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I am trying to set the RESET pin, which is set to high via internal pull up, of my MCU (3.3v) to low based on an external pin (3.7v). When the external pin is low, the RESET pin should be Gnd and if the external pin is high, the RESET pin should be disconnected (will get back to high because of the internal pull up). I am new to electronics by i think i found a good solution with a p-type MOSFET.

Source = RESET Drain = Gnd Gate = External pin (0-3.7v)

enter image description here

Right now i am struggling with the wide variety of MOSFET's out there (Farnell has 1000). Could anybody please help me to calculate the right values or give me other advice for solving that problem?

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  • \$\begingroup\$ Okay, thanks for the schematic. That won't work. It needs negative voltage to turn it on, and the body diode means it will sit always at 0.6V on the /RESET pin. \$\endgroup\$ Oct 2 '16 at 17:16
  • \$\begingroup\$ ok thanks for your help. So i will just use your schematics. The R1 is my external pin and Q2 my RESET pin right? \$\endgroup\$
    – perotom
    Oct 2 '16 at 17:21
  • \$\begingroup\$ Yes, that's correct. \$\endgroup\$ Oct 3 '16 at 0:17
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Rather than use discretes, I'd recommend to just use a digital inverter with 5 V tolerant inputs, but powered from the micro's 3.3 V supply.

If this is the only circuit driving the RESET pin, you can use for example 74LVC1G04.

If you need to do wired-OR logic to allow other circuits (like a mechanical pushbutton) to pull RESET low, you can use 74LVC1G06.

The cost of either of these devices is low enough that the placement of the part on the PCB is likely to cost more than the part itself (so that using a slightly pricier part to replace 2 or 3 cheaper parts pays for itself).

Edit:

I just noticed you wrote

When the external pin is low, the RESET pin should be Gnd and if the external pin is high, the RESET pin should be disconnected (will get back to high because of the internal pull up).

For non-inverting logic with open-drain output, try 74LVC1G07.

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  • \$\begingroup\$ Ok thanks i will take a look at that! I have to check if the external pin get's expcility set to gnd. \$\endgroup\$
    – perotom
    Oct 2 '16 at 17:59
  • \$\begingroup\$ Yes, a CMOS output is pulled to ground when low. \$\endgroup\$
    – The Photon
    Oct 2 '16 at 18:39
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Alternatively for non-inverting logic that swings from 3.7 to say 0.5v and the CMOS datasheet indicates VIH and VIL for 3.3 logic is 2.0 and 0.9V worse case (respectively) you can use a Schottky diode directly between the two interfaces to perform the reset.(with internal pullup)

  • Check and confirm both worst case output swings, and supply tolerances as well as RESET input VIH and VIL.
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You did not provide a schematic for the scheme you have in mind, but in any case it won't work. If you are using discrete parts you will need at least two transistors. Here is an example using BJTs, they could be replaced with complementary MOSFETs (or you could use a dual transistor with base resistors in a single package):

schematic

simulate this circuit – Schematic created using CircuitLab

When the input voltage is less than about 2.5V enough base current flows to turn turn Q1 on, which provides base current to Q2 turning it on and pulling the /RESET input low. R3 prevents leakage in Q1 from being amplified by Q2.

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  • \$\begingroup\$ Thanks! I added a schematics to my question. Do i really need that much components for such a simple task? Is there no easier solution? \$\endgroup\$
    – perotom
    Oct 2 '16 at 17:11
  • \$\begingroup\$ The open-drain single gate inverter @ThePhoton suggested is probably the easiest solution. I would consider the BJTs (there is a single package part with bot transistors and suitable resistors) only if the connection was going between boards and you wanted a bit more ESD immunity. \$\endgroup\$ Oct 2 '16 at 18:51

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