Why are we using test loads on linear drivers?

[Deleted member 8382]

well, am I pushing anyone? On this lasts posts I'm just asking for myself or for whoever who is interested! Do I need to tell you that this configuration wouldn't work for batteries? xd

 

[Eudaimonium]

Hey people why does the 'rate thread' bar up there suddenly appeared ( to me).
And it says: You have already rated this thread.
WTH ! I did not rate anything!
Sorry for off-topic but this is the first time this happened to me.

 

[Deleted member 8382]

Hey people why does the 'rate thread' bar up there suddenly appeared ( to me).
And it says: You have already rated this thread.
WTH ! I did not rate anything!
Sorry for off-topic but this is the first time this happened to me.

same here xd

 

[lasersbee]

well, am I pushing anyone? On this lasts posts I'm just asking for myself or for whoever who is interested! Do I need to tell you that this configuration wouldn't work for batteries? xd

Did I say you are pushing anyone....:thinking::thinking::thinking:

On this whole thread you were saying (by asking a rhetorical question)
that a Test Load was not needed for Linear Laser Drivers..

Some members here have stated that the safe way for most members
is to use a Test Load....

Whether you or anyone else uses a Test Load or not is that member's
choice...

The reasoning for or against using a Test Load has been clearly outlined
in this thread...
It is up to the individual Member to take that information to make a logical
decision on whether he wishes to use one or not...

That's all I said in the post above...

"To Use or Not to Use.... That is the question"

Sorry Albert.... I honestly don't know what you are trying to say..

Jerry

 

[Deleted member 8382]

You don't even need to use an LM317 Driver Circuit....

You can take your PS/Batteries and put a "calculated" resistor in series
with your DMM on Amps and check that you have the Current you want
to run your Laser Diode... without a test load...

The choice is ultimately up to the builder...

I understood that as a "other things could be done but we like it like this, stop it". Sorry again if I missunderstood you

 

[drlava]

So at this point, you don't need to use diodes at all for a test load on a LM317 based driver, but you need a load. I'll ask you directly (HIML9), what resistor to place in series with the MM to disipate some heat?

Do you have something against diodes?

You could use a 15 ohm 3W resistor for blu ray

8.2 Ohm 2W for red

 

[Hemlock_Mike]

Ohm's law assumes that everything is perfect (DC). Perfect connections, zero ohm wire and joints, no junction voltages, and everything linear and no AC.
It Ain't that way. Juction voltages on semiconductors are not linear and present unique problems. That's why we try to simulate the actual load with a "test load" prior to conecting "your precious" diode.

HMike

 

[Deleted member 8382]

@DrLava: If a resistor could make the job I wouldn't use diodes again xD So well, all in all, the only thing we can obviously avoid is to put an X amount of diodes for each diode. Let's say 6 diodes will work for all diodes?

@hemlock_mike: I'm very confused about this last post of yours. In case you really missed it the other 5 times I've said this, I'll repeat it for you We are talking about linear drivers based on the LM317 driver. This driver is based in the conversion of the LM317 constant voltage source on a current source by adding a resistor between the 2 pins where the LM317 develops the nominal voltage. At this point the other part of the load can change as much as they want without any change on the circuit current!

Yours,
Albert

 

[ossumguywill]

I havent been following everything, I thought I new quite a bit about this stuff, but all I know about thevennion stuff is that he pioneered the idea that any circuit can be reduced to a power source and a load... But anyway, the reason I use the test load is because for some reason my DMM cannot measure amperage very well (I think it has a blown fuse, but I don't really care enough to check). LOL so I have to use the 1 ohm resistor technique to get an accurate amp reading. so I have to use the 1n4001's voltage dropping properties (no idea how that works, gonna go look it up) to that I can measure voltage across the resistor without just measuring the output voltage of the driver (doesn't that just mean that higher voltage = higher "amps" reading?) I wish high school would do away with stupid time-waster courses like literature or my school's english and replace them with useful stuff like electronics or law. I have to figure all this out through the internet, and I probably sound like an idiot.
Oh well.

Will.

 

[HIMNL9]

okay, so this is at the end turning back xD

So at this point, you don't need to use diodes at all for a test load on a LM317 based driver, but you need a load. I'll ask you directly (HIML9), what resistor to place in series with the MM to disipate some heat? I mean a value that works for all the diodes we actually have

It's not so easy, sorry.

LDs are not linear components, for the first ..... and they cause a dropout, for the second ..... so, the things are a bit more complicates .....

Each model have an internal "equivalent resistance" that change with the current that you feed them, and a threshold line for conductivity (as any diode), where, under it, they acts almost as an open circuit (or better said, as a mid / high resistance circuit), and over it, as a voltage-dependent low resistance circuit ..... there's no single resistors that can emulate it with decent accuracy.

Other than this, using a resistor in series of your DMM, will false the reading, more than for the tolerance of an 1 ohm resistor in a dummy load ..... if you really want to use the less possible components, just use the 4-diodes-1 ohm-resistor load, still measuring the voltage through the resistor ..... if you want to measure it more accurate, change the resistor with a 10 ohm 0,5% 5W one ..... but not too much other to do, for this.

Or use a 10 ohm resistor and measure the voltage on it, you need a minimum 5W, better if 10W, resistor ..... no dropout, but the reading is still possible.

Ofcourse, in this case, for each mA, you read 10 mV, not just one ..... the measure is more precise, but the resistor dissipate 10 times more power.

You can also build a specific instrument, if you want, if you find one of those 4 ciphers mV units and a 10 ohm 10W 1% (or better) resistor

But no single resistors in serie to a DMM, sorry

 

[LuxIgnis]

I still believe the test load is useful to simulate voltage drop even on a DDL driver, so we can properly select the input voltage (battery).

ES. You can power the DDL driver with a 7V battery and set it at 300ma with a red diode. Then you go and attach a 8x blue ray diode... and... according to you, we'd still have 300ma, but it does not work that way. You can then change the input voltage to
9v-12v and then it sure works.

So is a diode-specific test load needed? I'd say it's useful.

 

[HIMNL9]

Uhm ..... i think there's a little confusion of terms

Is not that a diode-specific load is needed all the times that you need to regulate a driver (i never said this), cause, as Hallucynogenic have said, a regulation can be made also with just a resistor, or a DMM .....

Is just better use a load that cause the same, or little high, voltage dropout of the diode that you're planning to use, cause in this way, you can regulate the driver in the near-similar conditions of the ones in which it have to work, so you can see immediately if there's something wrong, this yes i've said ..... and also, for the non-steady hand problem

And, after all, a dummy load is not so high cost to build, and last you forever.

(at least, until you don't fry it trying to use it for regulate your 65A PSU for a IR bar ..... j/k )

 

[lasersbee]

Ohm's law assumes that everything is perfect (DC). Perfect connections, zero ohm wire and joints, no junction voltages, and everything linear and no AC.
It Ain't that way. Juction voltages on semiconductors are not linear and present unique problems. That's why we try to simulate the actual load with a "test load" prior to conecting "your precious" diode.

HMike

Very well put... even a layman can understand that...

Jerry