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

  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: equivalent

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!

  • \$\begingroup\$ 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. \$\endgroup\$
    – W5VO
    Commented Sep 11, 2017 at 17:44

1 Answer 1


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\$

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

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


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