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
    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\$

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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.