# Entrance resistance of a simple MOSFET circuit

1. Dec 26, 2013

### Jalo

1. The problem statement, all variables and given/known data

Find the internal resistance seen at Vin.

2. Relevant equations

3. The attempt at a solution

To solve the problem I started by ignoring the condensators and substituting the MOSFET by it's model. This resulted in the following circuit:

By applying Thenevin's theorem it's possible to find the internal resistance. To do that I applied a test voltage source, VT, at the terminals and I proceeded to find an expression that relates VT with the current i that passes through it.

By doing so I got to the result R = 2622 Ω. This appears to be wrong tho, since the solutions state that the resistance is 2.2kΩ.

If someone could help me I'd appreciate.

Thanks.

2. Dec 26, 2013

### Staff: Mentor

VDD is the DC power supply, and to AC signals it represents a very low impedance (i.e., a short-circuit) to ground.

The input resistance at a MOSFET gate is supposed to be very high, of the order of GΩ I think you'll find. (There is also some capacitance , which is always unavoidable.) The biasing resistors are parallel to this GΩ resistance, so the input resistance of this circuit amounts to just R1 // R2.

That's the easy part.

http://imageshack.us/scaled/landing/109/holly1756.gif [Broken]

Last edited by a moderator: May 6, 2017
3. Dec 27, 2013

### vk6kro

The model for a Mosfet usually shows a gate connection as an open circuit because the input resistance is close enough to infinite.

So, you would put R1 in parallel with R2 to ground and not have it go to the top of the current generator.

The positive voltage rail is regarded as ground for analysis purposes because it is assumed to be adequately bypassed.

You would then have a line going towards the current generator, but terminating in a small circle with the letter "G" (for gate) next to it.