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FrozenGate by Avery
"Glowtorch-type" mod (200mw 660nm red)
- Thread starter chimo
- Start date
Thank's chimo for cropping my picture and make it nicer 
Can someone that have better knowledge of electronics help in drawing the schematics from the picture?
Gazoo, I believe that the LD is retained by ring from the back, and I don't know what did they use to held the ring against the diode. There is no notch in the retaining ring for twisting the ring, etc. You maybe right that simple push can knock the diode out, but there is no way to know without destroying the LD. I just didn't want to destroy good diode, and need to figure out some other way.
Maybe later when I have good powerful IR diode ready to transplant in there, I will try to mess up with it
Anyone know what is the highest power IR diode available in 5.6mm form?
Can someone that have better knowledge of electronics help in drawing the schematics from the picture?
Gazoo, I believe that the LD is retained by ring from the back, and I don't know what did they use to held the ring against the diode. There is no notch in the retaining ring for twisting the ring, etc. You maybe right that simple push can knock the diode out, but there is no way to know without destroying the LD. I just didn't want to destroy good diode, and need to figure out some other way.
Maybe later when I have good powerful IR diode ready to transplant in there, I will try to mess up with it
Anyone know what is the highest power IR diode available in 5.6mm form?
Chimo, I believe it is a current regulation without PD (Photo Diode) feedback as well.
I manage to make it brighter and burn better by turning the pot clockwise. There is no way to find out exactly how much brighter though, not until I get the LPM-1 from Daedal group buy...
I manage to make it brighter and burn better by turning the pot clockwise. There is no way to find out exactly how much brighter though, not until I get the LPM-1 from Daedal group buy...
Bernhard said:Thank's chimo for cropping my picture and make it nicer
Can someone that have better knowledge of electronics help in drawing the schematics from the picture?
You are quite welcome.
It looks like a common current control scheme (you could probably find a version at SAM's laser page) . It's difficult to tell with the part numbers sanded off the chips, but I think it functions like this:
The chip marked DES is a transistor that "throttles" the current to the LD. The small part that looks like a transistor is likely a stable voltage reference. The larger IC is likely an op amp/comparitor (LM358?) that compares the voltage drop across the 1.5ohm sense resistor to a percentage of the reference voltage (through a voltage divider network - the 5k ohm pot allows some adjustment of the voltage divider). If the voltage drop is too low, the op amp provides more base current to the LD drive transistor, if it is too high, it reduces the base drive.
The large, 4.7ohm resistor is just a voltage drop resistor. It could be increased in size to allow for RCR123s. As it sits, that resistor is dissipating around 200mW! This is not a very efficient circuit.
Paul
I think you are spot on regarding the IC type, chimo. On my first one, I can read first and second digit perfectly, and read a small bit of third digit, whick looks like a number 3.
On my second unit, the first, second, third digit are barely readable, and the forth digit show up a bit and I'm very positive that it is a 5.
To be certain, we only need to guess the last digit and compared it against the reference IC database from Philips. Further, once the schematics of this circuit is available, we can compare it to the reference circuit design of this IC somewhere from Philips website.
On my second unit, the first, second, third digit are barely readable, and the forth digit show up a bit and I'm very positive that it is a 5.
To be certain, we only need to guess the last digit and compared it against the reference IC database from Philips. Further, once the schematics of this circuit is available, we can compare it to the reference circuit design of this IC somewhere from Philips website.
I'm just wondering, which resistor need to be change to allow bigger current?
Will bigger current cause higher strain to that poor resistor, or it is the other way around, that it will have to "resist" lower current because of the higher current being supplied to the LD?
Will bigger current cause higher strain to that poor resistor, or it is the other way around, that it will have to "resist" lower current because of the higher current being supplied to the LD?
Gazoo
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Bernhard,
Thanks for the explanation. I went back and looked at the pictures. Now I see what you meant by the ring that is holding the diode in. It does look like it was pressed in and would be a pain to get out. But I think it could be done if one needed to replace the diode.
Sorry I don't know what the most powerful infrared diode is in a 5.6mm can. I did a quick search and the highest I could find was 200mw cw. It almost looks from the pictures the housing could be modified to fit a 9mm diode.
If you changed the 4.7 ohm resistor to say a 3 ohm (for example) more current would flow through the circuit. The question is how much current can the driver board carry without damaging it.
Thanks for the explanation. I went back and looked at the pictures. Now I see what you meant by the ring that is holding the diode in. It does look like it was pressed in and would be a pain to get out. But I think it could be done if one needed to replace the diode.
