0
\$\begingroup\$

I am toying around with MOSFET to learn. I do not know how to use them, despite many futile earlier attempts, so I mash up some circuits to get a new starting point.

In the circuit below I have a driver (1EDN7512) and a logic level N-FET (AOSS32334C). According to the datasheet for the driver, when IN+ is HIGH and IN- is LOW, OUT will be high. This last HIGH goes to the gate, and my intention with that is to turn on the N-FET so "5V_IN" can go through and be seen on "5V_OUT".

Obviously I have missed something crucial here, because in the real world I measure 3.2VDC at "5V_OUT".

  1. To get what I want, should Q1 be a P-FET instead? (AOS21115C, for example)
  2. Will my circuit work like intended if I make the driver LOW instead of HIGH?

enter image description here

\$\endgroup\$
3
  • 1
    \$\begingroup\$ Did you not ask this question three years ago? \$\endgroup\$
    – Finbarr
    Commented Nov 28 at 9:53
  • \$\begingroup\$ When you say LOW and HIGH, do you mean changing your MOSFET position from high to low or inverting the logic to your MOSFET driver? \$\endgroup\$
    – winny
    Commented Nov 28 at 12:31
  • \$\begingroup\$ You have IN and OUT reversed for use as a switch. Otherwise, the source just follows the gate voltage (aka Source Follower = non-inverting) yet offset by Vgs where it can "almost" conduct ( high R) defined as the Vgs threshold = Vgs(th) or as I call it the near-off threshold. The conductivity increases by the squared difference above this threshold up to about twice the threshold. Use Source = IN as a fixed voltage i.e. 0V for Nch and Vdd for Pch. The gate change, Vgs to drain voltage change, Vds is always inverting like Base, Vbe to Collector Vce response in BJT's. \$\endgroup\$
    – D.A.S.
    Commented Nov 28 at 13:27

1 Answer 1

Reset to default
1
\$\begingroup\$

"Obviously I have missed something crucial here, because in the real world I measure 3.2VDC at "5V_OUT"." That is because your circuit operates in source-follower configuration. Your output voltage will be 5V_IN-Vgs(th) of your MOSFET.

  1. Yes, a P-FET would be very suitable for this application.
  2. Yes, but keep in mind that if your downsteam circuit you are trying to switch in and out will loose its ground connection. If it's an IC with connections elsewhere, many bad things will happen. If it's a solenoid, motor, LED or something else which has no connections elsewhere and no concept of ground, no issues.
\$\endgroup\$
2
  • \$\begingroup\$ 2) I don't think OP meant swithcing the GND but the need for inverting the driver output to enable the P-fet \$\endgroup\$
    – Justme
    Commented Nov 28 at 12:11
  • \$\begingroup\$ @Justme Ah, that could be the case, yes. Let's have OP clarify. \$\endgroup\$
    – winny
    Commented Nov 28 at 12:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.