High side driver with P channel FET

Will one of you be so kind as to double check my circuit here? This is feeding voltage from a battery to a switching buck regulator that supplies 5V to my Arduino and a few other things. Maximum current is 3A so max I'm moving 15W. Really much less than that, but going with max ratings to leave a margin. It's gotta be high side because of the way the grounds are connected later. That much I have been schooled on already.

The pin feeding it is not an Arduino pin, it's from a port expander. It has a 20mA max current.

Max Vgs for this P-FET is -16V. My supply can be from 7 to 18 depending on which batteries are hooked up. With the 1K and 330 I should be able to go to about 21V before I hit the max Vgs. And at the low end of my battery I'll still have about -5.25Vgs which should be just enough to get it turned on. I think I'm well within tolerance on everything else.

Or at least that's my novice opinion. I'd appreciate anyone who knows a bit more.

This may not be the best transistor, but it's what I have on hand. If it will work it will beat buying something.

With an emitter resistor, the base resistor does nothing, and can be removed (0 ohm).
There is ~4.3volt on the emitter, and ~4.5volt on the collector with a 5volt logic HIGH on the base.
Could add a zener between supply and collector, to be sure.
Like in the fourth diagram on this page.

Are you sure the buck converter doesn't have an enable pin (or can be accessed on the chip).
Then the whole mosfet circuit wouldn't be needed.
Leo..

Wawa:
Are you sure the buck converter doesn’t have an enable pin (or can be accessed on the chip).
Then the whole mosfet circuit wouldn’t be needed.

All it has is Vin, Vout, and two grounds. They were pretty cheap. The chip on it may have an enable pin but it’s tiny and I’m not going in after it. I can build this right next to the regulator and be done with it as long as I haven’t messed up my calculations.

I put the base resistor there to limit current a little extra. Wasn’t sure if it would help or not. I figured 4.3/330 is 13mA. That’s better than half my rating and I need next to nothing on the gate to turn on that FET so I figured I probably only needed microamps at the base. I guess 13 is less than 20 and it will save me a resistor to lose it.

This isn't exactly the one I have, the silkscreen is a little bit different on the back. But the parts and connections are the same.

Looks like pin 2 on that teensy little chip is Enable. Doubt I could get anything soldered to it though.

When your supply voltage is 7v, the MOSFET will not have a Vgs sufficient to turn on.

Have you considered a 5v relay ?


Pin 2, Enable Input. Pulling this pin below the specified threshold shuts the chip down. Pulling it up above the specified threshold or leaving it floating enables the chip.

larryd:
When your supply voltage is 7v, the MOSFET will not have a Vgs sufficient to turn on.

Help me understand. I'm not arguing, I'm just showing my work and asking for a correction.

At 7V I see Vgs will be (7*(1000/1330)) = 5.26V.

The gate threshold in the datasheet is -2.0V and what I get from figure 5 is that I'll be on enough to supply way more amps than I need. Where am I screwing up?

Sorry, should have posted this from the get-go. I know better.
https://www.mouser.com/datasheet/2/196/Infineon-I45P03P4L_11-DS-v01_00-en-785379.pdf

I'll assume we all know what a 2N3904 is.

.

Delta_G:
Doubt I could get anything soldered to it though.

This is a SOIC-8 chip. Not exactly tiny.
Just pre-tin a piece of hookup wire, and tack it on with a clean (no solder) fine soldering tip.
Leo…

If the voltage at the emitter is 4.5v and assume the Vce is .1v that leaves 7-4.6=2.4v(1k)Vgs

You need to look at RDS(on).

Drain-source on-state resistance RDS(on) VGS=-4.5V, I D=-25A - 13.1-18.7 mΩ
You need -4.5v Vgs
See figure 7

If you are switching on something which could have a high inrush current (buck converter is an example) it is often nice to control the rise time of the switching circuit to limit this.
Some purpose built load switches allow you to do this. Here is a simple example (it is a package consisting of nothing more than a p and n channel mosfet ) :

It doesn’t meet your specification (the max supply voltage is too low) but, from the data sheet, you see how it works with the addition of a capacitor and resistor and how you could apply it to your own circuit.

