I've read quite a lot of threads here about P-MOSFETS and how to drive them properly from IC but still have some questions.
Here's the schematic:


Some useful values from MOSFET datasheet
Vgs(th) -2..-4 V
Rds(on) ~0.07 Ohm
C ~1.5 nF

Schematic will be used to control a lot of common-anode RGB LEDs and current will be around 1A, so 60V/28A(max) MOSFET may be an overhead but that's what I got.
Can you please help me with the following
1. Will the schematic work as it is without NPN transistor controlling MOSFET gate?
2. Do I actually need current-limiting resistor to control Vgs or just the pull-up will do it?
3. Will MOSFET saturate on Vgs ~ -5V?
4. Will MOSFET close on Vgs ~ Vd - Vs = -1.7V?
5. How does the value of Pull-Up resistor affect how quickly I can switch MOSFET on/off (is 100Hz achievable?)

  • \$\begingroup\$ I have an equivalent circuit as the one in this question, except I’m using a 10k pull-up resistor and 12V on both the gate and source. The mosfet is conducting when my micro controller is powered off. Is that resistor value too high? \$\endgroup\$ Nov 29, 2017 at 17:53

1 Answer 1

  1. In general, simulating your design should answer that. Free tools like LTSpice or Qucs can be used for the simulation.
  2. The limiting resistor is good practice. The pull-up resistor does not limit current drawn from the microcontroller pin to the Gate. This current may spike beyond the MCU's pin maximum if directly connected, for charging the 1.4 nF input capacitance of the FET. A 180 Ohm resistor will be sufficient to limit pin current to under 28 mA, safe enough for many MCUs.
  3. No. The Transfer Characteristics graph indicates a Vgs of around -6 volts required to saturate this FET, for Vds = -30 Volts. For Vds = -5 Volts and Vgs = -4 Volts (not -5V, see next point), the On-Region Characteristics graph indicates a current of well under 2 Amperes.
  4. It should, as the rated minimum Vgs.th is -2 Volts, while the voltage divider formed by the 2 kOhm pull-up and the 500 Ohm current limiting resistor gives Vgs values of -1.36 Volts and -4 Volts respectively for MCU pin high and low. The problem you will face is at the pin=low part.
  5. Yes. The maximum switching cycle (turn-on + turn-off) for FQP27P06 is (45+380+70+190)=**685 nS**. Thus worst-case switching frequency is around 1.46 MHz (Vdd = 30 Volts, Rg = 25 Ohms). Allowing a hefty margin for the much lower Vdd = 5V and the 500 Ohm current limiting resistor, 15 KHz switching is definitely safe to aim for, much higher than the target of 100 Hz.
  • \$\begingroup\$ re MultiSim, I wouldn't call a 30-day trial a free tool. \$\endgroup\$
    – m.Alin
    Nov 22, 2012 at 11:53
  • \$\begingroup\$ @m.Alin Fair point, replaced it with a couple of other products instead. \$\endgroup\$ Nov 22, 2012 at 12:00
  • \$\begingroup\$ It was a very detailed answer. Just to make it clear - can I improve situation a bit by increasing pull-up value to, say, 20k? And one more thing - is it safe to rely on results obtained wih mentioned simulation programs even for simple circuits like this? \$\endgroup\$ Nov 22, 2012 at 19:24
  • \$\begingroup\$ @IgorRybalka You could use a pull-up of 20 kOhms and a gate resistor of 180 or 220 Ohms. That will increase the voltage swing of Vgs, resulting in a harder switching, and less heat generated in the MOSFET due to faster switching. \$\endgroup\$ Nov 22, 2012 at 21:06

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