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When trying to Plot the function

1/(1 + E^(((e + Quantity[-7.05, "Electronvolts"]) 
  *(Quantity[1, 1/("BoltzmannConstant")]))/T))

I get an incorrect plot looking likeincorrect Plot

However, if I first evaluate the points with Table and then use ListLinePlot it works as expected

correct Table@ListPlot

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    $\begingroup$ I'm not sure why you want to keep the units in the definition of f, but the following works: f[e_, T_] := 1/(1 + E^(((e + -7.05)*1)/T)). A good start when Plot returns an empty plot is to evaluate the function for some arbitrary values. Try and you'll understand why nothing plots. $\endgroup$
    – anderstood
    Mar 27 '18 at 20:27
  • $\begingroup$ If you really must have units: With[{T = 5000}, Plot[1/(1 + E^(((Quantity[e, "Electronvolts"] + Quantity[-7.05, "Electronvolts"])*(Quantity[1, 1/("BoltzmannConstant")]))/Quantity[T, "Kelvins"])) // Evaluate, {e, 1, 20}]] $\endgroup$
    – J. M.'s torpor
    Mar 28 '18 at 1:13
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    $\begingroup$ @anderstood Whenever Plot returns nothing I always evaluate the function for arbitrary values. That's why, after Plot failed, I replaced Plot with Table. As you can see, evaluating the function for arbitrary values in Table works. And when I ListPlot those arbitrary values, the function works. My question was why Plot fails when the function successfully evaluates for arbitrary values. $\endgroup$
    – Max Coplan
    Mar 28 '18 at 10:35
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I think a general fix might be using a wrapper function to add the units:

fp[t_] := f[Quantity[t, "seconds"]]
Plot[fp[t], {t, 0, 0.004}]
(* instead of *)
Plot[f[t], {t, Quantity[0.0, "Seconds"], Quantity[0.004, "Seconds"]}]

The problem seems to be that Plot does not always pass the parameter to f with units. It seems like Echos are able to show what happens:

Plot[Echo[f[Echo[t]]], {t, Quantity[0.0, "Seconds"], Quantity[0.004, "Seconds"]}]
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