Dummy Load for Laser Driver Development

Use a string of diodes, unless you plan to go really fast and the capacitance hurts. A Vbe multiplier might work as a variable diode.

But a resistor should be OK if it hits the same operating point as the laser.

Will your drive be DC or pulsed or pulsed fast?

John Larkin Highland Tech Glen Canyon Design Center Lunatic Fringe Electronics

If you want something better than a 5 ohm resistor dummy load, see the LM317 datasheet figure 8.8

As you did not specify your supply voltage, I don't know how close your 5 ohm resistor dummy load is to drawing

If you put the LM317 circuit in series with your 5 ohm resistor, you compute the current drawn by by 1.2/R1. So, for example, if R1 is 3.25 ohms the current drawn will be ~369 mA. This of course assumes a supply of enough "grunt" and within Vmax for the 317.

Some more detail: R1, at 3.25 ohms, will dissipate around half a watt. Use at least 1 watt. I don't know what you have on hand - I'd use power resistors - a 3 ohm in series with a .25 ohm.

Ed

Okay, many thanks. Yes, I know there were scant details provided but I only wanted vague suggestions I could maybe develop myself. The other idea I had was four diodes in series with a one ohm resistor so as to mimic the Vf of the laser diode. Fortunately I have a good selection of WW power resistors in my stash here.

The 4 diodes in series provide a 2.4 volt voltage drop, assuming .6 volts per diode. Call that Dd (Diode drop). The total voltage drop is the Diode drop (Dd) plus the drop across your 1 ohm R. So you need to know the current through R to compute its voltage drop.

Your R is 1 ohm. Your circuit looks like this: Vs---Dd---R---gnd. The voltage across R is Vs - Dd. Current through R (1 ohm) is found by I = (Vs-2.4)/1 = Vs-2.4 . That means I varies as Vs varies - I is not fixed. Thus we cannot say what the voltage drop is across R. That means the total drop cannot be established as equal your laser diode Vf using that circuit.

So in general, you need an active current limiting circuit for what you want to do. If we can get more specific - say a regulated supply of some specific or settable regulated output voltage, then we can use your circuit with a computed load resistance

Maybe you could post a schematic if I've misunderstood?

Ed

If you’re trying to get wide bandwidth, you need to include the wildly varying junction capacitance.

Cheers

Phil Hobbs

Cursitor Doom is an anonymous troll. His idea of what might constitute a gentleman is probably just as silly as the rest of his ideas.

Australians don't get all that excited about the social distinctions that distinguish English ladies and gentlemen from the lower orders. I worked in England for 22 years, and wasn't all that impressed by the kind of people who thought that they belonged to that particular class.

They weren't great at getting things done, and wasted a lot of time trying to create the right impression, rather than any kind of useful result.

Nothing about Cursitor Doom suggests that he has any right to call himself a gentleman - he's just a pretentious twit.

There are Dutch equivalents, and I got to know a few of them. The remnants of the Dutch regent class that have survived are rather more business-like than their English equivalents