Effects of "capping" voltage at higher diode currents?

I am curious if someone could explain what will happen if you were to "cap" or limit the voltage with rising current through a laser diode. Let me explain why I ask this a little more and then maybe my question will make better sense. When testing a diode I have seen several members here use a power source that regulates both voltage and current. On some of the testing it appears all that is being adjusted is the current (I.e. Upwards) and at higher currents the voltage starts to rise. It doesn't appear that the voltage is being increased by the tester yet the readout shows an increasing voltage. On these tests are they adjusting the voltage upwards or is this readout measuring the voltage change somehow caused by the diode at higher currents? Or another way of saying this is, does the voltage drop across a laser diode increase at higher currents? If so then this is the root of my question. It's based upon an assumption that may be incorrect the assumption I am making from watching these tests is that voltage drop across the diode increases as current increases. If this is the case and voltage drop across a diode does increase at higher currents then what would happen to the process if the voltage were capped somewhere below the typical drop for that current? Another example (remember my assumption could be totally wrong): let's say your testing a 445 diode and at 1a the drop across the diode is 4.5v, and then at 1.2a it rises to 4.6v, then at 1.4a it would rise to 4.8v (I'm using these numbers for demonstration, not saying this is what would be happening on a 445 diode). If this were the case and I regulated the voltage to 4.7v but turned the current up to 1.4a (that may have reflected a 4.8v drop) what would happen?

Again, my assumptions could be totally incorrect but from what I have seen it seems with certain diodes the voltage drop across the diode increases with higher currents. If I am wrong in this assumption then someone please explain why we see the need for higher voltages applied at higher currents.

The whole reason I ask this is to learn if there would be an advantage to regulating the current and voltage across a laser diode. Usually I build my own linear drivers from reverse wired LM3xx series voltage regulators. I could just as easily add voltage regulation before or after the current regulator if there was a derived benefit.

I understand the basics of why we generally use current regulation with LD's, due to their finicky nature to draw more current at higher temps and go into thermal runaway and they are current hungry, that at set voltages they can pass more or less current depending on varying factors. What I am less clear on is how they deal with voltage. Until now I've always thought of them as any other diode that has a characteristic voltage drop across it. Lately I am starting to see that voltage drop across a LD is more dynamic and may change with certain other variables. If this is truly the case then it would seem that regulating both current and voltage might improve stability and reduce potential adverse effects.

Really interested to see what someone has to say about this. Thanks agin for your assistance in advance.

Constant current is actually really just constant voltage, but not in a typical sense. Ohms law dictates that higher voltages through a certain resistance will increase the current flow. So, basically all a constant current driver does is adjusts the voltage into the load up or down until the desired current flows through it, and holds it there.

Realistically all a laser diode cares about is the overall power it has to dissipate. If you set up a constant voltage regulator right at the diodes sweet spot, it'd run just fine. However the catch is as resistance increases/decreases, and laser diodes not being a really ideal resistive load, a relatively small change in voltage could blow the diode up.

Say you set a constant voltage driver into a diode at 4.5V, the diode might pull 1.4A for example. Then maybe for some reason the resistance of the load decreases, the driver is still pushing 4.5V into it, but all of a sudden it's now pulling 2A. See where this is going?

Instead, if we regulate current, then the total power will always be the same. Regulating both voltage and current is redundant, regulating current will naturally regulate voltage too.

Sure, even with constant current the resistance of the load might change, but it'll always be a fixed current, there is no way for the diode to pull more than the driver is sourcing compared to driving it constant voltage. The total power into the diode will always be the same or less than the driver is supplying.