Sorry I don't know what the most powerful infrared diode is in a 5.6mm can. I did a quick search and the highest I could find was 200mw cw. It almost looks from the pictures the housing could be modified to fit a 9mm diode.
If you changed the 4.7 ohm resistor to say a 3 ohm (for example) more current would flow through the circuit. The question is how much current can the driver board carry without damaging it.
There are a few methods to get a higher current:
1. adjust the pot
2. change the sense resistor
3. change the voltage divider
With a higher current, the 4.7 ohm dropping resistor will dissipate more heat as well. What current are you planning on driving the IR diode at?
Paul
1. adjust the pot
2. change the sense resistor
3. change the voltage divider
With a higher current, the 4.7 ohm dropping resistor will dissipate more heat as well. What current are you planning on driving the IR diode at?
Paul
Bernhard said:
Bernhard,
Your calculation for power is correct, however, is "1W" a power consumption or optical output figure? The optical output will only be a fraction (very rough order of magnitude: 1/5 to 1/3) of the power consumed. Most of the power consumed is given up as heat.
The manufacturing/development push is to continue to creat more powerful packages with better efficiencies. The high-power LED industry has more than doubled efficiencies over the past few years.
Paul
Your calculation for power is correct, however, is "1W" a power consumption or optical output figure? The optical output will only be a fraction (very rough order of magnitude: 1/5 to 1/3) of the power consumed. Most of the power consumed is given up as heat.
The manufacturing/development push is to continue to creat more powerful packages with better efficiencies. The high-power LED industry has more than doubled efficiencies over the past few years.
Paul
Bernhard said:
I know there is something wrong in my calculation. That's why I keep wondering, where should the efficiency ratio fit in those figure? So assuming that the efficiency of LD is 25% (assuming from your data that it is getting the 780mW for 180mW output, and IR LD having the same efficiency lavel), it means the current will need to be 1600mA to drive 1W, is this correct? Don't know whether the circuit can withstand this, or it is capable of doing this with some mod...
The circuit seems well made and the pointer all together seems very sturdily put together. On the other hand... efficiency is very bad. I do not think I would leave the pointer on for too long for fear of the batteries blowing up! as it stands right now, I think I would be changing out the circuit on my pointer when it comes in. I don't like how this is being driven and I honestly think a better solution can easily be deduced. With a 6V supply, anything can be put in there. Effectively, a 4-stage current/voltage regulation circuit can be crammed on that board to make sure the diode doesn't see anything other than a ramp-up and ramp down at a very stable voltage and current. :-?
As about powering the IR diode with that circuit... I wouldn't go too far with that. It doesn't seem to be built efficient enough to supply that much and stay alive (the circuit or the diode).
Nevertheless, thank you both very very much for the pictures. They are very detailed, very clear, and gave me a very good idea of what to expect from the pointer that I will be getting soon
--DDL
As about powering the IR diode with that circuit... I wouldn't go too far with that. It doesn't seem to be built efficient enough to supply that much and stay alive (the circuit or the diode).
Nevertheless, thank you both very very much for the pictures. They are very detailed, very clear, and gave me a very good idea of what to expect from the pointer that I will be getting soon
--DDL
I did a little more schematic tracing...
Better shot with hot melt glue removed.
Pot side
Note that PD leg has been cut off
My shot at tracing the schematic. A real quick analysis: The pot wiper sets a voltage based on a fraction of the reference voltage. This voltage is compared to the voltage drop across the sense resistor. If the voltage across the sense resistor is too low, the second op amp provides more base drive current to the transistor to increase the current through the LD and the sense resistor. This circuit is likely less than 50% efficient with fresh batteries (it will become more efficient as the batteries die).
There are several vias that had to be traced by continuity checks. I upped the current from 208mA to 258mA.
Look what I picked up at a surplus electronics store (for those in Ottawa, that's "ADD Electronics"). The baggie in the background is 25-turn 10 ohm trimmer pots.
Better shot with hot melt glue removed.
Pot side
Note that PD leg has been cut off
My shot at tracing the schematic. A real quick analysis: The pot wiper sets a voltage based on a fraction of the reference voltage. This voltage is compared to the voltage drop across the sense resistor. If the voltage across the sense resistor is too low, the second op amp provides more base drive current to the transistor to increase the current through the LD and the sense resistor. This circuit is likely less than 50% efficient with fresh batteries (it will become more efficient as the batteries die).
There are several vias that had to be traced by continuity checks. I upped the current from 208mA to 258mA.
Look what I picked up at a surplus electronics store (for those in Ottawa, that's "ADD Electronics").
The baggie in the background is 25-turn 10 ohm trimmer pots. 
