Why are we using test loads on linear drivers?

[cistercian]

Why you should use a test load:

Lets say you decide to use a wire as a load and check the current, you want 100mA, it reads 100mA. So far so good. Next you solder your diode in place and power up, you notice the it isn't as bright as it should be... hmm.
What went wrong was that you forgot to check the input voltage, it was too low. But why it did work with the wire is that is has no voltage drop, and that was enough to get 100mA. And even if you did check battery the voltage after that so you made sure that it would maintain regulation you'd still get a false reading because you're measuring the battery without any load on it. By now you will be scratching your head quite a bit asking yourself why it doesn't work... Btw the 1.25V under operation is your receipt that it regulates properly.

It may be helpful here to point out the obvious:a good resistor has the same value regardless of the voltage applied to it.A diode junction does not, unlike a resistor a diode is a non linear device.IE the current flow can change dramatically with a tiny change in voltage.I also know some regulators will oscillate if operated under weird conditions.I would use a diode stack as a test load, and a very low voltage drop ammeter in series for final adjustment.

 

[Deleted member 8382]

Mmmm... That was a good point too. But well, at this point you just check the current at the entry of the driver and see that the current is lower than it was, so you know something is wrong

Anyway, I know a test load is usefully some times, but the whole point in the thread was that you can just use your multimeter in amp mode to set the current on a linear driver and that his method is even better than using a test load. A part from that, of course it's easier to just tell everyone to use them since it simplify the job for people that don't understand what they are dealing with

I know there's someone who won't believe it but the only purpose of this thread was to confirm my theory and to share it. It has been proven that many people didn't know that so I guess it wasn't a bad idea to post it.

From now I think I'll have to add a "" to every post lol

 

[HIMNL9]

Please correct me if i'm wrong, but seem that your basical question was not just "why we use test loads", but more probably "why we use these test loads with diodes and different voltage drops" ..... or i'm understanding wrong ?

Supposing i have understood right, then the answer require a specification, too ..... linear constant current drivers (like the ones that use LM317, as example), maintain a constant current through the load, as far as the input voltage is higher enough respect the output voltage, for have a difference bigger than the dropout of the regulator (plus something for the dissipations )

This means that, as example, if you power the driver with 12V, and the dropout of the regulator is 3V, no matter if you use an IR, a red or a BR diodes, cause, regardless the fact that the diodes requires 1,8V, or 2,5V, or 5,5V, the difference from VI and VO is always enough (1,8+3=4,8, 2,5+3=5,5, 5,5+3=8,5, from these to 12V, there's always a good difference)

But, if you power a linear driver with, as example, 2 li-ion cells (7,2V, nominally), when you use an IR diode, you still have a good difference (4,8V), when you use a red one still the difference is enough (5,5V), but if you connect a BR diode, the required voltage (8,5V) is greater than the one that you have available (7,2V), so you can never reach the current that you want, or also in case that the current that you need is low, it's still unstable.

So, the diodes are used for simulate the VF of a real LD (or, also better, a little bit more high) ..... in this way, you see immediately with the load if the driver is working good or not, without risk an LD ..... the "risk an LD part is due to a thing that, usually, peoples don't take in consideration, about linear drivers ..... let me do an example, with the BR thing .....

Suppose that you connect the two cells at their nominal value to the driver, and then set it with a PHR diode connected, measuring 100mA ..... the driver is NOT working in current stabilizer, in this moment, cause the voltage at the input is less than the one you need ..... so, practically you're regulating the 100mA only from the trimmer (and this mean that the trimmer can be in a position that, at regular power voltage, may give like 200mA, as example) ..... now, suppose that you change the batteries with a pair of fresh, just charged ones ..... that for li-ion cells, usually means 4,2V each, so, 8,4V ..... at this point, your driver is working almost at his set current (that was NOT the right one, before), so, 180 / 200 mA pushed in the poor PHR, that, probably, don't agree and say you "bye bye" in few seconds ..... and i'm sure that you agree with me that if this happens with a PHR, is not a big pain, but if the same thing happens with your new 8X, it's a bit different, right ?

