I have an N-channel MOSFET circuit like this, except there is a more detailed (still resistive) load circuit in place of the LED and R1.

enter image description here

I attach my oscilloscope probe to gate (no other connections to gate at that point), and ground to ground. I can see drain-side circuit voltage fluctuations (measured separately) on the scope (low current, low voltage).

I thought the gate was pretty well insulated from the drain side; I don't know but based on my currently limited knowledge of MOSFETs I am wondering if the MOSFET is fried.

  • 2
    \$\begingroup\$ You are really good at drawing!!! \$\endgroup\$ – perilbrain Sep 30 '12 at 16:21

You should never leave the gate of a MOSFET floating like that. Static charges can easily puncture the gate oxide, permanently damaging it. Tie a high-value resistor (1 megohm or so) between the gate and the source.

Yes, the gate is very well insulated from the drain/source, but there is a significant amount of capacitive coupling between them, which is probably what you're seeing on your scope.

  • \$\begingroup\$ +1 for static charges, I am always amazed how these MOSes acquire so much of static charges... \$\endgroup\$ – perilbrain Sep 30 '12 at 16:29

You say you are probing the gate, but you are watching the drain on the scope? How is this possible without a second probe on the drain?

Assuming this is actually what you have done, then since the gate is extremely high impedance and floating, then pretty much anything moving near it or touching it will cause the voltage at the gate to vary, and hence the drain-source current and drain voltage also. Touching the probe to it will certainly do this.

This is why you should never leave a FET gate floating (unless you are building a static detector) Add a pulldown resistor to ground (e.g. 10-100kΩ) to keep it off until you apply a turn on voltage.

EDIT - another possibility is that the drain has a signal present at it, and you are seeing this at the gate due to capacitive coupling between them - in this case Dave's answer fits.

EDIT 2 - It seems you are trying to implement something like this:

Touch switch

This circuit will stay on for a while after touching. On touching the finger bridges the +V and the MOSFET gate, charging the cap and turning it on. The high value resistor then gradually discharges the cap to turn it off again after a while.
You can change the resistor/cap values for a longer/shorter on time.

EDIT 3 - here is a simple Set/Reset circuit using 2 transistors:

SR Flip Flop circuit

To Set, apply the supply voltage to the S input (e.g. button between base and Vs. Vs could be e.g. 5V - 12V), to Reset, do the same for the R input. Output can be taken from either O point (one is inverted) and used to drive your alarm circuit. Transistors can be any general purpose NPN (BC337, 2N2222, etc)

Here are the waveforms:

Circuit Waveforms

Also, here is a simpler version, where you ground the inputs to switch states (R1,R2 = 1kΩ, R3,R4 = 10kΩ):

Alt Latch Circuit

  • \$\begingroup\$ (I just happen to know what is on the drain-side since I built/tested that circuit separately) Pulldown is a good suggestion - and I would do exactly as you recommend except I want this to be "touch on" - i.e. the same as the circuit called "FET touch switch" in Forrest Mims' electronics book. If I added the suggest pulldown, it would be a momentary-on, but I want continuous on after triggering so someone will investigate - it's for a level-alarm circuit. Is this what you mean by "static detector"? \$\endgroup\$ – Paul_A Sep 30 '12 at 15:47
  • \$\begingroup\$ @user192127 - I added a MOSFET touch switch circuit - is this similar to the one from the book? About the static detector - you can make a simple static detector with a FET with floating gate attached to "antenna", and resistor/LED. \$\endgroup\$ – Oli Glaser Sep 30 '12 at 15:59
  • \$\begingroup\$ Here's the what the book shows. I understand that this circuit may be a demo-only textbook circuit, and I am interested in better ways though I do want to understand what I am missing with this simple example circuit (it works on a breadboard with my real circuit in place of the LED/resistor - that's why I amd wondering if I fried the MOSFET) i269.photobucket.com/albums/jj56/primeq/touchMOSFET.png \$\endgroup\$ – Paul_A Sep 30 '12 at 16:02
  • \$\begingroup\$ Is that the circuit from the book? It's not a good one at all IMO - it will not hold it's value indefinitely and will be very susceptible to external noise. At the very least it should have a high value resistor, and a capacitor between gate and ground to filter noise and provide a decent amount of capacitance to hold the charge. This is okay I guess for a rough solution. If you want continuous on until switched off a latch would be the best solution - you can buy an IC or make one with a couple of transistors. \$\endgroup\$ – Oli Glaser Sep 30 '12 at 16:11
  • \$\begingroup\$ Oli, thanks for confirming what I had begun to suspect. The resistor/capacitor on the gate is probably what I will do as a pragmatic and simple solution. I would up-vote your answers if I had the mojo \$\endgroup\$ – Paul_A Sep 30 '12 at 16:19

There are two primary capacitances that will cause this for you. the first is know and gate/darin overlap capacitance and the second is gate/channel capacitance. The second capacitance will be noticeable if here is any conduction in the channel. So if the transistor is on then that will be the dominant capacitance. What you're seeing is normal.


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