Why use a dummy load?

[Xplorer877]

I've been out of the hobby for like a year but I'm back to make a 6x 405nm build.

Anyways I'm going to power my 6x SF-AW210 diode with a flex drive and a single CR123a cell. But my question applys to any driver I suppose.

First off, the potentiometers on drivers adjust the current output. So they must vary the voltage output to the load to achieve the current level set by the pot. So regardless of the resistance of the load, the driver must adjust the voltage to "force" the set current to flow. Correct?

So why then, when you use a dummy load do you have to measure the VOLTAGE across a 1ohm resistor and then calculate the current? Wouldn't it be easier to simply measure the current through the circuit? Furthermore, wouldn't it be even easier to ditch the dummy load and resistor altogether and just measure the current as it flows through your diode!?

For example, before I solder my diode to the driver I use test leads or alligator clips to hook up my diode with a ammeter in series with the diode. I turn the potentiometer all the way down and power up the driver. I'm planning on running my 6x diode at 190mA for a output of ~250mW. So as I slowly turn up the pot I look at my ammeter and once the current reaches 190mA I stop.

Wouldn't that be more accurate than a dummy load?

For me the issue is not making a dummy load. I have one built already. I'm just wondering why everyone is calibrating their drivers to a string of rectifier diodes when it's easy to tune it to the real thing

Get what I mean?

-Tony

 

[toogoofy]

Yo,
If I understood your question correctly I think this thread will give you your answer.

http://laserpointerforums.com/f38/another-driver-question-43482.html

Hope this helps,
Toogoofy

 

[Greenmechanic]

I am just repeating what's in the linked post but I am going to add my opinion;
Laser diodes are too sensitive to cap. discharges from the driver, to trust them to clip leads. Sure the method you talked about will work, as long as you don't lose connection for a microsecond. If you do, your diode is history.
Even a tarnished connector on the leads can cause this. Doing this with a PHR is one thing, (you are out $10) but with a a 6X or 12X, why risk that much money for an extra 60 seconds of work?
Also the pots. on some drivers are super sensitive. While trying to set the pot., turning it a hair's width extra can push an extra 50mA or more into the LD.
It boils down to "better to be safe then sorry", it's way to risky trying to set the current using the LD. IMHO

 

[Deleted member 8382]

Listen me

The Flexdrive is not linear and the current it supplies vary depending on the load. While what you said can be done on linear drivers (LM317, Rkcstr...), the ammeter load is nothing compared to a laser diode, so it's not an accurate way to do it. Also, you could break your ammeter!

 

[chipdouglas]

the pots are so sensitive a minute turn could go from 0 to max. and you arent actually measuring voltage, but milli volts across the 1ohm resistor. dummy loads guarantee that the pot is going to be set right the first time.

 

[Xplorer877]

Listen me

The Flexdrive is not linear and the current it supplies vary depending on the load.

But how could that be true? If the driver supplied different currents under different loads, then the current it supplies to the dummy load will be different then that it supplies to the diode.

I think it must work the other way around. Regardless of the load the driver will supply the same current, which means it must vary the voltage appropriately. In other words the pot does not adjust the voltage it adjusts the current.

Yes I agree with you all that these diodes are sensitive and expensive.

Even though the diode is attached to a driver Ohm's Law still holds. Which means that laser diodes must have a resistance (which will vary depending on their operating wattage). With the 6x SF-AW210 diode at 5.3 volts the diode will draw 210mA so it has a resistance of about 25.2 ohms.

I guess I'm just not understanding the way these drivers work. A 1 ohm resistor is a very small resistor. Just one volt will cause an amp of current to flow through it. So the flex drive would have to sullpy just milli volts to get a current of a few hundred milli amps. On the Flex Drive pdf file it says that the regulated output is between 2-5.5 volts. Thats why I'm not getting why the dummy load method works. Rectifying diodes have little to no resistance when the current is flowing through them.

-Tony

 

[Deleted member 8382]

You're completely wrong!

"But how could that be true? If the driver supplied different currents under different loads, then the current it supplies to the dummy load will be different then that it supplies to the diode."

No, that's why we use dummy loads that simulates exactly a laser diode.

I think it must work the other way around. Regardless of the load the driver will supply the same current, which means it must vary the voltage appropriately. In other words the pot does not adjust the voltage it adjusts the current.

You adjust the current, but only for that load. As soon as you change the load you'll have a different current

"Even though the diode is attached to a driver Ohm's Law still holds."

