pn junction diode

Say we have a silcon pn junction (diode) - i.e., a block of p-type on left, attached to a block of n-type semiconductor on right:

anode ------[ p | n ]------ cathode

Because of diffusion we get a barrier potention at the junction, which makes the n-side/cathode 0.7V higher than the p-side/anode:

- 0.7V + anode ------[ p | n ]------ cathode (*)

1) Now is there any way to "measure" this potential difference right from the diode using some instrument?

Now to get ride of the depletion layer (barrier) we need an "opposite" external voltage equal in magnitude to the 0.7V shown in (*) :

(barrier potential) - 0.7V + anode ------[ p | n ]------ cathode

  • 0.7V - (external voltage)

2) Now, why isn't the resulting voltage of the diode 0V (sum of barrier and external)? How come we only measure the external 0.7V using a voltmeter when the diode is forward biased?

3) Is there anything wrong with the thought process I've outlined above?
Reply to
Lax
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I sometimes just think of diode as a wacky current dependant resistor.

D from BC British Columbia Canada.

Reply to
D from BC

My 2 pesos: A diode is not a battery. The diode obeys i=isubs*exp(v/vt). Say you curve trace a diode. Well, you view it linearly, so looking on the curve tracer, you say the potential is 0.7V, but really that is only true at some current level. if you were the type to work the diode harder, you might say a diode drop is 0.75 V,

Think of the so-called junk buffer circuit. [PNP emitter follower followed by NPN emitter follower.] It doesn't have zero offset, yet it is a better example of attempting cancellation than the one you suggested.

Reply to
miso

Yes, but it can't just be ANY diode. It needs to be open to light. There is a threshold effect in photoelectric output that scales with the built-in potential.

Interestingly, ANY two metals take on a potential difference on contact, and that's why thermocouples work. But to measure such potentials inside a circuit one needs to control ALL the circuit's materials and temperatures. Doing the measurement by photoelectric emission is possible, there have been experiments with vacuum lathes (because a fresh metal surface is required, not a dirty exposed-to-air crust).

Reply to
whit3rd

whit3rd wrote: (snip)

(snip)

You might want to look into this belief a bit, since it is completely wrong.

--
Regards,

John Popelish
Reply to
John Popelish

No, not wrong, just a bit perverse. The little electrons redistribute quickly into charged layers after contact, but at the moment of contact, there IS a field at the interface.

Reply to
whit3rd

Have it your way.

Since you brought it up, please tell me more about how thermocouples work.

--
Regards,

John Popelish
Reply to
John Popelish

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Contact potentials are well known in condensed state physics. They can be quite high and have essentially unlimited current, but the naughty thing is getting that voltage out without cancelling it with another junction or series of junctions. A difference in temperature generates a low order difference, which is how thermocouples work.

Tim

-- Deep Fryer: A very philosophical monk. Website @

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Reply to
Tim Williams

Show me the phrase "contact potential" or "low order difference" in that article.

Here is the sentence that tells what makes thermocouples work, "The thermopower, or thermoelectric power, or Seebeck coefficient of a material is a measure of the magnitude of an induced thermoelectric voltage in response to a temperature difference across that material." Thermocouples are made of two materials that produce different induced thermoelectric voltages in response to the same temperature difference across those different materials.

The thermocouple output is the difference of those two thermal gradient induced potentials. It has nothing to do with anything produced at their contact surface. They don't even have to contact each other, They can be connected through any number of seriesed contact surfaces with intermediate conductive materials, as long as all those contact surfaces are at the same temperature. The contact surfaces produces no voltage.

--
Regards,

John Popelish
Reply to
John Popelish

Geez, there must be a lot of engineer-types on this blog. All the discussion is about externals, just phenomenology, not about the innards.

The density of allowed states in two dissimilar metals causes, on contact, a diffusion of electrons from the material with the lower density of allowed states, into the other (and in the case of PN junctions, two separate bands, the holes and electrons, BOTH undergo such diffusion). The charge transfer, of course, builds up a charge layer and causes an electric field (which eventually leads to zero net diffusion current).

The amount of that field minimizes the Helmholtz free energy (see Wikipedia - it's got a temperature term, and entropy which relates it directly to the density-of-states effect). The temperature dependence of the field, and the voltage drop that measures the field, makes it useful for thermometry.

