semic

Revamp lectures 13 and 14.

src/13_semiconductors.md

@@ -132,13 +132,14 @@ $$\frac{dE_h}{dp_h} = \frac{-dE}{-d\hbar k} = \frac{dE}{dp}.$$
 
 Finally, we derive the hole effective mass from the equations of motion:
 
-$$m_h \frac{d v}{d t} = +e (E + v\times B).$$
+$$m_{h,i} \frac{d v}{d t} = +e (E + v\times B).$$
 
 Comparing with
 
-$$m_e \frac{d v}{d t} = -e (E + v\times B),$$
+$$m_{e,i} \frac{d v}{d t} = -e (E + v\times B),$$
 
-we get $m_h = -m_e$.
+we get $m_{h,i} = -m_{e,i}$.
+An additional band index $i$ was introduced to clarify that the electron/hole mass relation only applies to particles **within the same band**.
 
 ## Semiconductors: materials with two bands.
 
@@ -146,10 +147,11 @@ In semiconductors the Fermi level is between two bands.
 The unoccupied band is the **conduction band**, the occupied one is the **valence band**. 
 In the conduction band the **charge carriers** (particles carrying electric current) are electrons, in the valence band they are holes.
 
+The occupation of the bands is dictated by the Fermi level. 
 We can control the position of the Fermi level (or create additional excitations) making semiconductors conduct when needed.
-
-Only the bottom of the conduction band has electrons and the top of the valence band has holes because the temperature is always smaller than the size of the band gap.
-
+The easiest way to control the Fermi level is through temperature by inducing thermal excitations. 
+In most semiconductors, the temperature is always smaller than the size of the band gap $k_b T \ll E_G$.
+As a result, only the bottom of the conduction band has electrons and the top of the valence band has holes.
 Therefore we can approximate the dispersion relation of both bands as parabolic.
 
 ![](figures/semiconductor.svg)
@@ -165,15 +167,19 @@ $$ g(E) = (2m_e)^{3/2}\sqrt{E-E_c}/2\pi^2\hbar^3$$
 $$ g_h(E_h) = (2m_h)^{3/2}\sqrt{E_h+E_v}/2\pi^2\hbar^3.$$
 
 Here $E_c$ is the energy of an electron at the bottom of the conduction band and $E_v$ is the energy of an electron at the top of the valence band.
-
 Observe that because we are describing particles in the valence band as holes, $m_h > 0$ and $E_h > -E_v$.
 
+Lastly, it is important to point out the notation that we adopt here. 
+In the above plot, $m_h \neq -m_e$ because it refers to two **different** bands: valance band for holes and conduction band for electrons. 
+In most literature, the band indices are neglected such that $m_{h,v} \to m_{h}$ and $m_{e,c} \to m_e$. 
+
+
 ??? question "a photon gives a single electron enough energy to move from the valence band to the conduction band. How many particles does this process create?"
     Two: one electron and one hole.
 
