# pn junction diode

Discussion in 'Electronic Design' started by Lax, Jan 10, 2008.

1. ### LaxGuest

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?

2. ### D from BCGuest

I sometimes just think of diode as a wacky current dependant resistor.

D from BC
British Columbia

3. ### Guest

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.

4. ### whit3rdGuest

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).

5. ### John PopelishGuest

whit3rd wrote:
(snip)
(snip)

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

6. ### whit3rdGuest

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.

7. ### John PopelishGuest

thermocouples work.

8. ### Tim WilliamsGuest

http://en.wikipedia.org/wiki/Thermoelectric_effect

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

9. ### John PopelishGuest

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.

10. ### whit3rdGuest

Geez, there must be a lot of engineer-types on this
blog. All the discussion is about externals, just

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.

11. ### John PopelishGuest

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.

12. ### whit3rdGuest

...   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.

13. ### neon

1,325
0
Oct 21, 2006
see boltzman constant for diodes drop. the forward voltage drop of any diodes is never constant varies with current and tenperature, it is an exponantional inpedance. it is assumed a drop of .6 to ,7v fully conducting but it could be much less it depends on the current flowing,

14. ### Phil HobbsGuest

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