From: cd@notformail.com
On Fri, 26 Dec 2025 21:01:34 -0500, ehsjr wrote:
>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
>
The power supply is sorted out, but before I apply it to an expensive
laser diode, I wanted to check I had the current in the right
ballpark. These fuckers are real current hogs!
I found this on the Odic Lasers website. They're suggesting something
curiously similar to my earlier idea: four diodes plus a 1 ohm
resistor:
https://odicforce.com/epages/05c54fb6-7778-4d36-adc0-0098b2af7c4
.sf/en_GB/?ObjectPath=/Shops/05c54fb6-7778-4d36-adc0-0098b2af7c4
/Categories/Background_and_Projects/Setting_the_Laser_Driver
--- SoupGate-DOS v1.05
* Origin: you cannot sedate... all the things you hate (1:229/2)
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