# PMOS Circuit Problem

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

2. Relevant equations •Id=K*(Vsg-Vth)2 •KVL

3. The attempt at a solution On part A, Haven't had much experience with PMOS. Used large signal model, to find Rs. Equivalent circuit:

We know that Vs=0. That means Vd=-7.5 V. If we say the voltage across Rs is Vdd-Vd, then that voltage is 22.5 V. Used 22.5V/Id and got 7500 ohms.

Used the Id equation to find Vsg to be 7.5 V

Vsg=Vs-Vg, and Vs=0, so Vg=-7.5

Now, we know there are 2 unknowns, so we need 2 equations. The first equation, we used voltage division to be •-7.5=15*(Rg2/(Rg1+Rg2))

Then, the second equation was using the small signal model. The input resistance was the Parallel between Rg1 and Rg2:

•3M=((Rg1)-1+(Rg2)-1)-1 •We disregard the 10km because it is so small in comparison

This is where the problem lies. If we continue, then it leads to a value with negative ohms, which is not possible.

Any help is appreciated!

• Any time you get a voltage that's greater than or less than your power supply, you should immediately be suspicious. In this case, your first assumption is incorrect. You know that VD = 0V, VS is the side of the transistor with the arrow.
– W5VO
Commented Sep 11, 2017 at 17:44

From $V_{SD} = 7.5V$ we can find $R_S = \frac{15V - 7.5V}{3mA} = 2.5\:\textrm{k} \Omega$

The PMOS gate-source voltage $V_{GS}$ at $I_D = 3\:\textrm{mA}$

Can be found from this equation:

$$I_D = \textrm{K}(V_{GS} - V_{TH})^2$$

solving this for $V_{GS}$ we get this:

$$V_{GS} = \sqrt{\frac{I_D}{K}} + V_{TH} = \sqrt{\frac{3mA}{0.2mA/V^2}} + 2V = 5.873V$$

So, the gate voltage needs to be $5.873$ volts lower than the source voltage.

$$V_G = 7.5V - 5.873V = 1.627V$$

Therefore you need a voltage divider that will provide this voltage to the PMOS gate.

So we have:

$$V_G= V_{DD} \cdot \frac{R_{G2}}{R_{G1}+R_{G2}} = 1.627V$$

$$R_T = \frac{R_{G1} \cdot R_{G2}}{R_{G1}+R_{G2}} = 3\textrm{M}\Omega$$

Solving this we have

$$R_{G2} = R_T \cdot \frac{\frac{V_{DD}}{V_G}}{\left(\frac{V_{DD}}{V_G} - 1\right)} \approx 3.3\textrm{M} \Omega$$

$$R_{G1} =R_{G2}\cdot \left(\frac{V_{DD}}{V_G} - 1\right)\approx 27\textrm{M}\Omega$$

As for the part $C$

$$C1 > \frac{1}{2 \pi F\cdot R} = \frac{1}{2 \pi 20\textrm{Hz}\cdot (10\textrm{k}\Omega +3\textrm{M}\Omega)} \approx \frac{0.16}{20\textrm{Hz}\cdot 3\textrm{M}\Omega} \approx 2.7\textrm{nF}$$

$$C_2 > \frac{0.16}{20\textrm{Hz} \cdot (2.5\textrm{k}\Omega+1\textrm{k}\Omega)} \approx 2.2\mu\textrm{F}$$