Poorman Mutli-Lux setup method

[download]

I found some interesting setup method to use the cheap driver board like 7135.
It can be used multi Seoul P7 or Cree,
like 2x li-ion to 2x LED or 3x li-ion to 3x LED.
It works quite well for me. Hope it helps you too.

The Cree XM-L have much lower Vin, 7135 does not work like 3x XM-L. (May-2012)


Each P7 can get 2.8A !!!
-----------------------------------------------------------------------------------------------------------


Each LED get 1.4A !!!


3 wires only.

[Bimmerboy]

Download - Very interesting. Can you tell us more about this setup?

[download]

The DIY multi-level driver is same as NetKidz & StefanFS had posted.

http://www.candlepowerforums.com/vb/...d.php?t=192677

http://www.candlepowerforums.com/vb/...d.php?t=195358

My wiring way to use it as multi-led only.
Use same driver to drive more than one LED.
The driver control one LED only, the other LED just follow the current.
Under one condition: 1x Li-ion or 3x NiMH to 1x LED.
Simple & Cheap.

[Bimmerboy]

After reading those two threads, plus looking at the above diagrams, I'm still confused!

Some questions:

Are you using a single 8 X 7135 board, or the controller/slave setup ?

How is it possible to put that much Vbatt into this? I thought the 7135 boards have a max input of 6V. Does your discovery mean it's 6V for each P7? If so, that's a pretty damn cool discovery.

In your first diagram, does the single P7 (connected to driver out) run in parallel or series with the other two P7's (connect to driver in)? I'm assuming series, because Vin would match the total emitter Vf, but I'm not understanding the diagram. EDIT: Think I figured it out... they'd be in series because driver in to driver out is in series. Correct?

Hoping to understand this setup, because it could be EXACTLY what I need to regulate 3 or 4 X P7 with 4 X Emoli.

Thanks, Download!

[download]

Bimmerboy,
I would suggest you try single 7135x4 board first, ( just half of current go though the LED ) muti-level or not, my setting is same as NetKidz & StefanFS wiring method 2.8A.

If you use 4 X Emoli, your driver connect 1x P7 and 3x P7 serial outside will be fine. ( 1 li-ion serve 1 LED only )

I tried & run for a while. It works quite OK for me.
eg: 3x P7 - The driver handle 1x li-ion power only, the serial P7 x2 share the 2x li-ion voltage, and all LED react the same function of the driving LED.

Hope this helps.

[Bimmerboy]

I think I'm beginning to see the light.

The extra P7's connected to the "driver in" side are providing the voltage drop so that the board itself only sees the Vbatt of one Li-ion. Do I have that correct?

If so, that means you can't use 4 X Li-ion for three P7's in an attempt to get full regulation throughout the entire runtime, but that cloud has a silver lining anyway, since you'd never know when the batteries are getting too drained. The light wouldn't dim until the batteries were being damaged. So, there's sort of a built-in protection here with a 1:1 setup... our eyes.

I definitely vote for this as "killer find of the month", maybe the year! A cheap, multiple P7 driver that seems to be working well. Totally surprised there's been no reaction to this.

Thanks again, Download. This sounds excellent!

P.S. The heck with half current... I'm going straight for 2.8A! I'm also thinking of using a D2DIM with this.

[Doh!Nut]

I am glad I found this, I was just about to post a question on the similar circuit, I had a look at the data sheet http://www.micro-bridge.com/data/ADD/AMC7135.pdf and saw that unlike most drivers that have four connections and the LEDout is not the same as the Batt Neg.

The AMC7135 has only three connections, with a common ground so whatever goes through the LED comes out the ground (and straight into the next LED )

This means that I could have a 3 P7 light running from 12v NiMh pack

Nick

[download]

Bimmerboy, You do see it, 4x P7 is very bright light.
D2DIM should work better for direct drive, only need more trick to wire extra control switch.
This wiring method is just easier only.

Nick, in my test result, the multi-level driver did not survive to drive more the 3x Ni-MH per LED in 3 x LED setting. (I took out all diode to gain more voltage)
I would suggest 3 P7 run 9 Ni-MH Max. only.

