0
\$\begingroup\$

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.

\$\endgroup\$
3
\$\begingroup\$

What I understand about potential difference

The basic electronic says a source voltage is the result of a potential difference between its poles.

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.

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 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.

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.

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.

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.

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.

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.

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.

\$\endgroup\$
  • \$\begingroup\$ great explanation. My brain still processing. Well, what if the negative pole of battery is +5v (considering its positive is +10v consequently generating an effective potential of 5v in this source) and the N pole of diode is connected to it forward? Shouldn't the negative side of battery(which is actually positive) attract the free eletrons from N pole of the diode? \$\endgroup\$ – ropbla9 Nov 16 '15 at 4:38
  • \$\begingroup\$ even in forward connection the negative pole of diode has different potentials and what I can't understand is why they don't attract themselves? \$\endgroup\$ – ropbla9 Nov 16 '15 at 5:12
  • \$\begingroup\$ let me rephrase the question. I already know what interests the movement of charges is just the difference of potential, but the potential of POLES exists too! So, assuming that the N side of the diode hypothetically would have a negligible negative potential and the negative battery side had -10 volts potential, I think there should be potential difference between the diode and the negative pole of the battery and thus obtain one reverse effect even if its forward connected. \$\endgroup\$ – ropbla9 Nov 16 '15 at 6:57
  • \$\begingroup\$ This really makes no sense to me. I think I'm going to die with a knot on my head, haha \$\endgroup\$ – ropbla9 Nov 16 '15 at 6:58
  • 1
    \$\begingroup\$ @ropbla9: The assumptions in your example don't make sense. We talk about voltages at certain points in a circuit as shorthand when it's clear what the reference is, but there's no such thing as 'potential of poles' in absolution; this is like measuring the length of a stick by only looking at one end. Likewise, saying the negative side of your 5 V battery is at +5 V is arbitrary and has no physical meaning. You also can't have a voltage difference over a direct connection. Connecting one diode terminal to a battery terminal means both are at the same potential, whatever you like to call it. \$\endgroup\$ – Marcks Thomas Nov 16 '15 at 11:58
1
\$\begingroup\$

A diode has no practical potential. If you put a volt meter across it when it's out of circuit there won't be a potential.

A diode is made out of a P-N junction. One side has free electrons (cathode) the other has free holes (anode). In the middle is a neutrally charged layer. The holes stole some of the electrons. That neutral layer keeps the rest of the electrons from filling the rest of the holes.

A diode has two modes of operation, forward biased and reverse biased.

Reverse biased

When a positive potential is applied to the cathode and a negative to the anode, the holes and electrons move away from each other attracted by opposite potential. No current flows except for a small leakage current.

Forward biased

When a low (0.1v) positive potential is applied to the anode and a negative to the cathode, the neutral layer starts to grow. As the voltage is increased there comes a point where the neutral layer can no longer separate the two sides and current begins to conduct. The exact voltage depends on the type of diode. Silicon diodes conduct at around 0.7v. Schottky diodes conduct around 0.2v.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.