No no no no no no and no! Ohm's Law only applies to linear electronics! When non-linear components come in, it just disappears. In other words, you can only apply ohm's law to ohmic materials, and diodes are not one of them.

from wikipedia

There are, however, components of electrical circuits which do not obey Ohm's law; that is, their relationship between current and voltage (their I–V curve) is nonlinear. An example is the p-n junction diode (curve at right). As seen in the figure, the current does not increase linearly with applied voltage for a diode. One can determine a value of current (I) for a given value of applied voltage (V) from the curve, but not from Ohm's law, since the value of "resistance" is not constant as a function of applied voltage. Further, the current only increases significantly if the applied voltage is positive, not negative. The ratio V/I for some point along the nonlinear curve is sometimes called the static, or chordal, or DC, resistance[18][19], but as seen in the figure the value of total V over total I varies depending on the particular point along the nonlinear curve which is chosen.

 

[Xplorer877]

I totally agree with you.

I meant Ohm's law holds for the 1 ohm resistor in the dummy load. I know that diodes don't fallow Ohms law that's why I said "Which means that laser diodes must have a resistance (which will vary depending on their operating wattage)". They have a non linear resistance that changes as more current flows.

It's the same thing with LEDs. You can't just measure the resistance with an ohm meter to figure out the voltage they should be run at.

However, I disagree with you about how the drivers are set.

You're saying that the current changes when you switch from the dummy load to the diode. Well doesn't that defeat the purpose of using it in the first place? Whenever people ask how a diode should be powered, we tell them a current. So it would make sense for a driver to vary the voltage to achieve that current. There is no way that a few rectifiers emulates any laser diode perfectly. Like you pointed out they are non linear. So different diodes have different resistance curves (I-V curve).

Look at this:

"The driver operates as a current regulated source, varying the output voltage above and below the battery voltage as necessary to meet the current output setting."

That's right off the Flex Drive PDF. The "current output setting" refers to the setting set by the pot.

Otherwise, you're implying that the pot adjusts only voltage. If that was true, then yes, the current would change under different loads.

Even though the driver regulates both current and voltage it is still bound by Ohm's Law, whether linear or not. A given voltage through a given load will cause a unique current to flow. Therefore, the pot must adjust either the voltage output or the current output but not both.

You cannot force 1 amp at 1 volt through a 10 ohm load, whether linear or not.

-Tony

 

[c4r0]

So why then, when you use a dummy load do you have to measure the VOLTAGE across a 1ohm resistor and then calculate the current? Wouldn't it be easier to simply measure the current through the circuit?

You don't have to calculate anything because when you have a 1 ohm resistor, the current equals voltage. This way of measure is used because then you don't have to open the circuit (and before that turn off the power, in case of driving a LD) when you want to connect the meter. You can just connect the voltmeter while the driver is working and read the measure as amps instead of volts. It's just more convenient.

 

[Benm]

I'm just wondering why everyone is calibrating their drivers to a string of rectifier diodes when it's easy to tune it to the real thing

Get what I mean?

I get what you mean. However, i think its always good practise to test any driver on a dummy load - wiring mistakes do happen, and it would be sad if you blew up an expensive laser diode due to that.

With linear (lm317 and such) current sources, you could argue that just connecting an amp meter will be a sufficient test. I still recommend using a dummy load though, especially when the expected voltage drop across the regulator isnt very large.

With a LM317 current soure, there must be 4.25 volts between input and output to operate within manufacture specifications. If you operate, for example, a blu ray diode from 2 lithium cells, you are below this limit and performance is not guaranteed, though it might work perfectly well in a practical setup.

 

[Deleted member 8382]

I'm gonna answer to each question below

I totally agree with you.

Fine

I meant Ohm's law holds for the 1 ohm resistor in the dummy load. I know that diodes don't fallow Ohms law that's why I said "Which means that laser diodes must have a resistance (which will vary depending on their operating wattage)". They have a non linear resistance that changes as more current flows.

Ah, I get you. We use that resistor avoid measuring current directly. It's easier to just measure the voltage on it instead of the current. However, you're free to measure the current directly if you're willing to do so.

It's the same thing with LEDs. You can't just measure the resistance with an ohm meter to figure out the voltage they should be run at.

Laser diodes, Light emiting diodes... They're all diodes at the end

However, I disagree with you about how the drivers are set.

