Where do I connect wires on this board?

TigerhawkT3

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It's the 350mA DX 20-pack.



All I really need to know is where to connect the LED +ve and -ve, as the battery will connect to the contacts on the board's underside.
 
the pics for the other drivers (mainly identical but with more of these black three feeted parts for more power) show:

red wire: middle contact
black wire: soldered at "Q2".

if red is "+" and black is "-" (as it should be), these are the points to use
 
I tried those spots, and they didn't work. I don't know if it was that none of the pads I tried were correct, or if the first one was wrong and blew out the board. Q2 does sound right for the -ve, but the middle contact (near the diode) connects to Vin, so it's not that. I'll probably register at DX and post my question there to see if anyone (perhaps "Kyle" himself) can answer.

I'm just glad these things were only $0.75 each. :)
 
If you click on the pictures at dealextreme you will see that "yellow" is right: positive in the middle and negative at Q2.
 
When all else fails, go to the spec sheet.



The pad at Q2 is LED-
The center contact on the back of the board is Batt+ and LED+
The outer Ring with all the vias is Batt-

The diode on board protects the 7135 from reverse polarity. Vdd should not draw much current.

Note that this driver completes the ground circuit.
 
Interesting - I've never heard of a board with a combined batt and LED terminal; most other have separate contacts for Vin, GND, LED+, and LED-. I'll try the "shared" setup with a new board, and we'll see what happens.
 
:crackup: Well, it turns out that I since I had the batteries in backwards in my fixture to get the polarity I wanted, the reverse polarity protection was preventing contact. :ohgeez: A solder blob later and some fiddling later, and I have a much brighter book light. Granted, it uses more power (around 400mA instead of 250), but the output is white and powerful.

And yes, LED+ connects to Vin, with LED- connecting to Q2. Thanks, everyone!

I've been doing some other mods lately as well, so I'll start a new thread with pictures in the Homemade/Modified section.
 
this looked too simple to be a driver board dont you think? an IC and a zener....wheres all the other components
I don't think it's a zener. Just an ordinary diode, as chimo said. And I think it's very clever the way they've rigged it to work with several of the reg chips, with solder bridges to connect the extra regulator ICs. Also, it's a lovely photo of the board TigerhawkT3.

Granted, it uses more power (around 400mA instead of 250), but the output is white and powerful.
Is that a measured 400mA?

Edit: BTW, you should add a www. to the start of your link.
 
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I don't think it's a zener. Just an ordinary diode, as chimo said. And I think it's very clever the way they've rigged it to work with several of the reg chips, with solder bridges to connect the extra regulator ICs. Also, it's a lovely photo of the board TigerhawkT3.
Thank you very much!

Is that a measured 400mA?
No no, just a very rough estimate. The DX page says it drives the emitter at 350mA, so at between 80-90% efficiency, it would be around 400mA draw from the cells. Very rough estimations, I know, but also very easy! :)

Edit: BTW, you should add a www. to the start of your link.
:thinking: Works fine for me.
 
No no, just a very rough estimate. The DX page says it drives the emitter at 350mA, so at between 80-90% efficiency, it would be around 400mA draw from the cells. Very rough estimations, I know, but also very easy! :)
You might misunderstand how it works. The current through the protection diode is only a few microamps, so unless you want to work things out to several more significant figures than is really necessary, it can mostly be ignored. Next, it's a linear regulator, which means that it sort of acts like a variable resistor, dropping the unwanted volts and dissipating that power as heat. The closer to the dropout voltage the supply voltage is the better the efficiency will be. Does that make some sense?

:thinking: Works fine for me.
Are you accessing this page thru www.candlepowerforums.com or just candlepowerforums.com?

Edit: See http://www.candlepowerforums.com/vb/showpost.php?p=2028873&postcount=1 point 9.
 
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You might misunderstand how it works. The current through the protection diode is only a few microamps, so unless you want to work things out to several more significant figures than is really necessary, it can mostly be ignored. Next, it's a linear regulator, which means that it sort of acts like a variable resistor, dropping the unwanted volts and dissipating that power as heat. The closer to the dropout voltage the supply voltage is the better the efficiency will be. Does that make some sense?
Indeed it does!... So, what's the upshot of that? Would it mean that the closer my Vin is to Vf, the better my efficiency? What if Vin is lower than Vf - what does it do then?
Are you accessing this page thru www.candlepowerforums.com or just candlepowerforums.com?

Edit: See http://www.candlepowerforums.com/vb/showpost.php?p=2028873&postcount=1 point 9.
I'm not going through "www." So, should I put a www. before each link I post, even if it doesn't appear to be necessary? For example, the above link works just as well as http://candlepowerforums.com/vb/showpost.php?p=2028873&postcount=1 (no www).
 
Indeed it does!... So, what's the upshot of that? Would it mean that the closer my Vin is to Vf, the better my efficiency? What if Vin is lower than Vf - what does it do then?
:tinfoil: OK... Vf being the operating voltage of the LED and Vin the supply voltage?

The spec sheet claims a 0.12V typical dropout voltage. That's the voltage that the Vout pin will be at. Maximum efficiency will be just before it drops out of regulation, when Vin = Vf + 0.12V. With a Cree at 350mA, 3.3V it'll be 96.5%. Roughly. With a supply at 4.5V it'll be about 73%, but that'll rise quickly as the battery voltage drops. With AAA cells it'll be up into the mid 80s within minutes.

When Vin drops under the required regulation level the LED and the regulator appear to do a tricky balancing act that depends a lot on the operating voltage of the LED for any particular current. I think it has the net effect is the current dropping a fair bit quicker than it would with direct drive. I'll do a bit more testing when I get some flatter batteries.

I'm not going through "www." So, should I put a www. before each link I post, even if it doesn't appear to be necessary? For example, the above link works just as well as http://candlepowerforums.com/vb/showpost.php?p=2028873&postcount=1 (no www).
People sometimes get annoyed if they go to a thread and find out they can't post because their browser or their cookies or something think it's a different server and they're not logged in. Edit: Besides, the solution to that "fault" was (IMPORTANT - please include www. in ALL LINKS) so it seems to be the accepted convention.
 
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That polarity protection diode is feeding the Vdd pin, which wants at least 3V. With 0.6V drop across the diode, Vin at anything under 3.6V could give problems.

With three NiMH cells at the base rate of 350mA it'll be OK, since if they drop under 1.2V each they don't have much left anyway. With alkaline, half their life is under 1.2V, or NiMH AAA cells at 1A will drop to about 1.1V each by the end of their life. In those situations it might be worth removing/shorting the diode if you plan to use the board with an alkaline battery.
 
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