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Lest assume we have the setup presented in the image.enter image description here

What would be the potential difference (not voltage drop) between points A and B? Why?

What effect will the pn junction barrier potential have on the potential difference between A and B?

According to this book (page 79) the barrier potential can't be measured with a voltmeter (I will get 0V reading) but it exists :).

So I assumed that potential difference between A and B is equal with the barrier potential of the pn junction even if I can't measure it. Is this correct? So B will be at a lower potential than point A?

Some people said that there is no potential difference between points A and B, but on the other hand if I measure with a voltmeter the potential at point A relative to the "-" of the battery I will get a 5V reading, when I measure the potential at point B relative to the "-" of the battery I will get a reading of 4.8V. So I am a little confused :). I am a beginner and I am trying to correctly understand what is going on.

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  • \$\begingroup\$ There will be no drop, because there's no complete circuit and no way for current to flow through the diode. If you measure a different drop between A and B relative to the battery negative terminal, it is because your meter is completing the circuit. \$\endgroup\$
    – The Photon
    Commented Jun 4, 2012 at 16:16
  • \$\begingroup\$ What about potential difference? Not voltage drop. \$\endgroup\$ Commented Jun 4, 2012 at 16:18
  • \$\begingroup\$ "Voltage drop" is an informal way to say "potential difference". Maybe someone being very pedantic will point out some theoretical distinction between the two terms. But in normal usage they're the same thing. \$\endgroup\$
    – The Photon
    Commented Jun 4, 2012 at 16:29
  • \$\begingroup\$ Potential and voltage drop usually have the same meaning when used in electronics. When used to mean different things you look at the definition you are using and pply it. Here "barrier potential is the voltage that would be necessary IF YOU APPLIED IT to cause the jucntion to conduct, even if you did not actually apply it. If you step out a a 10th floor apartment window you would change "potential" by about 30 metres - even if you don't do it. If you stand on the ground and look at the window the "potential" (gravitational head) is 10 metres. ... \$\endgroup\$
    – Russell McMahon
    Commented Jun 4, 2012 at 16:30
  • \$\begingroup\$ ... A 1N4004 silicon diode at about a microa,p will probably have ana actual drop of 0.3V (just guessing). Take that to say 1 mA and it will be about 0.6V., Take that to 1A and it will be about 0.8V+ . Figure 4-10 of your book on page 96 seems to cover this well. Note the rising V at no I until a certain V is reached. This is not quite true and the graph scale hides reality but it's close enough for pracyice. \$\endgroup\$
    – Russell McMahon
    Commented Jun 4, 2012 at 16:34

3 Answers 3

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You won't be able to measure it with a voltmeter, but there will indeed be a potential difference. In a P-N junction with no external voltage applied negative charges from the N-doped region will migrate to the P-doped region, and vice versa. The blue region will be negative charged, the red one positive. The difference in charge causes a 0.7V across the barrier. (Will depend on the donor and receptor concentration.)

enter image description here

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  • \$\begingroup\$ By "You won't be able to measure it with a voltmeter, but there will indeed be a potential difference" you are referring to a potential difference between points A and B? So even if I can't measure it there is a potential difference of approximately 0.2-0.7V (depending on the value of the barrier potential) between A and B. Right? \$\endgroup\$ Commented Jun 4, 2012 at 18:13
  • \$\begingroup\$ It is electrometry. The charged area of material is possible to detect without causing a current. The electrometer probe should be very tiny and sense with no contact from some distance. I beleive it is possible. \$\endgroup\$
    – user924
    Commented Jun 5, 2012 at 1:22
  • \$\begingroup\$ @stevenvh Did I understand correctly your answer? You are referring to a potential difference between point A and B; right? Thank you. \$\endgroup\$ Commented Jun 5, 2012 at 9:39
  • \$\begingroup\$ @Buzai - Yes. Look at the bottom graph. The potential difference is created across the junction, but doesn't change further on, either left or right, including the wires. \$\endgroup\$
    – stevenvh
    Commented Jun 5, 2012 at 9:43
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A PN junction creates a local potential field. The high concentration of majority carriers (holes in P, electrons in N) diffuse accross the junction because the concentration is much lower on the other side. However, that displacement of the carriers causes a electric field. A equillibrium is reached where the electric field pushes the carriers back to where they came from exactly opposite to the diffusion pressure. This has the side effect of clearing the junction region of minority carriers, which make the diode "off".

If a external E field is applied to counter the diffusion, then the carriers move back to their respsective sides, allowing the minority carriers to return, causing the junction to conduct again. This is how a diode turns "on".

Anyway, you can't measure the voltage caused by the local E field in the junction using a closed circuit because you don't have a way of doing so without additional carrier gradients that then make their own reverse E field. The net diffusion-caused E field will be zero in a loop.

However, it seems to me that it could be possible to measure the PN junction static E field another way. Suppose it is used in a vaccuum tube to deflect a electron beam. I haven't tried that, but I think it should work.

In any case, you can't get power from this without putting some in somehow. That is exactly how solar cells work. A solar cell is a big diode with the junction layer close enough to the surface so that ordinary light photons can penetrate it. A photon hitting the junction in the right place causes a loose electron, which is then accellerated by the field and can show up between the ends of the device as a current.

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  • \$\begingroup\$ So the depletion can exist at junction of doped and pure silicon as well ? what will be such device ? What will it do ? \$\endgroup\$
    – user924
    Commented Jun 5, 2012 at 1:18
  • \$\begingroup\$ @Rocket: Undoped silicon is not a good conductor (has high bulk resistivity), so not much. A more interesting case is P or N silicon joined with a regular conductor. That's exactly what a Schottky diode is. \$\endgroup\$ Commented Jun 5, 2012 at 12:46
  • \$\begingroup\$ Thank you :). I am not trying to get power out of a diode. Just trying to understand things better :) \$\endgroup\$ Commented Jun 5, 2012 at 15:36
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... when I measure the potential at point B relative to the "-" of the battery I will get a reading of 4.8V.

This example allows me to bring up a good point that some people don't realize.

By connecting your voltmeter between point B and the battery return, you are now creating a closed circuit. The input impedance of the voltmeter will be very high, but it isn't going to be as high as free space. High-end meters may be in the gigohm range - cheaper one, perhaps megohms.

It is possible that your meter is allowing sufficient current flow to create the small voltage drop that you're observing. Real diodes aren't like perfect switches, and will minimally conduct before hitting the knee and allowing high currents to flow. (The Shockley diode equation describes this behaviour.)

It's always important to remember that any measurement tool (meter, scope probe, etc.) may impact the circuit that you're measuring, especially high-impedance / floating circuits like the one in your example.

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