Calculating barrier height

[Paul]

Hi,

I'm trying to write a small program to design basic Schottky diodes.
I'm using the following equation to calculate the current produced by
a small applied voltage, far below Vt -->

I = A * Aast * T^2 * exp(-barrier/Vt) * exp(V/(n*Vt) - 1)

where barrier is the barrier height, A is the contact area, T is temp
in kelvin, Vt is thermal voltage, V is applied voltage, n is ideality
constant, and Aast is the Richardson constant -->

A = 4 * PI * q * m * k^2 / h^3

where m is the effective electron mass.

Everything seems fine except for my barrier height function, which is
-->

if(ntype) {
barrier = F - Eea;
} else {
barrier = Eg - F + Eea;
}

Obviously something's missing, because there's no dopant density in
the equations. An increase in dopant density will increase the
current. I'm thinking my barrier height equation is overly simplified.
Obviously the barrier height changes with dopant density. Does anyone
have the equation(s)? I'd appreciated it.

Regards,
Paul

 

[JosephKK]

Hi,

I'm trying to write a small program to design basic Schottky diodes.
I'm using the following equation to calculate the current produced by
a small applied voltage, far below Vt -->

I = A * Aast * T^2 * exp(-barrier/Vt) * exp(V/(n*Vt) - 1)

where barrier is the barrier height, A is the contact area, T is temp
in kelvin, Vt is thermal voltage, V is applied voltage, n is ideality
constant, and Aast is the Richardson constant -->

A = 4 * PI * q * m * k^2 / h^3

where m is the effective electron mass.

Everything seems fine except for my barrier height function, which is
-->

if(ntype) {
barrier = F - Eea;
} else {
barrier = Eg - F + Eea;
}

Obviously something's missing, because there's no dopant density in
the equations. An increase in dopant density will increase the
current. I'm thinking my barrier height equation is overly simplified.
Obviously the barrier height changes with dopant density. Does anyone
have the equation(s)? I'd appreciated it.

Regards,
Paul

Sounds like you need to look somewhere in sci.physics.*

 

[William Frensley]

Hi,

I'm trying to write a small program to design basic Schottky diodes.
I'm using the following equation to calculate the current produced by
a small applied voltage, far below Vt -->

I = A * Aast * T^2 * exp(-barrier/Vt) * exp(V/(n*Vt) - 1)

where barrier is the barrier height, A is the contact area, T is temp
in kelvin, Vt is thermal voltage, V is applied voltage, n is ideality
constant, and Aast is the Richardson constant -->

A = 4 * PI * q * m * k^2 / h^3

where m is the effective electron mass.

Everything seems fine except for my barrier height function, which is
-->

if(ntype) {
barrier = F - Eea;
} else {
barrier = Eg - F + Eea;
}

Obviously something's missing, because there's no dopant density in
the equations. An increase in dopant density will increase the
current. I'm thinking my barrier height equation is overly simplified.
Obviously the barrier height changes with dopant density. Does anyone
have the equation(s)? I'd appreciated it.

No, in this case the doping level drops out of the equation.
What really counts is the carrier density at and above the
actual energy maximum. You would think that lowering the doping will
push the Fermi level deeper into the bandgap, reducing the density at
the top of the barrier. But, the reduced doping will also reduce the
total band-bending (at any fixed bias voltage), reducing the real
physical barrier height. The change in
band-bending is logarithmic in the carrier density, but this is then
exponentiated by the Boltzmann distribution, resulting in a net factor
of one. The only density that counts is the conduction-band
density of states Nc (assuming n-type diode), and Nc is hidden inside
the Richardson constant (which is just Nc times the mean thermal
velocity of the carriers).

By the way, do not take the (barrier) = (metal work fn) - (semi
electron affinity) equation seriously. All the textbook authors believe
it, but they are not interface experts. In particular, it completely
fails for the most common semiconductors: Si, Ge, GaAs, InP. See
Kurtin, McGill, and Mead, Phys. Rev. Lett. vol. 22, pp. 1433-6 (1969)
for some of the evidence.

- Bill Frensley

 

[Paul]

No, in this case the doping level drops out of the equation.
What really counts is the carrier density at and above the
actual energy maximum.  You would think that lowering the doping will
push the Fermi level deeper into the bandgap, reducing the density at
the top of the barrier.  But, the reduced doping will also reduce the
total band-bending (at any fixed bias voltage), reducing the real
physical barrier height.  The change in
band-bending is logarithmic in the carrier density, but this is then
exponentiated by the Boltzmann distribution, resulting in a net factor
of one.  The only density that counts is the conduction-band
density of states Nc (assuming n-type diode), and Nc is hidden inside
the Richardson constant (which is just Nc times the mean thermal
velocity of the carriers).

By the way, do not take the (barrier) = (metal work fn) - (semi
electron affinity) equation seriously.  All the textbook authors believe
it, but they are not interface experts.  In particular, it completely
fails for the most common semiconductors: Si, Ge, GaAs, InP.  See
Kurtin, McGill, and Mead, Phys. Rev. Lett. vol. 22, pp. 1433-6 (1969)
for some of the evidence.

- Bill Frensley

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.

Please, doesn't anyone know how to do this?

Thanks,
Paul

 

[William R. Frensley]

Thanks, but that cannot be correct because the diode zero bias
resistance, Ro, significantly decreases with an increase in dopant
density.

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.

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.

Well known for exactly which devices?

- Bill Frensley

 

[Paul]

I take it that you have experimental data demonstrating this?

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 -->

http://www.ee.byu.edu/cleanroom/ResistivityCal.phtml

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 -->

http://www.ee.byu.edu/cleanroom/ohmic-schottky.phtml

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?

Regards,
Paul

 

[William R. Frensley]

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 -->

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:
http://origin-products.skyworksinc.com:81/docdownload.jsp?docID=NN30-DSN-V4Q5UIMLW2&tabname=TechLib
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
http://eesof.tm.agilent.com/docs/iccap2002/MDLGBOOK/7DEVICE_MODELING/2DIODE/DIODE.pdf

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.*

- Bill Frensley

 

[Paul]

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:http://origin-products..skyworksinc.com:81/docdownload.jsp?docID=NN30-...
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?  Seehttp://eesof.tm.agilent.com/docs/iccap2002/MDLGBOOK/7DEVICE_MODELING/...

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.*

- Bill Frensley

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.

Thanks,
Paul