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# Why are we using test loads on linear drivers?

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#### Deleted member 8382

##### Guest
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 xD

Isn't the Lm317 supposed to be a constant voltage source? We are making a constant current source out of it by applying the 1.25V constant voltage to any load to get any current following the Norton circuit scheme right?

At this point, no matter what you put in the test load right? I mean, why are we putting there diodes if we could just put the 1 ohm resistor directly!?

Yours,
Albert

#### ElektroFreak

##### New member
Truthfully, you don't even need a test load at all with a linear driver. You can just hook up the laser diode and measure current at the driver input. Since the driver itself draws only a negligible amount of extra current, the reading will be the same as the current reading at the driver output.

D

#### Deleted member 8382

##### Guest
I alredy know that. This is not what I meant anyway. I understand current is constant on the whole circuit, what I mean is that I don't understand why test loads are used at all to "set" the current xD

#### ElektroFreak

##### New member
Seems that what I just said would answer that question for you.. Who knows why? There is no good reason.

#### daguin

##### New member
Truthfully, you don't even need a test load at all with a linear driver. You can just hook up the laser diode and measure current at the driver input. Since the driver itself draws only a negligible amount of extra current, the reading will be the same as the current reading at the driver output.
With a linear driver, the test load is really just a safety check. Most of our "testing" is "input" related rather than output related. We use the test load to make sure the driver is set low enough so as to NOT blow the diode on power up.

When I make multiple modules, I simply start with the drivers turned down all the way for assembly. Then I turn each one up until the desired output is reached. Actual currents vary with diode efficiency.

Peace,
dave

#### nikokapo

##### 1
Because the test load simulates a diode and thus you can set the desired current for your diode by measuring it with a multimeter.

#### ElektroFreak

##### New member
With a linear driver, the test load is really just a safety check. Most of our "testing" is "input" related rather than output related. We use the test load to make sure the driver is set low enough so as to NOT blow the diode on power up.

When I make multiple modules, I simply start with the drivers turned down all the way for assembly. Then I turn each one up until the desired output is reached. Actual currents vary with diode efficiency.

Peace,
dave
One could also simply turn the pot on the driver all the way down before connecting the diode, and then once the diode is connected, attach a meter between the power source and the driver input and ramp the current up to where you want it to be, being careful not to overshoot.

@niko: But why even use a test load? It's very possible to set the current on a linear driver using the method I just described in the above paragraph and not harm the laser diode at all.

I've never used a test load with a linear driver, only switching or buck/boost.

D

#### Deleted member 8382

##### Guest
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.

I've read just so many tutorials where it's mentioned to use a test load to set the lm317 driver or the rkcstr. It just makes no sense!

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

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#### daguin

##### New member
One could also simply turn the pot on the driver all the way down before connecting the diode, and then once the diode is connected, attach a meter between the power source and the driver input and ramp the current up to where you want it to be, being careful not to overshoot.
You are right (and I do in many instances). However, in practice, many of the people building lasers do not have a very steady hand, they are impatient, the pots are small and sensitive, and sometimes our little pots take great leaps in current without warning.

"Overshooting" is much too easy to do for most beginning builders.

Peace,
dave

#### ElektroFreak

##### New member
^For sure. It is a wise safety measure all around, especially if you're using expensive diodes..

D

#### Deleted member 8382

##### Guest
My question is not about testing the current before installing the diode. My question is why do all tutorials tell you that you need X amount of diodes to simulate X laser diode when at the end it just doesn't matter!

Yours,
Albert

#### ElektroFreak

##### New member
I've wondered that myself in the past, and I believe it does matter with many drivers since sometimes you can't get an accurate current reading if current begins to flow at a lower voltage than the diode you're using requires. Why build separate test loads for each kind of driver? It's simpler just to build a load that works with any driver..

#### daguin

##### New member
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.
I've read just so many tutorials where it's mentioned to use a test load to set the lm317 driver or the rkcstr. It just makes no sense!
They are not "constant current" drivers. They are "variable current" drivers. The current is just "set" to stay at the current we pre-select. Not only does the current vary based on the position of the pot, it varies with the available and used voltage.

The test load is needed (as I have stated in other posts) as a "safety measure" to prevent accidentally over driving the diode (and blowing it) on power up. It is also needed to simulate the voltage draw of the diode. Given a constant voltage input and identical resistance in the driver, if you change the VF of the load, the current changes. That is why we use different test loads for different types of diodes.

Peace,
dave

D

#### Deleted member 8382

##### Guest
Daguin I'm so damn proud of what I'm just gonna do. I think I'll have to correct your for the first time in my life :crackup:

They are not "constant current" drivers. They are "variable current" drivers.
The lm317 is a constant voltage source. it applies 1.25V to the resistor you put on the adj pin. This way we get a constant current out of it. of course, if we modify this resistor or use pots we get different currents

The test load is needed (as I have stated in other posts) as a "safety measure" to prevent accidentally over driving the diode (and blowing it) on power up. It is also needed to simulate the voltage draw of the diode.
I know all this. What I meant was more likely why do we construct test by putting several diodes on them when a simple resistor would suit (maybe one or two diodes because the the driver would fail at so low ranges, but not to simulate the diode!)

Given a constant voltage input and identical resistance in the driver, if you change the VF of the load, the current changes. That is why we use different test loads for different types of diodes.
The problem is that you are not getting a constant voltage input. You are getting a constant current. So no, changing the load won't change to current (unless you go out of the driver limits)

Yours,
Albert

#### Meatball

##### New member
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.

#### daguin

##### New member
Daguin I'm so damn proud of what I'm just gonna do. I think I'll have to correct your for the first time in my life :crackup:
The lm317 is a constant voltage source. it applies 1.25V to the resistor you put on the adj pin. This way we get a constant current out of it. of course, if we modify this resistor or use pots we get different currents
I know all this. What I meant was more likely why do we construct test by putting several diodes on them when a simple resistor would suit (maybe one or two diodes because the the driver would fail at so low ranges, but not to simulate the diode!)
The problem is that you are not getting a constant voltage input. You are getting a constant current. So no, changing the load won't change to current (unless you go out of the driver limits)
Yours,
Albert
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

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