add a version of the plot with N_A=0

Closes #53 (closed).

docs/14_doping_and_devices.md

@@ -230,18 +230,26 @@ E_Fs = np.array(
 
 fig = pyplot.figure(figsize=(10, 10))
 ax1, ax2 = fig.subplots(nrows=2)
-ax1.plot(Ts, E_Fs)
-ax1.hlines([E_C, (E_C + E_V) / 2, E_D], Ts[0], Ts[-1], 'k', linestyles="--")
+ax1.plot(Ts, E_Fs, label=r"$N_A \neq 0$")
+ax1.plot(
+    Ts,
+    [equilibrium_E_F(T, N_C, N_V, N_D - N_A, 0, E_C, E_V, E_D, E_A) for T in Ts],
+    label="$N_A = 0$",
+    linestyle="--",
+    c="#1f77b4",
+)
+ax1.hlines([E_C, (E_C + E_V) / 2, E_D], Ts[0], Ts[-1], 'k', linestyles=":")
 ax1.set_ylim(-.1, 1.1)
 ax1.set_ylabel("$E_F$")
 ax1.set_xlabel("$T$")
 ax1.set_yticks([E_C, (E_C + E_V) / 2, E_D])
 ax1.set_yticklabels(["$E_C$", "$(E_C + E_V) / 2$", "$E_D$"])
+ax1.legend()
 draw_classic_axes(ax1, y=-.05, ylabeloffset=0.01)
 
 ax2.plot(Ts, N_V * np.exp((E_V - E_Fs) / Ts), label="$n_h$")
 ax2.plot(Ts, N_C * np.exp(-(E_C - E_Fs) / Ts), label="$n_e$")
-ax2.hlines((N_D - N_A), Ts[0], Ts[-1], 'k', linestyles="--")
+ax2.hlines((N_D - N_A), Ts[0], Ts[-1], 'k', linestyles=":")
 ax2.set_ylim(0, (N_D - N_A) * 3)
 ax2.legend()
 ax2.set_ylabel("$n$")
@@ -257,7 +265,7 @@ As we go from highest to lowest temperature, we observe several regimes:
 * **Intrinsic limit** . If the temperature is sufficiently large, then $n_i \gg |N_D-N_A|$ and therefore $n_e \approx n_h \approx n_i$. Additionally, if holes are heavier than electrons, then $E_F$ grows with temperature in this limit.
 * **Extrinsic limit**. As we decrease the temperature, we decrease the number of intrinsic carriers to the point where most of the charge carriers come form the fully ionized donors. As a result, the number of carriers stays approximately constant in this temperature range.
 * **Freeze-out limit**. Once the temperature is sufficiently low $kT \ll E_G - E_D$, we expect the electrons to "freeze away" from the conduction band to the donor band. The charge carriers still come from the donors, however, not all donors are ionized now.
-* **Zero temperature**. There are no charge carriers in neither conduction nor valence bands. The highest energy electrons are in the donor band and therefore $E_F$ should match the donor band.
+* **Zero temperature**. There are no charge carriers in neither conduction nor valence bands. If there are no acceptors, then the Fermi level goes to halfway between the donor and the conduction bands. Otherwise, the highest energy electrons are in the donor band and therefore $E_F$ should match the donor band.
 
 !!! check "Exercise"
     check that you can reproduce all the relevant limits in a calculation.