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An analytical estimate of diode’s depletion depth.
Gregory Prigozhin
August 6, 2007
Here I describe a simple way to estimate depletion depth and potential profile in the APS diodes made
on a high resistivity substrate.
Fig. 1 shows schematically a cross section of the diode array assuming that substantial voltage is applied
to the p+ regions to deplete silicon deep enough. When the depth of depletion becomes larger than the pixel
size, the boundary of the depletion region becomes flat. We can use this fact to make a very simple estimate
of the depletion depth. If one draws a box (shown by a dashed line in the Fig. 1) with its lower side aligned
n+
p+
n+
p+
n+
depleted n− silicon
undepleted n− silicon
Figure 1: Cross section of the diode structure and Gaussian box.
with the boundary between depleted and undepleted silicon and its upper side aligned with the metallurgical
boundary of the p − n junction near the surface of the device, then the Gauss’s law for the volume inside the
box can be written as follows. Assuming that such box is three dimensional and occupies N pixels and that
side walls of the box coincide with the center lines between pixels, where lateral component of the electric
field is zero, we evaluate total charge inside the box as qN Nd dApix . The electric field flux through the walls
of the box is nonzero only at the p+ region boundary. The area of this region is N Adiode . The n+ channel
stops for simplicity are shown to be deeper than the p − n junction, while in reality they are shallower. The
distortion of the results due to this is clearly negligible, because the field is very weak close to the n+ region
and the amount of charge involved is small compared to the total charge in the depleted volume. The electric
field flux integral is N ǫEj Adiode , where Ej is electric field at the junction (it is assumed to be constant over
entire diode area). The Gauss’s law states that the two quantity are equal, which leads to
ǫEj Adiode = qNd dApix
We need to determine depletion depth d, knowing the voltage V0 applied to the diode. The question is how
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E
E
Ej
linear portions,
same slope
Ej
x
x
d
d
a)
b)
Figure 2: a) Fiels as a function of depth in 1D case. b) Field as a function of depth at the center of circular
diode.
to relate V0 and Ej . In a simple one-dimensional case (when the diode occupies entire wafer) the derivative
dE/dx of electric field over depth x is proportional to the density of dopant in the depleted substrate, and,
hence, is constant. This means that if we plot E as a function of depth, the plot is a stright line. Since
voltage is an integral of electric field, V0 is an area under the triangle and is equal to Ej d/2 (see Fig. 2,a).
In the case when the diode occupies a small portion of a pixel the plot of E vs. x becomes nonlinear. The
plot of E vs. depth is schematically shown in Fig. 2,b. