Connect with us

Various questions regarding optocouplers

Discussion in 'Electronic Design' started by Michael, Jan 9, 2008.

Scroll to continue with content
  1. Michael

    Michael Guest

    Hi there - I'm working on finalizing component selectrion for a high
    side switching circuit (as discussed here a month or so back). I was
    recommended the FOD617 optocoupler, but I'd like to find a similar if
    not better surface mount part.

    I'm looking at using a darlington optocoupler. I think the Sharp
    PC452J00000F looks nice, with a 1000% current transfer ratio at 1ma.
    The datasheet for the part mentions a VCE saturation voltage (http://
    document.sharpsma.com/files/pc452_eJ.pdf, figure 8) of about 1V. What
    does a saturation voltage even mean with a phototransistor? I mean
    with a normal transistor that means a large percentage of the base
    current is not being amplified. Is it sort of the same with an
    optocoupler? Or is this just the lowest that VCE for the part to
    conduct current?

    Similarly - with just a normal transistor optocoupler, what would a
    VCE saturation voltage mean?

    Also, are there any major advantages between transistor optocouplers
    and darlington optocouplers? From what I can see, you have a much
    higher (~5-10x higher) current transfer ratio with a darlington, but
    you also have a higher VCE saturation voltage (about 1V as compared to
    about 0.1-0.2V). Is one kind faster than the other, or are there any
    other major differences?

    Thanks!

    -Michael
     
  2. Tom Bruhns

    Tom Bruhns Guest

    The saturation voltage is the voltage you'll measure at the collector,
    emitter grounded, at the specified coupler input current and collector
    current. "Saturation" means the same thing as in a bipolar transistor
    excited by injected base current. It doesn't matter a whole lot if
    it's an electrical base connection that injects charge, or photons
    that create it in the same region. It's the region of the Vce vs Ic
    curves, loosely, where the collector voltage changes very little with
    changes in collector current. Clearly, it's the input transistor of
    the darlington that's saturated, and the base-emitter drop of the
    output transistor adds to the saturation voltage of the input
    transistor.

    Optos with photodiode outputs are generally much faster (especially in
    turn-off) than transistor output optos, and the darlingtons are even
    slower. A key problem is that there's a lot of stored charge to get
    rid of. It's like if you connect up a bipolar amplifier transistor,
    drive the base hard enough to saturate it, and then disconnect the
    base and let it float. You'll have a long turn-off time to deal
    with. It helps to not allow the transistors to saturate: operate
    with a collector-emitter voltage greater than the saturation voltage,
    and at an appropriate current. See fig. 10 of the data sheet you
    referred to.

    Cheers,
    Tom
     
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day

-