• Dummy Load for Laser Driver Development

    From Cursitor Doom@cd@notformail.com to sci.electronics.design on Fri Dec 26 17:56:58 2025
    From Newsgroup: sci.electronics.design

    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
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  • From john larkin@jl@glen--canyon.com to sci.electronics.design on Fri Dec 26 11:04:44 2025
    From Newsgroup: sci.electronics.design

    On Fri, 26 Dec 2025 17:56:58 +0000, Cursitor Doom <cd@notformail.com>
    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

    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
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  • From ehsjr@ehsjr@verizon.net to sci.electronics.design on Fri Dec 26 18:36:57 2025
    From Newsgroup: sci.electronics.design

    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

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  • From Cursitor Doom@cd@notformail.com to sci.electronics.design on Sat Dec 27 00:35:47 2025
    From Newsgroup: sci.electronics.design

    On Fri, 26 Dec 2025 18:36:57 -0500, ehsjr <ehsjr@verizon.net> 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.
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  • From ehsjr@ehsjr@verizon.net to sci.electronics.design on Fri Dec 26 21:01:34 2025
    From Newsgroup: sci.electronics.design

    On 12/26/2025 7:35 PM, Cursitor Doom wrote:
    On Fri, 26 Dec 2025 18:36:57 -0500, ehsjr <ehsjr@verizon.net> 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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  • From Phil Hobbs@pcdhSpamMeSenseless@electrooptical.net to sci.electronics.design on Sat Dec 27 02:11:15 2025
    From Newsgroup: sci.electronics.design

    Cursitor Doom <cd@notformail.com> 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 yourCOre trying to get wide bandwidth, you need to include the wildly varying junction capacitance.

    Cheers

    Phil Hobbs
    --
    Dr Philip C D Hobbs Principal Consultant ElectroOptical Innovations LLC / Hobbs ElectroOptics Optics, Electro-optics, Photonics, Analog Electronics
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  • From Bill Sloman@bill.sloman@ieee.org to sci.electronics.design on Sun Dec 28 03:04:45 2025
    From Newsgroup: sci.electronics.design

    On 27/12/2025 4:56 am, Cursitor Doom wrote:
    Gentlemen (IOW not you, Bill),

    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
    --
    Bill Sloman, Sydney

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  • From Cursitor Doom@cd@notformail.com to sci.electronics.design on Sun Dec 28 14:26:15 2025
    From Newsgroup: sci.electronics.design

    On Fri, 26 Dec 2025 21:01:34 -0500, ehsjr <ehsjr@verizon.net> wrote:

    On 12/26/2025 7:35 PM, Cursitor Doom wrote:
    On Fri, 26 Dec 2025 18:36:57 -0500, ehsjr <ehsjr@verizon.net> 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-adc0-0098b2af7c4e.sf/en_GB/?ObjectPath=/Shops/05c54fb6-7778-4d36-adc0-0098b2af7c4e/Categories/Background_and_Projects/Setting_the_Laser_Driver
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  • From ehsjr@ehsjr@verizon.net to sci.electronics.design on Sun Dec 28 15:31:46 2025
    From Newsgroup: sci.electronics.design

    On 12/28/2025 9:26 AM, Cursitor Doom wrote:
    On Fri, 26 Dec 2025 21:01:34 -0500, ehsjr <ehsjr@verizon.net> wrote:

    On 12/26/2025 7:35 PM, Cursitor Doom wrote:
    On Fri, 26 Dec 2025 18:36:57 -0500, ehsjr <ehsjr@verizon.net> 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-adc0-0098b2af7c4e.sf/en_GB/?ObjectPath=/Shops/05c54fb6-7778-4d36-adc0-0098b2af7c4e/Categories/Background_and_Projects/Setting_the_Laser_Driver


    Thanks, now I get it! Your supply, in the circuit below, already
    has adjustable current limiting:
    Supply +---[D1]---[D2]---[D3]---[D4]---[R]---Gnd

    So what you want to do is connect a volt meter across
    the 1 ohm resistor and adjust the supply to provide a
    constant current through the resistor so the meter
    reads 370 mV. (Or whatever current you are after.)
    The current through the 1 ohm resistor will be the
    same number as the voltage reading.

    Ed


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