Old 11-15-2014, 06:27 PM #1
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Default step down from 12v BEFORE multiple LM317 drivers?

I have 8 LM317 laser drivers based on the rog8811 "It can be done" circuit.
These will be powered in parallel by 12v DC power.

Given the "ideal 7.2 voltage" suggested by rog8811, I'm wondering if it would be more efficient to have an initial upstream regulator to step down the 12v to 7.2v which would then feed my 8 drivers. By "more efficient" I mean generating less waste heat by my regulators overall.

My lasers are being used as short length trip wires, and as such I have dialed them down to low current, about 11ma, if that makes a difference.

Thanks for any advice.


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Old 11-15-2014, 06:32 PM #2
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Default Re: step down from 12v BEFORE multiple LM317 drivers?

Yes, it would be more efficient to step the voltage down before the LM317's. You can probably go lower than 7.4V even. Set it to a couple volts above the diode voltage.

Edit: Assuming you're using a buck converter to step down the voltage. It would be the same efficiency with linear regulators.

Last edited by ARG; 11-15-2014 at 06:58 PM.
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Old 11-15-2014, 06:50 PM #3
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Default Re: step down from 12v BEFORE multiple LM317 drivers?

If you use a linear regulator to drop the voltage from 12V to 7.4V , the overall heat generated will be the same just spread abit more out over the 9 regulators .

with out dropping the voltage , if you are running 11ma per regulator that's 3 volts over the diode ( ruff guess on red diode worst case ) , 1.216V over the sense resistor and then 7.8V over the regulator .

At 11ma that's 0.0858 W of heat per regulator

>>

With the 12V to 7.4V regulator ( assuming its linear ) , that's 88mA though it ( 8* 11mA ) so with a drop of 4.6V , that's 4.6*0.088 = 0.4048 W

With that in mind , it would then be 3V + 1.216V = 3.2V over regulators

so 3.2*0.011 = 0.0352 W pre regulator .

to be honest if you are using a linear regulator to drop it from 12V to 7.4V , unless you really need to reduce the tiny amount of heat generated in the LM317 then go for it but for a heat generation of 0.0858 W I don't think it would be worth it as all you are doing is removing around 0.05W of heat from each regulator and dissipating it in the 12V - 7.4V one


If its a switching regulator for the 12V-7.4V then it would be abit more efficient overall with around 0.035W dissipated by each of the 8 regulators.
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Last edited by DashApple; 11-15-2014 at 06:59 PM.
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Old 11-15-2014, 07:35 PM #4
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Default Re: step down from 12v BEFORE multiple LM317 drivers?

Thanks for the replies!!
Looks like the "ideal 7.2v" got transcribed as 7.4v in the replies, but it matters not.

So just to fill in the more obvious parts of ionlaser555's math...

For the 8 drivers using 12v directly scenario,
"At 11ma that's 0.0858 W of heat per regulator"

so for the 8 drivers:
8 * 0.0858 = 0.6864 W total

For the pre regulator 12v to 7.4v step down scenario,
"that's 4.6*0.088 = 0.4048 W" for pre regulator

"3.2*0.011 = 0.0352 W " per driver regulator

so for the 8 drivers:
0.0352 * 8 = 0.2816 W driver regulators combined

so for the pre regulator and 8 drivers:
0.4048 + 0.2816 = 0.6864 W total


So as both replies suggest, both scenarios generate the same total wasted 0.6864 W, assuming a linear regulator is used.

I would not be using a buck/switching regulator, nor am I trying to reduce a tiny amount of heat. I just wanted to make sure I'm not doing anything obscene.
I feel much more confident.

Last edited by muniorbust; 11-15-2014 at 07:35 PM.
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