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Barrier potential discrepancy

Kevin Weddle

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I thought I would give this new section a try. What about the diodes barrier potential. We all know that it measured with a meter and the polarity is that of the battery. I have read, though, that since the free electrons move from the N type to the P type, a negative potential appears in the P type while a positive potential appears in the the N type. This means we now have a potential that aids the battery while we measure a potential that opposes the battery. Any suggestions? I just take this for granted becuase I believe this is what happens right at the barrier point and that as you leave the barrier, there is a different polarity.

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Hi Kevin

In a pn junction in equilibrum, where p and n regions "touch", majority carriers from each region recombine with majority carriers from the other region. That is: free electrons from the n region recombine with holes in the p region and holes from the p region recombine with free electrons in the n region. A depletion region is thus created. In the depletion region there are no "mobile" carriers. The diffused holes leave behind uncovered fixed negative charges and similarly there are fixed positive charges left behind by the diffused electons. In this way, a separation of charges occurs, causing an electric field (- is in the p region and + in the n region). This electic field is called built-in potential and has the opposite polarity of the battery when you forward bias the junction. It is given by:
Vbi=VTln[(NdNa)/ni2 ]
Vbi is the built-in voltage
VT is the thermal voltage for Si (26mV/deg C)
Nd and Na is the concentration of doping atoms in the n and p region respectively
ni is the intrinsic carrier concentration in Si

This potential is constant, has a value of 0,7V approximately for Si diodes and normal doping (depends on doping and temperature), and you have to overcome it in order to have a current flow through the junction. This is practically what you measure and call forward voltage drop. When you bias the diode you do have an electron current and practically you create an electric field opposite to the built-in voltage. When the electric field you apply equals or exceeds the built-in voltage, your diode conducts.

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Just as a side note there is current before .7v is reached. The current is created by the movement of electrons as they are reducing the depletion region. I say reducing the depeletion region because when there is no voltage applied to the junction, the depletion region is wide. When you reverse the junction, the negative end of the battery repels electrons against the junction while the positve attracts electrons away from the junction. In order to get current, the electrons in the N type must cross the junction and combine with holes in the P type.

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Sure there is current before 0,7V! As you see in the equation the built-in voltage depends on doping (and temperature) so you can play with it to some extent by controlling the doping. Not to mention that electron-hole recombination is a statistic phenomenon and therefore when you say 0,7V it does't mean that there is no current at all at lower voltages.
As for the depletion region, it can not be significantly reduced or disappear. It exists there because you put two differently (p and n type) doped semoconductors together. And yes, it can be extended when you reverse bias the junction

You can take a look here to find a well written and more descriptive approach:

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.7v for silicon and .3v for germanium and you subtract this from the source.

If you want to be more exact in your answers, you can compensate for bulk resistance(which is the resistances of the p and n regions together) which is usually around .23ohms for si, but is usally negligible unless your striving for accuracy or depending on what your trying to accomplish.

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