New KCC Quiz AQQ299 about Opamp Open Loop Gain and LTspice

the solution for

20*log10(M) = M

can be found with the Lambert W function
which is a solution for x
of the function y = x*exp(x)
W(y) = x

with
x = -M*ln(10)/20

you get
y = -ln(10)/20

and with the Lambert W function
x = W(y)

you get finally
M = (-20/ln(10)) * W(y)
M = (-20/ln(10)) * W( -ln(10)/20 ) =~= 1.14028753461

which proves true with my calculator

20*log10(1.14028753461) = 1.14028753461

brgds
steve

PDF

branch -1 of lambert W function

gives another real solution

of M = 29.353070859629902

branch -1 of lambert W function

gives another real solution

of M = 29.353070859629902

You're still missing the third root Steve.  I only got two on the first attempt.  Note that question is about magnitude so the solutions must be real.  Logs of negative numbers are like heading North from the North pole!

yes i did not see that it was asked a soution for

abs(20*log10(M))==M

the third solution is

M = 0.90142340431067646240910539747071

i calculated it with matlab

here is the script

clear;clc;
format long
syms M

k=-1:0;
SK=(-20*lambertw(k, -log(10)/20))/log(10);

eq = abs(20*log10(M))==M;
SM(1)=vpasolve(eq,M,0.9);
SM(2)=SK(2);
SM(3)=SK(1);

format compact
sprintf("Solutions of M = ")
double(SM')

here is the output

ans =
"Solutions of M = "
ans =
0.901423404310676
1.140287534610586
29.353070859629902

looks nice with logarithmic M-axis

Bravo steve12 , you get it! Congratulations!

POSTSCRIPT

steve12,

I must say life’s a learning experience. Your answer introduced me to the Lambert Function. After a bit of research on it I gave it a try solving this problem. With a few substitutions I was able to express the problem in the form of

        ue^u = ±k

Using an online Lambert Function calculator, I found two answers from the W₀ branch and one from the W_-1 branch.

Thanks much,

AlB

i'm glad that you find it useful too

brgds steve