How to design a high-side switch without intermediate BJT/MOSFET?

Using Arduino General Electronics
1

Hello.

I have a microcontroller operating at 3.3V and a frequency inverter (Mitsubishi FR-S500) operating at 12V.

My overall goal is to pass a digital signal (0V/12V) controlled by the MCU to the inverter's input pin.

Problem arises because of the voltage difference of the two devices. It is obvious that I can't directly connect an output pin of the MCU since the voltage is not enough for the inverter to interpret it as a logical high, but I can switch a 12V circuit with the MCU.

Here is the schematic:

image
image932×529 4.53 KB

I replaced an output of the MCU with a switch (SW2), and the input pin with the label "Vin". Everything in the red box is encapsulated and I can't modify it.

So I want "Vin" to measure 12V when SW2 is closed to 3.3V, and 0V when SW2 is closed to GND.

However, in this example setup I used a N-channel MOSFET. Turns out, it will not work that way, since voltage at the source of the MOSFET will be 12V, and the voltage at the gate is at most 3.3V, which results in negative Vgs, while it needs Vgs to be at least 2.5V to turn the MOSFET on.

So N-channel is not the way. I tried to do it with P-channel.

image
image846×446 4.19 KB

But the voltage of 3.3V is not sufficient to turn the MOSFET off, because Vgs is at most -8.7V, and the cheap low-voltage MOSFETs I intend to use are opened with Vgs < -1.4V.

It's not that I want to use just MOSFETs. With BJTs the same problem arises for both PNP and NPN.

I saw some configurations using an intermediate NPN BJT/N-channel MOSFET activating PNP BJT/P-channel MOSFET. But I'm not sure if that's the only right way to do it. Two MOSFETs seem to be somewhat redundant.

My question is, what is the right way to design the switch? And are there ways to avoid using an intermediate BJT/MOSFET to drive another BJT/MOSFET?

Thanks in advance!

2

You MUST meet the circuit requirements. An option is a charge pump MOSFET high side driver chip, but why do you see the other standard approaches as a problem?

two MOSFETs seem to be somewhat redundant.

Don't be silly. A component that solves a problem is not redundant.

3 Likes
3

You have correctly described the problem. Vgs needs to be (about) zero to turn-off the MOSFET, and Vbe needs to be (about) zero to turn off a regular transistor.. This is why low-side switching is more common.

4

You need to use similar to this.

HighSideSwitch
HighSideSwitch788×1532 164 KB

The BJT in A or B can be replaced by a N channel MOSFET.

Or

Use a relay with an N channel MOSFET driver.

1 Like
5

Do you have the spec's of the input? What is the minimum input voltage at Vin so that it will read as high?
You could try to use a 4K7 resistor from Vin to 12V and pull Vin low/high with a NPN or Nchannel mosfet. Vin will be something like 11V but will probably work.

1 Like
6

Exactly. But in my case, it seems like I can't do a low-side switch.

7

What is the function of Q1, Q9, Q8 in these circuits?

8

It would be optional.

Q1, in circuit A, is there to prevent battery reversal situations.

1 Like
9

Thanx, that makes sense!
And has nothing to do with question of OP....

10

Correct

11

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