I have this circuit taken out from a book.enter image description here

The book says that :"The circuit as shown will work for positive signals up to about 10V; for larger signals the gate drive is insufficient to hold the FET in conduction (Ron begins to rise)"

What does this mean? That I can't switch a higher voltage than my gate voltage? And why would Ron rise if the input signal passes 10V ?

Thanks in advance.


You can't switch voltages higher than the gate voltage, no. You must always have the gate-to-source voltage (Vgs) greater than the transistor's threshold voltage if you want it to conduct well. If you want "pass a 10V signal" then you want the source and drain to be at 10V, which means that the voltage on the gate must be (at least) 10V plus Vth. Lower values of Vgs will result in higher drain-to-source resistance, so you really want the gate voltage to be much higher than the voltage you want to switch.

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    \$\begingroup\$ Ohh right, I was confused because for some reason I was thinking about the source being tied to ground in which case you can switch higher voltages than the gate voltage. But in this configuration Vd ~= Vs.. \$\endgroup\$ – Simon Maghiar Jun 14 '18 at 12:01

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