Let's see

You're saying that the current changes when you switch from the dummy load to the diode. Well doesn't that defeat the purpose of using it in the first place? Whenever people ask how a diode should be powered, we tell them a current. So it would make sense for a driver to vary the voltage to achieve that current. There is no way that a few rectifiers emulates any laser diode perfectly. Like you pointed out they are non linear. So different diodes have different resistance curves (I-V curve).

I didn't said the current is going to change from the laser diode to the dummy load. I said it changes trough different loads, so that we use a dummy load which emulates a laser diode very well. Not perfectly, but very well. You can test this yourself if you want, make the I-V curve for a dummy load and then for a Laser diode, they're basically the same. It would make a huge difference, however, to connect the driver to your MM directly, which is what you suggested in the first place

Look at this:

"The driver operates as a current regulated source, varying the output voltage above and below the battery voltage as necessary to meet the current output setting."

That's right off the Flex Drive PDF. The "current output setting" refers to the setting set by the pot.

The Flexdrive keeps a constant current on your diode for many input voltages. That's what it's designed for, I believe. I really don't know if it's regulating voltage or current, I don't know if the voltage stays constant if you change the load, so maybe it's neither one or the other thing. I really don't know about this last one.

Otherwise, you're implying that the pot adjusts only voltage. If that was true, then yes, the current would change under different loads.

Or that you're not adjusting any of them...

Even though the driver regulates both current and voltage it is still bound by Ohm's Law, whether linear or not. A given voltage through a given load will cause a unique current to flow. Therefore, the pot must adjust either the voltage output or the current output but not both.

Why not?

You cannot force 1 amp at 1 volt through a 10 ohm load, whether linear or not.

Obviously xD

-Tony

 

[charliebruce]

"But how could that be true? If the driver supplied different currents under different loads, then the current it supplies to the dummy load will be different then that it supplies to the diode."

No, that's why we use dummy loads that simulates exactly a laser diode.



You adjust the current, but only for that load. As soon as you change the load you'll have a different current

"Even though the diode is attached to a driver Ohm's Law still holds."

No no no no no no and no! Ohm's Law only applies to linear electronics! When non-linear components come in, it just disappears. In other words, you can only apply ohm's law to ohmic materials, and diodes are not one of them.

Sorry to say this but you're wrong. A test load doesn't accurately simulate a diode, it only provides an approximation of the way an LD behaves. A constant-current driver will work across a variety of different loads without needing adjustment - as long as that potentiometer stays in the same place, the current will remain constant too. The test load only ensures that the driver has something like a laser diode to push current through (vital for something like the flexdrive), that won't blow like a laser diode if you accidentally overshoot. Ohm's law always applies, that's why it's a law. All that it means with non-linear components (diodes, LDs, semiconductors), is that the resistance of components changes depending on other factors (resistance is not constant).

I didn't said the current is going to change from the laser diode to the dummy load. I said it changes trough different loads, so that we use a dummy load which emulates a laser diode very well. Not perfectly, but very well. You can test this yourself if you want, make the I-V curve for a dummy load and then for a Laser diode, they're basically the same. It would make a huge difference, however, to connect the driver to your MM directly, which is what you suggested in the first place

In response to your last post, the flexdrive is constant-current, meaning that as long as the pot stays in the same place, so will the current reading - you could use any number of diodes in a test chain and achieve the same current reading (within reason - obviously a chain of 100 would give too high a forward voltage). The load makes no difference to the current, provided it's within acceptable limits.

 

[Deleted member 8382]

Then I might have understood something wrong. Why are we making dummy loads for BR diodes and others for red diodes?

 

[charliebruce]

In ideal situations, there's no reason to use two different loads, but the main reason is to check whether the driver is capable of putting out the current at a higher forward voltage (5v for blu-ray, compared to 3v for red, for instance) - in some situations, a driver would manage to put out the same current, but only up to, say, 4v - so, it wouldn't be able to run a blu-ray at the desired current, only red. The best example of this actually happening, is a Rkcstr driver with too low an input voltage (Vin - Vdrop < Vf).

 

[Deleted member 8382]

Thanks for the info then. I thought drivers were set for a family of diodes each time lol

 

[Xplorer877]

Well I'm understanding it a little better now. It's not like I never used a dummy load I was just unclear about how/why it works. Admittedly it can get confusing. Semiconductor electronics is a bit beyond my knowledge so far. It should be fun when I get there (I'm studying to become an electronic engineer).

-Tony