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FrozenGate by Avery

LPC-815 red module powered by single Li-ion + linear driver

laserdudephil said:
There's something very wrong with this picture (schematic.) If you're not providing more than 4.3V then that 4.3V zener in the reference divider is not regulating. What that means in plain english is that the laserdiode power varies with battery voltage and that's no longer a constant current regulator.
Click to expand...
1. I cannot say if the part number is correct. What I've written there is what the SMD component database told me. It's in SOT-23 package, has 3 pins and "431" written on it. If you provide better information + datasheet about that component I'll fix it.
2. The part of your post which I highlighted in red is, in fact, very wrong. This driver (on the schematic) was confirmed, not only by me, to provide constant current at input voltage of over 2,7V.
laserdudephil said:
You should use this circuit with a minimum of 6V (2 x 3.0V Lith. cells)
Click to expand...
I'm leaving this experiment to you. I don't have unlimited supply of these drivers.

And by the way, what role do you think has the 0,33Ohm resistor? The one connected to the inverting input of the Op-Amp.
 





Ah, now I understand. According to NXP that would be a 4.3V zener. But what you probably have instead is a National Semiconductor LM431:
LM431 - Adjustable Precision Zener Shunt Regulator
which is a 2.5V reference.

The 0.33 Ohm resistor is a current sensor. The laserdiode current must pass through this resistor. This creates a voltage drop across the resistor which is compared by the op-amp to the reference voltage. The op-amp then ramps the voltage up or down to compensate.
-Phil
 
laserdudephil said:
Ah, now I understand. According to NXP that would be a 4.3V zener. But what you probably have instead is a National Semiconductor LM431:
LM431 - Adjustable Precision Zener Shunt Regulator
which is a 2.5V reference.
Click to expand...
Thank you ... LM431 seems a lot more logical compared to the 4,3V zener (which was non-sense). I fixed the schematic.
The bad part is that this makes the schematic a bit more difficult to replicate.
 
Actually National Semiconductor is very nice about sending out free samples. Don't wear it out thought. :shhh:
 
Hey i know this was probably already talked about, but honestly i dont have time to go through the whole thread. One of my lasers is the 50mW greenie for DX (the true green pointer) so i am figuring the driver is the same from the module you bought. My question: if i remove the present resistor, and replace it with a a 3ohm Res, then hook up a 20x diode (keep in mind i am not touching anything other than the resistor) will it produce over 200mA you think? And i have already tested the potentiometer....on this driver from DX, i cant seem to bypass it--its a set current flow i think form the factory. So any way, that was my question. Like i said, sorry if this was already talked about
 
oh my bad i didnt mean 3Ohm, i was thinking of another project as i was typing.lol. I havent looked at mine yet, but i am assuming the resistor is the same..so you tell me-what should i replace it with..only though if what i said is right ^^
 
creanzworld17 said:
Hey i know this was probably already talked about, but honestly i dont have time to go through the whole thread. One of my lasers is the 50mW greenie for DX (the true green pointer) so i am figuring the driver is the same from the module you bought. My question: if i remove the present resistor, and replace it with a ...
Click to expand...

Lets do the math. What is the reference voltage range in the circuit shown above. If the pot wiper is all the way at the top, it's 2.5V * 5000 / 25000 = 2.5 / 5 = 0.5V. If it's all the way at the bottom, the reference voltage is 0. So what current flows through a 0.33 ohm resistor which is dropping 0.5V? I=V/R 0.5/0.33=1.5A The circuit as shown would have a range of 0 to 1.5A. If the current sening resistor was 3.3Ohms, the range would be from 0 to 150mA. 'hope this helps. Don't assume that your circuit matches this schematic
and you'll be fine. Measure twice and cut once.
-Phil
 
creanzworld17 said:
My question: if i remove the present resistor, and replace it with a a 3ohm Res, then hook up a 20x diode (keep in mind i am not touching anything other than the resistor) will it produce over 200mA you think? And i have already tested the potentiometer....on this driver from DX, i cant seem to bypass it--its a set current flow i think form the factory.
Click to expand...
What do you mean? The potentiometer is not working ? Or glued ?
If this is not the case you can just set the current using the potentiometer without replacing the 0,33Ohm resistor. It will be a bit touchy to set the current so it's recommended to use test load.
If the pot is not working it will be hard to set the current at the desired value. laserdudephil explained pretty good how the resistor, you want to change, affects the circuit.
 
lazerov said:
What do you mean? The potentiometer is not working ? Or glued ?
Click to expand...

I am not quite sure. On my other drivers from DX, the potentiometers are accessable. But this one i am talking about, i cant seem to do anything to affect it. I say that, because all the lasers i get from DX, "i take it apart" and get to the pot, and usually amp it up a few more mW's. This driver i cant get to change. Kind of annoying, becasue like you said, i then cant do anything to increase or decreas mA ...and no, it is not glued, it can be turned left and right--it just doesnt do anything to the output of the laser
 
creanzworld17 said:
But this one i am talking about, i cant seem to do anything to affect it....
Click to expand...

I have to agree with Laserov. Your pot's probably shot. But it's got to be shot in a weird way. You see that 10k resistor next to the pot in the schematic? If the pot is shot -usually meaning that the wiper is not contacting the resistive strip,- that resistor in parallel with the resistive strip's (in the pot) 10k ohms equals 5k ohms. In other words, the circuit sees the pot as fully on. Since that would likely blow out your laser diode, I have a hunch that someone at the factory -if you can call it that- rather than replace the pot, soldered a different resistor across where that 10k one goes. Other than that I'm stumped.
 
@ Laserdudephil:
If the wiper wasn't touching the resistive strip, then the driver wouldn't work at all. I mean, it would have been only R3 (10k) connected to the non-inverting input of the Op-Amp, thus 0V there. There is some possibility that the pin of the potentiometer which should be connected to (-) is not properly soldered. This way you'll have some (dangerously high) output, but way less adjustable. We can only speculate on this, anyway.

@ Creanzworld17:
I recommend strongly against using the driver with the not working pot. If the pointer with that driver works fine, use it that way.
You can use a working driver without modification, but, as I already said, the current adjustment is going to be very sensitive. You can replace the power resistor with some between 0,51 - 0,68Ohm, if you want more precise adjustment. Originally the resistor should be 0,33Ohm (color code: Black,Orange,Orange + 2 other color bands). Don't forget that with a red LD you'll not be able to use 3V to power the laser. At least 3,6-3,7V are needed (one li-ion is fine). Also a red diode has different pin-out than IR one, so you should take this into account.
Btw, pics are not working for me, too.
 
lazerov said:
@ Laserdudephil:
If the wiper wasn't touching the resistive strip, then the driver wouldn't work at all. I mean, it would have been only R3 (10k) connected to the non-inverting input of the Op-Amp, thus 0V there. ...
Click to expand...

Absolutely right. I must have been low on sleep and thought that the wiper was tied to V-. However if the pot was open on the V- side, the adjustment would bring the noninverting terminal between 1/4 and 1/3 of 2.5V =
0.625V<Vref<0.833
1.89A<iLD<2.52A
which is clearly way too high. 'must be something else.

-Phil
 
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