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FrozenGate by Avery

DIY Homemade laser diode driver

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GreenLaser said:
hey guys will my diode driver work if i use a 1n4004 diode a 200 ohm pot and a 10ohm resistor instead of a 4ohm???
cheers
Click to expand...

Why don't you ask all your same Driver questions on the Threads you
have already opened asking basically the same questions...:thinking:

You seems to look like a kid running from room to room looking for
the same answers that were already answered... :undecided:

How many times are you going to ask the same questions....:thinking:
The answers are not going to change....:wtf:

how do i find out if they are 0.25% 0.5% 1% 2% so then i could find out the ohm value.
Click to expand...

Buy a $7.00 Digital Multi Meter......

http://cgi.ebay.ca/BRAND-NEW-CEN-TECH-7-FUNCTION-DIGITAL-MULTIMETER-/330430485277?cmd=ViewItem&pt=LH_DefaultDomain_0&hash=item4cef31131d



Jerry
 
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Does anybody know if this driver will work with the casio a140 diodes? I would like to run at about 1w. Any modifications needed?

Thanks
 
They will work up to 1.5 amps. Be sure to provide ample heatsinking and power resistors to handle the current though. Realistically there are much better drivers to choose from for those amounts of current, mostly omitting the 3+1.25V voltage drop required in this desgn.

If you want to drive a 445 from 12 volts this design will prove to be a good solution, but keep in mind that te voltage regulator will burn more power than the laser diode, and will require cooling to match!
 
Thanks for the reply,

benm do you have a link to a schematic for the better design you were referring to.
 
Hi

was thinking if i changed to a non electrolytic capacitor, would it discharge itself through the diode when the power is switched off?
 
Benm said:
They will work up to 1.5 amps. Be sure to provide ample heatsinking and power resistors to handle the current though. Realistically there are much better drivers to choose from for those amounts of current, mostly omitting the 3+1.25V voltage drop required in this desgn.
Click to expand...


Dear All,

From what I understand in order to calculate the V of the powersource needed to fuel said driver (from the original post) one has to follow the equation:

Vin = Vlaserdiode_drop + V1n4001_drop(0.7V) + Vresistor_drop(equals to Vref=1.25V) + Vlm317_drop(~3V)

So in order to drive a Laser diode at 4V folrward voltage drop


Vin = 4V + 0.7V + 1.25V + 3V = 8.95V

Am I correct? Does the drop across the resistors in the circuit which is equal to Vref=1.25V needs to be included in the calculation?

P.S.

This has been probably asked and has been answered before and I apologize if I asked it for a hundredth time in this thread... I tried to search through it with several keywords but would not come up with the info I needed. And looking through 54 pages of posts is juuust too much! So please forgive my n00bish approach.
 
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Yes, the drop on the resistor must be calculated together with all the others, too ..... and the drop of the LM317 is usually 2,5V, but calculating half volt more is always a good thing to do.

By the way, if you need a less dropout, the LM1117 (not 117), have a typical dropout of 1,2V, usual on load of 1,5V (in case you need some more extra voltage for a diode that requires a bit more, and you don't have it from the PSU ;)) ..... and is pin-to-pin compatible with the LM317.
 
HIMNL9 said:
Yes, the drop on the resistor must be calculated together with all the others, too ..... and the drop of the LM317 is usually 2,5V, but calculating half volt more is always a good thing to do.

By the way, if you need a less dropout, the LM1117 (not 117), have a typical dropout of 1,2V, usual on load of 1,5V (in case you need some more extra voltage for a diode that requires a bit more, and you don't have it from the PSU ;)) ..... and is pin-to-pin compatible with the LM317.
Click to expand...

Thank you!
 
You must also make sure your resistors and/or pot are rated to handle that kind of current...

A 1/2 Watt resistor or pot is rated for a max of 400mA's with an LM317: (1.25 volts at sense pin)

1.25 X .400 = .5

So if you wanted 1500mA's, you would need a 2 Watt resistor.

1.25 / 1250 = .001

So 1 Ohm 2 Watt resistor with an LM317 would give 1250mA's...
 
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Bluefan said:
Why the extra V1n4001_drop(0.7V)?
Click to expand...

The rectifier diode is part of a circuit and works as a safeguard in case you insert the battery the wrong way in :)
jayrob said:
You must also make sure your resistors and/or pot are rated to handle that kind of current...

A 1/2 Watt resistor or pot is rated for a max of 400mA's with an LM317: (1.25 volts at sense pin)

1.25 X .400 = .5

So if you wanted 1500mA's, you would need a 2 Watt resistor.

1.25 / 1250 = .001

So 1 Ohm 2 Watt resistor with an LM317 would give 1250mA's...
Click to expand...
Jayrob those calculations are not entirely correct I think. The Ohms law: I=V/R R=V/I V=I*R and you are doing: V*I :)

If you want to calculate the W that resistor should handle you should probably follow this: W = V^2/R or W = A^2*R

So for 0.4A on LM317 R = 1.25/0.4 = 3.125ohm and W = 0.4^2 * 3.125 = 0.5W

And if you want 1.5A R = V/I = 1.25/1.5 = 0.8ohm and W = 1.5A^2*0.8ohm=1.8W

LOL Nevermind i missed the obvious conversion W = VA hehehehehe I'm and idiot and you were right :)
 
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Hey, how do we calculate the ohms of the resistor we need? im not using a POT
im using a 240mw 660nm laser diode
 
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