This MOSFET IPP45P03 only needs 5V to turn on, so you can use a zener to protect its gate-source from over-voltage:

For slower switching scale up every resistor value.

In the diagram from post#0, there can never be more than 1000/330*4.35volt= ~13volt across the 1k resistor.
Leo..

larryd:
If the voltage at the emitter is 4.5v and assume the Vce is .1v that leaves 7-4.6=2.4v(1k)Vgs

I'm sorry. I don't understand. Again, still not arguing, but trying to find out why I'm wrong.

There's a 1000ohm and a 330ohm and 7 volts to drop. To me it seems like you'd drop 5.25 over the 1K and then drop 1.75 over the 330. So the emitter would be at 1.75V right? How do you get the 4.5V at the emitter?

Should I put the 330 resistor on the other side of the NPN? between the collector and the gate?

I'm terribly confused now. This is why I write code and don't design circuits.

If I can't make this work out and have to order parts then I'll order a better regulator with an enable pin. But was really hoping I could get this to work and get it built today.

The emitter voltage (when the transistor is on) is 4.5v, i.e. since the base is at 5v, there is 5v - .5Vbe = 4.5v at the emitter.

This means the 330R (emitter resistor) has 4.5v across it. That leaves ~ 7 - 4.5v = 2.5v across the 1k (if your supply is 7v).

As MarkT showed, see his circuit offering in post #9.

Also, you may want to consider using pin 2 (enable pin on the converter) to see if you can control the converter directly.

In general, when you have an emitter resistor, the emitter voltage will be the base voltage minus one diode drop, or about 4.35V with 5V at the base. But I'm not sure that's quite right in this case with current at the collector being so restricted. Taken to the limit, if nothing was connected to the collector, and with 3.3K resistors on both base and emitter, the emitter voltage might be more like 2.2V. Or in this case, increasing the base resistor might actually reduce the emitter voltage. I haven't really thought through this, but you can certainly experiment with it and see what happens.

But I think your suggestion of moving the emitter resistor up to the collector, and tapping the gate from the middle of the 1k/3.3k divider, and just connecting the emitter to ground, makes a lot of sense. Then your math works. But then you should use a much higher value base resistor so as not to waste power.

One possibility on the enable pin is to see if you can put your iron tip on it and use something to bend it up - lifting it from the board. Then it would be easier to solder a wire to it. Or even desolder and remove the entire chip, bend up the enable pin, then resolder the chip.

Why do you have an emitter resistor at all?

aarg:
Why do you have an emitter resistor at all?

To limit Vgs on the P channel mosfet to keep within its specification (that is, if you don't use the zener diode alternative solution) and possibly to control the slew rate.

aarg:
Why do you have an emitter resistor at all?

6v6gt:
To limit Vgs on the P channel mosfet to keep within its specification (that is, if you don't use the zener diode alternative solution) and possibly to control the slew rate.

Exactly. Because I'd have to order a zener to build it that way.

larryd:
The emitter voltage (when the transistor is on) is 4.5v, i.e. since the base is at 5v, there is 5v - .5Vbe = 4.5v at the emitter.

This means the 330R (emitter resistor) has 4.5v across it. That leaves ~ 7 - 4.5v = 2.5v across the 1k (if your supply is 7v).

As MarkT showed, see his circuit offering in post #9.

Also, you may want to consider using pin 2 (enable pin on the converter) to see if you can control the converter directly.

OK, I think I see. Tell me if I have this right now. I’m forgetting that there is a second source of voltage in this circuit. I’m treating the 7V as if all there is is two resistors to ground. But that isn’t the case, there’s a second voltage source connected between the two resistors and that’s why I’m not calculating the drop right.

I’ve got four of these regulators and I only need three. I think I’ll try to solder to that pin on one of them but I’m not the steadiest hand with that sort of thing. If I mess one up I gotta build like this for the others.

But thanks for helping me understand the circuit. To me right now it is more important to learn what I did wrong there than to find the best answer. It’s more about learning the thing I don’t know there. Really appreciate the help.

So when I'm calculating voltages around this circuit I should imagine this as an equivalent:

I think that was my fundamental misunderstanding.

Would it be different if that was an NFET instead of an NPN?

Not that I want to do that, but now I'm just trying to learn how this stuff works.