Setting Current via transfer function. Check me?

Discussion in 'Math' started by Zeppelin1007, Sep 13, 2015.

  1. Zeppelin1007

    Thread Starter New Member

    Mar 6, 2012
    Hey guys.

    This should actually be an easy one.

    Im playing with a PLD 5000 laser driver. Adding a diode to my cnc machine. It has a current monitor pin, as in you can read the outputgoing

    Anyways, it says, in order to set the current from a voltage reading, use the transfer function:

    V=I(LD) /2A/V
    I(LD) is Current of LaserDiode.

    My driving voltage is 12V. I'd l ike to hit around 1.5A

    Am i correct in guessing that 2(1.5)/12 = 0.25=V?

    Ie that if i read V=0.25 on my monitor pins, Im at 1.5Amps on the output?
  2. MrAl

    Distinguished Member

    Jun 17, 2014

    I think what you mean is:

    which when solved for Id gives us:

    and E is the measured voltage. So if you measure 0.25 volts then you have 0.5 amps.

    That's what it sounds like anyway, but you need to test this first anyway before connecting your laser diode. You can probably use two diodes in series with a small resistance like 0.1 ohms in series, and that way you can test the setup before the real laser diode is connected. You can measure the voltage of course, and you can measure the current by measuring the voltage across the 0.1 ohm resistor and using:

    but that's only when you use a 0.1 ohm resistor in series.

    It's always a good idea to test first anyway that way the real diode does not get blown out by accident.
  3. WBahn


    Mar 31, 2012
    I think you are not transcribing it correctly.

    What you have written is

    V=I(LD) /2A/V = [I(LD) /2A]/V = I(LD) / (2 A·V)

    Order and associativity of operations matter.

    It is probably

    <br />
V \: = \: \frac{I_{LD}}{2 \frac{A}{V}}<br />

    Rearranging this slightly, you have

    <br />
V \: = \: I_{LD} \( 0.5 \frac{V}{A} \)<br />

    The driving voltage and the monitoring voltage are two very different things. You apply whatever driving voltage is needed to obtain the monitoring voltage that corresponds to your desired current. If you want 1.5 A of current, then your monitoring voltage will need to be 0.75 V. This is NOT the voltage that you drive the device with. The monitoring voltage is responding to the current that is actually flowing -- the voltage of the power supply that you have available doesn't enter into it.

    If you track the units you will see that this cannot be correct. But, since no one cares enough about being correct to bother tracking their units....

    Nope. If you read 0.25 V on the monitor output, then you will have a diode current of 0.5 A, since you have 2 amps of diode current for each 1 volt at the monitor output.