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I know this question sounds silly, as if there was a potential difference a current would be created when the terminals are connected together and this would mean energy has come from somewhere.

The reason I ask this though is that from my understanding of the depletion region and built in potential of a diode it seems like if you connected a voltmeter across the whole diode it would show the value of the built in potential.

This is explained in the image below:

pn junction under equilibrium bias

At first, electrons flow from the n type to the p type because there are a higher concentration in the n type, and holes do vise versa. This is called the diffusion current. The first electrons and holes to cross the pn boundary are the ones which are closest to it; these carriers recombine when they meet each other and are then no longer a carrier. This means there is a depletion region of no carriers near the pn boundary. because electrons have left the n type material, and holes have left the p type material, there is a surplus of positive and negative charge on the n and p side of the pn boundary respectively. This causes an electric field that opposes the diffusion current, and so no more electrons or holes cross the boundary and combine. In short, only the electrons and holes near the boundary combine, because after they have done that an electric field is formed that prevents any more carriers from crossing. The current due to this electric field is called drift current, and when in equilibrium this will equal the diffusion current. Because there is an electric field at the boundary (pointing from the positive charge to the negative charge) there is an associated voltage. This is called the built in potential.

If you sample the electric field at each point along the diode from left to right, you would start with 0 in the p region because there are an equal number of protons and electrons. As you approach the depletion region you would see a small electric field pointing back towards the p region, caused by acceptor impurities which now have an extra electron (due to recombination) and therefore now have a net negative charge. This electric field would increase in strength as you get closer to the boundary, and then die away as you get further away.

This electric field means there is a voltage, as shown in graph (d). The p side is at an arbitrary potential, and the n side is at a potential higher than this because there is an electric field between them. This means there is a potential difference across the depletion region; this is known as the built-in potential.

But why when I connect a volt-meter across the whole diode will I not see this built in potential?

enter image description here

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I've found an answer on wikipedia but I do not understand it at all. After 3 years of studying EE and classes studying electromagnetism and maxwell's equations I thought I understood what voltage was. Turns out I don't :( –  Blue7 Apr 14 at 5:37
    
Heck, that's an intimidating Wiki page. I'm going to have to read it again in the morning :) If it makes you feel better, I've been an EE for a decade, and have a good physics background, but I didn't know this... –  bitsmack Apr 14 at 6:47
    
Because of the law "conservation of energy". Otherwise, we would had had an infinity energy source by simply placing billions of diodes on a silicon chip. –  hkBattousai Apr 14 at 9:05
2  
Consider that a voltmeter does not measure the electric field itself. Ask yourself, "Self, if it doesn't measure the electric field, what does a voltmeter actually measure, and why do we use it rather than a real electric field meter?" –  Adam Davis Apr 14 at 13:48
    
possible (actually an essential) duplicate of Internal difference in a diode –  Alfred Centauri Apr 14 at 14:13

3 Answers 3

up vote 7 down vote accepted

I think, the answer is relatively simple. Do you know the working principle of a "Schottky diode", which is based on a semiconductor-metal junction? Now - what happens if you connect a voltmeter (or any other load) across the diode? You create two Schottky junctions which exactly compensate the diffusion voltage inside the pn diode. Thus, no voltage can be measured. With other words: You cannot use the diffusion voltage to drive any current through an external load.

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It seems like the answers vary on different versions of this question, but I like this answer best. And I don't know the working principle of a schottky diode, can you please explain or link a simple explanation? What happens when you connect a p or n type material to a normal conductor? Another question, is the wikipedia link that I mentioned in the comments nothing to do with the answer? –  Blue7 Apr 14 at 17:53
    
AS I have mentioned, it is a metal-semiconductor junction. See wikipedia under "Schottky diode". –  LvW Apr 14 at 19:43
    
I've been reading about metal-semiconductor junctions, and I'm now getting a better understanding of why you can't measure the built in potential. Just to clarify though: Is there an electric field across a metal-semiconductor junction? –  Blue7 Apr 14 at 22:56

If you had an electrostatic voltmeter with a resistance much higher than your D.U.T. Series Resistance, which is possible, but the diode leakage would have to be equally high to prevent discharging the Static Potential.

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It is a very nice curiosity question! Same question came up to me when I was in my second year. But until I came across the Threshold voltages in Transistors and PN junction voltage drops, the picture became little clear.

You are absolutely right (last paragraph), because there is a change in potential due to electric field in the depletion region, there is higher potential from n-type side and negative potential from p-type side, making the intrinsic potential difference built up. That is why, to allow the current to flow through diode (PN junction) you would need higher potential from P-type and n-Type such that their difference is larger than the intrinsic potential difference which is in opposite direction to applied voltage across diode. This is what we call forward biased diode! I am sure you know this basics. Now lets go to the real question ->

If you were to probe your virtual Digital voltmeter exactly at the two depletion boundaries then I am sure you would see the voltage difference there, but its quite impossible to do with the regular multiplier. I am sure there are ways that semiconductor companies have special probes to sense these voltage differences. But if you were to measure the disconnected diode from your regular multimeter (same this is taken in consideration when you simulate it in LTSPICE that the probing is done at the ends of the diode not internally). Basically, your Graph (D) has this answer it self. Graph shows that both ends of diode have no electric field present. since the Electric field is conservative, and two diode ends (ends of P and N type materials) have no charge and electric fields at the ends are cancelled due to diffusion, as a results there is no electric field present at after the diffusion region ends, that means their difference is also 0 and measured voltage difference is 0 V as well. Hope this helps!

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