This is the schematic: P-MOSFET high-side switch

For this switch setup to work, time delays even up to 50 ms is acceptable.

How do I calculate/estimate resistor values?

  • R1: Gate pull-up for P-MOSFET. (defaults to P-MOS being OFF)
  • R2: I don't know why it's needed or whether it's needed at all. (comment if you know.)
  • R3: Limiting MCU output pin current (when charging the N-MOSFET gate capacitor)
  • R4: Providing some resistance while the manual momentary-switch is held down. (limiting in-rush current before the P-MOSFET turns ON hard.)
  • C1: making the P-MOSFET turn-on-time longer.

Datasheets images: (first two are of 45N03 N-MOSFET)

45N03 Max Ratings 45N03 Characteristics

IRF7416 Max Ratings IRF7416 Characteristics

  • \$\begingroup\$ You have 15k pull down for your N-FET, which is very reasonable. If you set the pull up R1 to 15k, what value (range) do you need to set R2 to in order to satisfy Vgs(th)? With that value, what switching speed do you end up with and is that reasonable in your application? If yes, are all power dissipation figures ok for the resistors? \$\endgroup\$
    – winny
    Jun 1 at 18:55
  • \$\begingroup\$ @winny I assume R2 is conducting only when discharging P-MOS gate into ground. but why not remove R2 ? \$\endgroup\$ Jun 1 at 19:07
  • \$\begingroup\$ Try/simulate it! \$\endgroup\$
    – winny
    Jun 1 at 20:18

R1 and R2 form a voltage divider to give the correct Vgs to the PFET when the NFET is on. If you remove R2, then you can potentially exceed Vgs of the PFET and damage it.
In your circuit, since you are only at 12V and Vgs max is 20V, you could probably eliminate it. In that case choose R1 to limit current through the drain-source of the NFET.

For R3, 1k-10k will work. Unless you are driving fast hard signals, like for motor drive, there is a lot of latitude for a "switch".

R4 depends on the rest of your circuit. Use ohm's law, \$\large\frac{Vin-Vout}{R}\$

For C1 I would simulate it to see. I think it could vary based on source impedance, load impedance etc. If you keep R2 and connect C1 to the gate, instead of ground, then it's an RC circuit which can be used to calculate the Vgs at time \$t\$. ie. you can control the turn on ramp.

  • \$\begingroup\$ "choose R1 to limit current through Vds of the NFET". Vds of the NFET is 25V. how R1 affects that ? \$\endgroup\$ Jun 1 at 20:52
  • \$\begingroup\$ Vds can withstand 25V and not conduct. When you turn on any FET you have current flow through the drain to source. That current needs to be controlled, otherwise you burn up the FET. I guess I should have wrote \$Ids\$ \$\endgroup\$
    – Aaron
    Jun 1 at 20:55

Note that your R1 could be wired to the Drain of the N-MOSFET. The P-MOSFET can take the full 12V on its gate. R2 then only serves to slow down the turn on of the P-MOSFET. You can make R2 as low as 0 Ohms.
R1, only needs to flow enough current to defeat the leakage in the N-MOSFET (500uA) at some small voltage. (<Vgsth of P-MOSFET ie 1V)
1V / 500uA = 2K
That will work when very hot. Note that 12^2/2K is the power in the resistor.

  • \$\begingroup\$ "That will work when very hot" . what being hot? you mean when ambient/junction temps are high? \$\endgroup\$ Jun 2 at 9:07
  • \$\begingroup\$ The numbers I used were for the junction temperature being 175C. Look in the data sheets for more about leakage and temperature. \$\endgroup\$
    – Ken Smith
    Jun 2 at 22:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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