<jupyter_start><jupyter_text>Summary from last class Instrinsic charge carrier concentrationThe concentration of thermally-generated electron-hole pairs varies (lots!) with temperature, and is:$$n_i = B \cdot T^{3/2} \cdot \exp\left(\dfrac{-E_g}{2 k_B T}\right) \;\; \mathrm{\/cm^3}$$  DopingAdding impurities to the crystal from Group III or IV elements (replacing an Si atom with another like B or P) also adds extra electrons or holes to the structure.Impurities which add extra *free electrons* are called **donors**, with concentration $N_D \;\; \mathrm{cm^{-3}}$.Impurities which add extra (free) *holes* are **acceptors**: $N_A \;\; \mathrm{cm^{-3}}$. Even when doped, a semiconductor crystal has no net charge$$\sum \text{charges} = q \left( +N_D + p - N_A - n \right) = 0$$This, combined with the fundamental relationship$$n\cdot p = n_i^2$$ Allows us to calculate the concentration of *free electrons* and *holes* for any doping condition (always choose $+$ version):$$n = \dfrac{(N_D - N_A) \pm \sqrt{(N_D - N_A)^2 + 4 n_i^2}}{2} \; \text{ and } \; p = \dfrac{n_i^2}{n}$$ or, solved the other way:$$p = \dfrac{(N_A - N_D) + \sqrt{(N_A - N_D)^2 + 4 n_i^2}}{2} \; \text{ and } \; n = \dfrac{n_i^2}{p}$$ **Q:** Are the above two sets of solutions equivalent? **Q:** Does it matter which set to use depending on which of $N_D$ or $N_A$ is larger? When we dope in practice, the impurities $N_D$ and/or $N_A$ are in the range $10^{14} \text{ to } 10^{21} \; \mathrm{cm^{-3}}$, which is much greater than $n_i$ at normal temperatures.  This results in us using the following approximations all the time:$$\text{when $N_D$ is largest:  } n \approx N_D \text{  and  } p \approx n_i^2/N_D$$$$\text{when $N_A$ is largest:  } p \approx N_A \text{ and } n \approx n_i^2/N_A $$$\rightarrow$ Notice how only the minority carrier concentrations are affected by temperature (holes for n-type doping, electrons for p-type doping).  Conclusions* $n_i$ varies lots with temperature* We can change the concentration of free electrons ($n$) and (free) holes ($p$) by **doping**.* Minority carrier concentrations are greatly affected by temperature.  Majority carriers not so much.  Charge movement in semiconductors* Drift - constant velocity proportional to E-field* Diffusion - movement from high to low concentration  Drift current: Physics phundamentals* An electron in an electric field experiences a force.* This force causes the electron (which has mass) to accelerate.* Why does this not therefore cause an *increasing* current in a material? $\rightarrow$ the electrons are scattered by the solid -- always accelerating but constantly getting knocked off track.![From: https://ecee.colorado.edu/~bart/book/book/chapter2/ch2_7.htmfig2_7_2](fig/zeghbroeck_fig2_7_2.png) resulting in an *average velocity* which is constant and proportional to the applied E-field$$v \propto \vec{E}$$ The constant of proportionality is called **mobility** ($\mu_n$ for electrons and $\mu_p$ for holes).  Its units must be $\mathrm{\frac{cm^2}{V\cdot s}}$.For silicon:* $\mu_n = 1350 \;\;\mathrm{\frac{cm^2}{V\cdot s}}$* $\mu_p = \phantom{1} 480 \;\;\mathrm{\frac{cm^2}{V\cdot s}}$ Recall that $\mathrm{V(olts)} = - \int_a^b \vec{E}\,dx$.For holes: $\vec{v}_h = \mu_p \vec{E}$ $\rightarrow$ same direction as $\vec{E}$ field vector.For electrons: $\vec{v}_e = -\mu_n \vec{E}$ $\rightarrow$ opposite direction as $\vec{E}$ field vector. Current flow in a bar:![Voltage source across a conductive bar.](fig/current-flow-bar.png) Start with current due to electrons:![Electron current](fig/electron-current.png) Substitute with electron velocity from earlier: $\vec{v}_e = -\mu_n \vec{E}$ (see the double-negative!), then divide both sides by the cross-sectional area $W\cdot h$.$$J_n = \frac{I_n}{W\cdot h} = \mu_n \vec{E} \cdot n \cdot q$$$J_n$ has units of $\mathrm{A / cm^2}$: *current flux*.  In an unfortunate naming convention, everyone else calls this term electron **current density**. By the same reasoning, we can find the hole **current density**$$J_p = \left(\text{hole velocity}\right) \left(\text{charge density}\right)$$$$J_p = \left(\mu_p \vec{E}\right) \left(p \cdot q\right)$$ The total **current density** is then$$J_{total} = J_n + J_p$$$$J_{total} = q \left( \mu_n n + \mu_p p \right) \cdot \vec{E} \;\; \mathrm{A/cm^2}$$  Mobility changes with doping :(![From: <NAME>, Principles of Electronic Devices.](fig/zeghbroeck_fig2_7_3.png)https://ecee.colorado.edu/~bart/book/ What causes this second-order effect?* More doping means a less uniform crystal and more opportunities for scattering.* But (free) charge density increases faster than mobility decreases, so conductivity still increases.  Velocity saturation![Jaeger, Microelectronic Circuit Design](fig/jaeger-velocity-saturation.jpg) At large E-field strengths, the drift velocity does not continue to increase (light speed is $3\times 10^{10}\;\mathrm{cm/s}$).  The velocity (therefore current) remains constant and no longer varies with E-field (voltage).  In a circuit context, the device changes from behaving like a resistor to more like a current source.  Velocity saturation is very common in modern integrated circuits.  For example, a $130\,\mathrm{nm}$ chip process uses a $1.2\,\mathrm{V}$ supply voltage, giving internal E-fields of<jupyter_code>print('%2.2g V/cm' % (1.2 / 130e-9 / 100) )<jupyter_output>9.2e+04 V/cm
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