So, also if a dummy load is not needed if you have a steady hand and are working with input voltage high enough, is always better to use a dummy load with a dropout voltage near or a bit high to the real LD, when you are working with low or near-the-limit input voltages (or if you're working near the end of the trimmer track and you're not sure that your hand is steady enough ) .

BTW, there's also a secondary reason, for use a dummy load, not so important as the first one ..... if you drop some voltage through the diodes, and dissipate it with them, is all power that not become dissipated from the LM317 (the power dissipated from linear regulator is given from the difference from the Vi and the VO, minus the dropout, multiplied for the current) ..... you can also use just an 1 ohm resistor, instead a full dummy load, example (again ), VI = 12V and current set to 300mA, if with the load simulating a BR, plus the dropout (8,5V), the power dissipated from the LM is 1.05W (12-8.5)*0.3 , with an 1 ohm resistor only, the power dissipated from the LM is 2.7W (12-3)*0.3 ..... and, cause the TO220 container is based on a 1W (maximum 1,5W) dissipation, your driver become burning hot and go in self protection each few seconds (if not just die )

Sorry for the long post, and i hope i understood correctly your question

 

[Deleted member 8382]

@HIML9: I get the point, but if you are simulating the diode I assume you already know the voltage drop on the laser diode, so it's all simple maths to know what the input voltage needs to be! Btw, probably you are right about the thread title, but I think I clarified it trough the post

 

[HIMNL9]

@HIML9: I get the point, but if you are simulating the diode I assume you already know the voltage drop on the laser diode, so it's all simple maths to know what the input voltage needs to be! Btw, probably you are right about the thread title, but I think I clarified it trough the post

Perhaps i also jumped over some posts, i'm sorry about that .....

About the math, yes, it's enough for calculate ..... but, if you're not sure that your driver is working good, using a dummy load remain a good way for not risk your LD ..... and using one that reproduce the VF of your LD, is still a good way for see if all your assemby works good in some extreme conditions (VI near the requested one, when batteries start to discharge, as example )

 

[Deleted member 8382]

Okay, but listen to what you are saying. The load helps to know if your whole circuit works ok in extreme conditions. I never said not, the post was about why using the test load to set the current on a linear driver

 

[HIMNL9]

I said also this .....

So, also if a dummy load is not needed if you have a steady hand and are working with input voltage high enough, is always better to use a dummy load with a dropout voltage near or a bit high to the real LD, when you are working with low or near-the-limit input voltages (or if you're working near the end of the trimmer track and you're not sure that your hand is steady enough ) .

said in poor words, when you are in the last pair of mm of the trimmer course, and if your hand is not steady enough, you risk to send 500mA to the diode, instead 200, turning the trimmer half millimeter too much

I suppose it's a reason good enough, when you're regulating the current with a 6X or an 8X connected ..... or not ?

 

[Deleted member 8382]

Don't get offended but I'm getting the same answers over and over. When I say that a test load is not needed to set the current on a linear driver, I mean that I would use the MM on amp mode directly or the 1 ohm resistor as load, NOT THAT I WOULD SET IT DIRECTLY WITH THE LASER DIODE!

I wouldn't like to look hostile, but please, read the whole thread because It's the third or forth time that I say this, each time to a different member...

 

[daguin]

Don't get offended but I'm getting the same answers over and over. When I say that a test load is not needed to set the current on a linear driver, I mean that I would use the MM on amp mode directly or the 1 ohm resistor as load, NOT THAT I WOULD SET IT DIRECTLY WITH THE LASER DIODE!

I wouldn't like to look hostile, but please, read the whole thread because It's the third or forth time that I say this, each time to a different member...

This is why, in Academe, we do this kind of discussion, once a year, over drinks, in a strange city. The written word is very frustrating.

