Problem with the LM317 DDL

Hi,

I just finished assembling the DDL Driver based on the schematic images on Rog8811's site.

I am having readings that do not agree with the instructions.

I am not getting the 1.25V reference voltage across the middle pin of the 317 to after the resistors.

I am getting the full input voltage at the output from the driver circuit, where the LD will sit.

I am confused about the two diagrams, the assembly, and the testing diagrams.

I think there might be a discrepancy between the two diagrams. on the assembly diagram it shows the middle and last pin on the pot to be commoned, where as on the circuit test diagram it shows the first and middle pins to be commoned.

And on the Veroboard layout the resistor is before the pot with the first and middle pins commoned.

Am I reading these diagrams incorrectly?

Which one is correct?

Heres a link to his page just incase,

Laser driver - It can be done

if you need me to circle the parts of the circuit that I think are different, I'll do that and post them here later on tonight.


thanks for any insights you can offer..

Pictures will help immensely so we can see your wiring.

I see this problem come up a lot, and I haven't been active on this site for very long! (I was registered 4years ago but never active, and forgot that account).

Anyway, I think the problem stems from people who don't normally look at schematics not knowing exactly what the schematics mean. To alieviate this I have drawn up a schematic with the common LM317 Constant Current Driver configuration on it twice;
1) Shown how it normally is, with explanation of where the different voltages are.
2) Shown arranged differently but electrically the same. It may look very different to the layman's eyes, but it is really exactly the same. This time it should eliminate errors commonly run in to by those who "just replace the lines on the schematic with wires". A common mistake is to put the load between OUT and GROUND when it should really be between ADJUST and GROUND.





EDIT: Also, I wanted to say that you do NOT need to use all 3 pins on a pot!!!! I see this all the time and there really isn't much call for it. When using a Pot as a VARIABLE RESISTOR and not as a VOLTAGE DIVIDER you only need TWO PINS. The inner pin (the "wiper") and one of the outer pins. You can choose which outer pin you want based on which way you want the pot to turn to increase resistance. With the part you turn facing up at you and the pins facing towards you, if you use the left and center pin, resistance will increase when you turn it clockwise. If you use the center pin and the right pin, resistance will decrease when you turn it clockwise.

So, if you wanted to use a pot on a constant current driver like for a laser diode, you could use the center pin as "in" and the right pin as "out" of the pot, and when you turn the knob to the right (clockwise) it will increase the current being supplied.

Ignore the output voltage. It's a constant current driver. You need to run it with a load to measure the current. Your ammeter itself can serve as a load.

Hello, heres a photoshop sketch of my understanding of the three diagrams.



I built my driver same as the assembly diagram.

my confusion is whether or not the other two are electrically the same.


@Sigurthr: thanks for the tips about pots, I did not know that. I am very amateur with electronics of this nature. by your description, I should be able to break the common between the middle and third pin, and make the middle pin the "out" and that should solve everything.

Heres a macro of my build sofar, compactness was my prime concern, hehe.



it's for a 20mA 405nm, my resistor is a 12.5ohm, so by this, my driver should have a max of 100mA output.

unless anyone disagrees, I'm going to follow my understanding of Sigurthr's advice.

thanks!


p.s. I think old CAT5 makes great solid copper wire stock.

I personally never like using parallel diodes for reverse polarity protection. Even though it will shunt most of the current in the event of a polarity reversal, there will still be a reversed voltage across the diode that way, and that is enough to kill some LDs. Just put it forward biased in series with the LM317's IN pin, the silver strip pointing towards the LM317 and away from the positive battery terminal. That will provide reverse polarity protection as well. You will incur a 0.7v drop though, so you would need to provide 3.7V higher than your laser diode's Vf. Switch it out with a schottky if you can.

I would also add a second cap before the LM317 for input filtering. IMO input filtering is more essential than output coupling with these type of drivers.

All the drawing will work, but depending on which orientation you have the pot, it may need to be turned the opposite direction for a current increase.

Sigurthr said:

I personally never like using parallel diodes for reverse polarity protection. Even though it will shunt most of the current in the event of a polarity reversal, there will still be a reversed voltage across the diode that way

Click to expand...

Yeah, <1V. The vast majority of laser diodes can handle <1V reverse voltage.

Thanks for the help all, I hope to get some time tonight to make the modifications and retest the circuit.

Cyparagon said:

Ignore the output voltage. It's a constant current driver. You need to run it with a load to measure the current. Your ammeter itself can serve as a load.

Click to expand...

Just thinking about this. if the diode is a 405nm, so requiring ~4.5V, and I'm measuring the entire 10V that I'm putting into the driver across the test load, would that not mean that the LM317 is not taking its 3.7V ? could this 10V at the output burn the diode?

or is it that the LM317 will allow nothing through it at all until the threshold Voltage of 3.7V is reached, then it all flows through at a regulated current?

so the 10V measured at the diode location means nothing because the current is regulated so low?

please forgive me if this question is ignorant to common electronics knowledge.

Jimmy996 said:

Just thinking about this. if the diode is a 405nm, so requiring ~4.5V, and I'm measuring the entire 10V that I'm putting into the driver across the test load, would that not mean that the LM317 is not taking its 3.7V ? could this 10V at the output burn the diode?

or is it that the LM317 will allow nothing through it at all until the threshold Voltage of 3.7V is reached, then it all flows through at a regulated current?

so the 10V measured at the diode location means nothing because the current is regulated so low?

please forgive me if this question is ignorant to common electronics knowledge.

Click to expand...

The reason it is showing the full 10V is because your circuit is WRONG.



I had mentioned this a few posts ago:

Sigurthr said:

A common mistake is to put the load between OUT and GROUND when it should really be between ADJUST and GROUND.

Click to expand...

thank you so much Sigurthr!

I'll go fix this up.

Anytime! I know electronics can be daunting to the beginners, and simple mistakes like this are hard to spot by the untrained eye.

Glad to help!

P.S. if you notice I fixed the pot set up as well in the drawing, hehe. If you're gonna do it, might as well do it right!

There are basically two modes of operation for power supplies. Constant voltage, and constant current. Constant voltage supplies keep the voltage constant, and will supply whatever current is required to keep the voltage stable. Constant current supplies keep the current constant, and will supply whatever voltage is required to keep the current constant.

You can't measure the current of a constant voltage supply (unless there is a load), because that leads to a massive draw in current. Similarly, you can't measure the voltage in a constant current supply (unless there is a load), because that leads to a large surge in voltage .

Trust me. Ignore the voltage reading. Just hook up your ammeter to the output. Measure the current, not the voltage.

Jimmy996 said:

and I'm measuring the entire 10V that I'm putting into the driver across the test load

Click to expand...

Cyparagon, you are completely correct. However, I thought he WAS measuring with an appropriate load.

Yep, I'm using the simulated load of 6, 1N4001 in series with a 1hom resistor.

Across the resistor, the voltage drop was measuring as expected. ( according to rog8811's instructions) between 10mV and 98mV as the pot gets turned.

anyhoo, Im going to rebuild the circuit as Sigurthr correctly re-drew up for me, and report back with my findings.

thanks again everyone!


Edit:

Sanity Check..

heres a Pinout image of the LM317

According to this, I do have it set up correctly..

http://www.afrotechmods.com/reallycheap/UPS/LM317.jpg

yes/no ?

Sorry, I missed that part.