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This is a very theoretical question and mainly just for my understanding for current flow.

Let's say I have a DC voltage source. I use two metal wires to connect my DC voltage source across a piece of intrinsic/undoped/pure silicon semiconductor. Everything is at room temperature.

enter image description here

My understanding of the metal wire is that:

  • It has a very low bandgap energy and thus the covalent bonds in the valence band that the atoms of the metal form with each other break extremely easily and hence we have a lot of electrons in the conduction band, available to flow, just waiting for a voltage to be applied to start moving!
  • Because of this, it can be thought of as having a 'low' resistance - current flows easily!

My understanding of the piece of intrinsic/undoped/pure silicon semiconductor is that

  • Its bandgap energy is somewhere between that of a good metal conductor and that of a insulator, hence a semiconductor. At room temperature, some of its covalent bonds will too break and thus some electrons will be available for conduction, just waiting for a voltage to be applied.
  • The number of electrons available for conduction is significantly lower than that of the metal conductor. Hence, for this example, I think of the pure silicon as having a 'medium' resistance.

From my understanding of basic circuit theory, I draw the equivalent circuit as follows, enter image description here

From this, I can see that the current flowing in the circuit is limited by the medium resistance of the piece of silicon.

My question is understanding how this current is limited from a atomical level? The metal has excess free electrons, right? Why can't these excess free electrons travel through the lattice? Why do we have to rely on the limited number of free electrons in the silicon piece?

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  • \$\begingroup\$ Unfortunately your models of "resistance" aren't particularly useful. I'm not sure your background, but you'd probably be well-served reading some more detail on semiconductor physics... it's a large topic that takes many quite a while to understand that I don't think is a great fit for this website. Here's a resource from a quick web search. \$\endgroup\$
    – Shamtam
    Apr 5, 2021 at 22:12
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    \$\begingroup\$ And to complicate all this, there's a thing called a "Fermi level" (of which I cannot remember all the details) that bends the bandgap of the semiconductor where it's in contact with the metal; this changes the behavior at the interface in complicated ways, up to and including making the junction rectifying (as with a Shottky diode). So you can't even be assured that the metal + junction + silicon + junction + metal assembly will act like plain old resistors in series. \$\endgroup\$
    – TimWescott
    Apr 5, 2021 at 22:42
  • \$\begingroup\$ I see. I have a side question - do EEs know semiconductor physics in a lot of detail? \$\endgroup\$
    – jrr76994
    Apr 5, 2021 at 23:33
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    \$\begingroup\$ Some do. EEs that specialize in semiconductor devices are often the world experts in semiconductor physics. Topics surrounding semiconductor physics (and quantum mechanics) are some of most common PhD dissertations in EE. \$\endgroup\$
    – KD9PDP
    Apr 5, 2021 at 23:37

3 Answers 3

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Good question - there's a lot to unpack here, it's not as straightforward as it appears! Before answering your questions, a little bit of quantum mechanics is needed to understand it.

First:

My understanding of the metal wire is that: It has a very low bandgap energy and thus the covalent bonds in the valence band that the atoms of the metal form with each other break extremely easily and hence we have a lot of electrons in the conduction band, available to flow, just waiting for a voltage to be applied to start moving!

I'd correct that a little: metals don't have a bandgap energy at all, at least not as we'd think compared to semiconductors (they have bands, just not near the Fermi level). The Fermi level (the energy level where there's a 50% chance of an electron state being occupied) is in the middle of a band for a metal, which is why there are so many electrons available to carry current. In an undoped semiconductor, it's in the middle of a band gap - far away from any electron state, so there are not many carriers available (unless you make it very hot). When you dope the semiconductor, you move that Fermi level closer to a band edge, increasing the likelihood of an electron in the conduction band, or the lack of an electron (hole) in the valence bad. If you highly dope a semiconductor, you can drive the Fermi level into the conduction band - making it very metal-like.

