Our tour through semiconductor physics will focus on the main features that are relevant to a circuit designer — those that affect the behavior with respect to voltage and current. The foggy valleys where the underlying quantum-mechanical properties that form the base of these features will be left for a different, more focused, exploration.
| Symbol | Units | Name |
|---|---|---|
\(T\) |
\(\mathrm{K}\) |
temperature |
\(E_g\) |
\(\mathrm{J}\) or \(\mathrm{eV}\) |
material bandgap energy |
\(B\) |
\(\mathrm{cm^{-3}K^{-3/2}}\) |
material constant |
\(k_B\) |
\(\mathrm{\dfrac{J}{K}}\) or \(\mathrm{\dfrac{eV}{K}}\) |
Boltzmann constant |
\(q\) |
\(\mathrm{C}\) |
elementary charge |
\(n_i\) |
\(\mathrm{\dfrac{\#}{cm^3}}\) |
intrinsic charge carrier density |
\(n\) |
\(\mathrm{\dfrac{\#}{cm^3}}\) |
free electron density |
\(p\) |
\(\mathrm{\dfrac{\#}{cm^3}}\) |
(free) hole density |
\(N_D\) |
\(\mathrm{\dfrac{\#}{cm^3}}\) |
donor doping density |
\(N_A\) |
\(\mathrm{\dfrac{\#}{cm^3}}\) |
acceptor doping density |
\(\mu_n\) |
\(\mathrm{\dfrac{cm^2}{V \cdot s}}\) |
electron mobility |
\(\mu_p\) |
\(\mathrm{\dfrac{cm^2}{V \cdot s}}\) |
hole mobility |
\(D_n\) |
\(\mathrm{\dfrac{cm^2}{s}}\) |
electron diffusivity |
\(D_p\) |
\(\mathrm{\dfrac{cm^2}{s}}\) |
hole diffusivity |
\(q \left(N_D + p - N_A - n\right) = 0\)
\(n \cdot p = n_i^2\)
\(n = \dfrac{\left(N_D - N_A\right) \pm \sqrt{\left(N_D - N_A\right)^2 + 4 n_i^2}}{2}\)
If \(\left(N_D - N_A\right) \gg 2 n_i\), then we can approximate \(n \approx \left(N_D - N_A\right)\).
\(p = \dfrac{n_i^2}{n}\)
\(p = \dfrac{\left(N_A - N_D\right) \pm \sqrt{\left(N_A - N_D\right)^2 + 4 n_i^2}}{2}\)
\(n = \dfrac{n_i^2}{p}\)
Similarly, if \(\left(N_A - N_D\right) \gg 2 n_i\), then we can approximate \(p \approx \left(N_A - N_D\right)\).
The basic physics of an atom’s structure tells us that electrons are negatively charged and protons are positively charged. Current in amperes is moving charge in coulombs per second. Therefore, current in electronics can be due to electrons moving or protons moving.
Before moving on, here is a reliable saying:
If your protons are moving, you have bigger problems.
Electronics with semiconductors is frequently called “solid-state” for a good reason --- our materials are (all) in the solid state of matter!
Protons moving means your material is in a liquid or, heavin forbid, a gaseous or plasma state. Surely you recognize that this cannot possibly end well :)
Electrons moving is totally fine. In fact, electrons moving in free space between electrodes is fundamentally what is happening in vacuum tubes. The vacuum just ensures that the electrons don’t collide with other (air) molecules while in motion.
This is why the field is called electronics!
Moving positive charges in the form of ions does happen in processes like electrolysis used in electroplating, for example, but that bumps up against chemistry and is out of our scope for now.
Negatively-charged electrons in motion are our only currents. Keep reading for how we fake having positive charges also moving inside semiconductors.
An intrinsic semiconductor is a material that has no added impurities. Its electrical properties therefore are determined only by the properties of the material itself.
The concentration of thermally-generated electron-hole pairs varies (lots!) with temperature, and is:
It is much more common when calculating numbers to need ni2:
The constant B is derived from material properties related to allowable energy levels an electron can occupy in a crystal.
The law of mass action relates the concentrations of free electrons and holes under thermal equilibrium:
This equation holds for all semiconductors. See Wikipedia: Mass action law (electronics) for a short description and more references.
| Semi. | Eg [eV] | B (×1015) [cm-3K-3/2] |
|---|---|---|
Si |
1.12 |
5.2 |
Ge |
0.66 |
1.66 |
GaAs |
1.42 |
0.356 |
GaN |
3.49 |
|
SiC |
3.26 |
Adding impurities to the crystal from Group III or IV elements (replacing an Si atom with another atom such as B or P) also adds extra electrons or holes to the structure.
