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Why are we using test loads on linear drivers?

jayrob

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I think it is as Dave tried to explain, the test load also simulates the voltage demand of a particular diode...

As you know, if the supply voltage is not enough, then the circuit will not regulate the current properly. So a test load is needed, in conjunction with a proper input voltage. Both together will allow you to properly set the current.

And this is just a good practice, since we use boost circuits as well...
 
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Hallucynogenyc

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Definitely there's something wrong with you today. You are now linking me to pictures on how to use the LM317 as a current source. Did you ever read what I posted? I'll quote it for you:

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 add a 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).
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.
And to respond the las question, as I just told you:

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.
 

ElektroFreak

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^no dude, there's nothing wrong with me, and I'm done with you.

Those are PDF datasheets NOT PICTURES, and those pages contain the basic "DDL" circuit printed right on them, which you would know if you had looked. Adding capacitors and diode is just common sense from an electronics standpoint, and therefore do not need to be drawn into the circuit diagram. Now, if all you plan on doing is telling all of us that we don't know what we're talking about when all we're trying to do is help you understand this stuff, then you can take that attitude and shove it where the sun don't shine, friend. That's trollish BS..

You started this thread with a question, not a lesson on how to use regulators. Frankly, I don't need an electronics lesson from you.
 
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Hallucynogenyc

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@Jayrob:I know the diodes simulate the voltage drop of the certain diode we are testing, but again, since the LM317 is being used as a constant current source, the voltage drop on the load won't change anything.

I invite you to test this yourself. Set a DDL driver to for instance 100mA. Then put a red diode on it, then a bluray, an infrared and if you want even a random value resistor. The current won't change!

@EF:Oh my god. Really. Oh my god. Didn't you just told me to specific places in the datasheets in which pictures of the circuits were illustrated? I am 200% sure of what I'm saying. Trolling is a whole different thing from defending a "scientific" theory with arguments. Have you ever heard the myth of the cave? I suggest you to think about it. Please. And I mean seriously please, read my last posts again.
 
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MarioMaster

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The differing diode counts is for simulating different diodes. Red have about half the voltage drop of BR and IR diodes are even less than red. Depending on the battery source or available power, a test load is quite advisable to make sure the voltage source doesn't droop too much when the proper load is applied.

Some builds here have the battery voltage already close to the regulation limit so checking the operation status of the LM317 before installing the diode is a good idea. You could set the current too high due to a drooping battery, and then blow the diode when fresh batteries are inserted.

Also, the more load you put on the LM317 the MORE heat it generates. You're getting confused in that an equal current load with more VOLTAGE will decrease the heat output being the linear driver that it is.

I'm sure there's also language barrier problems as well. Your intent is unclear.
 

drlava

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Just saw this thread, is seems there is some confusion on this issue, even in life-members here :)

It's a good question Hallucynogenic, you are right to ask. In fact, in an ideal world you COULD set the current on a DDL/rkcstr driver simply with a multimeter in amps mode in place of the diode and test load. You are correct also, in an ideal world, it would not matter the load voltage for this driver, the output current would remain the same. David, this is because the LM317 does not operate as a resistor, it has active feedback with high gain in order to maintain the 1.25V across its out and adj pins.

This results in a nearly equivalent output current whether the load is 0.5V or 6V, as long as the battery voltage is > Vdo+1.25V+Vload.

The Flexdrive works similarly, within the range of 2.2V to 6V output, at one pot set point, the output current will be virtually constant. Again this is by design through sampling and amplification, is not a violation of Ohm's law.

The simplest explanation for the original question has already been given: It's a good practice, and, with the built-in 1 ohm resistor it makes current measurement easy for any type of driver.
 
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Hallucynogenyc

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I might be wrong with the heat thing, but not with the others one. Just to make totally sure I just went and tested it myself.

Mariomaster, I'm totally sure we have multiple language problems, but since there's no other solution than keep learning English I guess you will have to try to understand my English as it is now xDD

Read this I posted above:

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
Isn't this totally true?
 

ElektroFreak

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Just saw this thread, is seems there is some confusion on this issue, even in life-members here :)

It's a good question Hallucynogenic, you are right to ask. In fact, in an ideal world you COULD set the current on a DDL/rkcstr driver simply with a multimeter in amps mode in place of the diode and test load. You are correct also, in an ideal world, it would not matter the load voltage for this driver, the output current would remain the same. David, this is because the LM317 does not operate as a resistor, it has active feedback with high gain in order to maintain the 1.25V across its out and adj pins.

This results in a nearly equivalent output current whether the load is 0.5V or 6V, as long as the battery voltage is > Vdo+1.25V+Vload.

