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I'm attempting to connect an infrared laser diode to a laser diode driver circuit. I'm inexperienced with electronics and am having difficulty interpreting the technical information of my parts.

The laser diode is a ARIM ADL-78051TL. Its datasheet is available here.

The driver circuit is a iC-WK iCSY WK2D. This component seems to be part of a series, and has two datasheets. One specific to this version here, and one that details how the chip on the circuit works here.

My first questions concern how these parts should be connected to each other. I've compiled diagrams from the datasheets to illustrate my questions:

Laser Diode and Driver Circuit diagram.

As is visible in the image, I don't understand what "common pin laserdiode" and "2" are meant to connect to. Googling "common pin" indicates it has some relation to ground, but I didn't find a definitive answer. I suspect that the "2" pin on the laser diode is meant to go to ground, since pin 1 is for the photo-diode and pin 3 is for the cathode, but the datasheet doesn't explicitly mention this.

My other questions are concerned with the amount of voltage demanded by the diode and supplied by the driver. I've created a diagram with some tables from the datasheets, where I've highlighted the rows my questions are about:

enter image description here

My first question concerns supply voltage, with the relevant information highlighted in red. The driver seems to be supplying any diode connected to it with 2.4V to 6V, depending (I think) on how much the regulator on the driver has been adjusted. The diode datasheet mentions that the diode will only work with a voltage up to 2.2V.

Will this mean I need to reduce the driver's output voltage by at least 3.8V, to ensure the diode isn't damaged?

My second question concerns operating current, and has it's relevant information highlighted in blue. The driver datasheet mentions it can supply a current up to 90mA. The diode datasheet indicates a maximum permissible current of 40mA. My understanding is that If I correctly limit the voltage to 2.2V, I won't have to worry about too much current. Is this correct?

I hope that I've stated my questions in a clear and understandable manner, and would be overjoyed if any of you can help me.

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  1. "Common pin laserdiode" refers to the terminal shared between the photodiode and the laser diode (Pin 2 in your Laser Diode Diagram). The correct connection is shown in the datasheet you linked:

    In this block diagram you can see that:

    • The LD pin (Pin 1 in your Laser Diode Diagram) is connected to the LDK terminal of the integrated circuit present in the driving module;
    • The PD pin (Pin 3 in your Laser Diode Diagram) is connected to the MDA terminal;
    • The common node between the laser diode and the photodiode is connected to the LDA terminal;

    The other datasheet describes the whole module (including external components):

    enter image description here Here you see that:

    • LDK is connected to the LK pad of the board;
    • MDA is connected to the PD pad;
    • LDA is connected to the C pad;

    So you have to connect C (driver module) to 2 (laser diode).

  2. The voltage interval you read in the module's datasheet is referred to the driver's supply, it doesn't mean these voltages are applied to the laser.

  3. The current absorption is determined by the load impedance (the laser diode's resistance). In principle, as the diode may not be able to limit the current (due to its "low" resistance), a resistor should be put in series with it in order to prevent a "high" current from damaging the device. Anyway, in the driver module I see a 4.7 kΩ (plus a 10 kΩ trimmer) is already mounted between the photodiode and ground so there shouldn't be any concerns about the current (actually you can expect it from a laser diode driver module).

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    \$\begingroup\$ I've assembled the parts per your answer and everything is working perfectly. Thanks a lot for your help :) \$\endgroup\$
    – lars
    Commented Jan 4, 2022 at 21:14
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    \$\begingroup\$ You're welcome :) \$\endgroup\$
    – Yuri
    Commented Jan 13, 2022 at 11:12

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