-
Dopants which add extra free electrons are called donors, with concentration \(N_D \;\; \mathrm{cm^{-3}}\).
-
Dopants which add extra (free) hole’s are acceptors: \(N_A \;\; \mathrm{cm^{-3}}\).
Even when doped, a semiconductor crystal has no net charge. So let’s add up all of the extra charges present in our doped crystal:
The symbols of equation \(\eqref{eq-sum-charges}\) are familiar, but we need to carefully look at the signs of each term:
- n
-
Density of negatively-charged free electrons, including both thermally-generated and extras contributed by donor dopants
- ND
-
Density of donor dopants. The sign is positive because a donor dopant also contributes a net extra proton along with its extra free electron.
- p
-
Density of positively-charged (free) holes, including both thermally-generated and extras contributed by acceptor dopants
- NA
-
Density of acceptor dopants. The sign is negative because an acceptor dopant is short a proton in addition to being short an outer-shell electron (a.k.a. hole).
The above charge balance equation, combined with the law of mass action fundamental relationship:
allows us to calculate the concentration of free electrons and holes for any doping condition (always choose the \(+\) version of the quadratic formula because the density number must be non-negative):
or, solved the other way:
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. Therefore we use the following approximations all the time:
Notice how only the minority carrier concentrations are affected by temperature (holes for n-type doping, electrons for p-type doping).
As usual, we lead with the punchline — results first, then back-fill with its creation story.
| Drift |
constant velocity proportional to E-field |
| Diffusion |
movement from high to low concentration |
Two mechanisms of movement with two types of charge carriers yields four types of current in a semiconductor.
| electrons | holes | |
|---|---|---|
drift |
\(\phantom{-} q \cdot n \cdot \mu_n \cdot \vec{E}\) |
\(\phantom{-} q \cdot p \cdot \mu_p \cdot \vec{E}\) |
diffusion |
\(\phantom{-} q \cdot D_n \cdot \dfrac{\mathrm{d}\, n}{\mathrm{d} x}\) |
\(- q \cdot D_p \cdot \dfrac{\mathrm{d}\, p}{\mathrm{d} x}\) |
-
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?
Think about this question, then <click to reveal>
The electrons are scattered by the solid — always accelerating but constantly getting knocked off track.
Notice in this video during the Plinko game (around 3:00), the camera pans downward at an approximately constant rate: YouTube: The Price is Right former biggest Plinko win primetime.
This results in an average velocity which is constant even though an individual electron is always accelerating.[1]
The average electron velocity is proportional to the applied E-field.
The constant of proportionality is called mobility (μn for electrons and μp for holes) and must have units of \(\mathrm{\frac{cm^2}{V\cdot s}}\).
For silicon, these values are around:
-
\(\mu_n = 1350 \; \mathrm{\frac{cm^2}{V\cdot s}}\)
-
\(\mu_p = \phantom{1} 480 \; \mathrm{\frac{cm^2}{V\cdot s}}\)
Recall that the electric potential difference that we commonly name by its units of volts is only and truly the path integral of the electric field. Fortunately, the E-field is a conservative field, so the result of the integration only depends on the end points:
- For holes
-
\(\vec{v}_h = \mu_p \vec{E}\), movement in the same direction as the \(\vec{E}\) field vector.
- For electrons
-
\(\vec{v}_e = -\mu_n \vec{E}\), movement in the opposite direction as the \(\vec{E}\) field vector.
Imagine a bar of silicon
Let’s begin by considering the current that flows due to electrons.
We know the electron (average) velocity, the density of (free) electrons, and the geometry of the bar.
|
Note
|
Notice the double negative! Each negative has a different origin even though the net result is a positive sign. |
Let’s normalize this into a current flux by dividing by the cross-sectional area W·h.[2]
|
Note
|
Jn has units of \(\mathrm{A / cm^2}\) which is the units of a flow per unit area or 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
The total drift current density is then
Finding the current that you would measure with an ammeter from this expression merely requires multiplying by the cross-section area of the bar.
But… look back at Voltage source connected across a (semi)conductor bar. and see that we are applying a voltage across the ends of the bar (using an ideal voltage source) — what is \(\vec{E}\) ?
An easy-ish way to remember what to do is to recall the units of the electric field: volts per meter. We get our voltage back by multiplying by meters, or the Length of the bar.
This only works if the cross-section area is uniform along the length of the bar.
Think about what would happen if the middle of the bar was necked down to a smaller area. KCL forces the current to be constant so therefore {insert thinking here}.
What causes this second-order effect? Does this mean that conductivity decreases with more doping?? <think first, then click to reveal>
-
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.
