Could someone please explain the function of the MOSFET in the following bi-direct level conversion circuit:

Image from: http://www.rocketnumbernine.com/2009/04/10/5v-33v-bidirectional-level-converter/

What would happen if it was gone and the source and drain wires were just shorted?

 

Rotate the image anti-clockwise by 90 degrees, your confusion will come down by a great deal.

Anywas the gate of the MOSFET is driven by 3.3v to drive 5volt ckt.

 

Let me know if this is correct:

When L1 is driven low, VGS would goto 3.3V and thus the FET would be on and H1 would see a low. If LI is high (or left floating as the case might be), VGS ~= 0 and the FET is off meaning H1 would be pulled high.

If that's correct, how does it work the other way? I got the impression that if H1 was driven low, then L1 would go low too. Not sure I see how though...

 

When H1 is driven low, provided that the FET has a suitably low threshold the FET channel will be conducting, (in the inverse direction), so pulling L1 down.


When H1 goes high however, L1 will follow it up at first, but the FET channel will stop conducting as the source potential rises towards that of the gate. The pull-up resistor then takes L1 up close to 3.3V, but no further.


At least that's my best guess!