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I tried to make a NOT gate with N type mosfet and everything seemed good when the gate had voltage, however I found strange things when the gate was open and that I put test components between the source and drain.

My voltmeter indicates a voltage of 1.8V between drain-ground, but a voltage of 1.6V between drain-source, even if nothing is going on at the gate. Then, I expected to have the remaining 0.2V at the 1M test resistor and/or parallel led I put between source-ground, but both were indicating 0V.

What am I missing ?

This mosfet's order is: gate, drain, source.

schematic

simulate this circuit – Schematic created using CircuitLab

schema

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closed as unclear what you're asking by Matt Young, JRE, Daniel Grillo, Andy aka, Transistor Mar 10 '16 at 22:24

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    \$\begingroup\$ Edit your question, hit the button to draw a schematic, draw your schematic. \$\endgroup\$ – brhans Mar 9 '16 at 13:32
  • \$\begingroup\$ I suspect you have the same problem as this guy. \$\endgroup\$ – The Photon Mar 9 '16 at 16:38
  • \$\begingroup\$ I added the schematic of the part in question \$\endgroup\$ – MB101874 Mar 9 '16 at 17:39
  • \$\begingroup\$ Is your 1.8V source polarity correct? What's R2 for? \$\endgroup\$ – Transistor Mar 9 '16 at 18:20
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I found strange things when the gate was open... a voltage of 1.6V between drain-source, even if nothing is going on at the gate.

The Gate in a MOSFET is very well insulated, so when 'open' it can easily pick up a static charge. Your 'strange' readings may have been caused by the FET getting enough Gate voltage to partially turn on.

When you added the 1M resistor the Gate would have discharged to 0V and then the FET would have turned off, so its Source voltage should also have dropped to 0V (since it is connected to Ground via a low value resistor).

Another possibility is that you connected a 'test component' which had significant voltage drop at very low current. Most multimeters have an input resistance of ~10M, so if you measured the Drain-Source voltage without a pull down resistor the LED could get enough current to drop 0.2V.

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  • \$\begingroup\$ I see. I assume instead of the multimeter at D-S, I could put an equivalent 10M resistor in parallel, and try to measure the 0.2V at the LED ? \$\endgroup\$ – MB101874 Mar 9 '16 at 17:51
  • \$\begingroup\$ Yes. And I see now that the resistor across your LED is 1M, not what I thought (150 Ohms), so 1.6V would be the expected measurement even if the LED wasn't present! \$\endgroup\$ – Bruce Abbott Mar 9 '16 at 20:07

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