It hopefully makes sense that the charge velocity can’t increase so much as to exceed the speed of light, so this is the obvious speed limit. (Light speed is about \(3\times 10^{10}\;\mathrm{cm/s}\))
The velocity (therefore current) approaches a lower limit and no longer varies linearly at large E-field strengths (voltage). In a circuit context, this means that the device changes from behaving like a resistor to more like a constant current source.[3] Velocity saturation is very common in modern integrated circuits.
A \(130\,\mathrm{nm}\) chip process uses a \(1.2\,\mathrm{V}\) supply voltage, giving internal E-fields of
>>> print('%2.2g V/cm' % (1.2 / 130e-9 / 100) )
9.2e+04 V/cmwhich is well within the velocity saturation regime according to Drift velocity versus E-field. From Jaeger, Microelectronic Circuit Design.
The total diffusion current density is then
Look back at Four currents in a semiconductor (A/cm2) and match the terms we’ve just worked through. Only one term has a negative sign, hole diffusion, be careful to not make sign errors!
The goal of these subsections is to develop a model of the pn junction that is useful for circuit design and analysis. We will end up with a relationship between current and voltage, and another expression for junction capacitance (charge storage).
Imagine taking two separate bars of silicon, one doped to be n-type (ND) and the other doped to be p-type (NA). Push these bars together so that there is a single bar where the middle abruptly changes doping types and levels.
At the moment of contact, there is a huge concentration gradient for both holes and electrons. The p-type region with a majority of holes is right next to the n-type region with very few holes and vise versa.
|
Note
|
The n and p sides are reversed compared to previous figure. This is to match our other reference materials. |
This large gradient causes the holes at the edge of the p region to flow via diffusion to the low concentration side and electrons at the edge of the n region flow via diffusion the opposite direction. Because of the opposite signs for charge polarities and movement direction, the net diffusion current is in the same direction, which is to the left in Initial diffusion currents at the junction..
After a long time, this diffusion will eventually stop because the electron and hole densities have become uniform:
|
Warning
|
NO !!! We have forgotten about the protons that are in this material! |
If your protons are moving, you have bigger problems.
Think about the Group IV dopant atom (say phosphorus) on the n side of the junction. That extra electron (and extra proton!) is in a region of many free electrons and next to a region with few. So it will tend to move (diffuse) to the right towards the p region.
This particular free electron came from the 5th outer valence electron of the phosphorus dopant atom. Therefore, when it diffuses away, the 15th proton in the nucleus is left without a charge-balancing mate. We now have a +1 positively-charged phosphorus ion.
Similarly, when the hole contributed by the Group III dopant (say boron) on the p side diffuses to the left, we remember that the hole was filled with an electron. This electron fills the 4th outer valence “slot” and completes the covalent bonding structure in that area. Don’t forget that boron is short a proton that matched the now-filled hole. Therefore the region has a net −1 negatively-charged boron ion.
Both of these types of ions cannot move because the charges are either a) part of the nucleus, or b) a bound electron. This means that we have:
Charges
In
Spaaace [4]
or called the depletion region because the volume is eventually depleted of free charge carriers.
Now that there are separated charges, Gauss' law, \(\nabla \!\bullet\! \left(\epsilon \vec{E}\right) = Q\), implies that there is a non-zero E-field inside the depletion region. This field is maximum at the boundary between the n and p type doping and, for this example, points to the right.
Recall that the diffusion currents are to the left and the drift current from the E-field is opposite, to the right. Because this semiconductor bar is not connected in a circuit, KCL forces the current at every x location to be zero. The junction is in equilibrium and the drift and diffusion currents must be equal and opposite.
|
Caution
|
NO ! (again). The currents must be balanced for holes and electrons individually, or it makes no physical sense. |
Take the free electrons equation and substitute their definitions:
Integrate and divide both sides by n
The left side is the potential difference developed across the depletion region and is given the symbol \(V_0\). Also, recall Einstein’s relation, \(\frac{D}{\mu} = \frac{k_B T}{q}\), and substitute
We usually are only concerned with the magnitude of V0; notice that swapping the fraction so the majority carriers term (nn) is in the numerator cancels the leading negative sign. Recognize, for the next section, that the higher potential is on the n side, at location b, given the E-field direction. A similar process can be done to find V0 from the hole currents.
One final set of substitutions using our (good) approximations of majority and minority carrier concentrations
where \(V_T\) is called the thermal voltage.
\(V_T = \frac{k_B T}{q}\) is something you should memorize.
-
It is 25.85{mV} at 300{K} (26.85°C)
Both 25{mV} and 26{mV} are used in practice for hand calculations at “room temperature.” You will also see 40 = 1/0.025 in equations. Because VT is usually inside an exponent, the numerical result of a calculation can be sensitive to which approximation you used. Therefore it is best to state your assumptions instead of assuming the reader rounds the same direction as you do.