Why are we using test loads on linear drivers?

[Deleted member 8382]

They are used as a precaution. What if the driver is bad? It is better to find that out while using a test load, rather than the actual diode. The small rectifier diodes are used because they have a similar voltage drop to the laser diode we want to simulate. Since diodes have very little to no resistance, the voltage drop across the load of the driver, is going to directly affect the amount of current that the driver is allowed to provide. I hope this helps.

it actually makes me laugh how people will answer without reading anything.

 

[Meatball]

I DID read this entire thread so far. If you only want a couple people to post here, then say so.

Sorry for offering an answer or two to your question.

 

[ElektroFreak]

it actually makes me laugh how people will answer without reading anything.

^Dude, your questions have been vague to say the least. Maybe instead of venting your linguistic frustrations on the people who are trying to answer you, you should rethink what you are trying to ask and make it as understandable as possible. I didn't have a clue what you were asking until halfway through this thread.

 

[Deleted member 8382]

Perhaps we are suffering from a difference in labels. The LM317T is NOT a "driver." It is one component of a driver. The "driver" includes the potentiometer as well. Variable resistance (via the pot) is part of a driver. The driver's purpose is to allow variation in current (and to smooth out and limit spikes). The current from a driver is adjustable. In order for it to be a "constant current" driver, you would have to eliminate the pot. Maybe it would be better for understanding for me to say it is an "adjustable current" driver. We do not "drive" diodes with an LM317T alone. The LM317T may be a constant source, but a "driver" (with a pot) is adjustable and variable.

If you change the load, and the resistance and voltage input remains the same, the current changes.

Someone else will have to come in with the formula, but you cannot change one variable in an equation and have the result remain unchanged.

Peace,
dave

You are not talking about the lm317 driver. you are talking about the DDL driver. The lm317 driver/chip/circuit is just a simple constant current voltage source that can be easily turned into a constant current source by putting a resistor on it.

Deadel just purposed to adda a capacitor to compensate the possible (not dangerous IMO) battery spikes, a diode to protect from a bad battery insertion (not needed for me) and a pot in order to change the current later withour soldering (not needed if you do not plan to change it).

So yes, I'm driving my phr with just a 10 ohm resistor and a lm317 chip (and it have many hours on it).

The formula is easy. V=R·I (incredible eh?)

The voltage by the lm317 is 1.25V, put a 10 ohm resistor on it and yep, you just got that the current is 125 mA. And since this current is coming from a constant voltage source, it's constant too.

constant current= won't change despite the circuit it is on.
adjustable current= you can change it somehow.

Do the test by your own if you want. the current is not going to change if you change the load.

Yours,
Albert

 

[Cyparagon]

EYE know what you mean. Don't quote me on this, but I believe the lm317 can be used as either voltage regulator:

OR as a current regulator:

I'm not sure which the DDL is.

 

[Deleted member 8382]

@EF/meatball: sorry but I am really frustrated because no one understand what I mean and I thought it was obvious. isn't it? Sorry then if it isn't!

Until now the "test loads simulate a laser diode" explanation had always made sense for me but after having some electronics classes I'm having doubts now

niko answers

Because the test load simulates a diode

I say:

wait. it seems no one is listening to me xd

the lm317/rkcstr are giving CONSTANT CURRENT. This means the current will be the same no matter if you put there a laser diode, a resistor, a test load or just a wire.

Meatball answers:

They are used as a precaution. What if the driver is bad? It is better to find that out while using a test load, rather than the actual diode. The small rectifier diodes are used because they have a similar voltage drop to the laser diode we want to simulate. Since diodes have very little to no resistance, the voltage drop across the load of the driver, is going to directly affect the amount of current that the driver is allowed to provide. I hope this helps.

Am I really the only one who feels like if he had not read? Again, sorry if you did...

 

[pullbangdead]

It's been recommended for all noobs because some drivers absolutely require a full load (the flexdrive).

Do YOU want to explain to every single new person that comes on this board why they need a test load for the flexdrive and don't need one for the rkcstr? I don't. So for the sake of sanity, and the sake of every driver a noob buys, it's much easier to say "every driver needs a test load.

If we didn't recommend "always use a test load to set current", some noob would show up yelling at people because he broke his flexdrive when he didn't use a testload, "because it said not to for the rkcstr, and aren't they the same?!?"

It's just for consistency.

 

[Deleted member 8382]

EYE know what you mean. Don't quote me on this, but I believe the lm317 can be used as either voltage regulator:

OR as a current regulator:

I'm not sure which the DDL is.

See the first one is just using the current from the first resistor in order to create a voltage drop. If you change this second resistor you change the voltage. The capacitors are just to avoid spikes etc...

It's all about Norton-Thevennion conversions!

 

[Deleted member 8382]

It's been recommended for all noobs because some drivers absolutely require a full load (the flexdrive).

Do YOU want to explain to every single new person that comes on this board why they need a test load for the flexdrive and don't need one for the rkcstr? I don't. So for the sake of sanity, and the sake of every driver a noob buys, it's much easier to say "every driver needs a test load.

If we didn't recommend "always use a test load to set current", some noob would show up yelling at people because he broke his flexdrive when he didn't use a testload, "because it said not to for the rkcstr, and aren't they the same?!?"

It's just for consistency.

I understand that as a reason. But it's not needed for me haha. Is this really the only reason?

Anyway, it's just that I find it strange that is not mentioned anywhere...