All the discussion has been on Seebeck effect, tabulated thermocouple voltages, etc. That isn't an explanation at all, just a description of the apparatus. And, no simple apparatus is gonna tell you about the thermocouple voltage of a SINGLE junction of two metals. All your meters work only on complete circuits (which will have TWO junctions or more).

The main way this relates to electronic design is in the microvolt range, at DC. No feasible amplifier technology will ever give you zero errors at low frequency, because neither the factory calibration fixture nor the normal circuit environment is so uniform in temperature or controlled in wiring composition that the thermocouple voltages won't randomize (or worse, couple) to make input 'offset error', usually in the few-microvolts range. The best you can do, is to make the circuits very small, with minimum power dissipation, and wrap thermal blankets around 'em.

Reply to
whit3rd

With a Usenet title of "sci.electronics.design", ya think?

Keep digging. That hole isn't quite deep enough. We can still see you.

Provably not true by simple experiment with a thermocouple and a millivolt meter. The only point in having the second thermocouple is to remove the temperature dependence of the meter connections on that measured voltage.

Damn, you can dig two separate holes, simultaneously.

At this rate, you could disappear from view, permanently, in just a few days, in two holes, at the same time. Of course, you could go back and actually learn about how thermocouples work and fill these holes back in. But I am not betting that will happen any time soon. Shame, really.

--
Regards,

John Popelish
Reply to
John Popelish

=2E.. =A0 And, no simple apparatus

When I think of this experiment, I see metal A, joined to metal B, with copper leads from A to a meter movement, to B. So there is an A-B thermocouple, a B-copper couple, and a copper-A couple. At thermal equilibrium, A, B, and the copper all take on different potentials, but no current flows in the loop. Current only flows if the couples aren't at thermal equilibrium.

Reply to
whit3rd

This isn't a blog, it's a USENET NEWSGROUP that was created for engineers to discus design and theory. You are accessing it through the half assed 'Google Groups' HTML interface.

If it's over your head try: news:sci.electronics.basics or

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for all lowly google group users.

--
Service to my country? Been there, Done that, and I\'ve got my DD214 to
prove it.
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Reply to
Michael A. Terrell

Let me come to the guy's defense a bit...the physics of the Seebeck effect really is more or less the same as that of the depletion region in semiconductors--a balance between diffusion and drift.

Consider two pieces of metal, e.g. calcium (work function of about 2 eV) and platinum (work function about 6 eV), both of which are electrically neutral (same number of electrons and protons). Electrostatics is conservative force, i.e. in a static field, if you take an electron around a closed loop, its energy doesn't change. If you rip an electron out of the calcium and take it out to a large distance, it'll cost you 2 eV. If you now let it fall onto the platinum, it'll release 6 eV. (A fine point: the energy released is actually the electron affinity of the material, not the work function, but the two are essentially identical for partly-filled bands, i.e. conduction electrons in metals.)

Sooo, when you touch the Ca to the Pt, electrons start flowing across the boundary until the two Fermi levels line up. But that results in a big boundary layer of free charge inside the Pt, enough so that the voltage drop across the charge layer equals the difference in the Fermi levels of the metals. (This is not quite the same as the difference in the work functions because the electrostatic image potential, which is part of the work function, depends on the surface charge density of the metal.)

In a metal, these layers are very thin because the density of free charge is so high, which makes the Debye shielding length very short.

The Seebeck effect arises from the different temperature coefficients of the two processes, diffusion and drift. As you increase the temperature, more electrons drift into the Pt, which changes the potential.

If you have two junctions back to back, the common mode temperature cancels out by symmetry, but in the presence of a temperature difference, the cancellation isn't complete, and a thermocouple voltage can be measured via an external circuit.

The chemical potential and free energy business is just a physicist's or chemist's way of describing the result--there's a variational principle in thermodynamics that systems seek the state of minimum free energy (This is either the Helmholtz or Gibbs free energy, depending on whether you're keeping the number of particles constant or not.) It's much easier to calculate that way than having to keep the kinetics in view all the time.

Cheers,

Phil Hobbs

Reply to
Phil Hobbs

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