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jermstur
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  • Hi. jermstur.

    I'm still a amateur, too.:yh:

    If you use 5 of 1N5405 and 0.2ohm at 1.2A like this:
    flexdrive(+)--1N5404--1N5404--1N5404--1N5404--1N5404--0.2ohm--flexdrive(-)

    1. the minimum wattage of resistor should be 0.288W.
    [ P = I * V = I * (I * R) = 1.2 * 1.2 * 0.2 = 0.288 W ]

    2. total voltage drop
    = Vdrop of 1N5404 + Vdrop of reistor[ V = I * R ]
    = 0.8 * 5 + 1.2 * 0.2
    = 4.24V

    3. the current of your test load should be 5 times of the voltage measured at your DMM.
    V = I * R = I * 0.2
    I = 5 * V

    It is difficult to simulate exact Vdrop because there is some variation between laser diodes, and Vdrop of 1Nxxxx is 0.7-0.9V fixed step, not a spectrum.

    I tested with 1ohm. It's simple V=I.

    And remember one thing. At near 1A, 5-pin chip of flexdrive should be heatsinked.

    I wish this may help your 445nm build.
    Good luck.

    SHIN
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