Simple methods find one extremum, which is a maximum:

In[2]:= f = F /. D -> 1. 10^-24 d // Simplify;

In[3]:= df = D[f, d] // Simplify;

In[4]:= extremum = d /. (sol = Solve[df == 0, d, Reals][[1]])

During evaluation of In[4]:= Solve::ratnz: Solve was unable to solve the system with inexact coefficients. The answer was obtained by solving a corresponding exact system and numericizing the result. >>

Out[4]= 3154.42

In[5]:= d2f = D[df, d] // Simplify;

In[6]:= (* extremum is a maximum *)
d2f /. sol

Out[6]= -2.05399*10^-8

In[7]:= (* with value *)
f /. sol

Out[7]= -1.88228

In[8]:= Plot[f, {d, extremum/10, 10 extremum}]

Are you sure that your expression has a minimum? It seems to me that the infimum is attained as D tends to -Infinity. The structure of your expression is clearer if treated this way:

coeffs = Cases[F, _Real, All] // Union;
subst1 = Thread[coeffs -> Table[c[i], {i, Length[coeffs]}]];
c0 = (coeffs[[-4]]/12);
subst5 = Map[Reverse, 
   Cases[Rationalize[subst1[[1 ;; 9]]], Except[_Real -> _]]];
subst2 = Thread[
   Table[c[i], {i, 10, 13}] -> c[0]*Rationalize[coeffs[[-4 ;;]]/c0]];
subst3 = subst1 /. subst2;
F2 = Simplify[F /. subst3 /. subst5]
subst4 = c[0] (-D + c[2]) :> Log[x];
F3 = FullSimplify[F2 /. subst4]
F4 = F3 /. Map[Reverse, subst1]
Plot[F4, {x, -1.98^(1/15), 5}]

where x == Exp[c[0] (-D + c[2])] is a new variable.

Your expression involves high powers of the parameter. I think that's why you're getting overflow. Also, note that you're actually using NMinimize because your expression does not have infinite precision.