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FrozenGate by Avery

DIY Homemade laser diode driver

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i got the whole circuit built but in the group bag of potentiometers at radioshack the lowest was a 1k so would that work, i would just turn it realy slightly
 





How many turns on the pot? 1 turn could be iffy, 25 turn would be OK'ish, don't forget to put a low ohm fixed resistor in series with it just in case though.

Regards rog8811
 
Ok, 2 questions...
1) What is the difference between the lm317 and the lm317t (is lm317t useable)?
2) What type of modifications do I need to perform to this driver in order to make it Blue-ray worthy?
 
I notice this setup uses 6volts, but a rechargable cr123 for DX outputs 3.6volts. Using 2 you get 7.2 volts, and when full charged 8 to 9 volts! suggestion to regulate this down? My diode takes 2.7 volts at 150mA. Also in the schematic where would you put the switch so that the LD doesn't blow up? to prevent the capacitor from doing something like that couldn't you just put a bleader resistor in?
 
guyfrom7up said:
I notice this setup uses 6volts, but a rechargable cr123 for DX outputs 3.6volts.  Using 2 you get 7.2 volts, and when full charged 8 to 9 volts!  suggestion to regulate this down?  My diode takes 2.7 volts at 150mA. Also in the schematic where would you put the switch so that the LD doesn't blow up?  to prevent the capacitor from doing something like that couldn't you just put a bleader resistor in?
Click to expand...

I know exactly what you are thinking, cos i was thinking the same way before bugging Daedal and Gazoo with a lot of questions..

You do not need to regulate the voltage down. The circuit needs 6V OR MORE. It will automatically regulate the voltage down to a value, where the current will allways be exactly what you set it to.

You could feed the circuit up to 37V and the voltage on the LD would stay the same.. The LM317 would heat up much more tho, since it would turn all the excess power to heat.


This circuit, as it is, is perfect for battery applications, as long as the voltage is ABOVE 6V. Below 6V the current would start dropping, so if you would reduce the battery voltage before this circuit, as the batteries would discharge, the voltage would drop and so would the current, below a certain point.

Using two rechargable lithiums, you would get almost 8.4V when they are full, and 6V when they are empty. This means the current on the LD would be constant over the entire battery capacity.



EDIT: Oh, and the switch should be put between the battery and the LM317. It doesn't matter if on the + or the -, the effect is the same. But for some reason i always put it on the + side.

The capacitor is there to protect the LD. It can only kill it, if it gets disconnected from the LD and charged. As long as you make sure the circuit is OFF and short the capacitor BEFORE soldering it to the LD, it can not kill it. After that it is recommended to never remove it from the LD.
 
Can't really read all of 34 pages without wasting huge ammounts of time so;
It would be normal to read ~5.4v across the diode, and it's fine to connect a LD to this high a voltage as long as the current it controlled?
 
Chris. said:
Can't really read all of 34 pages without wasting huge ammounts of time so;
It would be normal to read ~5.4v across the diode, and it's fine to connect a LD to this high a voltage as long as the current it controlled?
Click to expand...


I don't fully understand your question. 5.4V is not normal for a red laser diode.


This circuit regulates the current by regulating the voltage to the LD. Even when the LD heats up and it's resistance decreases, the current will stay the same, because the circuit would drop the voltage a bit.

Therefor the current ALLWAYS stays constant, as long as the imput voltage is ABOVE 6V. The voltage also stays constant, if you discount the variations from the heat. You do not have to worry about voltage, you just have to set the correct current..


It is also smart to test the circuit before trying it on a LD, by attaching an array of LEDs to it. This way you could measure the voltage and current and see, if it works fine. The voltage would NOT be the same as later on the LD, since the LEDs have a different forward voltage. But as long as you make sure the current is regulated, you don't need to worry about the voltage.

Then, before soldering the LD to the circuit, you have to make sure it it off, and short the capacitor!

Even safer is, if the capacitor is shorted and soldered to the LD directly.


I don't know how it is possible to measure 5.4V on the LD. Maybe it is damaged. Maybe something is wrong with the circuit and it killed the LD. Where exactly are you measuring this? What is the current?
 
Chris. said:
Can't really read all of 34 pages without wasting huge ammounts of time so;
It would be normal to read ~5.4v across the diode, and it's fine to connect a LD to this high a voltage as long as the current it controlled?
Click to expand...

Hmm, after reading your question again, i think i know what happened...

You powered the circuit ON without a LD on it, right?

Then you measured the voltage where the LD should be, am i correct?


In this case, it is most likely your circuit is doing just fine, but it can not regulate the current, since there is nothing for the current to flow through. This is why you see such a voltage.

If you soldered the circuit together correctly, it will work perfectly. Powering the circuit ON without a LD is not very safe, as this charges the capacitor to this high voltage (5.4V). This could kill the LD when you connect it.

Power the circuit OFF and short the capacitor, to discharge it!


Then, you can test the circuit on an array of a few LEDs, to make sure the current is regulated.


After that power the circuit OFF, short the capacitor and solder the LD to it and do not remove it again.
 
That's exactly what i'd done, thanks!
(I'd also left out the capacitor as I was just taking readings so no worries there)
Working on breadboard atm... I'm about to transfer this to vero-type stuff.
So, I'll try this with an LED.
I'm just worried as, e.g. a battery is not affected by a light load and will always read in the range of voltages you want to see.
 
Jeez, maybe we should make a step by step DIY website for all the noobs that do not feel like reading the whole forum. That way we can leave this area for questions other than "will the diode work if I put 5V through it? and will a cd burner work?". I will take this idea into consideration depending on the time I have. Daedal, I have to say, good job making this driver. It has gotten a lot of people involved.

Maybe we should give out LPF oscars or something like that. ;D ;D
 
Chris. said:
I'm just worried as, e.g. a battery is not affected by a light load and will always read in the range of voltages you want to see.
Click to expand...

200-300mA is not really a high current for batteries. The voltage will not be affected. Even at a much higher current the voltage would drop only a little on a full battery..


But i suggested to test it on an ARRAY of LEDs. One LED can only handle 20mA. If you solder 5 LEDs together, you can test the circuit up to 100mA and so on..


But from what you said so far it seems it's working.. It is much better to solder everything together tho, so you don't have to worry about loose contacts, that could even kill the LD (like if the capacitor get's disconnected from the LD even for a moment with the circuit ON)..
 
aaronX987 said:
Jeez, maybe we should make a step by step DIY website for all the noobs that do not feel like reading the whole forum. That way we can leave this area for questions other than "will the diode work if I put 5V through it? and will a cd burner work?". I will take this idea into consideration depending on the time I have. Daedal, I have to say, good job making this driver. It has gotten a lot of people involved.

Maybe we should give out LPF oscars or something like that. ;D ;D
Click to expand...


It's called instructables, he should of made it into an instructable.
 
To minimise the current initially, should the potentiometer be at its minimum or maximum value?
 
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