Measuring the built-in potential of a pn junction

[Mayan Fung]

TL;DR

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?

 

[Dale]

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

 

[anorlunda]

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.

 

[Mayan Fung]

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

 

[nasu]

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.

 

[Baluncore]

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?

 

[Dale]

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.

 

[nasu]

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.

 

[Mayan Fung]

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.

 

[Dale]

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

 

[alan123hk]

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