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If an nmos which has the gate and drain connected to VDD, and the source connected to a grounded capacitor, the nmos will start conducting and the capacitor will start charging as long as VDD > VTn (the threshold voltage of the nmos). The gate is connected to VDD and the source is connected to a grounded capacitor so VGS = VDD - Vx (Voltage across capacitor).

Because the nmos is in conducting mode it will act like a resistor, so a current can flow from VDD to the the capacitor and charge it. This should keep charging until Vx equals VDD, but apparently it doesn't. In steady state, the capacitors final charge value is VDD-VTn.

Why does this happen?

The threshold voltage (Vtn) concerns the voltage that the gate needs to be at before a conducting channel between the source and drain can form. Why would this cause the capacitor not to charge to VDD when the capacitor isn't connected to the gate?

The picture below shows the circuit which I am talking about. enter image description here

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The voltage that must be greater than \$V_{TN}\$ is the voltage from the gate to the source, \$V_{GS}\$, not just the voltage at the gate. Once the source voltage rises to \$V_{DD} -V_{TN}\$ the voltage from the gate to the source will just equal \$V_{TN}\$. Any increase in the source voltage will cause \$V_{GS}\$ to be less than \$V_{TN}\$ so current flow will stop. Therefore, the voltage at the transistor's source can't rise above \$V_{DD}-V_{TN}\$ (based on this simple model).

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  • \$\begingroup\$ Thanks for your answer, this has helped a lot. I just have one question though, why is the VGS The voltage that must be greater than VTN? I always assumed it was the voltage between the gate and body (B), and since the body of an nmos is connected to ground, this is equal to VDD. This makes more sense in my head because the inversion region in the p material is caused by an electric field between the source and the body. \$\endgroup\$ – Blue7 Apr 17 '14 at 0:06
  • \$\begingroup\$ @Blue7: No, the "body" of the MOSFET is usually connected to the source, not ground. \$\endgroup\$ – Dave Tweed Apr 17 '14 at 0:59

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