Yes, another LM317 thread

[Abray]

I apologize if this has been covered anywhere. I searched, researched, and looked through all the first page threads and couldn't find anything similar, and now i'm stuck.

I just purchased the popular A 140 445nm diode, and my driver is having issues. Before deciding on a permanent host for it and all, I decided to at least get it running on some kind of a circuit. I made some blu-rays before so i had plenty of LM317s around.... I know they aren't the most efficient, but all I want right now is to see this beauty shine (at full potential).

I'm using the cookie cutter daedal circuit. 9V battery, LM317, 47mF 16 volt cap, diode for reverse current protection, and 3x 3 ohm resistors in parallel on the LM317. This should be supplying over an amp of current to the diode, but not surpassing the limits of the LM317, correct?

Here are my results...

I tested with a test load to make sure my circuit wasn't too messed up, ended up getting way lower current than hoped (about 130 mV was measured, so translates to about 130 mA assuming my voltage drop is reasonably close).

The LM317 and diode MIGHT've gotten warm when i left the circuit running for very long, but it also could've been my touching it too much, idk. Either way, insignificant heat.

Battery voltage BEFORE tests: 8.4 V

Ammeter hooked up in series with diode circuit turned on for about 5 seconds: ~220mA

Voltage across diode: ~3.84V

Basically, what am I doing wrong here. Why aren't I getting the "correct" (expected, desired) amount of current through the diode OR test load?

I know some basics about circuits, but nothing too complex...

Thanks for any help!

 

[jimdt7]

You can use these two calculators, they are very helpfull, i use them everytime i make an lm317 circuit.

LM317 Voltage Calculator

LM317 Current Calculator


Hope this helps you solve your problem !

Could you mind posting a picture of your circuit ?

 

[Abray]

I actually used that current calculator when I was putting the circuit together haha. And if this circuit should ultimately be regulating the current, is there any reason to calculate the voltage? It should be getting enough from the battery how it is set up, and we also tried putting multiple batteries in series so we actually had 21 volts....but same current across diode, and still no heat on any components.

I can't take a picture RIGHT NOW, but maybe in a couple hours when my roommate gets home. So it IS possible we have something on the breadboard messed up. I checked it tons of times though, and my roommate's been working on it with me too and he hasn't noticed anything wrong. I might try putting it together a different way, or swapping out certain components.

 

[Cyparagon]

A 9V battery cannot provide enough current for this diode.

 

[chipdouglas]

What cyparagon said.... The 9v batteries are not high current. They arent even good for 100ma builds.

 

[Abray]

Thank you! I figured it would be some stupid little problem like this haha.

Just out of curiosity, why can't it provide enough current? And is there any way to tell before hand which batteries can and can't?

EDIT:

I know a little about circuits, but this is obviously something i\I haven't learned about yet haha.

 

[Wolfman29]

Basically, the size of the battery dictates how much current it can shoot out compared to other batteries of the same chemistry. However, most people don't know that 9Vs are actually composed of 6* AAAA batteries, which are tiny little things... they can't supply that much current because they are so small!

 

[Abray]

Alright, I'll see if there's any way I can generate enough current for this circuit (stupid greedy LD!). Is there any "standard" set of batteries to get this much current at this voltage?

Also, is there anyway to calculate how much current batteries could put out, or is it more of a trial and error thing?

 

[Wolfman29]

There are some ways to calculate it, but it's all phenomenological, so it doesn't matter anyway. A good rule of thumb for li-ion batteries, the ones we use here, is roughly two to three times the current capacity (labeled as mAh) in current.

If you used two fully charged li-ions, that's 8.4V at max charge and most of them can handle 1.25A current draw except the smaller ones.

 

[Ghostchrome]

phenomenological

:slinks of to google:

 

[Wolfman29]

Basically, it means that the equations are determined by measurements anyway and not by some fundamental laws, so there is no point in using equations to determine other values, when you can use the values used to determine the equations instead