SPI Isolation with Open Drain MISO
I see a few designs where a schottky diode is placed with its cathode facing the output on the isolated/uC side of the isolator and then another pullup before going to the uC. Can anyone explain to me what the point of that would be?


It sounds like you are looking at this schematic from the above datasheet....



If we look at a single channel, it looks like this...







simulate this circuit &ndash; Schematic created using CircuitLab

Here, diode D1 converts the push/pull output of the isolator to effectively an open collector output. 

Here are the possible cases for the above circuit with MCU being an open-collector pin and OUTPUT being a push-pull pin:

MCU: Pull Low, ISOLATOR: Pull Low

No problems. 

MCU: Hi-Z, ISOLATOR: Pull Low

No problems. Current flows though R1 and D1 into OUT, the MCU sees a low. 

MCU Hi-Z, ISOLATOR: Drive High

No problems. R1 pulls MCU input line high, MCU sees high. 

MCU: Pull-low, ISOLATOR: Drive High

Yikes. If the two directly connected, then they would fight it out and one would burn up. But adding D1 prevents current from flowing from the high OUT into the low MCU. 

R1 is needed for the case where OUTPUT is high and the MCU is not pulling down. In this case, R1 will pull up so both see the line as high. 

Make sense?