Cheers!

[jeffosborne]

From the AMC7135 data sheet:

Supply voltage range 2.7V ~ 6V

You are putting 10.8 volts nominal on a device that likes only up to 6 volts.

Have you TRIED the circuit you are proposing?
Anybody read the data sheets?

Warm regards,
Jeff O.

[download]

Both of wiring method I am using right now.
It works fine for me. Just share this little finding.

[Doh!Nut]

I am a little surprised that did not work with 4 NiMh per LED, but was half thinking about the circuit I was considering where the output of one driver went to the Vin (or Vdd) of the next stacked driver. That would be within the voltage specs but would be wasting lots of power.


The worst case for downloads circuit above is that you have low Vf LED, say 3.2V

So a fresh 12v Pack gives 14+v
14v minus two times 3.2v gives 8.1v accross the driver - way off spec. You would have to drop two or three cells to bring it back down to spec. (as download said)

Four P7 (Gulp)

14.5v - (3*3.2v) = 4.9 accross driver (worst case)
Running condition
12.5 - (3 * 3.7v) = 1.4v accross driver and LED- ie it wont work .
Even if the driver did not drop any voltage accross itself and the voltage was devided equally that would leave 3.1v accross each LED, not enough for 350ma.

If I want to use this in a torch with existing 12v NiMh pack maybe I will have to get a circuit, and find the actual vf of the 3 P7s and put in a small resistor.

N

[download]

I mean the multi-level circuit IC will not survive, not AMC7135, 7135 just overheat, cutdown the power & flash. It will not die easily.

You may try this:

(P7 + 7135 with multi-level) Serial (P7 + 7135) Serial (P7 + 7135) -> 12V

Each 7135 board have diode to drop the voltage a little, should work now.

[OldNick]

Download

This is very clever. I believe you have tried this and it works, but I am completely puzzled.

Simple circuit, but I don't get the principles. What does that board do that allows you to connect the Bat (+) to the two lower LEDs (-) that are not driven by the board (the "slaves")? It's effectively in series with the battery and boosting the volts? Or is it sharing the volts between the driver/LEd combo and the slave LEDs?

Why do the lower ("slave") LEDs follow the current of the two that are actually driven?

Also, if you used a single board instead of the parallel version, I assume you could use the board to drive, say, just 2 Crees "in series" and not the paralleled version you have.

Given that I actually do not want to drive my Crees at 1400mA, how do you use this? Do you have a multi-stage board, and not use high, or do I use one of the lower-current boards (which I think have only 3 of the ICs on them rather than 4)

I would appreciate any help from you (or anyone) about this.

Nick (the second)

[Mr Happy]

This is very clever. I believe you have tried this and it works, but I am completely puzzled.

...

I would appreciate any help from you (or anyone) about this.

I did not see this thread earlier, but it looks like a very creative use of the driver.

It works because the 7135 is a linear current regulator (in other words it is like a fancy variable resistor in the circuit). Because it is a linear regulator, the current everywhere in the circuit is the same, just as if it were a simple resistor.

In the first circuit then, all three P7's see whatever current the 7135 is set to, for example 2.8 A.

Now let's look at the voltages. When fully charged, three Li-ions in series produce about 12 V. On the supply side of the 7135 we have two P7's in series. Each of these will drop their Vf, which for the sake of example we might assume to be 3.5 V. Therefore the voltage seen on the input side of the 7135 will be 12 - 2 x 3.5 = 5 V. We therefore are within the allowable input range of not more than 6.0 V. Lastly, the 5 V input provides enough margin to drive the P7 on the output, assuming the same Vf of 3.5 V.

Out of curiosity, we can take the worst case when the battery is discharged and the P7's happen to have a higher Vf of 3.7 V. Close to discharge the battery will have a voltage nearer to 9 V than 12 V. From this we subtract 2 x 3.7 V giving 1.6 V. Clearly 1.6 V on the input to the 7135 is nowhere near enough to drive the output LED, so all the P7's will have shut off long before the battery gets this low.

As a result, the setup will not be able to extract all the charge from the battery, but on the good side it will protect against over discharge and prolong the life of the cells.

