- #1
Mayan Fung
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- 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?
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?