Desperatly looking for driver (updated kennan driver?)

[waTom]

I hope this is the right thread as it is about the kennan driver (mainly).

I would like to use one for a one led flashlight that should be powered by two cr123a batteries but also be able to take 18650.

The description says "Total input voltage must be 1-3V higher than output voltage" but it doesn't say what happens if not. Does the driver simply stop working or would it be direct-drive?

[uk_caver]

The description says "Total input voltage must be 1-3V higher than output voltage" but it doesn't say what happens if not. Does the driver simply stop working or would it be direct-drive?

With an (older) Kennan I have, if running a single Cree off 3xNiMH (~3.45V getting to the Kennan due to the presence of a series diode for reverse-voltage protection), I was getting ~270mA draw from the cells, and presumably about the same current going to the LED.
At that current, there was 0.27V drop through the Kennan circuitry.

[waTom]

Thanks!

[kosPap]

double post, sorry...

[kosPap]

Hi guys? How about this version of the Keenan board?



link to supersize image

this one is missing the black part next to the inductor...

I cannot read it any further and I wonder if the - wire is soldered properly. It seems like it connects both the R200 and 622 parts...

[uk_caver]

R1 and R2 are connected anyway (see circuit diagram in post #53)

In the circuit you posted a picture of, R6 is present as a 6k2 and R7 is missing - that does a permanent pull-up the Chip Enable input (not sure why they used a resistor there, rather than a shorting link - maybe to make it easier to add external pull-down for on/off control?)

Also, R4 is a 1R0 resistor and R5 has been replaced with a capacitor, presumably to form the 1R0/10u (or similar) series combination, as recommended in the PT4105 datasheet for use at >=12V input.

That looks like a smart thing to do if dispensing with IC1 - I think I'll probably do it myself on some Kennans I'm using in a battery charger I'm working on.

[kosPap]

R1 and R2 are connected anyway (see circuit diagram in post #53)

In the circuit you posted a picture of, R6 is present as a 6k2 and R7 is missing - that does a permanent pull-up the Chip Enable input (not sure why they used a resistor there, rather than a shorting link - maybe to make it easier to add external pull-down for on/off control?)

Also, R4 is a 1R0 resistor and R5 has been replaced with a capacitor, presumably to form the 1R0/10u (or similar) series combination, as recommended in the PT4105 datasheet for use at >=12V input.

That looks like a smart thing to do if dispensing with IC1 - I think I'll probably do it myself on some Kennans I'm using in a battery charger I'm working on.

friend you lost me here....in laymans terms was that an improvement of the Kennan board style? What do you mean by the>=12V input reference? taht the board it is optimized for that voltage range? What do you mean by the term pull up?

also R1&R2 are connected but doesn't things change if the led input wire is placed between them? (are they in a series or parallel?). Also in another board the is a solder excess taht tuches the adjacent chip pin (If my eyes don't fool me)

thanks, Kostas

[zzonbi]

"friend you lost me here....in laymans terms was that an improvement of the Kennan board style?" I think so.

"What do you mean by the>=12V input reference? taht the board it is optimized for that voltage range?" Yep, IC takes up to 20V input voltage.

"What do you mean by the term pull up?" It means a 'hanging in the air' input
voltage is pulled up to a higher voltage so as to see a logical 1 signal. It could be pulled down as well, eg by connecting to ground.

"also R1&R2 are connected but doesn't things change if the led input wire is placed between them? (are they in a series or parallel?)." Things are already changed by C1. They are neither series, nor parallel.

"Also in another board the is a solder excess taht tuches the adjacent chip pin (If my eyes don't fool me)" Eyes could fool, better take a multimeter and verify. And be sure you press the probes rather hard to make a good contact.

Looks like they listened and finnaly put that RC snubber. I agree with you uk_caver, this new version looks better. Perhaps they didn't shunt R6 to make it easier to add an R7 such as the chip would disable when batteries discharge lower than a desired level (the supposed role of the missing IC). Your version is more probable though. Now if only my Kennan would work, because I don't think they intend to send me a replacement :-/

Do your Kennans make a hissing sound?

[uk_caver]

friend you lost me here....in laymans terms was that an improvement of the Kennan board style? What do you mean by the>=12V input reference? taht the board it is optimized for that voltage range? What do you mean by the term pull up?

