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Old 11-16-2007, 02:57 PM #1
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Default More power from the IR diode?

I recently made an LMT317 based controller as described here, and used a 10 ohm resistor.
The ~650nm diode is very bright but lacks in terms of burning power, so I decided to try the IR diode from the same drive.
Supprisingly it burns and smokes far more on the same 10 ohms of resistance
What's going on? Ive desoldered it now as I'm pretty worried about killing my eyes but.. > I wanted a RED burning laser not an invisible one

The drive spec was 8x dvd RW, 32x cd RW....

I'm thinking of finding a ~1 ohm resistor and using the pot to do all the work? (Its currently at 0 for both diodes)
With just 10ohms of resistance would my current be 3v/10ohm = 300mA which seems way to high?


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Old 11-16-2007, 03:12 PM #2
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Default Re: More power from the IR diode?

Are you sure you are feeding the circuit enough voltage? The drop over an IR diode is a bit lower than for a red, so that might explain the difference in radiometric power.

Apart from that, i doubt the IR would burn -that- much better at equal current, since they are not much more efficient.
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Old 11-16-2007, 03:21 PM #3
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Default Re: More power from the IR diode?

It's being 'fed' from 4 AA batteries at ~ 6v.
Should I try a 9v battery maybe?
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Old 11-16-2007, 03:48 PM #4
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Default Re: More power from the IR diode?

Chris, re-read the DIY driver thread. As well as some of Gazoo's posts about the 317. I don't think you're quite understanding how it works.

The sense resistor (Between the Vout and Adj pins) is used to control the current.

1250 / sense resistor rating in Ohms = current in mA

If you use a resistor and a pot (which is a variable resistor) in series, size the fixed resistor + the lowest resistance from the pot to give you your max current.

Example:
I wish the max current to be 400mA.
1250mV / 400mA = 3.125 Ohms (Google 'Ohms Law')
My pot at its minimum reads 1.5 Ohms (Some will be higher, some will be Zero! Make sure you check that it can't bottom out even for a moment if you twist it HARD or press on it)
3.125 - 1.5 = 1.625
2 parallel 3.3 Ohm resistors give me 1.65 Ohms.

So I need 2 parallel 3.3 Ohm resistors in series with my pot as this will let me drive at any current up to 397mA.

As for the power source, 4 AA alkaline cells won't last long at all, the voltage will sag to the point where the current to your LD starts dropping in just a few minutes. I like 6 NiMH cells, 4 E2 Lithium (Too expensive for me) if space is a concern, or 2 rechargable Lithium cells if you really want to cut down on the space.

My new favorite it 3 or 4 NiMH cells, an AMC7135, and a resistor parallel to the LD. But for more info look at Gazoo's work on the 7135s
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Old 11-16-2007, 04:06 PM #5
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Default Re: More power from the IR diode?

Quote:
Originally Posted by a_pyro_is
Chris, re-read the DIY driver thread. As well as some of Gazoo's posts about the 317. I don't think you're quite understanding how it works.

The sense resistor (Between the Vout and Adj pins) is used to control the current.

1250 / sense resistor rating in Ohms = current in mA

If you use a resistor and a pot (which is a variable resistor) in series, size the fixed resistor + the lowest resistance from the pot to give you your max current.

Example:
I wish the max current to be 400mA.
1250mV / 400mA = 3.125 Ohms (Google 'Ohms Law')
My pot at its minimum reads 1.5 Ohms (Some will be higher, some will be Zero! Make sure you check that it can't bottom out even for a moment if you twist it HARD or press on it)
3.125 - 1.5 = 1.625
2 parallel 3.3 Ohm resistors give me 1.65 Ohms.

So I need 2 parallel 3.3 Ohm resistors in series with my pot as this will let me drive at any current up to 397mA.

As for the power source, 4 AA alkaline cells won't last long at all, the voltage will sag to the point where the current to your LD starts dropping in just a few minutes. I like 6 NiMH cells, 4 E2 Lithium (Too expensive for me) if space is a concern, or 2 rechargable Lithium cells if you really want to cut down on the space.

My new favorite it 3 or 4 NiMH cells, an AMC7135, and a resistor parallel to the LD. But for more info look at Gazoo's work on the 7135s
I see what you mean now, current is controlled via the 1.25v drop between the Vout and Adj pins.
So I have a max of 125mA which seems fine for an 8X actually, I'm going to stick another 10ohm resistor in parallel to give me 250mA max. Thanks =]
I was really just supprised by the power of the IR diode, I thought it would be far less powerfull.
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Old 11-16-2007, 04:22 PM #6
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Default Re: More power from the IR diode?

No prob.
With my 8X, 140mA was just right. It didn't get too hot, and I could run it for 2 minutes or so. Alas, it died at 175mA. but that doesn't mean anything for your diode, just be careful as I think 250mA would not be kind to an 8X diode.
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Old 11-16-2007, 06:59 PM #7
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Default Re: More power from the IR diode?

Found a 6 ohm resistor for now, the IR LD is still going @ 208mA Cuts reasonable(~3mm) holes in bin liners (no lens, close range).
I always expected the IR diode to be weaker than the red LD, and it may be, haven't re-soldered the red LD yet.
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