 

[Meatball]

@EF/meatball: sorry but I am really frustrated because no one understand what I mean and I thought it was obvious. isn't it? Sorry then if it isn't!

Hey I know what you mean. I was simply trying to give a complete answer of my own, instead of building up on what had already been said, since it was still unclear. Don't assume someone didn't read just because they are replying to the first post, instead of the last one.

 

[ElektroFreak]

You are not talking about the lm317 driver. you are talking about the DDL driver. The lm317 driver/chip/circuit is just a simple constant current voltage source that can be easily turned into a constant current source by putting a resistor on it.

Deadel just purposed to adda a capacitor to compensate the possible (not dangerous IMO) battery spikes, a diode to protect from a bad battery insertion (not needed for me) and a pot in order to change the current later withour soldering (not needed if you do not plan to change it).

So yes, I'm driving my phr with just a 10 ohm resistor and a lm317 chip (and it have many hours on it).

The formula is easy. V=R·I (incredible eh?)

The voltage by the lm317 is 1.25V, put a 10 ohm resistor on it and yep, you just got that the current is 125 mA. And since this current is coming from a constant voltage source, it's constant too.

constant current= won't change despite the circuit it is on.
adjustable current= you can change it somehow.

Do the test by your own if you want. the current is not going to change if you change the load.

Yours,
Albert

Daedal didn't propose anything. The DDL circuit is included in EVERY Lm317 datasheet. It is the standard LM317 current limiting configuration. We just happen to use it to power LDs.

Look, man.. you're taking kind of a hostile tone here, so I'm pretty much done answering your questions. If you have this all figured out, why the hell are you asking us? So you can jump all over us for trying to help you? you can keep that stupid bullshit.

 

[daguin]

Do the test by your own if you want. the current is not going to change if you change the load.
Yours,
Albert

Then you are artificially limiting the current by having a "limit" on available voltage (and current) at input.

I don't even need to do the experiment. I know basic math.

voltage at source (drawn through) resistance = current @ voltage draw (probably + heat)

If you change ANY component in the equation, you cause a change in some other component of the equation.

If your voltage at source AND your resistance remain constant AND you change the voltage draw, the only component left to change is the current (and heat).

Peace,
dave

 

[Deleted member 8382]

@EF: wtf, I wasn't pretending to jump on everyone. I was just trying to know why isn't this mentioned anywhere. Now it seems like it's not mentioned because most of the people here don't know it so I'm trying to explain. You really didn't need to go like that. I really have no reason to go hostile with Dave and trust me that there was not any hostile intention on that post. Also, the DDL circuit is not included in every datasheet. The only thing included in the datasheet is the suggestion to use the constant voltage as a current source.

@Dave: I think you're forgeting what the lm317 does when you think in the equation. It's supplying the diode from the batteries. The extra power is transformed into heat. The more load you put on it the less heat it's going to dissipate, until the point in which it will need more voltage from the batteries, keeping the current constant.

Really, if you put one or two laser diodes the current will still be the same!

 

[jaycey]

When I first came here I didnt have a clue about drivers, measuring current, test loads etc.

I started by building the DDL driver, but thought 'hey I can just turn the pot till it lights up' pop went the first diode, pop went the second diode, pop went the third diode.
Now if you go back to my posts then I was using diodes from Plasmon UDO drives.
If only I knew then what I know now I could have saved them and sent 1 out to someone for testing (I suspect they are similar to a PHR)

Anyway, it was the use of a test load and multimeter that showed me where I was going wrong, the difference between 80mah and 280mah was around a millimeter on the pot.
The pots (100ohm) are extremely sensitive and as soon as I learnt how to measure the current I took it up slowly and have never had another blow on me.

Now I dont understand what you are saying, because I dont have much electronics knowledge, but I know that what has been demonstrated SIMPLY, works for people who are new to this hobby and want to have a go themselves at building a laser driver that is adaptable enough to power the majority of diodes we use.

If you come up with a simpler solution then great! but hey, its pretty simple as it is

 

[Deleted member 8382]

Dear jaycey,

When you test the current on a DDL driver, you were probably told that you had to use 4 diodes for a red laser diode simulation, 6 diodes for a bluray diode simulation, etc...

This is only needed when testing advanced drivers like the Flexdrive. For the DDL, for instance, if you just put the 1 ohm resistor to measure the current it will work no matter if the diode is red, infrared, bluray, or X-ray (joke).

I'm not saying test loads are not needed, only that their construction concept is wrong.

This is what I'm trying to prove

 

[ElektroFreak]

The current-limiting circuit is in the following LM317 datasheets:


http://cache.national.com/ds/LM/LM317.pdf

^bottom of page 19

http://www.datasheetcatalog.org/datasheet2/c/0hj5dxz6qa1kdqxjigl5zpok4iky.pdf

^page 6 figure 10

http://www.datasheetcatalog.org/datasheet2/a/0s922lq6p7gwsc2hhzx6z4e3pl3y.pdf

^bottom of page 9

http://www.datasheetcatalog.org/datasheet/texasinstruments/lm317.pdf

^bottom of page 5

and perhaps more, but I'm done looking. It's just common sense to add a capacitor or two for transient and ESD protection and maybe a diode for reverse polarity protection.

There is nothing more annoying IN THE WORLD than someone who asks a question, and then proceeds to tell everyone who answers that they don't know what they're talking about. It's rude to say the least. It's like I just said: If you've got this all figured out already, why in the HELL are you asking us?