I'm sorry for the length of this post, but I don't find it easy to describe things concisely. (and I hope space forgives my modification of the excellent diagram from post #53)

The 8-pin chip on the board (the PT4105E) has an input called 'CE' (chip enable), which is used to turn the chip on and off. If this pin is pulled low (eg connected to the 0V input or similar, that switches the chip off. If this pin is connected to the positive supply, it turns the chip on.

The PT4105 datasheet is at:
www.ihonlynorth.com/forums/attachments/fabrication/1696d1210005394-led-tech-4x4s-pt4105e.pdf

In the original Kennan boards (which were sold last year), this CE input pin was connected to the positive voltage supply to the chip (ie the input from the battery), so the chip was always on.

In the newer Kennan boards (sold recently, and as described in post #53), the combination of IC1 and R4, R5, and R7 turned the PT4105 off if the battery voltage dropped below a certain voltage.

In the board you were asking about, a resistor was used to 'pull up' the CE input (effectively, to make the voltage at the CE pin high enough to keep it permanently on, whatever the battery voltage was, similar to the way the original boards work.

What is an improvement depends a lot on how someone plans to power the circuit.
In some situations, having a low-voltage shutdown (like the New Kennan boards) can be good (basically, if you're running off Li-Ion cells, and the shutdown voltage is appropriate to save the cells from overdischarge).

However, in other situations, having a shutdown voltage can be either a problem if it's set too high for the supply you're using (such as 4xNiMH?), or meaningless if it's set too low. For example, a shutdown voltage of 6V wouldn't protect your cells if you were running off 4xLi-Ion cells in series.
In those situations, the old Kennan or the board you showed a picture of would be better, or at least as good.

Page 6 of the datasheet for the PT4015 suggests that if the input voltage is 12 Volts or above, a certain resistor and capacitor in series are connected across the input to help keep the circuit stable. The datasheet suggests that that extra resistor+capacitor is is probably not necessary if running from batteries, which is presumably why the Kennan boards don't include them.
However, on the board you have, this resistor and capacitor are added.

This could be useful if the board is going to be run from some kind of power supply other than batteries, such as if running LEDs off a 12-18V power supply powered from AC mains, and it isn't going to cause any problem or waste of power even if the board is run from batteries.

The circuit diagram for the circuit you showed seems to be:

You can compare this with the one made by space in post #53, which I used to generate this diagram from.

also R1&R2 are connected but doesn't things change if the led input wire is placed between them? (are they in a series or parallel?). Also in another board the is a solder excess taht tuches the adjacent chip pin (If my eyes don't fool me)

R1 and R2 are already connected, so there's no problem if the solder for the output wire flows and touches both R1 and R2, since that isn't making any new connection between components on the board.

Looking at the datasheet, there are some PT4105 circuits where the connection between R1 and pin 8 of the 4105 needs to use a resistor - these circuits are where some dimming control is being used.

However, if the PT4105 is running at a fixed output, a direct connection between the 'output' end of R1 and pin 8 of the PT4105 can be used. In that situation, it doesn't matter if some solder does bridge across from the R1/R2 connection to pin 8 of the PT4105, since that will only result in R2 being shorted out, which won't affect operation.

[kosPap]

Thanks for the explanation. Somehow I got it...and i can safely & efficiently use 6V with this

Now, would you prefer this flavor over any other version of the board???

[uk_caver]

Now, would you prefer this flavor over any other version of the board???

Personally, it doesn't matter too much me.
Given the way I use them, I don't need the low voltage cutoff in any applications and though I might sometimes have a use for the the RC circuit, that is only in applications where I have plenty of space to add it externally to the board if necessary.

So far, in various projects, I've been using Kennan boards I bought last year, (like the first picture in post #29), which didn't even have a place to put the low-voltage cutoff chip.

I just looked in my component store and checked out the newer Kennan boards I bought a few months ago, and they're just like the boards you showed a picture of, without IC1, and with an RC circuit on the input.

[zzonbi]

This latest version does not have an external circuit for low voltage shut-down. Both R6 and R2 are redundant, they can be shorted.
I am wondering how this circuit can be modified to get a constant voltage supply. Perhaps getting the feedback from a voltage divider only would do that?

[CM]

Good thread. wish I found this earlier.