I understand that if you set up a constant voltage across an LD the current can change up or down, in fact I mentioned this above. This is essentially why we don't hook a battery up directly to an LD, the LD can (and will) run away so to speak, and allow more current than desired to pass. I understand Ohm's law very well too. On rethinking my question as it applies to Ohm's law I think I can rephrase it much simpler: does an LD's resistance change with increasing current? Back to the issue of having a constant voltage across an LD and the current through it, if the circuit follows Ohm's law (no reason to suspect it shouldn't) then the only way current can increase across an LD at a set voltage is for the resistance of the LD to decrease. So basically, as an LD heats up it's resistance decreases, more current is passed, power is dissipated in the form of heat, more heat equals less resistance, equals more current through the diode, and the cycle continues into so called "thermal runaway".

When I think of the word redundant I think of more than one identical attribute. I'm not sure I would say having both voltage and current regulation is redundant because they are not the same. Possibly the end result as it relates to an LD is the same but I am not yet convinced of this because of the example both of us have referred to with regards to thermal runaway. If strict voltage regulation were redundant (I.e causing the same effect as) in regards to current regulation then we shouldn't worry about thermal runaway or changing current through an LD. Yet we do. Unless this is a factor of imprecise regulation to the degree that a very tiny change of voltage causes a huge increase in current, and our "regulation" is not regulation in the strictest sense.

Voltage drop across an LD implies resistance in its simplest terms, at least as I understand the dynamics of an LD. Going back to Ohms law, for the current to change across an LD at a given voltage the LD must change it"s resistance unless there is something peculiar about LD's that allow them the ability to operate outside this law. I can't I angina one.

I guess I should rephrase one of my other questions as well: watching these diode tests on the forum, when using a power source with the voltage and current readouts I see the voltage going up as the current and output goes up. Is the voltage being turned up by the tester along with the current, or is this just a measurement of the voltage across the diode at a given current.

If it is a measurement than then it would appear resistance is increasing with increased current (I.e. Ohms law dictates for increased V then R would increase). This would appear to be the opposite of the thermal runaway effect in which you would expect to see V drop.

I guess it should be asked directly too here: what are the characteristics of an LD with regards to current and internal resistance?

Again, remarks all very appreciated.

the voltage is a measurement of the Vf.

If you pump a certain current through a diode, that diode will drop a certain Vf. If you put a certain voltage across the diode, it will draw a certain current. As your logic tells you, this follows Ohms law, except that the diode's equivalent resistance varies with temperature, much as you posited in your above post.

For a small change in voltage, you get a large change in current. Again, you did reason this out in your above post. I think that it being easier to "fine tune" larger increments in currents rather than minute changes in voltage, coupled with the thought that power output scales with relation to current on a more 1:1 basis than with relation to Vf (not really, but just easier to think of it that way), is the main reasons why we regulate the current to control diode output.

Jmillerdoc said:

Voltage drop across an LD implies resistance in its simplest terms, at least as I understand the dynamics of an LD. Going back to Ohms law, for the current to change across an LD at a given voltage the LD must change it"s resistance unless there is something peculiar about LD's that allow them the ability to operate outside this law. I can't I angina one.

I guess I should rephrase one of my other questions as well: watching these diode tests on the forum, when using a power source with the voltage and current readouts I see the voltage going up as the current and output goes up. Is the voltage being turned up by the tester along with the current, or is this just a measurement of the voltage across the diode at a given current.

If it is a measurement than then it would appear resistance is increasing with increased current (I.e. Ohms law dictates for increased V then R would increase). This would appear to be the opposite of the thermal runaway effect in which you would expect to see V drop.

I guess it should be asked directly too here: what are the characteristics of an LD with regards to current and internal resistance?

Again, remarks all very appreciated.

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A lot of the videos you will see on diodes usually are running as constant current at the same time they monitor the voltage because as the current though the diode increases the voltage drop over the diode will also increase as well due to the NP junction .

Wouldn't the voltage drop be a function of the NP junction as The series resistance of a laser diode it tiny , I have a 500mW 670nm diode with a series resistance of 0.25 Ohms , but its running current is 1.1 Amps , yet it has a voltage drop of 2.5 Volts .