Electronic – Using a voltage divider and an N-channel MOSFET to switch an Arduino input pin from a EL2008 Beckhoff digital output module

You have received a large amount of useful input related to the question that you asked and it will be of assistance to others who read these answers in future.

However, you have been wasting people's time and confusing your self because you stated what you thought you needed to do to fix your problem instead of telling your people what your problem is. While there is some overlap the answers that have been given mainly relate to things that you are not trying to do. While they do somewhat address what you are trying to do, the diagram that you provided would make almost no sense in most contexts and is NOT doing what it appears to be doing.

Lesson: "Tell us what you actually are trying to do and we will tell you the best way to do it".

Real question: See Maxim DS1822 Data Sheet -
PAGE 5 - POWERING THE DS1822 and
page 6 SUPPLYING THE PARASITE-POWERED DS1822 DURING TEMPERATURE

In the related diagram below Vpu is a "weak pullup" and the FET is a "strong pullup".
When the Vdd pin is grounded, power supply energy may be provided via the DQ line and is stored in an internal capacitor Cpp (C parasitic power). During most portions of operation the "parasitic" feed provides enough current Ipp at an acceptable voltage to power the IC. During some operation Ipp is inadequate and the iC must either be powered via Vdd or via a higher current source (see data sheet page 5). During these high current operations the FET is turned on to provide extra supply current. This low resistance power feed clamps the bus high and presents it being used for signalling by other ICs on the bus, so the 'strong pullup' is enabled only for as long a period as required.

SO:

You do need a FET for pullup, you do need a high side FET, this need is most simply met with a P channel FET - all as advised by others.

As Vmicrocontroller (Vmcu) is >= V1_wire_bus, the FET is not being used as a level converter but as a high side power supply switch.

Choosing a MOSFET:

Connecting a suitable P Channel MOSFET as shown in the diagram will fill the need. Many FETs will do the job.

Rdson / On resistance: MOSFET must have low enough on resistance = Rdson for the task.
A MOSFET that dropped 0.1V at 2 mA would probably suffice
Rdson = Vdrop / Iload =
= 0.1v/2 mA = 50 Ohms.
You'd have immense difficulty buying a P Channel FET with Rds = 50 Ohms = normally available ones are typically 50 to 5000 times BETTER (lower Rdson) ie 1 Ohm down to say 10 milliOhms.ie ANY P Channel MOSFET that meets other specs will have an OK Rdson

Gate operating voltage = Vth or Vgsth:
Vth or Vgsth should be << Vcpu.
ie the μP (microprocessor) should easily drive the MOSFET.
A 3.3V μP will JUST operate a MOSFET where Vth = 3V.
Operation will be better at Vgsth = 2.5V
and better again at 2V. Lower again does not hurt.

Vds_max > say 10V is OK - 20V or 30V better. > 30V OK.
Ids_max is so low as to be met by anything.

The horrible BSS184 - datasheet here is 20 cents in 1's at Digikey and does the job well enough. Digikey and others have many more that will do a better job - but not needed here.

There are two main sets of characteristics quoted for pins, absolute maximum ratings and typical ratings.

The HC-06's datasheet says it will accept input at 3V - 4.2V with 20mA - 40mA

Those are absolute ratings. The pin is rated to handle 3V to 4.2V and 20mA - 40mA current. In a normal circuit, digital input pins will draw very little current, as they are CMOS logic gates. There might be a pull-up resistor, that connects the pin internally to Vcc, or a pull-down resistor, that connects the pin internally to GND, which will increase the current draw.

Without the pull-up or pull-down resistor enabled, the pin has very high input impedance. Your proposed one resistor circuit looks like this:

^{simulate this circuit – Schematic created using CircuitLab}

You effectively have a voltage divider with the other resistance being the input impedance of the HC-06 input pin.

So the voltage on the RX pin with 5V on the Arduino TX pin would be:

$$ V_{RX} = 5V \frac{R_{really\,really\,big}}{R_{line} + R_{really\,really\,big}} \approx 5V$$

This is outside the maximum voltage rating.

Solution

The solution is to use some kind of level shifter. The resistors in the divider should be high enough not to violate the Arduino maximum current output and much less than the input impedance of the input pin. They also must be low enough that the input capacitance of the pin doesn't 'smear' the signal too much (slew rate). You can think of the capacitance as resisting the change in voltage, so sharp inputs start to get rounded off: (image taken from http://www.johnloomis.org)

For you quoted figures we would then have:

$$ V_{RX} = 5V \frac{\left(\frac{1}{20K} + \frac{1}{ R_{really\,really\,big}}\right)^{-1}}{10K + \left(\frac{1}{20K} + \frac{1}{ R_{really\,really\,big}}\right)^{-1}} \approx 5V \frac{20K}{10K + 20K} \approx 3.3V$$

In Arduino land, people would by a logic level shifter board for this purpose. One common one has a voltage divider for 5V TX to 3.3V RX, and a transistor for 3.3V TX to 5V RX. Others have transistors both ways, so that the voltages can be different from 5V and 3.3V as using a voltage divider the ratio is fixed, and to effectively have a very low output impedance on the TX pins and thus avoid slew rate problems.