[2xTrinity]

This latest version does not have an external circuit for low voltage shut-down. Both R6 and R2 are redundant, they can be shorted.
I am wondering how this circuit can be modified to get a constant voltage supply. Perhaps getting the feedback from a voltage divider only would do that?

Basically what this chip does is behave like a 200mV regulator. It tries to ensure that the voltage seen at pin 8 (FB) is 200mV. There is a sense resistor between the LED negative, and ground, to convert the output current into a reference voltage. The latest Kennans delivered have had a 0.2 ohm resistor as the sense. There will be 200mV dropped across that sense resistor when current trhough the LED is one amp, thus, it behaves like a 1 amp current regulator. Increasing the sense resistor value will decrease the regulated output current.

If you ground your LED (or whatever you're driving with the board) directly, and remove the sense resistor, you can wire up a voltage divider to pin 8. I'd set up a voltage divider using a 1k-ohm resistor between the feedback pin and ground, and then a 100k linear trimpot between the voltage output, and your feedback pin. This way, by adjusting the trimpot, you'll be able to achieve any voltage output you want all the way up to 20V (the max the board can handle).

[TorchBoy]

If you ground your LED (or whatever you're driving with the board) directly, and remove the sense resistor, you can wire up a voltage divider to pin 8. I'd set up a voltage divider using a 1k-ohm resistor between the feedback pin and ground, and then a 100k linear trimpot between the voltage output, and your feedback pin. This way, by adjusting the trimpot, you'll be able to achieve any voltage output you want all the way up to 20V (the max the board can handle).

That's not very dissimilar to the approach taken in the first "Dimming Control" diagram on page 5 of the datasheet. They still have the LED connected to ground via R2, though, which would mean for that particular arrangement VFB was never shorted to ground. I don't think there's any advantage for the LED.

Using either of those voltage divider setups, if your input voltage varied the current regulation would change, right?

[2xTrinity]

That's not very dissimilar to the approach taken in the first "Dimming Control" diagram on page 5 of the datasheet. They still have the LED connected to ground via R2, though, which would mean for that particular arrangement VFB was never shorted to ground. I don't think there's any advantage for the LED.

Using either of those voltage divider setups, if your input voltage varied the current regulation would change, right?

In the voltage divider I specified, the feeback resistor is bypassed so there is no current regulation, only voltage regulation, which is what the poster I was replying to wanted.

Using the "dimming control" on the datasheet, increasing the reference voltage in effect causes the output current to decrease linearly (I have tested the board, and yes the relationship between current out and voltage in is remarkably linear IF the feeback resistor is used). The value of the feeback resistor still determined the "maximum" output current (eg, externally applied voltage = 0).


I haven't tested the board extensively in an open-loop setting at all (eg, LED or load directly grounded, and FB not conncted in any way to the output). But I imagine that the relationship between input voltage and output current would no longer be linear, and might even vary based on factors such as Vf shift due to heating in the LEDs etc.

[TorchBoy]

Thanks for that clarification, 2xT. I guess I should revise what I said about advantages for the LED with R2 in there - staying in current regulation instead of voltage regulation would be an advantage in most situations.

It's good to know the board is so controllable.

[manne]

The layout has been changed again:

http://www.candlepowerforums.com/vb/...64&postcount=1

R2 and R6 is 3k on mine.
The diode has become bigger format (like first kennan) while the output capacitor is smaller.

[cytoe]

Anyone use this board to power 3 x 3watt led's (cree, etc) using a 14.4 Li-ion pack? I've had 3 boards die w/in a couple of minutes of power on. The original boards worked fine. These failing boards are the new design, but not the latest. Trying to figure out if they are worth ordering again to try out. The next best option would probably be a BuckPuck...but it costs more

[Brlux]

I played with the Zetex zxld1360 a while ago for the next generation of my home lighting setup. They are great little switchers, the best and worst part of them is the control pin. It can be driven with roughly a .3-2.5V DC voltage to control the brightens but the thing I didn't like is it didn't go low enough for my liking before they would just shut off. If configured with the .1 ohm sense resistor the lowest LED current I could get before they would go in to shutdown was about 200mA. If you want to go any lower then you have to use a PWM on the control pin but it has to be below 500 Hertz so in my opinion the frequency would be annoying. You could go with a higher value sense resistor like a .2 ohm and you would be able to get it down closer to 100mA and still have your 1A drive capability when feeding 2.5V on the controle pin but then you are sacraficing efficiency. They are very small 5 pin SOT-23 and a bit pricy but have a wide input range up to 30V and are very efficient. I was getting around 94% range when driving 3 rebels form 12V. Unfortunatly my project could not work with the minimum output limitations. They may still find their way in to other projects of mine down the line though.