The Flexdrive works similarly, within the range of 2.2V to 6V output, at one pot set point, the output current will be virtually constant. Again this is by design through sampling and amplification, is not a violation of Ohm's law.

The simplest explanation has already been given: It's a good practice, and, with the built-in 1 ohm resistor it makes current measurement easy for any type of driver.
^There you go Albert, there's your answer. It's already been given in this thread, but there it is again. Here's a quote where I said basically the same thing:

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..

@EF:Oh my god. Really. Oh my god. Didn't you just told me to specific places in the datasheets in which pictures of the circuits were illustrated? I am 200% sure of what I'm saying. Trolling is a whole different thing from defending a "scientific" theory with arguments. Have you ever heard the myth of the cave? I suggest you to think about it. Please. And I mean seriously please, read my last posts again.

Whatever, man..

Look, so far in this thread EVERY poster has tried to help you understand. Each time you say we're wrong. At this point, as far as I'm concerned, you can learn this stuff the hard way, just like I did. Go to college or Google it.
 
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Hallucynogenyc

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AT THE END SOMEONE WHO UNDERSTANDS ME, THANKS GOD!

hey DrLava, you say in a ideal world only. why? I mean, I think it's even more precise to do it this way! 6 diodes are 4.9V drop, not 5. Just this little imprecision is higher than any other produced by my method right?

Thanks for explaining it the right way: "this is because the LM317 does not operate as a resistor, it has active feedback with high gain in order to maintain the 1.25V across its out and adj pins."

You all even made me go and test this lol
 

jayrob

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Just saw this thread, is seems there is some confusion on this issue, even in life-members here :)

It's a good question Hallucynogenic, you are right to ask. In fact, in an ideal world you COULD set the current on a DDL/rkcstr driver simply with a multimeter in amps mode in place of the diode and test load. You are correct also, in an ideal world, it would not matter the load voltage for this driver, the output current would remain the same. David, this is because the LM317 does not operate as a resistor, it has active feedback with high gain in order to maintain the 1.25V across its out and adj pins.

This results in a nearly equivalent output current whether the load is 0.5V or 6V, as long as the battery voltage is > Vdo+1.25V+Vload.

The Flexdrive works similarly, within the range of 2.2V to 6V output, at one pot set point, the output current will be virtually constant. Again this is by design through sampling and amplification, is not a violation of Ohm's law.

The simplest explanation for the original question has already been given: It's a good practice, and, with the built-in 1 ohm resistor it makes current measurement easy for any type of driver.
This is exactly what I did a couple of years ago when I first started building DDL drivers...

I tested current using only my DMM, and it worked just fine. But I learned that it is not the best way to do it. It could also harm the meter. My meter has safety stops built into it, but it is still not the best way to do it.

It is just good practice to use a test load. A test load is not expensive. And we use other types of drivers. Along with various supply voltages.

That's the answer to why... And let it still be recommended to all new ones learning to build! :)
 
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Hallucynogenyc

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Then you could just have told me from the very first:

"yes you are right but we are using them because we are lazy to make different loads for different drivers"

:crackup:
 

daguin

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I said it in another thread and I repeat it here.

Al seems to be in a bad mood lately

Peace,
dave
 

drlava

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hey DrLava, you say in a ideal world only. why? I mean, I think it's even more precise to do it this way! 6 diodes are 4.9V drop, not 5. Just this little imprecision is higher than any other produced by my method right?

I haven't seen your method, but the closer the load is to the actual diode voltage, the more precise the measurement will be, even if the worst case error is only a percent. Additional error creeps in because the 1 ohm resistor commonly used is not a precision resistor. If you have a calibrated multimeter, the most precise way to set the current would be to use a dummy load in series with your multimeter in amps mode, thus bypassing the 1 ohm sampling resistor error.
 

jayrob

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That's the way I do it...

Just measuring mA's and not using the 1 ohm resistor.

Also, most meters have two inputs for the plugs. It is better to use the higher rated input plug for the probe. The 10A plug, not the 400mA plug. The 400mA max input has some kind of resistance in there that can affect the accuracy of the reading...
 
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Hallucynogenyc

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When I said my method, I meant just using the multimeter in amps mode ;) Isn't that way easier and precise than mounting 6 diodes and a resistor? As you just said, people resistors aren't precision ones, the diodes are not even making the correct voltage drop...

@Jayrob: Mmmm... Interesting. Why is the other plug made for then?
 

drlava

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When I said my method, I meant just using the multimeter in amps mode ;) Isn't that way easier and precise than mounting 6 diodes and a resistor?
the answer:
the closer the load is to the actual diode voltage, the more precise the measurement will be
The 400mA range may use an ohm or two, the 10A range usually uses under 1/2 ohm sense resistor.
 




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