# Built-in potential in pn junction

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• EmilyRuck
In summary, the energy of an electron in a solid structure is always negative, including the E_c and E_v levels in the band diagram of a pn junction. When the junction is built and thermal equilibrium is reached, the depletion region creates an electric field and a potential barrier. The potential barrier, represented as \phi_i, is calculated using a positive value for q, but from the point of view of an electron, the p-region is actually at a higher potential. The vacuum level, which represents the energy of a free electron outside the crystal, is not depicted in the band diagram of a pn junction but is affected by the built-in voltage.
EmilyRuck
Hello!
The (potential) energy of an electron in a solid structure is always negative; also the $E_c$ and $E_v$ levels (conduction band and valence band limits) are negative, in the band diagram of a pn junction.
When the junction is built and thermal equilibrium is reached, the depletion region creates an electric field and a potential barrier $\phi_i$.
1) First of all,

$\phi_i = \displaystyle \frac{kT}{q} \log \left( \displaystyle \frac{N_D N_A}{n_i^2} \right)$

where $k$ is Boltzmann's constant, $T$ is the absolute temperature and $q$ is the absolute value of the electron charge. The energy step due to this built-in potential is $q \phi_i$, but why a positive $q$ is used to calculate this step? Given a voltage difference $\Delta V = \phi_i$, the corresponding potential energy difference should be $\Delta U = q \Delta V = - |q| \Delta V$ in the case of an electron.
2) Let's suppose anyway that $q \phi_i > 0$.
In the conduction band of the n-region, just the charge carriers with an energy $E = E_c + q \phi_i$ can cross the junction and diffuse to the p-region. So, I would say that the p-region, with respect to the n-region, is at a higher potential.
In order to have a reference image, let's consider this document, page 11. Why in the last plot, which shows $V$ as function of position, the n-side is at a higher potential and the p-side at $V = 0$? (Consider $V_A = 0$, that is the case of thermal equilibrium)
3) What about the vacuum level? It is never depicted in the band diagram of a pn junction; but how is it affected by the built-in voltage?
Thank you anyway for having read.

Emily

From the point of view of electron what you say is right p-side is at higher potential. But it is a convention to consider from the point of view of a positive test charge. What do you mean by vacuum level you mean at absolute zero. At absolute zero all the bonds will be in tact and there will not be any free charge to create potential difference or depletion region.

EmilyRuck
Let'sthink said:
From the point of view of electron what you say is right p-side is at higher potential. But it is a convention to consider from the point of view of a positive test charge.

Ok, thank you.

Let'sthink said:
What do you mean by vacuum level you mean at absolute zero. At absolute zero all the bonds will be in tact and there will not be any free charge to create potential difference or depletion region.

By "vacuum level" I mean the energy of a free electron outside the crystal, as stated http://ecee.colorado.edu/~bart/book/book/chapter2/ch2_3.htm, par. 2.3.3.2. Anyway, maybe the vacuum level is bent like the energy bands across the junction due to the electric field.

## What is built-in potential in pn junction?

Built-in potential in pn junction refers to the electric potential difference that exists between the two sides of a pn junction when there is no external voltage applied. It is caused by the diffusion of majority charge carriers from one side to the other, resulting in a depletion region where there is a lack of charge carriers.

## How is built-in potential in pn junction created?

Built-in potential in pn junction is created during the fabrication process of a pn junction. When a p-type semiconductor and an n-type semiconductor are brought into contact, electrons from the n-type material diffuse into the p-type material and holes from the p-type material diffuse into the n-type material. This creates a depletion region where there is a lack of majority charge carriers and results in the formation of a built-in potential.

## What is the significance of built-in potential in pn junction?

The built-in potential in pn junction plays a crucial role in the functioning of semiconductor devices. It determines the direction and magnitude of current flow when a bias voltage is applied to the pn junction. It also affects the barrier height and the width of the depletion region, which in turn affects the device's electrical characteristics such as voltage breakdown and capacitance.

## How is built-in potential in pn junction affected by temperature?

The built-in potential in pn junction is affected by temperature. As temperature increases, so does the diffusion of charge carriers, causing the depletion region to shrink. This results in a decrease in the built-in potential. On the other hand, at lower temperatures, the depletion region widens, increasing the built-in potential.

## Can the built-in potential in pn junction be modified?

Yes, the built-in potential in pn junction can be modified by using a process called doping. By introducing impurities into the semiconductor material, the concentration of majority charge carriers can be altered, thus changing the width and magnitude of the built-in potential. This is commonly used in the fabrication of semiconductor devices to control their electrical characteristics.

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