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help with simple ddl driver

bmw

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I made a simple ddl driver out of a diode (for revers voltage protection), a 16v 47uf capaciterfor voltage spikes),and a resistor(for current setting). I am going to use a loc diode and i want to run it 400ma. what value of resistor should I use? Will it work without a lm317?
 
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bmw

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i am not using a lm317 right now. do i have to use one. I was trying to do it with out a lm317 so i could use 2AA.
 
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i am not using a lm317 right now. do i have to use one. I was trying to do it with out a lm317 so i could use 2AA.
It would help if you let us know what you are using besides a
Rectifier Diode and a Capacitor and a Resistor...

Only those parts will not regulate the current to a Laser Diode
over the battery voltage variations over time...:cryyy:

Jerry
 

rhd

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lol

It's tough to make an LM317 DDL circuit, without an LM317 ;)

Can you have a BMW without having a vehicle?

** Philosophy hurts head **
 

bmw

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lol i feel stupid:eek::banghead::oops: . well If i have a lm317 can I still use 2AA?
BTW, B.M.W. is my initials lol :D
 

rhd

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Nope, 2AA won't give you enough voltage.

With an LM317, to drive a LOC, I suspect you would need 4, maybe 5 AA batteries.
 

bmw

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ok i tried a 2.7 Ohm resistor but I only get 350ma. why is that?
 
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2.7 ohms will give you around 470mA, possibly more. Not good. It'll kill your diode fast.

Use 3 ohms, or 3.3 ohms. Also, use lithium/lithium ion batteries. AAA's won't cut it. Sometimes, neither will AAs (bad quality, and/or empty ones).
 

bmw

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i am using 4AA and a 2.7 Ohm resistor only gives me 350ma
 
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You are not listening.

First of all, You're not feeding enough voltage to the driver. Well, it's there on the limit, but batteries could be drained a little.

Second, forget primary batteries. If what you're using is not rechargeable, it'll cost you A LOT OF MONEY! Trust me, I know firsthand.

Third, use 3.3 ohm unless you want your diode to die anywhere from instant to 1 minute runtime.
 

bmw

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i am using alkaline AAs and i get about 6.5v. So according to Eudaimonium that is not enough voltage and or current to power this circuit properly. so if I power it with 2 3.6v AAs it should put out 400ma with a 3.1 Ohm resistor?
 
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i am using alkaline AAs and i get about 6.5v. So according to Eudaimonium that is not enough voltage and or current to power this circuit properly. so if I power it with 2 3.6v AAs it should put out 400ma with a 3.1 Ohm resistor?
If you mean, power them with two 14500 batteries (which are lithium-ion equivalents of AA size cells), yes it will work.

Fully charged lithium ion cells give off 4.2V of surface charge, too. Just something to keep in mind.
 




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