The above looks okay. Note that I completed the second equation (red bit).Ok so i get:
Node 1: (Vs-V1)/R - (V1-2Vx)/R - (V1-Vo)/R = 0
Node 2: (V1-Vo)/R + Is - Vo/R = 0
I think you have a sign issue with the first expression of the pair. The second looks okay.Which leaves me with V1 - Vo = Vs
and -V1 + 2Vo = Is*R
Have I done anything wrong?
The above looks good for the open-circuit case (finding Vo). You should point out at this point that the above expression for Vo is also Vth (the Thevenin voltage).Ok so I have tried attempted the theory used from gneill.
My final equations are:
V1 + Vo = Vs
-V1 + 2Vo = IsR
Vo = (Vs + IsR)/3
V1 = (2Vs - IsR)/3
... and since Vo is shorted at this point ...And short circuit current equates to:
Io = Is + (V1-Vo)/R
Egads! You're right. Very careless of me, and good catch on your part.I didn't get those equations. I think the problem is the node marked "2Vx" in reply #8 should have been labeled as "-2Vx".
I went through this quickly just now and I got the same answer for R_Thv as I previously got using the "test source" method outlined in reply #12.
Once you've fixed up the above, solve the first pair for Vo in terms of Is and Vs.uart, using replacing the centre node with -2Vx I get the following equations:
5(V1) -3(Vo) = Vs
-V1 + 2Vo + IsR <----- not an equation
Io = Is + (V1 - Vo)/R <------ Solving for short circuit current, Vo should be zero
Rth = Vo/Io
Using the initial values set in comment #5 do you get Rth = 2.9kohms? I instead simply end up with the value of R
Yeah I think you've got it now. Once you fix those mistakes pointed out by gneill (the "not an equation" just needs to be set equal to zero) then you should definitely get the correct answer. (BTW. The answer turns out to be a fairly simple fraction of "R").Of coarse! Vo is not the same when the output is shorted, what an oversight! I didn't even pick that up from the hint that you tried to give me.
Thanks, I dont think i could possibly get this wrong now (But you never know)