Question

Bit of hairy algebra question here. Question attached on next post.

Answer 1

I have this equation:\[v_{in} = v_{gs} +\frac{R_s}{2}k(v_{gs} - V_T)^{2}\]and I need to solve for vgs - Vt. Any tips?

Answer 2

The solution is:\[v_{gs} - V_T = \frac{\sqrt{2R_sK(v_{in} - V_T) + 1} -1}{R_sK}\]

Answer 3

\[\Large v_{i n} = v_{gs} +\frac{R_s}{2}k(v_{gs} - V_T)^{2}\]\[\Large v_{i n} - v_{gs} =\frac{R_s}{2}k(v_{gs} - V_T)^{2}\]\[\Large 2\left(v_{i n} - v_{gs} \right)={R_s}k(v_{gs} - V_T)^{2}\]\[\Large {2\left(v_{i n} - v_{gs} \right) \over R_sK }=(v_{gs} - V_T)^{2}\]\[\Large v_{gs} - V_T =\sqrt{2\left(v_{i n} - v_{gs} \right) \over R_sK }\]\[\Large v_{gs} - V_T ={\sqrt{2\left(v_{i n} - v_{gs}\right)} \cdot \sqrt{R_sK} \over R_sK }\]\[\Large v_{gs} - V_T ={\sqrt{2R_sK\left(v_{i n} - v_{gs}\right)} \over R_sK }\]I'm not sure where that \(-1\) and \(+1\) is doing there though...

Answer 4

That was my problem. At first, I thought, well, let's try this: and did it like that. But the lecturer started by subtracting vT on both sides and then subtracting (vin - vT). He gets to:\[\frac{R_s}{2}K(v_{gs} - V_T)^2 + (v_{gs} - V_T) - (v_{in} - V_T) = 0\]

Answer 5

Then magically come up with the solution. I didn't understand why to go on this kind of trouble.

Answer 6

Perhaps to get \(v_{gs}\) only one on side. In the solution I wrote up there, we have a \(v_{gs}\) in the square root also,

Answer 7

Ah, got it. My mistake. It's not vin - vgs on the solution. It's vin - Vt

Answer 8

So yeah. Maybe it would be clearer if I had asked as the last equation. How to solve for vgs - Vt from there?

Answer 9

\[\Large \frac{R_s}{2}K(v_{gs} - V_T)^2 + (v_{gs} - V_T) - (v_{in} - V_T) = 0\]This looks like a quadratic. I would try using the quadratic formula with \[\large a=\frac{R_s}{2}K\]\[\large b=1\]\[\large c= -(v_{i n}-V_T)\]

Answer 10

Thanks @KingGeorge. I kinda was on the right track but I thought I had to complete the square first.

Answer 11

You're welcome.