The second circuit can be analyzed in a similar way, except that now two LED's have been put in parallel. The trouble here is that if the two LED's have different Vf values, the current will not be equally distributed between them. Worse, if one gets hotter than the other the imbalance will tip further in the unequal direction.

It is not typically advised to put LED's in parallel because of this problem of current balancing. It may work but it's not a robust arrangement.

[download]

Hi OldNick,

I updated the diagrams for protected 18650 li-ion only, this is exactly what I am actually using.

The 7135 board transfer the extra voltage to heat before feeding LED. (like 4x 7135 ~1.2A to LED)

Because the Voltage of my Li-ion and LED Vin is close. So the extra voltage of multi-Li-ion is still OK for 7135. But not much.
If you get ultra low voltage of LED, result may be different.

In actual finding, 3x P7 diagram, the (two lower LEDs) share most the voltage of 2x li-ion, the driver take care 1x li-ion with little extra, when the muli-level driver switch dim, it limited the current of whole circuit, so all the led dim as same level.

In 4 lux diagram, mainly use 2x Li-ion, it need 2x LED in serial to share it.
Parallel extra LED are just share the current of 8x 7135, not a must in 3x 7135.
If you use 2x LED only, wire LED like diagram, (no more extra parallel LED) one of the LED use (3x 7135) for 2x Li-ion is fine. Each LED got ~0.9A.


Mr Happy,
You said better than me.
In the real world, when the battery voltage lower than LED, driver run as direct driver and LED will run lower Vf too. It will drain all the battery dry till protected circuit works.

Hope this helps!

[OldNick]

Ok. Thanks for the replies from both of you. My brain needed a kick.

I am OK with electrics, but just coming up to speed on these sort of current regulators, as distinct from voltage regulators.

I get the shared current. The 7135s actually allow a current by raising and lowering voltage then. As you say, a smart resistor. So as long as the volts left over from the "free" LEDs is within the volts of the 7135s, all stages will work. Within that range, the 7135s share the voltage across the circuit to maintain the current.

But for all of me, the two "free" P7 LEDs (and the free ones in the Cree cct) are in reverse polarity. The Positive of the battery goes to the Negative side of the LEDs.

Sorry if I am missing something (apart from the functioning grey matter)

Nick

[Mr Happy]

But for all of me, the two "free" P7 LEDs (and the free ones in the Cree cct) are in reverse polarity. The Positive of the battery goes to the Negative side of the LEDs.

I think you are right. I noticed that too, but I didn't want to make my post any longer than it already was.

The positive (anode) side of the LED should be facing the positive (+) terminal of the battery. The LEDs do all appear to be backwards in the diagrams. The red and black coloring of the wires seems a bit out of place too...

[OldNick]

Oh good...for me anyway.

I reckon the LED that is actually driven by the board is OK for polarity. It's just that the connection to the centre + of the board is "blocked" by the board. It's difficult to draw that bit without either what we see or having the LED connected to both centre and edge, or look that way.

Download : Care to clear this up? As far as I am concerned, the idea is brilliant whether you have got it wrong in the drawing or not, and if the drawing is correct, then I need help to understand.

Nick

[download]

, really sorry, corrected the diagram uploaded.

[OldNick]

, really sorry, corrected the diagram uploaded.

Bad luck, mate. Thanks for that. Still clever stuff. I am going to work with this. Sort of a $2 buckpuck! <G>

I give you an , not a <G>

Nick

[OldNick]

Hi OldNick,

I updated the diagrams for protected 18650 li-ion only, this is exactly what I am actually using.

Download

Are you saying it will not work with AAs?

You had them there at first and that is what I was going to use (I have them). But if they are no good, then I will have a think.


Nick

Nick

Nick <G>

[Mr Happy]

It will work with any batteries as long as you match the voltages correctly. Use my earlier post to give you an idea of how to work it out.

[07accordEX]

i have a quick question if you are using 1x li-ion per p7 how come the its 1x li-ion per 2x crees?

how would i wire it to use 4x li-ions for 4 crees?

[OldNick]

It will work with any batteries as long as you match the voltages correctly. Use my earlier post to give you an idea of how to work it out.

