One problem is scaling the x values, because they are close to floating point precision. Plotting your data and the fit on a linear rather than logarithmic x-scale also suggests an appropriate model. I multiplied the x values by 1E+9, and was able to get a decent fit with a single-exponential:
eqn = @(par,x) par(1) - par(2).*exp(x/par(3));
that after some experimenting, yielded:
fit_par =
268.8501e-006 248.9514e-006 -561.3801e-003
The actual value of fit_par(3) would then be -561.3801e-012.
A two- or three-exponential model may be difficult to fit, so I would experiment by successively subtracting two-parameter exponential terms similar to the initial term (with those initial parameter estimates for the initial set).
Since this appears to be an heuristic exercise, I also suggest using nlparci at this three-parameter and each subsequent step (sequential subtraction of two-parameter exponential terms). While you get a good fit and the confidence intervals remain significant (bounds do not include zero), continue adding to the third set of exponential terms. If at any step any bounds include zero, the model is over-parameterised and does not represent your data. You would then have to accept a simpler model.
