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03-31-2011, 06:08 PM #33
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Cyparagon
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Re: LM317 Driver DIY

^Yes, your design is good to go.

Quote:
 Originally Posted by rhd I don't think he can EVER get 1.5A. The circuit he referenced here has the pot in SERIES with the resistor. He's never going to escape the 11 Ohm R1.
R1 is .89Ω. The 11Ω I stated earlier was with the pot at full.

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03-31-2011, 06:20 PM #34
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rhd
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Re: LM317 Driver DIY

Quote:
 Originally Posted by Cyparagon R1 is .89Ω. The 11Ω I stated earlier was with the pot at full.
Ok, but I was replying to Jufran's statement:

Quote:
 Originally Posted by Jufran88 R1 = current limiting resistor = 11 ohms R2 = potentiometer = 100 ohms
R1 + R2 = 111 ohms when the pot is at full, and 11 ohms (plus a small minimum resistance of the pot) when the pot is at it's lowest.
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03-31-2011, 07:12 PM #35
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Cyparagon
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Re: LM317 Driver DIY

Oh I see. Yeah, that setup would be a big pile of useless
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03-31-2011, 07:24 PM #36
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Jufran88
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Re: LM317 Driver DIY

Ah, sorry for the mass confusion. I was just using those values as an example. But OP has .89 ohm set as the current limiter. When you calculate for maximum output you don't include the potentiometer in series.

Ok, here another example, not the values OP has.

I have set up an LM317 with a 1 ohm resistor in series with a 1K ohm pot (it was the lowest value potentiometer they had at radioshack without getting those bigger rheostats). I set up my test load for a red diode (Vf = 3V) since I don't have a power supply to achieve the 9V required for a 445nm LD or 405nm LD so instead I used 2 18650 batteries in series which should give me 7.2V

So, to calculate maximum output you need:

1.25/1 = 1.25A so that should be the value or get as close to it as possible.

but if you add the potentiometer in series:
1.25/(1+1000) = .0012A = 1.2mA

You only calculate the pot in series when your looking for mimimum output.

As you can see I'm outputting .909V or 909mA which is close to its maximum at 1.25A. Probably due to voltage sag in my batteries. However, if I had a bench PSU it would probably achieve the intended value.
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