Electric field throughout a semiconductor

Question

Suppose I have a semiconductor bar of length 2mm and I am applying a potential difference of 2V across it. Will the magnitude of electric field be same throughout the bar OR will it be different at different lengths(example different for x=0.5mm and x=1mm) ?

Answer 1

In a bar of ideal semiconductor with one side grounded and the other at some nonzero voltage you will have a constant electric field through the bar.

Another answer addresses the band structure of semicondcutors, which seems unnecessary to answer this question, but since it was brought up I will comment on it as well. The electric field present in a semiconductor is proportional to the derivative of the conduction and valence band edges. This is guaranteed to be a constant when there is no net charge. If there is net charge the field will change across the semiconductor and your bands will bend. In your example of an ideal semiconductor there is no net charge anywhere within the semiconductor and therefore the electric field must be constant.

In a practical block of semiconductor you will have to apply your bias somehow, and that is usually through a metal-semiconductor contact. These introduce some diode behavior and will change the field a bit.

Answer 2

Will the magnitude of electric field be same throughout the bar

No!

When we think about a semiconductor as semiconductor, we consider the electric field as superposition of two: the external field plus the periodic potential given through the fact that it's a regular pattern of atoms.

I don't know at which stage of your studies you are. So, the following might be something that you only learn in your second or third year of EE:

When you're looking at what makes semiconductors be semiconductors, you typically start modelling things a bit further down, and come up with a lattice of atoms. In the case that you don't apply an external voltage, that leads, with the cores being regularly placed, to a periodic structure in electric potential.

If you apply an external electric field (by applying a voltage), you indeed get a linear ramp, but that is superimposed with the periodic structure described above.

Then you realize (Ok, physicists at the end of the 19th century realized for you) that electrons are just particles that are observable "consequences" of a standing wave. Waves are described by the Schrödinger equation, which has a couple of free parameters; especially the electric potential.

You then look for a solution that fits to your periodic potential, because that's given by the position of atoms in your semiconductor lattice.

These solutions thus give us the state in which our atoms can be in. We call these collection of states "electronic bands". And if your lattice allows for two bands at the same position, but separated by some voltage, then electrons can only be in either of these bands, and need energy to cross the band gap. By superimposing an external electrical field, we can "slant" the whole band gap diagram and allow for electrons to flow from one band to the other. That's where the semiconducting behaviour comes from!

---

If this sounds nothing like you've learned so far, don't worry – it's on the curriculum of every EE program I've seen so far, and they'll be preparing you with the math you'll need in time; that's why you need to study things like Fourier transforms, a couple solutions to differential equations, a couple of things about probability densities and so on.

Answer 3

No, it depends on resistivity & electrical conductivity. Generally, magnitude of the electric field generated by a point charge with a magnitude charge.

And, about semiconductor bar, it depends on electric field.