# Are these resistors in series?

We had this question as a part of a small signal analysis question on the Final. This circuit I drew below is the DC analysis which is the first step pre-AC analysis which will have a totally different figure. This NMOS has a feedback resistor and then a resistor between the feedback resistor and the DC source. I didn't know how to solve it because the circuit implies the resistors are neither in series nor in parallel. We were required to find drain current, but regardless:
1) Are these resistors in series or parallel or neither?
2)How can we find the gate voltage in this case?
3)Can we find the Thevenin equivalent of cutting the circuit at the gate and drain?
Kindly don't make redraw the circuit using a small signal model since we want to do only DC analysis, and using the large signal model that contains a diode make the problem worse • Yes the resistors are in series. Yes, you need to analyse with and without Q1, to see how Q1 acts as a load. – Sparky256 Apr 17 '18 at 0:40
• For more complex circuits you can't always solve them by using series/parallel equations. You can come up with a set of simultaneous equations that includes the effect of the mosfet, and them solve the simultaneous equations. – immibis Apr 17 '18 at 0:51
• I'm afraid the trick is that 30M ohm resistance is kind of open wire :( – Joe Apr 17 '18 at 1:12
• Shame on you being rude! I don't know what timezone you are in, but here you asked the question in the middle of the night. Also, both sparky256 and immibis has given you fairly good pointers to how to solve this, so cut the crap – MrGerber Apr 17 '18 at 5:15
• Joe, you've asked the question rather poorly. Do you mean all three resistors, or just R2 and R1? What is your next step once you know which ones are in series? Also, it is quite likely no one is answering because you are so far off track that they don't want to go through a whole semester's worth of material to straighten you out. – JRE Apr 17 '18 at 5:18

If the current from the supply is I: The gate current = 0. Therefore $I - I_{ds}$ will be the current through both R2 and R1. They are hence like series.
Assuming MOSFET is in saturation, $$I_{ds} = K(V_{gs}-V_{th})^2$$ Also: $$V_{DS} = V_2 = (I - I_{ds})(R_1+R_2)$$ Also: $$12 = IR_3+(I - I_{ds})(R_1+R_2)$$ Also: $$V_{GS}= V_{DS}\frac{R1}{R_1+R_2}$$