Semiconductors

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On Semiconductors

There a bit of information I think is lost about semiconductors. There are really two types.

Intrinsic Semiconductors: These are pure, unchanged semiconductors. In any situation the charge of the material is neutral, so any electrons getting knocked in or out of their conduction bands will always create/leave behind a hole. You may ask why this is different than basically any other material. Wouldn't with just a piece of plastic there would be holes left behind when the electrons jump off? The answer is YES! The difference is that the difference between semiconductors and a piece of plastic is that as semiconductors heat up, more electrons jump out into the conduction band and the ease to jump into the conduction band is much easier than a piece of plastic. The term "semi" here in "semiconductor" is really just a term that means its less conductive than pure metal, but more conductive than an insulator like a piece of plastic.

To illustrate this further, let's take a bit of germanium (semiconductor). The resistivity of undated germanium at room temperature is approximately $0.47 ohm-meters$ while copper's is $1.68 \times 1 0^{- 8} ohm-meters$ . Intrinsic germanium is about 28 million times more resistive than copper at room temperature.

And just for fun if you increase the temperature of a semiconductor it becomes more conductive while metals become less conductive - this is because they're conductive for different reasons - semiconductors are conductive because of the atoms jumping to the conduction band. Metals are conductive because of just free electrons sitting around so when they heat up the vibrating lattice will make it harder for the electrons to move. Practically an intrinsic semiconductor will never become as conductive as a metal no matter how much you heat it up because it'll just melt or the lattice will break down but it's fun to illustrate the point.

Doped Semiconductors: As soon as they're doped all bets about $n = p = n_{i}$ are off. You're in a situation where $n ≫ p$ or $p ≫ n$ .

So because semiconductors are typically so resistive, the majority of practical applications they have to be doped. The resistance drops a ton when doped for example heavily doped germanium, resistivities can range from $1 0^{- 2}$ to $1 0^{- 3}$ ohm-meters and as we mentioned above copper is somewhere around $1.68 \times 1 0^{- 8}$ ohm-meters.

The reality is that this simple property of almost being a conductor allows them to interact in all kinds of cool ways like conducting when theres light or becoming more conductive when temperatures change.

Intrinsic Carrier Concentration $n_{i}$ .

The next bit after understanding semiconductors is basically asking "well how semi are those conductors." Well it depends on material. So take this table:

Material	$E_{g} (e V)$	$B (c m^{- 3} K^{- 3/2})$
Silicon (Si)	1.1	$5.23 \times 1 0^{15}$
Gallium Arsenide (GaAs)	1.4	$2.10 \times 1 0^{14}$
Germanium (Ge)	0.66	$1.66 \times 1 0^{15}$

Where $E_{g}$ is the bandgap energy for the electrons to jump to the conduction band (in electron volts) and the valence band. The lower this is the easier it is for the electrons to jump up and start conducting electricity, and $B$ is just scaling the interaction with temperature. It reflects how the number of intrinsic carriers change with temperature.

For sanity, the $n_{i}$ is talking about the number of charge carriers per unit volume, typically $c m^{- 3}$ . A value like $1.66 \times 1 0^{15}$ typically represents that many electron/hole pairs per cubic cm.

Calculate Intrinsic Carrier Concentration

$n_{i} = B T^{3/2} e^{\frac{- E _{g}}{2 k T}}$

and remember $k = 8.617 \times 1 0^{- 5}$

Extrinsic Carrier Concentration

Intrinsic semiconductors are pure and have an equal number of electrons ( $n$ ) and holes ( $p$ ). Hence, in intrinsic semiconductors, $n = p = n_{i}$

n-type Semiconductor: Doping with donor atoms (like phosphorus in silicon) introduces extra electrons. In this case, $n ≫ p$ or $p ≫ n$ .

This creates a new equation which is true for both intrinsic and extrinsic semiconductors which is $n \times p = n_{i}^{2}$ . This is true by the law of mass action. And remember p and n are charge carriers per unit volume, typically $c m^{- 3}$ .

Lets talk about doping concentration units

$N_{D}$ : This represents the concentration of donor impurities in the semiconductor. Donor impurities are atoms that, when added to a semiconductor, "donate" an extra electron to the conduction band. When a semiconductor is doped with donor atoms, it becomes an n-type (negative-type) semiconductor because of the excess electrons. Common donor atoms for silicon include phosphorus (P) and arsenic (As).

$N_{A}$ : This stands for the concentration of acceptor impurities. Acceptor impurities are atoms that "accept" an electron, creating a hole in the valence band. When a semiconductor is doped with acceptor atoms, it becomes a p-type (positive-type) semiconductor because of the excess holes. Boron (B) is a common acceptor atom for silicon.

In the context of the statement "If $N_{D} ≫ n_{i}$ then $n \approx N_{D} "$ — it means:

If the concentration of donor atoms ( $N_{D}$ ) is much greater than the intrinsic carrier concentration ( $n_{i}$ ), then the majority carrier concentration of electrons (nn) will be approximately equal to $N_{D}$ . In simpler terms, in a heavily doped n-type semiconductor, the number of free electrons is approximately equal to the number of donor atoms.

