From: ehsjr@verizon.net   
      
   On 12/26/2025 7:35 PM, Cursitor Doom wrote:   
   > On Fri, 26 Dec 2025 18:36:57 -0500, ehsjr  wrote:   
   >   
   >> On 12/26/2025 12:56 PM, Cursitor Doom wrote:   
   >>> Gentlemen (IOW not you, Bill),   
   >>>   
   >>> I've got a bunch of green laser diodes which are specified for 370mA   
   >>> current draw. I've been using a straight 5 ohm WW resistor rated at   
   >>> 10W as a dummy load, but it's crude and inaccurate. Is there something   
   >>> better I should be using? I've got some 2W blue ones to do later on as   
   >>> well so something which could be adapted for those would be a plus.   
   >>> Ideally something which mimics the knee you get as it starts to   
   >>> conduct.   
   >>>   
   >>> Cheers,   
   >>>   
   >>> CD   
   >>   
   >> 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   
   >> 370 mA when connected directly across the supply.   
   >>   
   >> 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.   
      
   There's a problem with that as I understand what you said. I   
   suspect I'm not understanding what you have in mind. Here's   
   what I see as the circuit from what you said:   
      
   Supply +---[D1]---[D2]---[D3]---[D4]---[R]---Gnd   
      
   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   
      
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