Various questions regarding optocouplers

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

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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

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