Therefore:

p = n_{i}^{2} / N_{D} if donor is dominant n = n_{i}^{2} / N_{A} if acceptors is dominant

All extrinsic relationships hold true for intrinsic carrier concentration its just that sometimes we can forget about the minority carrier

This is true for extrinsic and intrinsic: $n_{i} = B T^{3/2} e^{\frac{- E _{g}}{2 k T}}$

This is true for extrinsic and intrinsic: $n \times p = n_{i}^{2}$ and $n = p = n_{i}$ is a special case of the first equation.

What you'll notice this describes though is that for example the dominant carrier can be significantly greater than the intrinsic carrier ( $N_{D} ≫ n_{i}$ ) and the opposite is true for the opposite doping. What is this saying physically? Physically, the intrinsic carrier concentration sits as an "intermediate" value because it's governed by the probability of a rare event (thermal excitation across a large energy gap) compared to the more "guaranteed" carrier introduction by dopants.

The important bit here is that the intrinsic carrier concentration, $n_{i}$ , is inherent to the semiconductor material itself and is determined by its atomic properties, particularly the bandgap. It represents the number of carriers (electrons and holes) present in the pure semiconductor due to thermal excitations. This concentration doesn't change due to doping, but the presence of dopants can vastly increase the total number of free carriers, overshadowing the intrinsic concentration in terms of magnitude.

This is why the $\approx$ is so critical in $n \approx N_{d}$ or $p \approx N_{A}$ because the intrinsic property DOES change during doping it's just incredibly small, so we end up in a situation where the intrinsic carrier property is basically decoupled from the doping.

Current in the semiconductor

We have two types of currents

Drift Currents

Electron drift velocity $v_{d n} = - μ_{n} E$ where $μ_{n}$ is electron mobility and $E$ is electric field

Electron drift current density: $J_{n} = - e n v_{d n} = - e n (- μ_{n} E) = + e n μ_{n} E$

Then of course you can do the same with holes

Hole drift velocity: $v_{d p} = + μ_{p} E$ where $μ_{p}$ is the hole mobility

And of course hole drift current density: $J_{p} = e p v_{d p} = e p (+ μ_{p} E) = + e p μ_{p} E$ .

Then of course summing them gives total drift current density

$J = e n μ_{n} E + e p μ_{p} E = σ E = \frac{1}{ρ} E$

The total conductivity

$σ = e n μ_{n} + e p μ_{p}$

And rember $\frac{1}{ρ}$ is the resistivity.

So thats a lot of equations but we can break this down conceptually. Remember that drift velocity is the velocity of charged particles inside the electric field $E$ . In this case the velocity is quite straightforward. Lets look $v_{d n} = - μ_{n} E$ . The hard math is basically done for us with this equation, its basically saying that the speed that charge carriers move $v_{d n}$ is influenced linearly by $E$ at proportion of $μ_{n}$ . And this is super literal, the unit of $v_{d n}$ is just $m / s$ .

That sets the groundwork for the context equations. Now for current density, its fairly easy to see that to know the # electrons passing through some 2d surface area literally is:

The speed of the electrons $\times$ the number of electrons $\times$ the charge of each electron (to convert to current).

And just for fun remember that the elementary charge = $1.602 \times 1 0^{19}$ .

And then obviously because both holes and electrons can carry charge, the current is both of them summed together!

A big note here is that a lot of complexity is hiding here. The composition of the currents change as you look across the diode, the

Diffusion Currents

Electron diffusion current: $J_{n} = e D_{n} \frac{d n}{d x}$

Hole diffusion current: $J_{p} = - e D_{p} \frac{d p}{d x}$

What do these mean? Well they're pretty straightforward actually. Let's take $J_{n}$ which is of course the "current density." It represents the amount of current (due to the movement of electrons) per unit cross-sectional area of the semiconductor. Physically, this is a measure of how many electrons are moving across a given area in response to concentration gradients. The unit of current density is $A / c m^{2}$ , which describe a current per unit area. $D_{n}$ is the diffusion coefficient for elections which describe how rapidly electrons spread out, it's unit is $c m^{2} / s$ .

It's a bit hard to reason about these equations because they describe a "snapshot" of the diffusion at some given moment.

Total Current

Really simple, the way you find the total current in the semiconductor is you just add the previous equations in both drift, diffusion, hole and electrons!

Current: $j = j_{n} + j_{p}$

And then break it down more to each diffusion and drift component

$j_{n} = q μ_{n} n E + q D_{n} \frac{\partial n}{\partial x}$

$j_{p} = q μ_{p} pE - q D_{p} \frac{\partial p}{\partial x}$

Notice theres no time domain...get rekt! You can use Fick's second law for a time domain but I don't think thats in scope.