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

[Watson A.Name - 'Watt Sun']

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
http://www.belza.cz/ledlight/ledm.htm.

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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[Kevin McMurtrie]

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
http://www.belza.cz/ledlight/ledm.htm.

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.

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

 

[Watson A.Name - 'Watt Sun']

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.

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?

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.

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.

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

--
@@F@r@o@m@@O@r@a@n@g@e@@C@o@u@n@t@y@,@@C@a@l@,@@w@h@e@r@e@@
###Got a Question about ELECTRONICS? Check HERE First:###
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My email address is whitelisted. *All* email sent to it
goes directly to the trash unless you add NOSPAM in the
Subject: line with other stuff. alondra101 <at> hotmail.com
Don't be ripped off by the big book dealers. Go to the URL
that will give you a choice and save you money(up to half).
http://www.everybookstore.com You'll be glad you did!
Just when you thought you had all this figured out, the gov't
changed it: http://physics.nist.gov/cuu/Units/binary.html
@@t@h@e@@a@f@f@l@u@e@n@t@@m@e@e@t@@t@h@e@@E@f@f@l@u@e@n@t@@

 

[Kevin McMurtrie]

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.

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

 

[Watson A.Name - 'Watt Sun']

[snip]

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

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

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.


--
@@F@r@o@m@@O@r@a@n@g@e@@C@o@u@n@t@y@,@@C@a@l@,@@w@h@e@r@e@@
###Got a Question about ELECTRONICS? Check HERE First:###
http://users.pandora.be/educypedia/electronics/databank.htm
My email address is whitelisted. *All* email sent to it
goes directly to the trash unless you add NOSPAM in the
Subject: line with other stuff. alondra101 <at> hotmail.com
Don't be ripped off by the big book dealers. Go to the URL
that will give you a choice and save you money(up to half).
http://www.everybookstore.com You'll be glad you did!
Just when you thought you had all this figured out, the gov't
changed it: http://physics.nist.gov/cuu/Units/binary.html
@@t@h@e@@a@f@f@l@u@e@n@t@@m@e@e@t@@t@h@e@@E@f@f@l@u@e@n@t@@