Here's a quote from Igor explaining why the difference. But it probably doesn't matter which plug you use with a linear driver. Boost drivers are a different story. So just out of good practice, I always use the 10A plug...@Jayrob: Mmmm... Interesting. Why is the other plug made for then?
DMMs measure current by measuring voltage across an internal resistor (called "shunt"). Just like we sometimes put a 1 ohm resistor and measure voltage across it.
On top of that, the long wires of the DMM itself introduce a resistane, which the current has to flow through. And all that requires some extra voltage, which is lost, by the time current gets from one probe, through the DMM, to the other probe. This is called a voltage drop, and depends on HOW MUCH current is flowing through a resistance (the higher the current, the higher the voltage drop across a resistance -> U = R x I). If you use a second DMM, you could measure this voltage drop across the first DMM.
So, the DMM introduces a resistance into the system, the resistance causes a voltage drop - the driver gets a bit less voltage than the battery actually has. If a switching driver gets less voltage it will compensate by drawing more current, to keep power (P = U x I) the same. It's the only way it can supply a constant output...
If you would measure the voltage drop across the first DMM with another DMM, you could then calculate how much less the actual current draw in the laser is. But since the 10A (or 20A - whatever the highest) range uses a thick metal rod as a curent measuring shunt, the resistance introduced is small, so the measurement is quite close to the actual current in the build.
The reason this effect is not noticable with linear drivers is, that they draw a constant current (not constant power), as long as they get enough voltage. Putting a DMM in between reduces the voltage the driver gets a bit. But a long as it is still enough, the driver will keep drawing a constant current.
If however, a linear sytem would be barelly above the minimum voltage required for regulation, adding a voltage drop could reduce the voltage to bellow the minimum, and cause the driver to drop out of regulation, while it could still regulate for a while, if there was no DMM in it's way.
At a certain point, a linear driver might still get just enough voltage to regulate when measured through the 20A range, but drop out when measured in 200mA range.
Due to the different behavior of linear drivers, the effects of the voltage drop are harder to notice there, and since this voltage is low, the above situations are unlikelly to happen when testing a linear driver setup.
A situation where the difference in DMM's resistance between the two ranges is very noticable is in the many cheap chinese unregulated green lasers. There, the drivers are out of regulation most of the time, and act as complicated resistors. Measuring the current with the high range on the DMM will show a much higher current than the low range on the DMM!
Another situation is, if you were to make a laser with nothing but a pot between the diode and the driver, set the current with the DMM in-line and then remove it. The actual current in the laser could be so much higher, that it woud kill the diode!
Hey Albert....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
If you can't draw a mental picture in another persons mind....I meant the DDL circuit or the simple current regulator using a lm317. I think I was explicit enough!
How would you use the lm317 as voltage regulator for laser diodes? I mean, the difference of voltage from 100 to 150mA on a laser diode is very small, isn't it? (aren't they working as a Z-diode?)
You have sounded like a smarta$$ to some of the members on this Thread and that wasn't called for...
Just because you had "some electronics classes" doesn't make you an "expert" even if you think you are....
Did I? I really don't think I ever pretended to be an expert on it lol.so don't tell me that it can't be done...
This is my answerthey do have those things in Spain, don't they
What do have Spain to do with all this?
Quote:I freaking hate it when people ask a question, and then tell you you don't know what you're talking about
What I answered was not what I was responed about. And in fact the responses were wrong and so I tried to correct them. and you know I was right with the corrections I made.
From this point you just starting to flame me at no reason, telling me that you linked to images and not pdfs, when you just pointed me to the pdf page 19 at the botton, in which there was an image, just to make me look stupid.
Then you tell me that the DDL circuit is on the datasheet. I'll tell you something. This is the DDL circuit.
And this is what appears on the datasheets:posted him an imgae
That was all I was trying to tell you. But you didn't listen. just atacking me. It's Ok. Bye.posted him an imgae
I'll never understand people like you. You jumped on me wrongly and won't go back no matter what happens.
Ah, sorry for not being born in your country, I guess that was my biggest sin for you.