I want to use an n-channel mosfet to turn the supply from a battery on/off. I understand that typically you'd connect the positive side of the battery to the load, and the MOSFET would go between the load and ground. However, for my board I need to have the switch be between the ground plane and the negative terminal of the battery, not between the load and the ground plane. I've attached a simple schematic to show what I mean. The PIC controlling the gate of the MOSFET has it's own 3.3V supply and everything is tied into a common ground. Will this work the way I want it to? I already have the n-channel MOSFETs lying around which is why I want to use them.

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The load goes something like this, and you need a logic level MOSFET.

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Bill
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I understand your schematic is the way it's normally done. I'm wondering if there's a way to have the switch be between the negative terminal of the battery and ground.

 

Where's the switch supposed to be in your schematic? (I assume you are talking about a push-button to turn the circuit on, or are you referring to the MOSFET itself as the switch and it's being controlled by the PIC?)

Is the switch supposed to power up the circuit including the PIC? Then the PIC keeps the MOSFET on until it needs to be turned off?

We need a bit more information, particularly about the switch and where it fits into the circuit.

But, if you're referring to the MOSFET itself as the "switch", yes, the circuit as you posted will work (the drawing of the MOSFET is a bit confusing, the gate is hard to distinguish). I've done the same thing in a circuit to act as a self-power-up button. Press the button and it turns on the MOSFET which powers up the load (including a microcontroller) which in turns keeps the MOSFET on via one of its I/O pins. When the MCU wants to cut power, it sets the pin low, turning off the MOSFET.

 

When I say "switch" I'm talking about the MOSFET itself. Here's a better schematic of what I'm trying to do. The PIC has its own 3.3V supply. The PIC will also be used to measure the voltage of the battery, which is what the voltage divider formed by R1 and R2 is for.

Right now I have the drain of the MOSFET connected to ground and the source connected to the negative terminal of the battery. The PIC (and thus the gate) can only swing from 0 to 3.3 V. Vgs threshold of the MOSFET is .9V. I'm not sure this will work since I don't know what voltage the negative terminal of the battery will be when the MOSFET is "off".

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It will work if you connect the source and the battery (-) terminal to ground, then disconnect the drain and load from ground and connect them together. This way the source is always at ground (same level as the PIC's ground), and thus Vgs will always be either 0V (PIC line low, MOSFET off) or 3.3V (PIC line high, MOSFET on). The load would be grounded through the MOSFET when it's on.

I attached an edited version of your schematic.

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Which of course connects it just like my first post

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Bill
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Thanks for the help but maybe I am not being clear. The load connects to the ground plane on my board, which makes it difficult to get the MOSFET between the load and ground. That is why I wanted to put the MOSFET between ground and the battery.

I hooked up a little test jig to see if my original schematic would work and it doesn't. With the gate pulled to ground the MOSFET doesn't turn completely "off". Looks like I'll have to either use separate ground planes to get the MOSFET between load and ground or use a high side p-channel MOSFET.

 

I guess that's why it's usually done that way, because it works.

Floating the source on a MOSFET makes controlling it difficult, since it depends on the voltage between the gate and source.

What's the load? How hard is it to isolate from your board's ground plane?

 

A relay would work in your original configuration.

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Bill
Smart Kits build Smart People
http://www.blueroomelectronics.com/

 

Shouldn't be too hard to isolate the ground planes. I was just hoping I could do it this way to make things a little easier.

I realize a relay would work but they are bigger. I actually haven't been able to find a relay that is as small as the MOSFETS I have. The MOSFETs I have in hand are a SOT-23 package that can handle up to 6.5A and cost about 50 cents.

 

your circuit will work of course - with the load always supplied - no matter, if there is a signal from the PIC or not.

(since the PIC won't "see" the ground connection of the MosFet, and the MosFet gate doesn't "see" the ground connection of the PIC.)

If you want to use separate grounds use an optocoupler and wire the circuit as suggested by blueroomelectronics.

Boncuk

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I believe you can, but you would need to have some gate drive circuitry that is connected on the neg battery side of the FET to develop gate drive. That part would not be switched on and off by the N-FET.

 

The easiest solution for this original poster is the same as the schematic developed in the thread "Stupid FET Question". Use two FETs, one P and one N channel, to control the + supply to the load. Leave the load - grounded.

Stupid FET Question

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