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Having trouble with what I thought would be a simple idea (I'm sure it is, but I know next to nothing).

What I want: a cable from a 12 V battery to a 12 V water pump, with a float switch (500 mA max).

Where it is now, auto resetting 4 A breaker, float switch, P-type MOSFET and a pull-up resistor (10-ish kΩ).

+ to pump (shown here as a lamp), to drain
source to -
gate to float switch to - (with the pull-up to +)

Works great, but were I to implement this (oh yeah, its being built as a cable), the MOSFET becomes hot when "switched off".

Is there a simple solution I'm missing or am I just kidding myself and this is a stupid idea/will never work?

Any tips would be greatly appreciated.

schematic

simulate this circuit – Schematic created using CircuitLab

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4 Answers 4

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You have the P-channel MOSFET in a source-follower configuration. Move it to the other side of the pump.

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Try this instead using a P channel transistor for switching like this it needs to be configured as a high side switch. D1 is needed to dissipate the inductive spike formed when M1 turns off.

schematic

simulate this circuit – Schematic created using CircuitLab

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You have the drain connected to a fixed potential, and the load connected to the source. This configuration is called "common drain", or "source follower", in which the source tends to follow the gate in potential, but a few volts (\$V_{GS(TH)}\$) different.

Using the MOSFET model you show, with \$V_{GS(TH)}\approx 3V\$, NODE1 will be either +12V or +3V, which is not the full on/off switching behaviour you require.

You need the load on the drain side of the MOSFET:

schematic

simulate this circuit – Schematic created using CircuitLab

This is "common source", and will behave very differently. NODE1 will now be either 0V or +12V, with the load receiving nothing, or the full battery potential difference across it.

You should also protect the transistor from dangerous voltages produced by inductive loads like motors, by placing diode D1 across the load, as shown.

However, this does not explain the heating of the transistor, while off. If the MOSFET is damaged, that could cause heating. Perhaps you have long cabling, picking up noise, and keeping the MOSFET partially conductive. If so, this can be mitigated with a capacitor:

schematic

simulate this circuit

This seems a little complicated. You could achieve the same result using a relay:

schematic

simulate this circuit

With the right relay, you can drive the coil with 12V and 100mA or so, but have many amperes of current available for the load.

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  • \$\begingroup\$ Thanks a lot to everyone, I'll give that a shot. I had thought of the relay idea (which i would usually use) but since this is for a bilge pump in a wet bike and there is a high possibility of getting wet i moved away from the mechanical relay for something i could seal into something resembling a straight cable. I tried a solid state relay but it must not be very well isolated trigger to load and it didn't work at all as i was expecting. \$\endgroup\$
    – Steve
    Commented Feb 27 at 4:32
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You have the mosfet on the wrong side, you have it on the low side of the load. This won't work because the current will go through the body diode, and this won't turn off, it's a diode. You need to put the fet on the other side of the load.

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