I can not explain my question concisely as I am beginner and I have difficulties with chemical and mathematical terms, so I will discuss my logic from the beginning foolishly.
What I understand about potential difference
The basic electronic says a source voltage is the result of a potential difference between its poles.
With that I understand that a battery of 1.5V has a positive terminal with 1.5 volt charge and a negative terminal to 0 volts of charge.
The nature of the material excess electrons on the negative pole tend to go for the positive pole until they neutralize. When this happens the battery will have 0.75V at each pole and the potential difference will be 0v. Correct? I am not sure.
What I understand about diode
A diode is a semiconductor electronic component divided into two poles, one negative and one positive so with a potential difference between them; It is separated by a resistive layer between the poles.
When a diode is connected to battery directly (negative diode with negative source) etc, free electrons on the negative side of the battery repel the electrons free from the negative side of the diode to destroy the resistive layer and the current flow through the circuit.
What I can not understand
Let's say I have a source with 0 volts in its negative pole and a diode with -7 volts in its negative pole. Logically, the electrons on the negative side of the diode should go to the negative side of the source and thereby increase the depletion layer in the diode even when polarized directly to the source.
Another example … let's say I have a diode with upside potential of 5V connected to the positive side of the battery that has 20 volts potential. In this scenario, why the electrons of the diode will not meet positive side of the supply to the increasing depletion layer?
I think it got a little confusing but I hope someone understands me and can help me.
Best Answer
It is more practical to consider a voltage source as an element having electric potential energy, which is manifested through a potential difference between its terminals, and has the ability to move electric charge.
The Volt is the unit used to measure electric potential or potential difference. To measure amount of electric charge, the Coulomb is used.
The fact that a battery is 1.5 volts, means that the potential difference across its terminals is 1.5 volts, and is a measure of the battery capacity to do the work of moving electric charge, or that is, generate an electric current.
The diode is not divided into two poles, but it consists of semiconductor material N type and P type These materials are obtained from pure silicon, by adding impurities (doping), which provide electrons (N type) or generate holes (P type).
This device doesn't presents a potential difference between its terminals, ie, if we connect a voltmeter to the terminals, there will not measure voltage. The associated quantum effect is the Potential Barrier. Basically this is a distribution of the electric potential energy in the crystal lattice formed by silicon, which acts as a barrier that prevents the passage of electrical charges, whether these holes or electrons.
The connection that you describe, is called forward bias diode. Nothing in the internal diode is "destroyed". What happens is that the potential energy of the battery is greater than the energy implicit in the potential barrier, which, it is reduced, allowing the passage of electrical charges.
I guess you're talking about a 0 volt battery with the positive terminal and -7 volts on the negative terminal. In this case the only thing that interests us is that the difference in potential corresponding to this battery is 7 volts. Then in the application with the diode, the only thing we care about is whether it is forward bias or reverse bias.
Charges of equal polarity repel, so free in the N type materials electrons are repelled when connected to the lower potential pole. In this case, it is forward biased. Different polarity charges attract, so that the free electrons in the N-type material are attracted when connected to the pole of higher potential battery. In this case it is reverse biased.