I believe you have your answer. In perfect world, you are right. However, when building lasers (especially with inexperienced builders) it is a good idea to use the test loads. That's why we do it.

If the question has become more focused than "Why are we using test loads on linear drivers?", I recommend that you start a new thread with the more specific question posed.

If the discussion is about electronics theory rather than on the justification of the use of test loads, your audience will be much more focused and many of the non-electronic engineering type minds will stay out of it.

Peace,
dave

 

[HIMNL9]

Don't get offended but I'm getting the same answers over and over. When I say that a test load is not needed to set the current on a linear driver, I mean that I would use the MM on amp mode directly or the 1 ohm resistor as load, NOT THAT I WOULD SET IT DIRECTLY WITH THE LASER DIODE!

I wouldn't like to look hostile, but please, read the whole thread because It's the third or forth time that I say this, each time to a different member...

LOL, no, i don't get offended at all ..... and you're right, is always possible to regulate the current using only the 1 ohm, or just only the DMM, theorically ..... but there is a problem .....

let me copy and paste (i'm too tired for rewrite it again, LOL)

BTW, there's also a secondary reason, for use a dummy load, not so important as the first one ..... if you drop some voltage through the diodes, and dissipate it with them, is all power that not become dissipated from the LM317 (the power dissipated from linear regulator is given from the difference from the Vi and the VO, minus the dropout, multiplied for the current) ..... you can also use just an 1 ohm resistor, instead a full dummy load, example (again ), VI = 12V and current set to 300mA, if with the load simulating a BR, plus the dropout (8,5V), the power dissipated from the LM is 1.05W (12-8.5)*0.3 , with an 1 ohm resistor only, the power dissipated from the LM is 2.7W (12-3)*0.3 ..... and, cause the TO220 container is based on a 1W (maximum 1,5W) dissipation, your driver become burning hot and go in self protection each few seconds (if not just die )

 

[Eudaimonium]

Hey people hear my opinion...
The LM317 regulator chip, as far as i know, was never meant to be laser diode regulator. That is why we even need to put a spike protection capacitor over the diode...
Well, b'cuz it is never meant for so sensitive piece of electronics, it was supplied with a reference diagram suggesting a ... pot.
Now first of all, pot serves it's purpouse, but it is, by no means, good for adjusting the current for a laser diode (ESPECIALLY if the diode is already attached).
Why, IN THE NAME OF THE LORD, wasn't there a simple sticky thread with nothing more than a table as follows
TYPE ||| CURRENT||| RESISTANCE
SCC | | | 200mA | | | 6 ohms
LCC/LOC || 400mA || 3 ohms
PHR | | || 100/125mA | 10-12 ohms
6x BR || | 200mA | | 6 ohms

You know what I mean? And OFC a text explaining that the resistance goes in between of OUT and ADJ pins... Why for the name of Chuck Norris ( ) would you use the pot ? First, resistor is like, 10 or 15 times cheaper than a pot IMHO.
Next, the tuning step of driver building is thrown out the window.
Significant amount of time saved:
-to set the pot
-to measure current everytime after resseting
-to get new diodes after blowing out old ones with half of mm error in tuning
Significant amount of money saved:
-resistor cheaper than potentiometer
-MUCH less diodes blown out
Significant amount of nerves saved:
-not beeing able to set desired current
-not blowing out diodes
- NOT GETTING EF AND HALLUCINGENIC ENEMIES WITH THIS THREAD.

Why don't somebody simply explain that the schematic with 2x 10 ohm resistors, recifier diode, and a pot is overcomplicated?
All you need: resistor, 317 chip, cap.
End of story.
The table should serve it's porpouse to those who acctually need to look it put, meaning new guys in this hobby, that are still not familiar with currents for each type of diodes and so on...

