-privacy@sonic.net mentioned...

I tried o get it to restart below .9V but it wouldn't. I can go down to .6V but only if I start at .9V or more. I really didn't try it at above 3V, but to that point, the more voltage, the less current; as you say, it regulates the LED power. But what is the secret to getting the whole circuit to do this at ten times the LED power output?

I thought about that two transistor driver. I saw some circuit like that, and it said that if the supply V goes above 1.2V, the two transistors start to conduct, and you got a problem big time. Here, the rest of the circuit limits the current. But is there an advantage to using two driver transistors? Why not just one.

The next thing I'll play, er, experiment with is the 100k resistor, to see how it affect the output. I might try to change to a red LED to see what happens. Back to the lab, Boris.

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Watson A.Name - 'Watt Sun'

The 100K resistor is some negative feedback so that the transistors are in a high gain analog state before the oscillator starts. It will start without it if the power ramps up quickly.

The complementary pair produces a much better drive at high frequencies than one transistor, a bias resistor, and a pull-down resistor. You can adjust L and C so that the oscillator runs in the MHz region and you'll still get a razor sharp waveform. The downside is of course that it blows up at 2V :)

Reply to
Kevin McMurtrie

In article , snipped-for-privacy@sonic.net mentioned... [snip]

Oddly, when I lower it to 50k the total current is reduced and the LED puts out a bit less.

Then why not change it to a totem pole like the output stage of an audio amp. Use the complementary pair, but have the emitters drive the output stage. It won't go down to as low a voltage, but since the circuit is bootstrapped, it might work.

I lowered the 100 ohm resistor down to 50 ohms. The total current almost doubled and LED current went from 4.4 mA to 7.8 mA. This is a bit more like what the LED current should be. I would expect the LED current to be around 20 to 25 mA.

With the lower value resistor, the regulation isn't as good. The LED current and supply current increase as the supply V goes up from 1 to

2V, then it drops as the circuit regulates from 2V on up to 3V. But even so, it's putting out a substantially greater amount of light, and that's a good sign. I checked the freq and it's running at 58 kHz.
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