# Measuring the built-in potential of a pn junction

• I
• Mayan Fung
In summary, the reason why we cannot probe the built-in potential across a diode with a voltmeter is because the Fermi level of the n-type and p-type semiconductor is aligned, resulting in a voltage reading of 0. This is due to the formation of Schottky barriers at the two semiconductor-conductor junctions, which cancel out the built-in potential. However, if the diode is not connected in a circuit, it is not possible to measure a voltage difference. This is because a voltmeter requires energy to operate, and without a power source, there is no energy available to drive the meter. Additionally, the basic equation of a diode states that when no voltage is applied, no current will flow,
Mayan Fung
TL;DR Summary
Can we measure the built-in potential of a diode with a voltmeter?
I am thinking about the reason why we cannot probe the built-in potential across a diode with a voltmeter. Obviously, a diode is not an energy source, so it is impossible for it to show a voltage reading. After doing some research, I found some explanations and some questions about them.

1. The Fermi level splitting is what a voltmeter is measuring. The Fermi level of the n-type and the p-type semiconductor is aligned. Therefore, the voltage reading will be 0 across a diode.
2. When the diode is connected to the voltmeter, there are three junctions in total. Two semiconductor-conductor junctions at the two ends and the pn junction. The two semiconductor-conductor junctions form Schottky barriers which cancel the built-in potential exactly. Therefore, the voltage reading is 0.

I think that the two explanations are equivalent in the sense that the formation of the Schottky barriers is indeed the alignment of the Fermi level between the conductor and the semiconductor. However, if I separate the n-type and p-type semiconductors so that no pn junction forms, does it mean that I can measure a voltage difference if I connect them to the voltmeter?

Wait, we can measure the potential across a diode. Why do you think that you cannot? Am I misunderstanding what you are saying?

Delta2
Real voltmeters draw a current and dissipate energy. Because the OP mentioned that a diode is not an energy source, he concluded that a voltmeter must read zero.

Dale said:
Wait, we can measure the potential across a diode. Why do you think that you cannot? Am I misunderstanding what you are saying?
I mean I thought we can measure the built-in potential by directly connecting the diode with a voltmeter

Dale said:
Wait, we can measure the potential across a diode. Why do you think that you cannot? Am I misunderstanding what you are saying?
I think he means a diode which is not connected in any circuit. Just the diode and the voltmeter.

Delta2
nasu said:
Just the diode and the voltmeter.
Unless it is an electrometer, the voltmeter will consume some energy. Where will the energy come from to drive the meter? Maybe from the temperature of the diode, or from incident light?

nasu said:
I think he means a diode which is not connected in any circuit. Just the diode and the voltmeter.
Oh, that would be weird. Yes, I had assumed that the diode would be in a circuit with some sort of power source, so I didn't see why a voltage couldn't be measured.

Baluncore said:
Unless it is an electrometer, the voltmeter will consume some energy. Where will the energy come from to drive the meter? Maybe from the temperature of the diode, or from incident light?
I did not comment on the OP's idea, just tring to guess what it was. I did not say that you will measure something in these conditions.

Yes, I also found the idea of a diode, which is a passive component, showing a voltage reading strange. However, normal diagrams discussing the pn junctions usually shows a potential difference, just like the one here. It seems that from these diagrams, it is reasonable to measure the potential difference.

Sure, and you can measure that difference when it is part of a circuit where it is forward or reverse biased.

This is probably because according to the basic equation of a diode, when the voltage applied from the outside is zero, no current flows. So no matter how big the internal resistance of the instrument is, no voltage will be generated due to the flow of current, so the voltage measured by the instrument is zero.

https://en.wikipedia.org/wiki/Shockley_diode_equation

## 1. How is the built-in potential of a pn junction measured?

The built-in potential of a pn junction can be measured using a variety of methods, including capacitance-voltage (CV) measurements, current-voltage (IV) measurements, and photovoltaic measurements. These methods involve applying a voltage or light source to the pn junction and measuring the resulting capacitance, current, or voltage.

## 2. What factors affect the built-in potential of a pn junction?

The built-in potential of a pn junction is affected by the doping concentrations and types of the p and n regions, the temperature, and the material properties of the semiconductor. Additionally, any external bias or illumination can also impact the built-in potential.

## 3. Why is it important to measure the built-in potential of a pn junction?

The built-in potential of a pn junction is a key parameter that affects the behavior and performance of electronic devices such as diodes, transistors, and solar cells. By accurately measuring the built-in potential, engineers and scientists can better understand and optimize the functionality and efficiency of these devices.

## 4. What is the significance of the built-in potential in solar cells?

The built-in potential plays a crucial role in the operation of solar cells. It creates a barrier that separates the p and n regions, allowing for the efficient separation of photo-generated carriers and the generation of a current. The magnitude of the built-in potential also affects the open-circuit voltage and overall efficiency of the solar cell.

## 5. How does the built-in potential change with temperature?

The built-in potential of a pn junction decreases with increasing temperature. This is due to the increase in carrier concentration and mobility in the semiconductor material. As a result, the barrier height decreases, and the built-in potential decreases. This phenomenon is important to consider in the design and operation of electronic devices, particularly in high-temperature environments.

• Electrical Engineering
Replies
1
Views
902
• Electrical Engineering
Replies
12
Views
1K
• Electrical Engineering
Replies
14
Views
2K
• Other Physics Topics
Replies
2
Views
2K
• Quantum Physics
Replies
6
Views
1K
• Atomic and Condensed Matter
Replies
1
Views
1K
Replies
7
Views
2K
• Atomic and Condensed Matter
Replies
7
Views
760
• Introductory Physics Homework Help
Replies
18
Views
2K
• Chemistry
Replies
6
Views
2K