I believe something like this would save not only things mentioned above, but also:
-time needed for a laser hobbyist to explain stuff to newbie (i had a friend ho wanted to assembe that, he started asking me where can he get the pot and so... I just said : 'Hell with the pot! Grab 6 ohm resistor to replace 2x 10 ohms and a pot, and also ignore the rectifier diode... At first he was confused how can the circuit work with parts missing! Imagine how my keyboard suffered, when I managed to explain how sh't works, over IM ) He had no knowledge of such electronics BTW. Now he does

Does everyone understand what am I saying? Simple table! We could have avoided all of this mess, rivalship between those two, 3 page thread where actually nothing is said.

 

[Deleted member 8382]

yeah well, the whole problem is that most of the people don't understand how do the LM317 chip works, so we created an universal method for everyone and every diode. But anyway, I think there should be a thread explained all this in case someone wants to learn it or in case someone wants to save his time. I don't think such a table is necessary if you understand how this works (and it's very easy). V=R·I, where V is always 1.25V, so just find R with the desired current!

The pot is good if you want to use the same driver for different pots or you want to get currents like "150mA" (although I then use parallel associations). Anyway, I've been running a phr at 150mA with just a 8 ohm resistor WITHOUT diodes or capacitors and it works great.

Also, IIRC the capacitor is not to protect from the LM317 spikes, it's to protect from the batteries spikes (at least that's how Deadel justified it's use). I think the LM317 has almost no spikes.

@HIML: I suggested the 1ohm resistors for those who only have voltmeters, but discart this one. What if you use the amp mode only. I've done it and nothing happened! This also avoid the incorrections on the measurements due to the 1 ohm resistor tolerance!

Yours,
Albert

 

[Eudaimonium]

Hmm... I simply use the dummy load to check id the current is alright, however I never tried bare amp mode on DMM ...
But if you had such table handy, and the forumly itself beside it, imagine how much trouble it would save, not only blowing out diodes because of the pot, but also later the frustration for the blown diodes, because you fogure out: that you never needed the stupid pot.
In my 4 years of building lasers, I've blown like, 20 red diodes and some of IR...
Only because I was following da stupid vid on youtube:
Turns the pot until it glows very brightly
And how I'm suposed to know where is the diodes limit?!
I believe that 18 out of that 20 would be lasing even now if someone told me to replace the pot with 6 ohm resistor.
Lemme see you raise hands...
How many diodes were blown by turning the pot while diode was soldered on and power applied?

You know what I mean?
I maybe seem to deviate from original topic: Why do we use test loads on linear circuit?
It has become habit of measuring the current on the load, WHILE ADJUSTING THE POT.
IMHO, this thread and all the trouble would be avoided if someone said:
"Use THIS resistance to replace pot" before suggesting the usage of 317 regulator chip as LD regulator. The diagram provided in the datasheets wasn't made to be LD driver. If we use 317 as LD driver, we do not follow the diagram that does not show the LD driver-suitable circuit!
Even more! If you manage to set the current with trimmer (measured on the test load) you have to know that trimmer and all mechanically movable electronic components are first to degrade, leading to shorter MTTF, therefore reducing reliability of the whole system. Using fixed resistance FTW, and good match count

 

[Deleted member 8382]

there's already a thread to collect data about all diodes, so all you would need is to check the current you want for your diode and then apply the formula to get the resistor you need. IF you still want to use a pot, adjust it with the MM on resistor mode to get the resistance you need, and then atach it to the chip.

Easy and safe

 

[Eudaimonium]

there's already a thread to collect data about all diodes, so all you would need is to check the current you want for your diode and then apply the formula to get the resistor you need...

:thinking: That is what I've been saying all along...

But never use a potentiometer... You can set it (if you manage to set it precisely) but resistance could change by anything: SMD variants can change only because they were soldered on PCB, and as I said, pots in general are very unreliable in long shots, because of mechanical principle of operation...
Just take a look to see if your can on diode is 2 mm, 4mm or no can, and grab a resistor accordingly! I'd gladly pay ONE resistor same amount of money I'd have to pay for a pot: just for avoiding the fuss of tuning it!
But it is just me...

 

[Deleted member 8382]

Yeah, I completely agree with you on that too.