[jtr1962]

You could go with a higher value sense resistor like a .2 ohm and you would be able to get it down closer to 100mA and still have your 1A drive capability when feeding 2.5V on the controle pin but then you are sacraficing efficiency.

You're actually not sacrificing all that much efficiency in exchange for a much greater dimming range. If you're driving 3 LEDs then your total Vf is around 10V. The 0.2 ohm sense resistor only adds 0.1V of overhead at 1 amp drive current. This decreases the efficiency by about 0.1V/10V or 1%. A small price to pay IMO. That being said, I wish the ZXLD1360 was designed maybe for 5V on the ADJ pin = 100% output. 0.25V then would allow dimming to 5% while still allowing use of an 0.1 ohm sense resistor.

If using PWM then 500 Hz shouldn't be noticeable at all to the eye although it might still give a stroboscopic effect with things like fan blades.

[gillestugan]

New version again, this time a beautiful green PCB






The R2 and R6 are now 1k.
Seems to be running pretty cold and nice. 1.01A out before I modded it
I did the pot mod described earlier in this thread and made some efficiency measurements when running one Q5 off 12V. Its really great, to have dimming down to a few mA.
Here are the numbers:

0,87A 81%
0,50A 83%
0,25A 90%

[Doh!Nut]

Thanks for the update

I have a few of the version 2 keenans
Does anyone know if they can be driven in parallel to get 1400ma?

I have seen this done with two sandwich shoppe drivers.

A second question would be is it possible to wire them in a way that you could switch one off for a high/low. I ask this because I have tried it before using a double throw switch powering a buckpuck and a bucktoot. it would only run for few seconds before cutting out - I assume this was the open circuit protection cutting in because some current was flowing back through the unpowered driver. Is this likely or was it something else

Thanks
Nick

[Spykooo]

Hello Everybody!

I'm French and I don't speak English very well.

With friends, we made MTB Light with P4 Seoul.

For regular courant I choose this component on Kaidomain.

Today I tested the assembly. It's ok, the LED ignite.

But, I mesured the exit current without LED and I find 900mA????? It's normal?

I test with beeper my welding and enter (-) LED and (-) power the beeper is ok!!!! Where is my R1? It's like I jump R1 with my welding.

Please Help!

For amateur we can see my first realisation here : http://spykooo.unblog.fr/2008/06/14/...rage-pour-vtt/

[uk_caver]

900mA is normal if your driver has an 'R200' (0.2Ohm) for R1

The PT4105E chip used on the Kennan driver has a feedback voltage of 200mV, +/- 10%. With an R200 for the current-sensing resistor R1, that gives a current of about 1 Amp, +/- 10%.

(The earlier Kennans used an R270, which gave a current of ~740mA)

Since R1 is a very small resistance a continuity tester would probably react to it the same way as it would react to a short-circuit.

[Spykooo]

OK ok I'm happy for your answer!

I wanted 750mA and I have 900mA...
If I replace R1 (0.2ohm) by 0.27ohm I find 750mA?

You confirme my idea for the continuity tester.

Thanks for your help.

[gillestugan]

yes, and the driver will run cooler and more efficient.

[cytoe]

Can anyone confirm that this new driver revision can run 3xled from a 14.8v Li-Ion pack? Sort of lost confidence in this driver after toasting 4. Seems the last "new" design sucks.

[2xTrinity]

Can anyone confirm that this new driver revision can run 3xled from a 14.8v Li-Ion pack? Sort of lost confidence in this driver after toasting 4. Seems the last "new" design sucks.

The reason it's toasting is Kai, or whoever assembles these boards, dropped the sense resistor from ~0.25 ohm, to 0.2 ohm, effectively raising the current from 750mA to 1A. This was done to make the chip "better" and more powerful. The thing is, the PT4105 regulator chip really shouldn't be run higher than 750mA continuously as its efficincy drops and it can excessively heat up. If you replace that sense resistor, you will be fine to run any sort of setup you want.

[cytoe]

Thanks for the info. Do you know where I can find these resistors? The local RadioShack doesn't seem to carry such small resistors.