Clear[id, kn, vt, vdd, vo];
id = kn*(vi - vt)^2/2;
vo = vdd - id*rd;
Solve[D[vo, vi] == -1, vi]
Output:
{{vi->(1+kn rd vt)/(kn rd)}}
I want to simplify the result with vx=1/(kn*rd)
. It should be vx+vt
. So how to replace with kn*rd
with 1/vx
?
I try:
Simplify[vi /. Solve[D[vo, vi] == -1, vi][[1]], kn rd == 1/vx]
The result is:
1/(kn rd) + vt
Not vx+vt
.
The version is 12.0.0.0.
Simplify[vi /. Solve[D[vo, vi] == -1, vi][[1]], {kn rd == vx}]
givesvt+1/vx
(not as you saidvx+vt
$\endgroup$$Version
of Mathematica are you using? $\endgroup$vx=1/(kn*rd)
,different fromkn rd == vx
. $\endgroup$So how to replace with kn*rd with vx
? And that is what I used. I did not use the code they had. But what they wrote before it. May be that was a typo then. $\endgroup$