Logic-level, high-side, P-channel MOSFET switch

Question

I want to use a MOSFET for high-side switching of the DRV425 fluxgate as it has no enable pin, and has quite a high power consumption (150 mW @ 5V -> 30 mA):

Before the P-channel MOSFET WTM2301 there will be a signal inverter SN74LVC2G04 for reasons not important for my question.

I only have 3.3 and 5 V available on my board and want to switch the MOSFET with a 3.3 V signal. The DRV supply is 5 V. The MOSFET datasheet mentions "P-channel", and from my understanding it can be used for high-side switching, however, the test circuit looks like low-side switching in the datasheet.

Can anybody provide some guidance on the chosen parts/ the circuit? Before manufacturing the board it would be great to get some feedback from experienced people. To be honest, I am confused with the datasheet and not sure this will work.

Answer 1

A cheap solution would be the following (NMOS + resistor). If you are ok with some current consumption when the PMOS is ON, you can try it.

Case1: PMOS is OFF

Case2: PMOS is ON

You can increase R1 further and reduce current further but should make sure that the current is able to beat the IDSS of M3 and pullup M2 gate to 5V when M3 is OFF. Another potential risk is if there are sharp voltage transients of large magnitude expected on 5V supply, the transistor M2 could turn ON momentarily (when it is supposed to be OFF) if R1 is very high.

A level shifter would not have any current consumption. So, it is definitely better if you can afford it.

Answer 2

First, as per @Andy aka comment you need to swap S and D terminals of the MOSFET.

Second, your inverter is powered at 3.3V so its output voltage is at most 3.3V.

When you need to turn on the load, you feed 1 at the inverter input, its output is pretty much near 0V so it is -5.0V in respect to the drain. The transistor is as on as it is possible. Good.

When you need to turn off your P-MOSFET, you will feed zero at the input of the inverter, the MOSFET's gate will be 1.7V below the drain potential.

The threshold voltage of WTM2301 is -0.4 .. -1.0V so your MOSFET will be somewhat conducting. For a 30ma load this will be "on" for any practical purpose (see the datasheet).

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You need a level shifter.

Since you happen to use an inverter, you may simply use an inverter powered by 5V bus. Most of 5V MOS logic series them will be happy with 3.3V logic levels at the input.

Answer 3

The MOSFET as you have shown it is upside down. You need its source S at the top, connected to the positive supply.

When switching high-side with any P-channel transistor, to switch off the load (flux gate IC), the potential at the transistor's gate must be very close to its source, +5V in this case. To switch on, using the MOSFET you propose, the gate will need to be significantly lower than \$V_S - V_{GS(TH)} = 5-1 =4V\$, say, 3.5V at most. In other words you require gate potentials significantly above or below +4V.

A 74HC logic device powered from +3.3V is not capable of providing that (its output will be 0V or +3.3V), so some kind of translation from 0/+3.3V to <+3V/+5V is necessary.

You have a few options. In all circuits below, M1 is your WTM2301 MOSFET, and G1 is one of the 74LVC2G04 inverters:

^{simulate this circuit – Schematic created using CircuitLab}

Above, we shift the output of the inverter upwards by using diodes to develop some offset voltage, 2V (A) or 3V (B). This provides M1's gate with potentials of 3.0V and 4.8V (A), or 2.0V and 5.0V (B).

I include option A only for completeness; I don't recommend it, since all other solutions I present here will provide a full 5V "high" signal.

More "clean" solutions, getting M1's gate as far as it's possible to get from +4V given a +5V supply, require an active approach:

^{simulate this circuit}

In option C, M2 is low-side, and can be switched directly from G1's output. For 0V and +3.3V inputs, M2's drain goes between +5V and 0V, perfect to drive M1's gate most effectively, but the signal is inverted.

Option D is a design popular with "bidirectional level shifter" modules like those from ADAFruit, but it has the advantage over option C that it does not invert. It also uses the least components, which is nice.

Driving a MOSFET gate requires a lot of initial current to charge and discharge gate capacitance. You'll notice that in all these designs R1 is present to reduce this load on G1's output.

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Update

If you wish to use the TXU0104 in the manner you showed us, that's OK. However, since you are clearly in the market for options, and since the TXU0104 seems a bit overkill, I assume you are free to explore other ideas too.

The simplest by far is to use a TTL level compatible inverter or gate, which will trigger correctly for outputs from 3.3V CMOS devices like your ESP32, but produce 0V/5V outputs perfectly suited for the MOSFET gate. By using such a device, level translation and inversion is performed in the same unit.

Any 74HCT gate will do this. I don't think you need schmitt trigger inputs, since the ESP32 outputs will not be slow or noisy. You can use any regular inverter, such as those you find in a 74HCT04, or 74HCT2G04. You could even use NAND or NOR gates configured to invert, like those in the 74HCT00 and 74HCT02.

If you really need schmitt trigger inputs, then devices like the 74HCT2G14 and 74HCT132 are perfect. All the following circuits will level-translate and invert simultaneously:

^{simulate this circuit}

It's always useful to have a few NORs and NANDs available for other tasks, but whatever gates you don't use in the packages, don't leave their inputs unconnected. Always tie unused inputs to ground or the positive supply.

As long as the 74HCT device has a supply of +5V, you can apply inputs of 0V/5V or 0V/3.3V, and you will get outputs of 0V/5V. That means you can use these "HCT" things between your existing "LVC" output and the MOSFET, if you wish.

I'm a bit embarrassed that I didn't think to suggest this solution in the first place. Big oversight on my part, sorry.

Answer 4

The ESP's output can be raised slightly from 3.3V using a resistive divider to the 5V rail. In this way, its output can be translated from 0–3.3V to, for example, 3.0–4.3V.

Simulation link here.

Alternatively, you can retain the inverter (at a 5V supply) for a more decisive switch-on of the MOSFET (and therefore more output current):

Simulation link here.