Hi MrHappy

I can work it out OK, once I grasped the idea of the "smart VReg-for-current" that the 7135 is.

I was asking _because_ it looked OK, but Download had specifically removed the AAs from the drawing.

Nick

[Mr Happy]

I was asking _because_ it looked OK, but Download had specifically removed the AAs from the drawing.

I was just confirming. No dig intended

[Mr Happy]

i have a quick question if you are using 1x li-ion per p7 how come the its 1x li-ion per 2x crees?

how would i wire it to use 4x li-ions for 4 crees?

Here's a quick answer:

Two Li-ions will supply from 8 - 6 V depending on the state of charge. The Crees in the diagram are the high output R2 variety, and these tend to have a higher Vf, maybe 4 V or more at 700 mA. So you do the calculation in the same way; take the 8 V from the battery, subtract the 4 V Vf, and therefore the regulator sees a 4 V input. The numbers are within range, so the circuit works.

My feeling is that with that setup there is little margin of excess voltage. If the Cree R2s happened to have a Vf of 4.2 V, it would be very difficult to get maximum brightness out of them, especially as the battery discharged below 8 V.

Also, as I mentioned above, it is not usually a good thing to put LEDs in parallel. Unless you carefully match the Vf's, you may get uneven current distribution.

[OldNick]

i have a quick question if you are using 1x li-ion per p7 how come the its 1x li-ion per 2x crees?

how would i wire it to use 4x li-ions for 4 crees?

You don't need more Li-Ions to run more diodes, just more boards in parallel. However it's not the best scenario.

The drawing shown is actually _two_ 1400mA boards in parallel, to give 2800mA to drive a P7. Each board has 4 * 7135 regulators. Each 7135 gives 350mA to give a total of 1440mA per board.

The Cree LEDs only need (actually) 1000mA to drive to max. Therefore the circuit as shown needs two crees in parallel to accept the current, even when the input voltage from the cells could overdrive one. However as shown it will still overdrive the 2 Crees. It would work OK with the hungry old K2 Lumileds.

For crees I would use 2* (3 * 7135) boards instead, to give the required 2000mA for two in parallel.

So the battery usage is to give the required voltage, not to drive more LEDs.

As Mr Happy said, running LEDs in parallel is risky unless they are very well matched. I would prefer to use a single 1000mA board, driving 3 LEDs in series, and have the same setup for any more LE chains. You can drive them all from the same battery pack.

Obviously more batteries in parallel will give a longer runtime.

If you are still confused let us know.

Nick

[OldNick]

I was just confirming. No dig intended

Naaah! None taken. I appreciate your explanations. Better too much than the usual snappy one-liners

Nick

[download]

OldNick, I updated diagrams are what I am using actually.
It does not mean AA not work.
3x AA is very closed to 1x li-ion 18650 in many way.
It should work, if you match it like 3x AA = 1x Li-ion.
But I did not test it yet like 4x li-ion equal or not to 12x AA.

07accordEX, 2rd setup is: parallel 2 Cree in 1x Li-ion only
They just share max current like 2.8A /2 = 1.4A per Cree.

4x li-ions for 4 crees = 1x (4x7135) driver connect 1x LED or P7 and serial 3x LED or P7 outside like diagram 1 driven by 4x Li-ion.

You guys type faster than me.

[OldNick]

OldNick, I updated diagrams are what I am using actually.
It does not mean AA not work.
3x AA is very closed to 1x li-ion 18650 in many way.
It should work, if you match it like 3x AA = 1x Li-ion.
But I did not test it yet like 4x li-ion equal or not to 12x AA.

07accordEX, 2rd setup is: parallel 2 Cree in 1x Li-ion only
They just share max current like 2.8A /2 = 1.4A per Cree.

4x li-ions for 4 crees = 1x (4x7135) driver connect 1x LED or P7 and serial 3x LED or P7 outside like diagram 1 driven by 4x Li-ion.

You guys type faster than me.

Ok. Thanks for that.

I may type faster but your post says it better. I misunderstood the question. As you say, you can add more cells in series, as long as you have a new LED for each 3.6-4 volts of battery pack.

Nick