On Nov 29, 5:36*am, William Frensley <[Only registered users see links. ]> wrote:

Thanks, but that cannot be correct because the diode zero bias
resistance, Ro, significantly decreases with an increase in dopant
density. Please try the equations and you will see that the predicted
zero bias resistance does not change with an increase in dopant
density. Obviously, because there's no dopant density in the
equations. Something's wrong.

The aforementioned equation determines the current relative to an
applied voltage. The current *must* increase with an increase in
dopant density because it is well known that an increase in dopany
density decreases Ro.

I take it that you have experimental data demonstrating this?

If so, that is an indication that you are not measuring thermionic-
emission current. The diffusion-current model has a weak
(square-root) dependence on the doping level. That model has not
seen the light of day for many years. You will probably have to
go back to Crowell and Sze, Solid-State Electronics vol. 9, pp. 1035-48
(1966) to get the detailed information.

Before doing so, do yourself a favor and make sure you know what your device
is actually doing. Get a high-quality electrometer, that can measure
nanoamps. Measure the I(V) curve over the range of zero to a few hundred
millivolts (up into the current range of some microamps). Try plotting
the data on both linear and semilog scales. If the curve is a straight
line on the semilog paper, you are really looking at barrier-limited current.
If it is more nearly straight on the linear paper, what you are seeing is
parasitic conduction, probably along the surface. The diode equation will
never model this, but a shunting resistor will.

On Nov 30, 1:03 pm, "William R. Frensley" <[Only registered users see links. ]>
wrote:

I own two types of diodes made by Skyworks Inc., where one diode
(SMS7630) is heavy doped with Ro = 5.4K ohms, and another diode
(SMS7621, same diode except less dopant density) with Ro = 675K ohms.

I don't know how dopant density affects the depletion region
resistivity, but outside the depletion region there's a decrease in
resistivity with an increase in dopant density -->

Conductivity = (n * un + p * up) * e

Resistivity = 1 / conductivity

where n is electron carrier density, p is hole carrier density, un is
electron mobility, up is hole mobility, and e is charge.

Here's an online calculator -->

[Only registered users see links. ]

I don't have the equation for resistivity in the depletion region.
There are less carriers in the depletion region, but is it 100%
depleted? Could the dopant density make a significant difference in
depletion region resistivity? I understand that an increase in dopant
density decreases the depletion *width*, which in turn decreases the
diodes resistance. Although, what about resistivity?

Also, according to semiconductor mathematics, the barrier height
depends on if it's n-type or p-type. Barrier height changes Ro. The
type makes a huge difference in barrier height -->

[Only registered users see links. ]

For example, n-type Silicon with Aluminum contact is 0.2eV, while the
p-type is 0.92eV. So, what if it's just one atom more n-type, and then
you add two p-type atoms to make it p-type material? Does that
suddenly change the barrier height from 0.2eV to 0.92eV?

Paul wrote:
Once again, depletion-layer resistance is *not* a meaningful concept
in semiconductor device physics. The current transport mechanism
is barrier-limited, not ohmic.

I presume you have the data sheet for your devices: [Only registered users see links. ]
This gives the SPICE model parameters for both of these devices.
Since you are obviously approaching this from the circuit side, why
not just use the SPICE model? See [Only registered users see links. ]

The answer for R(V=0) is right in the data sheet. This is the small-signal
resistance which means it is derived by differentiating the I(V) curve:
1/R0 = dI/dV = qIs/kT at V=0, or R0 = 0.026V / Is.
For the SMS7630 device, Is = 5E-6A so R0 = 5.2 kOhm.
For the SMS7621 device, Is = 4E-8A so R0 = 650 kOhm.
(Your source was probably using kT/q = 0.025 V)

Now, why is there a difference in Is? The data sheet makes no claims
that these are the "same diode except for doping." In fact, in order
to get these differences in Is and therefore turn-on voltage, the
manufacturer is almost certainly fiddling with the barrier height.
This can be done by choice of metal, interfacial layers of different
sorts, thermal processing (annealing or sintering) and probably other
tricks known only to the process engineers in this technology.

The standard process-line technique for measuring barrier height
(as opposed to research-lab techniques requiring high-vacuum surface
characterization) is to measure the I(V) curve, fit Is, and then
apply the thermionic emission model in reverse to find \phi_Sb.
Of course, this requires knowing the diode area. Since these are
microwave devices, let's assume 10 square microns. Solving
Is = A^* T^2 exp(-\phi_SB / kT),
I get barrier heights of 0.40 eV for the SMS7630 and 0.46 eV for
the SMS7621. These are quite plausible.
*Again, the work function minus electron affinity formula, and tabulated
values of those quantities, are not at all reliable.*

On Dec 3, 9:26*am, "William R. Frensley" <[Only registered users see links. ]>
wrote:

Yes, it would be great to just work with spice parameters, but that
does no good if you want to predict a diodes properties beforehand. So
far I'm seeing semiconductor mathematics that's approximations, at
best. Like you said, taking the work functions and electron affinity
is not reliable. I've seen two equations to calculate Vbi. One is for
n-type and another equation is for p-type. I just don't think it's
like a switch, when the material has more n-type dopants then all of a
sudden there's a huge change in Vbi and barrier height.

BTW, I've read through a lot of Crowell and Sze book. I didn't see the
answers to my questions there either.