Second: You are asking about the resistance of a material, and trying to compare the resistance of metal to semiconductor. The resistance is derived from a term called "resistivity" times material length divided by area. You are correct that resistivity (\$\rho\$) is a function of carrier density, which is higher in a metal than an undoped semiconductor. But other things also matter, such as temperature and carrier mobility. Carrier mobility in turn depends on things like the quantum mechanical effective mass of a carrier in the semiconductor, and how scattering the semicondcutor is. The effective mass is basically "how much mass does an electron appear to have based on the relationship between its wavefunction's momentum in a crystal lattice and its energy?" Holes tend to be "heavier" than electrons.

Ok, with that background we can get to your questions:

My question is understanding how this current is limited from a atomical level?

Assuming you care about the atomic level (and can ignore material area and length), current is limited by the availability of carriers (e.g., semiconductor doping, semiconductor temperature), how scattering the material is (semiconductor crystal purity and temperature of a metal), what the effective mass of a carrier is (an electron in a metal typically weighs less than an electron in a semiconductor conduction band typically weighs less than a hole in a semiconductor valence band).

The metal has excess free electrons, right?

If a metal is electrically neutral, it has just the right amount of free electrons. That is - it has as many free electrons as there are + metal ions. If it had excess, it would have a net negative charge.

Why can't these excess free electrons travel through the lattice?

They can and do. The problem is getting in to the lattice. The electrons in the metal need a state in the semiconductor for them to be able to transition in to. They either have to quantum mechanically tunnel through the bandgap barrier (e.g., Schottky contact), or the semiconductor has to be doped such that the electrons in the metal have a suitable state without tunneling (e.g., ohmic contact). See: https://en.wikipedia.org/wiki/Metal%E2%80%93semiconductor_junction

Why do we have to rely on the limited number of free electrons in the silicon piece?

This is the same question as "how come a resistor limits the current through a diode?" Since the same current has to flow through all elements in series, one highly resistive element will "limit" the current through all the other elements in series. So the same number of electrons will go through both the semiconductor and the metal, it's just that the semiconductor will have a bigger voltage drop and effectively limit the total current through the entire series (assuming everything is set up correctly so that current can flow between the metal and semiconductor properly in the first place)

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  • \$\begingroup\$ Fantastic answer! There was a lot more behind that than I expected :) \$\endgroup\$
    – jrr76994
    Apr 6, 2021 at 0:10
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NO actually pure Silicon is an insulator and needs doping to make it a good conductor with a forward bias current.

Although technically due to electron and hole mobility values they are intrinsically semiconductors, but very poor ones without doping in superheated toxic gas chambers like arsenic, where they obtain high ratios of conductance with forward & no bias.

“As conductivity of pure silicon is quite low, hence it is not suitable as a circuit element for electrons and is doped with small amounts of other elements. This doping ensures that the conductivity of silicon is substantially increased and its electrical response is adjusted by controlling the charge and number of activated carriers.”

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  • \$\begingroup\$ Ok, I understand your point. Let's say the silicon had some doping where it would be able to conduct a fair amount of current but not as much as a metal like copper. How would electron interaction take place at the interface between the copper and doped silicon? Really, what I am trying to ask is that, there are a huge number of free electrons in the metal, why can't they just flow through the doped silicon as fast and in the same quantity as they were flowing in the metal? Why must the current be limited by the number of free electrons in the doped silicon? \$\endgroup\$
    – jrr76994
    Apr 5, 2021 at 22:28
  • \$\begingroup\$ My limited understanding is the conductor interface to a dielectric be it a linear insulator (capacitor) or non-linear (semi) is the determined by cross-sectional area and conductance of the interface material to become the ESR, bulk resistance but only partly the Rce and RdsOn...AFK back to painting... \$\endgroup\$ Apr 5, 2021 at 22:33
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Let's assume the silicon block starts suddenly to allow electrons which are trying to get out along the minus wire to flow along the block and sink to the +wire.

But how in the hell the electrons which just had got out of the minus wire into the Si-block would know where to rush? If the Si-block were a good conductor its own electrons would flow towards the +wire and that would leave a +charged path that the intruders would follow with no own effort. That's called electric influence.

But the electrons in the Si-block move nowhere, they are bound to their orbits in the crystal structure. Thus nobody makes strong enough electric field for the intruders, the field strength stays approximately the battery voltage divided by the width of the Si-block. The intruders accumulate near the intrusion point and soon their common field stops more electrons coming in. => No current.

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