Why can't diodes with built-in potential be used as batteries?

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Diodes like the 1N4001 and zener diodes have built-in potential due to thermal energy, but this potential opposes current flow rather than enabling it. When connected to a circuit, the built-in potential does not provide sufficient energy to light an LED, as it results in zero electromotive force around the circuit. The discussion clarifies that while there is a potential difference between the p and n sides of a diode, it is not usable for generating current in a closed circuit. Instead, energy generation requires overcoming this built-in potential, which is not feasible with standard diodes. For energy extraction, alternative methods like thermoelectric generators are suggested.
yinx
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Hello,

i was wondering for 1N4001 diode or zener diode has some built in potential, why can't they be used as batteries? I have some diodes lying around and I tried connecting to an LED, but it doesn't light up.

I know this question seem dumb, any help will be appreciated!

thanks!
Yinx
 
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yinx said:
Hello,

i was wondering for 1N4001 diode or zener diode has some built in potential, ...

Why do you believe they have "built in potential"?
 
pantaz said:
Why do you believe they have "built in potential"?

doesn't diodes have built in potential (aka contact potential)??

yinx
 
The built in potential of a diode is a result of thermal energy. At a temp above absolute zero, 0K, the silicon crystal lattice vibrates. These vibrations have energy which separates electrons from the parent atoms leaving holes behind. This is thermal generation of electron-hole pairs, ehp.

The electrons, e-, & holes, h+, move through the Si & cross the junction. On the n-side of the junction, holes accumulate & recombine w/ electrons, vice versa on the p side. This accumulation of charges results in a local E field near the junction, & the integral of E over the distance is the potential.

In order to forward bias the diode, i.e. establish forward current, the built in potential must be considered. It has a polarity which tends to oppose current flow. The source powering the network must supply a potential large enough to overcome this potential.

A good semiconductor physics text covers this in detail. It has diagrams which make it clear. That should be where you look for your answers. If you need clarification on what I've said, that would be fine.

Claude
 
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The term built-in potential is frequently used in electronics and isn't contact potential.
Even though there is such a potential difference between the P and the N side, when we close a circuit no current flows. We feel disappointed but Thermodynamics feels very happy (if it were a person) because we haven't been able to get energy out of nothing.
When we consider a closed circuit we realize the built-in potential isn't the only potential difference to account for. When you do it properly, you end up with zero electro motive force around the circuit and no current flows.
However, we can build a thermal engine. Simply heat a thermocouple with a flame and close the circuit with another thermocouple submerged in ice water. You'll notice current flows and you get work as in any thermal engine.
 

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