microcontrollermosfetswitching
I'm currently using this circuit to control my load from MCU logic HIGH
why just not do like this ?
circuit ? It's safe for my MCU ?
I use NDP6020P is a Vgs(th) max -1V my VCC is 3.7V.
I know in the last circuit when MCU is LOW my load is "ON" instead the first circuit "LOW" = "OFF"
Regards
With 1 meg from the gate of Q1 to GND, and SENS25% at 50k, Q1's Vgs will be 4.76 V, while with SENS25% at 600k, Vgs will be 3.125V.
That means that Q1 is almost certainly capable of being fully turned on - albeit somewhat slowly - by SENS25%, but R2 will prevent that from happening because the higher Vgs goes, the higher the current through R2 goes, which will cause the voltage dropped across it to increase, raising the voltage on the source, partially defeating the rise in Vgs.
A solution might be to wire your circuit like this:
Where, when Q1 is OFF, 5V across R1 R3 R4 will make Vout equal to a "1" into your Raspberry Pi, and when Q1 is ON, the drop across the LED and Q1's Rds(on) will make Vout equal to a "0".
If not, I think you'll need some extra circuitry or a different way to go about it.
Good question. You are wise to ask.
This is not recommended. Sometimes MCU inputs specify that a certain amount of current can be injected through the protection diodes. You will have to check the MCU you eventually settle on. Select a pullup that will insure you stay under the maximum allowable current. But it would probably just be easier to add an additional NPN or N-channel (BSS138) to pull down the gate of the PMOS. Then there is no problem and you are free to choose any MCU, because it will never "see" 12 V.
Edit: I was only thinking about damage to the MCU. But @abdullah kahraman raises another equally important point. The diode current will also cause the PMOS to turn on. So this just plain won't work.
Best Answer
You can drive an mosfet directly from the GPIO things. But you need to make sure it will work:
This is for an N-channel. That is easy.
There are extra difficulties with a P-channel. Since the voltage to turn the P-channel off should not exceed the maximum level of the GPIO. This will typically be a maximum of 5V on five volt tolerant IO.
Meaning switching a p-fet directly on GPIO is not possible when switching more than 5V.
In case of your NDP6020P, the Vgs(th) graph looks like this:
When Vgs is < 1 Volt, the fet is not conducting. The pullup to Vcc will do this.
When Vgs is > 2 Volt, the fet is conducting. Shorting gate to ground will do this.
The absolute gate voltage may not exceed the maximum GPIO voltage.
Typical circuit, including MCU internals:
simulate this circuit – Schematic created using CircuitLab
As you see. The 5V, or switched load voltage, may never exceed Vcc/Vdd of the MCU!