Can anyone explain the meaning of this circuit? Why is the resistor connected in parallel to the diode? (obviously they have the same voltage drop), why the diode connected to the voltage source in that way? I guess it's a basic question so bare with me... enter image description here

Edit: The 3 wire module on the right side is a connector to an external fan. Pin 2 of this connector is connected to 12V (which is not shown).

  • \$\begingroup\$ I've no clue what the right-hand, 3-wire module is. But the diode looks like it is there to handle the case where that node rises above the 5 V rail. Quick path to remove charge -- probably much faster than leaking through the 10 k resistor. \$\endgroup\$ – jonk Dec 10 '16 at 7:55
  • \$\begingroup\$ It looks like you've attempted to obscure some info that would provide useful information. \$\endgroup\$ – Spehro Pefhany Dec 10 '16 at 8:05
  • \$\begingroup\$ the 3 wire module is a connector to fan. Sorry for not pointing that out. \$\endgroup\$ – roy.me Dec 10 '16 at 8:22

The 10k resistor is a pull-up, makeing sure that the MOSFET is turned on by default.

The port on the right is probably a connector and pin 3 is an open collector which allows you to turn off the MOSFET by pulling it low.

The diode is a clamp diode to make sure the gate voltage doesn't exceed 5 V, probably due to miller effect though the MOSFET. At small signal levels, this is neglectable but perhaps the desiner wanted protection against ESD/excessive drive voltage from the outside. You'll see them commonly even when not fully needed.

That MOSFET in turn is an inverter and overall it looks like a 5 V TTL logic to 3.3 V translator which connects via a connector and assumes open-collector output on the sender. Output to your 3.3 V system is the middle wire on the left, the source of the MOSFET.

  • \$\begingroup\$ Thank you very much. What will be the voltage on the gate? 4.3V? Why is the resistor connected in parallel? \$\endgroup\$ – roy.me Dec 10 '16 at 8:26
  • \$\begingroup\$ The volage on the gate will either be 0 or 5 V. Note tge diode direction. The leftmost resistor is the pull-up for the inverter stage. \$\endgroup\$ – winny Dec 10 '16 at 8:47
  • \$\begingroup\$ Ok, so if you can, shed some light on the esd protection behaviour... \$\endgroup\$ – roy.me Dec 10 '16 at 8:58
  • 2
    \$\begingroup\$ This is not a good ESD protection. It is only an over voltage clamping protection. If the input voltage to the gate exceeds 5V, the diode will clamp it to around 5.7V. However, if the voltage is negative (and ESD can have positive or negative polarity) the diode will not clamp it. \$\endgroup\$ – Claudio Avi Chami Dec 10 '16 at 9:55
  • \$\begingroup\$ You are right. Thank you. I did not think of that. \$\endgroup\$ – roy.me Dec 10 '16 at 11:27

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