# MOSFET transistor

Circuit a) presents the circuit of polarization of a MOSFET and b) the equivalent model of small signal on $g_m$ is the transconductancy of the dispositive, and $r_0$ an equivalent resistance defined by $r_0=V_A/I_D$.

Data provided:

• $V_T=1.5V$
• $C_{OX}W/L\mu==0.25 mA/V^2$
• $R_L=10$ kOhms

A) Determine the quiescent operation point of the transistor B) Calculate the voltage gain of the small signal in the common source configuration as seen in figure b).

After doing my research I found that the quiescent point is the intersection between the load line, which I found applying Kirchoff as $I_D=(15V-V_{DS})/R_D$ and the intensity curve given by $I_D=C_{OX}W/L\mu*(V_{GS}-V_t- V_{DS}/2) V_{DS}$, both thought as plots of $I_D(V_{DS})$.

The problem, if I am not wrong yet, is that we don't know the value of $V_{GS}$. Maybe I could use $g_m=\Delta I_D/(\Delta V_{GS})$...In the circuit appear $g_m, r_o=V_A/I_D$. In the equivalent circuit we could calculate some equivalent resistances, but we don't know the explicit values of $g_m, V_A, v_i, r_o$...

I do not intend to get my homework done, but to get any possible kind of hint to go on with the resolution, and then post the complete answer so others like me can also learn from this example. Thank you for the help in advance.

• This may sound really stupid but Vgs (in your small signal ac analysis) is Vin – Andy aka May 19 '15 at 18:36

For the quiescent point, $V_{GS} = V_D$ since there is no current flowing into the gate of the MOSFET, and there is a capacitor blocking any current flowing into $R_C$.
And since you have an expression for $I_D$ you can write $V_D = 15V - I_D*R_D$; from there you should be able to solve this.