The discussion revolves around the built-in voltage of a pn junction diode and why it cannot be measured with a voltmeter when the diode is isolated. Participants explore the nature of this voltage, the role of current flow, and the concept of contact potentials at the junctions.
Participants express varying viewpoints on the nature of the built-in voltage and its measurement, with no consensus reached on the explanations provided. The discussion remains unresolved regarding the nuances of contact potentials and their implications.
There are limitations in understanding the relationship between built-in voltage, current flow, and contact potentials, as well as the definitions of terms used in the discussion. Some assumptions about the behavior of the diode in isolation are not fully explored.
Why does it not act like a voltage source? There are +ve and -ve charges present. Isn't it like an electric dipole?anorlunda said:Because the voltage is there only when a current flows. The voltmeter won't cause a significant current flow.
cnh1995 said:Why does it not act like a voltage source? There are +ve and -ve charges present. Isn't it like an electric dipole?
That's what I read somewhere but didn't understand fully. Could you please elaborate? What exactly is the contact potential?marcusl said:A more precise answer is because the contact potentials at the p-metal and n-metal junctions exactly counteract the built-in potential.
Did you hear already about a Schottky diode? This a junction between metal and semiconductor - and exactly such a junction exists (necessarily) caused by the contact material at both ends of the pn diode. And the diffusion voltages at these junctions compensate the diffusion voltage across the internal pn junction.cnh1995 said:That's what I read somewhere and didn't understand fully. Could you please elaborate?