MOSFET switch replacement not working

Question

I'm using a PNP MOSFET to replace a physical button. When I ground the lead I've attached to bypass the switch, the action I need activates (a notification light comes on - it's the activator button on this barcode scanner), so I know I'm working with the right lead.

In the same test, when I insert a DMM into the loop, I see that when pressed, there's about 60uA going through it.

My circuit is as follows:

^{simulate this circuit – Schematic created using CircuitLab}

The gate voltage goes from 5V to 0V when expected, and when I put a DMM between the drain of the MOSFET and ground, I can see ~55uA when the MOSFET is activated. However - the indicator light doesn't turn on.

Any ideas of what I should look for?

The external input is the circuit described in this question, but with a P channel instead of an N channel, as my sensor output goes high when it needs to be off, and low otherwise.

Here's the mosfet I'm using: http://www.diodes.com/datasheets/ZVP4424A.pdf

Answer 1

Measure the MOSFET source (connected to scanner) when the MOSFET is ON.
You'll find it does not pull down to ground as the MOSFET needs a voltage of about Vgsth to just start to turn on. You are probably getting 1+ Volts source to ground. source-

Driving an NChannel MOSFET , source to ground, drain to scanner should work. The drive sense will be inverted. This can be fixed if needs be.

If you wish to use the P Channel FET try a 100k scanner in to ground to see if it turns on. If not try lower R till it works. Then pull scanner input UP with MOSFET, source to V+, drain to scanner in. Again, polarity reversed.

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Magic :-):

This circuit converts

• A High / Ground signal at the MOSFET source

into

• An open_circuit / ground output at the MOSFET drain..

The external input voltage swing and current sinking ability has not been specified.
The circuit below will convert a 0 to V1 signal to 0 to V2 where V2 > V1, OR to a 0 /Open circuit output.

In this case when Vin is high, when (Vdd-Vin) < Vgsth_FET the FET will be off.
When Vin = 0, if Vgsth_FET < Vdd then FET will be on.
The input line needs to be able to sink the current from S1 - which is said to be low show it should work.

N Channel FET.
Switching polarity as original. Input low = on. Gate to V+, drain to scanner, source to external input.

^{simulate this circuit – Schematic created using CircuitLab}