How to modify a constant current LED driver 700mA into 350mA?

Is it possible to modify a constant current LED driver 700mA 5W (230V input) to use with 4x1,2W 350mA Diods??

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This is just a guess however.

Going over these circuits a number of times it would appear they have short circuit protection but not current regulation.

Try these circuits on the output.

The first circuit is fixed the second two are dimmable.

Since 4 x 2w = 8 W, that's significantly more than the device is designed to handle, no.
Allecc (author)  steveastrouk4 years ago
Sorry for missunderstanding but it should be 4 diods each being 1.2W 350mA diods.
IF you can get to the circuit board it MIGHT be possible, take some decent photographs and post them here.
Allecc (author)  steveastrouk4 years ago
Took a few pictures but i dont have the greatest camera, hope you can see. The first 2 are of the driver i first asked about, but i was wrong "Pout" is only 3W. This is supposed to be non-dimmable.

The other driver i got is bigger, 11W 660mA, this is supposed to be dimmable on PRI side with a leading/trailing edge dimmer.

Got a bunch each of these. What i would like to know is if i could modify the current they put out or increase the W they can handle?

Is this the right place to have a continuous discussion or should  i take it somewhere else?
Allecc (author)  Allecc4 years ago
Here are the specifications i could see.
I recommend looking at the data sheet for that IC (TNY266)
I found a copy of it here:

In the circuits given in the data sheet, it looks like the feedback is by way of pin  4, also called EN/UV,  and that current out of that pin is usually fed through an optocoupler,so I am guessing its the current through the LED on the other side of that optocoupler that is actually controlling pin 4.  The datasheet for the optocoupler visible on your board, I found here,

Anyway, it might be worthwhile to sort of map out the circuit on the side of the transformer that gets connected to your LED, and see if you can figure out how that circuit is sensing current, and how it is feeding that signal to the LED in the optocoupler.

One thing to look for is if any of those small valued resistors, like maybe the 0.27 ohm one, are wired in series with your LED, and thus that resistor and your LED share the same current.  If so, that might the current sensing mechanism you are looking to discover.

Final note:  That driver might work with 1 LED, but maybe not with 4 in a series. The specs on its little plate said Uout was 4-5 VDC.  The drop across one white LED is about 3.6 V, and the drop across 4 of them is 4*3.6 V = 14.4 V.  Also do the math for the power dissipation, and you'll discover that 4 white LEDs pulling 350 mA, P = 14.4 V *0.350 A= 5.04 W

Well the sense can't be by current passing through the LED, so we have to assume there's a little part we can't see near that optoisolator.

Its a clever little thing.
Allecc (author)  steveastrouk4 years ago
I could be a bit smarter, I would like the ability to dim the lights. But it is small, only 35x38x15mm. Not to bad going from 230VAC to my led.

I got another which is about the same size thou that is both dimmable on primary side and can handle 350,500,700mA, 12, 24VDC, THAT is a clever little thing.
OK.  I drew a diagram of my guess
of the current sensing works.

The component values are based on the values Allecc wrote on this picture:

Now I am guessing this 1.3 ohm resistor has the same current flowing through it as the Big White LED, and with a current of 700 mA flowing through it, the drop across that 1.3 ohm resistor would be 0.7*1.3 = 0.91 V.   That same voltage is across the 550 ohm resistor in series with the LED inside the optocoupler. 

Guessing that the forward drop across the optocoupler LED is about 0.6 V, gives a current of (0.91-0.6)/550 = 0.56 mA,  which is enough to call it "on". 

Also guessing that the optocoupler LED will not turn on at all, when the current through the 1.3 ohm resistor is less than 0.6/1.3 = 0.46 A  = 460 mA

So I think the answer to how to de-power this thing, to change it from a 700mA regulator to a 350 mA regulator, is to double the size of that 1.3 ohm resistor.  Make it 2.6 ohms, and then the optocoupler LED will switch on at approximately half the current level it did before.

I mean that's if this thing is actually wired that way.
Allecc (author)  Jack A Lopez4 years ago
Im realizing this is way over my head. I'll get some books and do some more study on my own before i get back to here. Please continue this discussion here if you want, i would still love to read it even thou i barely understand half of it.
I cant choose a best answer, you've all been great! Thanks allot and see you around here :D
Adjusting the power down should not be a problem.

The things I have marked here, the exact markings would be helpful.

Photos of the other side of the circuit boards would be helpful.

Without knowing the values of these things this is just a guess, the components marked guess in the picks are resistors and should be the ones to change to a pot to make the drivers dimmable.
Bear in mind this is a off-line converter - so what we're looking for can only be on the secondary of the system.

What you have tagged as a "regulator" is actually the triac firing the primary of the transformer, under the control of signals passed by the opto-coupler. Presumably, they have a sense resistor in parallel with the opto, and "measure" the voltage there.
Without part numbers and better picks of both sides of the circuit board, you really can’t say for sure what is what or does what, at this point everything is a guess unless you have the same board and you can read the part numbers.

The one you say is a triac from the picks could just as easily be a 4pin IC mosfet or a transistor

This is a similar power supply.

One is marked K3563 from experience I know that is a mosfet when I look it up on this site


I get 2SK3563 a 35 watt N channel mosfet.

The other is marked BA50BC0 a 1.0 Amp 5 volt voltage regulator.

And the little one that looks like a small transistor is marked TL1431C is an adjustable programmable voltage reference.

the optocoupler is a PO123

The mosfet, optocoupler and voltage reference is a circuit protection.

You just can’t tell whats what without the part numbers
No, I read the chip number and downloaded the data sheet - did you ?
Cant see them in the picks.
 I was able to get a better look at the 8pin IC it is a TNY266PN an off line switch a fault protection IC. There is no regulator on the small circuit board so I doubt you can change its output. The EL817 is an optocoupler. Go to this sight to get the data sheets http://www.maxim4u.com/
Increasing power isn't possible, too many things need uprating. We need to look closely at the components around the OUTPUT of the circuits to change the current.
Allecc (author)  steveastrouk4 years ago
I'll do that and get back to you. Thanks alot so far Ice/Steve!

Also if you (or anyone) have some good sources of information around this subject you could point me to, so i could do some more study myself, that would be appreciated :D
Yes somewhere in the circuit you will find this or something like it change the resistor.

The bottom voltage regulator on the right is used to regulate the current to about what you want.
555 tested.png
iceng4 years ago
I take it that your driver is not easily accessible to change.

First thoughts come to mind would be to place 2 LEDs in series
in parallel with two other es for a total of four.

And to guarantee the two series LED strings share the
current at 350ma each you will need to place a resistor in series with
each string.

Or place sub current regulators in series with each two series LEDs.

Allecc (author)  iceng4 years ago
Its not installed yet so right now its very accesible but once i but it in place, not so much. Is there any modification i can do inside it if i can acces it? Solder and swap some parts out?

Your solution sounds like a good one. Might be a stupid question but if i use 4 identical LED's in a 2+2 string, why do i need a resistor in there too. Shouldnt the current split equally between the 2 strings?

You'd think that, but you'd be wrong: Leds don't share current evenly, and the hotter one gets, the more current it draws, taking it from the other one, which gets cooler and draws less current. Pretty soon you are in a race to the top, and one LED burns out, closely followed by the second.
@1watt the leds should be putting out a fairly high temperature, and if all mounted to the same heatsink very close to one another, they might be able to survive by complementary heating of the lower current devices until a stasis is reached; but in all likelihood it would runaway. I'm surprised these days at how often high power leds ar used in parallel with few consequences