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Problem

Determine the state (cutoff, linear, saturation) of both transistors in the following circuit. enter image description here \$K = 0.4 \frac{\text{mA}}{\text{V}^2}, \: R_{D1}= 1\text{k}\Omega, \: \: R_{D2}=0.6 \text{k}\Omega, \: \: V_{DD}=5\text{V}, V_{to}=2\text{V}\$.

My attempt

From the schematic it's easy to see that \$V_{GS1} = V_{GS2}\$. My idea was to use load line analysis to analyze the transistor current and voltage behavior. I intend to find the operating point, or Q-point for both transistors, and see what region they operate in. Here is my work so far with my calculations on the left and my graph on the right of the paper.

enter image description here

However, I realized I don't know how to identify the Q-point at all. The definition of the Q-point is just the operating point for the circuit with zero input signal, but that doesn't help me in this case.

Can anyone help me out and tell me how to find the Q-point? And also if my idea of solving this problem is okay?

Maple calculation in response to Andy aka's answer

enter image description here

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  • \$\begingroup\$ If you did some more complex maths to find Vg (i.e. solving the quadratic equation after equating Vd to Vg and solving iD from Vcc and Rd) would you be able to tell if Vg was putting the FET in saturation or otherwise? \$\endgroup\$
    – Andy aka
    Aug 20 '20 at 12:55
  • \$\begingroup\$ Your amendment to the circuit is incorrect because now, the gates would float and everything is indeterminable. \$\endgroup\$
    – Andy aka
    Aug 20 '20 at 13:04
  • \$\begingroup\$ @Andyaka I have edited the circuit once more. I am not sure how to arrive at the quadratic equation you are mentioning. Can you give me some more help? \$\endgroup\$
    – Carl
    Aug 20 '20 at 13:34
  • \$\begingroup\$ I can leave an answer that shows how you find Vgs (2.0853616 volts) but where would you go from there? \$\endgroup\$
    – Andy aka
    Aug 20 '20 at 13:45
  • \$\begingroup\$ Put together K, Vto. Vdd, Rd1 and find the equation for Vgs and therefore V(Rd1) and I(Q1). (Come back to Rd2 and Q2 later) \$\endgroup\$ Aug 20 '20 at 13:45
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From a comment: -

If you did some more complex maths to find Vg (i.e. solving the quadratic equation after equating Vd to Vg and solving iD from Vcc and Rd) would you be able to tell if Vg was putting the FET in saturation or otherwise?

From another comment: -

I am not sure how to arrive at the quadratic equation you are mentioning. Can you give me some more help?

I can leave an answer that shows how you find Vgs (2.0853616 volts)

If you expanded the formulas (bearing in mind that \$V_G = V_D\$), you have you'd get this: -

$$V_G^2 - 2V_TV_G + V_T^2 = \dfrac{i_D}{k} = \dfrac{5}{k\cdot R_D} - \dfrac{V_G}{k\cdot R_D}$$

Therefore: -

$$V_G = \dfrac{2V_T - \frac{1}{k\cdot R_D}±\sqrt{\frac{1}{k^2\cdot R_D^2}-\frac{4V_T}{k\cdot R_D}+4V_T^2 - 4V_T^2+\frac{20}{k\cdot R_D}}}{2}$$

$$V_G = \dfrac{2V_T - \frac{1}{k\cdot R_D}±\sqrt{\frac{1}{k^2\cdot R_D^2}-\frac{4V_T}{k\cdot R_D}+\frac{20}{k\cdot R_D}}}{2}$$

If I plug in the values for k (0.0004), \$R_{D1}\$ (1000 ohm) and \$V_T\$ (2 volts) I get: -

$$V_G = \dfrac{4 - 1.25 ±\sqrt{6.25-20+50}}{2} = 3.7603986447 \text{ or -2.2603986447}$$

Clearly only the positive value is applicable for the MOSFETs.

Can you take it from here?

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  • \$\begingroup\$ First of all, thanks for your answer. The equation you are setting up only holds if we assume that the transistor operates in the saturation region, right? I guess that's a fair assumption, since Vds = Vgs. But your result of Vgs = 2.085 V is confusing me though. I have solved the quadratic equation with Maple, and I get that Vgs = 3.760 V, which must be the operating voltage for both transistors. I upload a picture of the calculations in Maple. \$\endgroup\$
    – Carl
    Aug 20 '20 at 15:07
  • \$\begingroup\$ @Carl I used a k value of 0.4 and not 0.0004. I expect that if you changed my numbers in the equation to reflect this you'd get Vg = 3.76 volts. The point of the help you requested was to show the process. \$\endgroup\$
    – Andy aka
    Aug 20 '20 at 15:09
  • \$\begingroup\$ But K has the units \$ \frac{\text{mA}}{\text{V}^2} \$, so I multiplied with 10^(-3) to compensate for the microamps. \$\endgroup\$
    – Carl
    Aug 20 '20 at 15:11
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    \$\begingroup\$ I realized how to solve the problem now. Thanks for the help, Andy. \$\endgroup\$
    – Carl
    Aug 20 '20 at 15:21
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    \$\begingroup\$ @Carl I've corrected my answer with the right value of k. \$\endgroup\$
    – Andy aka
    Aug 21 '20 at 13:29

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