 ## Semiconductor density of states and Fermi level
 
-### Part 1: Intrinsic semiconductor
+### Intrinsic semiconductor
 
 ```python
 E = np.linspace(-3, 3, 1000)
@@ -205,29 +211,36 @@ draw_classic_axes(ax, xlabeloffset=.2)
 4. Apply approximations to simplify the equations (this is important!).
 5. Find $E_F$ and concentrations of electrons and holes
 
-Applying the algorithm:
+Applying the first two steps of the algorithm:
 
 $$n_h = \int_{-E_v}^\infty f_h(E_h) g_h(E_h) dE_h = \int_{-E_v}^\infty\frac{(2m_h)^{3/2}}{2\pi^2\hbar^3}\sqrt{E_h+E_v}\frac{1}{e^{(E_h+E_F)/kT}+1}dE_h$$
 
-$$n_e = \int_{E_c}^\infty f(E)g_e(E)dE = \int_{E_c}^\infty\frac{(2m_e)^{3/2}}{2\pi^2\hbar^3} \sqrt{E-E_c}\frac{1}{e^{(E-E_F)/kT}+1}dE$$
-
-We need to solve $n_e = n_h$
+$$n_e = \int_{E_c}^\infty f(E)g_e(E)dE = \int_{E_c}^\infty\frac{(2m_e)^{3/2}}{2\pi^2\hbar^3} \sqrt{E-E_c}\frac{1}{e^{(E-E_F)/kT}+1}dE.$$
 
-Simplification:
-Fermi level is far from both bands $E_F-E_v \gg kT$ and $E_c - E_F \gg kT$
+Note that whenever calculating the hole dependent quantities, we replace all the relevant physical quantities with their hole equivalents.
+Since the hole energy is opposite $E_h = -E$, we replace the Fermi energy $E_F \to -E_F$ and the bottom of the valance band by $E_v \to -E_v$ in the integration limits.
 
-Therefore Fermi-Dirac distribution is approximately similar to Boltzmann distribution.
+In the third step, we need to solve the equation under charge balance $n_e = n_h$.
+The equation is not a pleasant one and cannot be solved analytically unless an approximation is made.
+Therefore, the fourth step assumes that the Fermi level is far from both bands $E_F-E_v \gg kT$ and $E_c - E_F \gg kT$.
+As a result, the Fermi-Dirac distribution is approximately similar to Boltzmann distribution:
 
-$$f(E)_{e/h} \approx e^{-(E_{e/h}\pm E_F)/kT}$$
+$$
+f(E)_{e/h} \approx e^{-(E_{e/h}\pm E_F)/kT}.
+$$
 
-Now we can calculate $n_e$ and $n_h$:
+Now we can move to the last step and calculate $n_e$ and $n_h$:
 
-$$n_h \approx \frac{(2m_h)^{3/2}}{2\pi^2\hbar^3}e^{-E_F/kT} \int_{-E_v}^\infty\sqrt{E_h+E_v}e^{-E_h/kT}dE_h =
-N_V e^{E_v-E_F/kT},$$
+$$
+n_h \approx \frac{(2m_h)^{3/2}}{2\pi^2\hbar^3}e^{-E_F/kT} \int_{-E_v}^\infty\sqrt{E_h+E_v}e^{-E_h/kT}dE_h =
+N_V e^{E_v-E_F/kT},
+$$
 
 where we used $\int_0^\infty \sqrt{x}e^{-x}dx=\sqrt{\pi}/2$ and we defined 
 
-$$N_V = 2\left(\frac{2\pi m_h kT}{h^2}\right)^{3/2}$$
+$$
+N_V = 2\left(\frac{2\pi m_h kT}{h^2}\right)^{3/2}.
+$$
 
 We see that holes are exponentially activated into the valence band. 
 
@@ -236,16 +249,22 @@ We see that holes are exponentially activated into the valence band.
 
 Similarly for electrons:
 
-$$n_e = N_C e^{-(E_c - E_F)/kT},\quad N_C = 2\left(\frac{2\pi m_e kT}{h^2}\right)^{3/2}$$
+$$
+n_e = N_C e^{-(E_c - E_F)/kT},\quad N_C = 2\left(\frac{2\pi m_e kT}{h^2}\right)^{3/2}.
+$$
 
 
 Combining everything together:
 
-$$n_h \approx N_V e^{E_v-E_F/kT} = N_C e^{-(E_c-E_F)/kT} \approx n_e$$
+$$
+n_h \approx N_V e^{E_v-E_F/kT} = N_C e^{-(E_c-E_F)/kT} \approx n_e.
+$$
 
 Solving for $E_F$:
 
-$$E_F = \frac{E_c + E_v}{2} - \frac{3}{4}kT\log(m_e/m_h)$$
+$$
+E_F = \frac{E_c + E_v}{2} - \frac{3}{4}kT\ln(m_e/m_h).
+$$
 
 An extra observation: regardless of where $E_F$ is located, $n_e n_h = N_C N_V e^{-E_g/kT} \equiv n_i^2$, where $E_g=E_c-E_v$ is the band gap of the semiconductor.
 
@@ -255,6 +274,25 @@ $n_i$ is the **intrinsic carrier concentration**, and for a pristine semiconduct
 > $$n_e n_h = n_i^2$$
 > is the **law of mass action**. The name is borrowed from chemistry, and describes the equilibrium concentration of two reagents in a reaction $A+B \leftrightarrow AB$. Here electrons and hole constantly split and recombine.
 
+### Conduction
+
+Earlier, we deduced that empty and filled bands provide no current. 
+We finish the analysis by considering partially filled bands of an intrinsic (pristine) semiconductor.
+To calculate the current, we utilize the Drude model and sum the electron and hole contributions:
+
+$$j = -n_e e v_e + n_h e v_h $$
+
+$$ -m_e v_e /\tau_e = -eE;\quad -m_h v_h /\tau_h = eE.$$
+
+We see that despite opposite velocity signs for electrons and holes, they carry electric current in the same direction:
+
+$$ \sigma \equiv \frac{j}{E} = \left(\frac{n_e e^2 \tau_e}{m_e}+\frac{n_h e^2 \tau_h}{m_h}\right) = n_e e \mu_e + n_h e \mu_h.$$
+
+We know that for intrinsic semiconductors, the hole/electron densities are $n_e = n_h = n_i \propto e^{-E_G/kT}$.
+Therefore, it is possible to measure the band gap of an intrinsic semiconductor by looking at the temperature dependant conductivity $E_G \approx d \ln \sigma / d [kT]^{-1}$.
+Additional information can be obtained using Hall effect.
+However Hall effect is much more complex in semiconductors since only the current in the direction perpendicular to the applied electric field must vanish.
+This, however only means that the electron current is opposite of the hole current in that direction, not that the electrons and holes move parallel to the applied current.
 
 ## Exercises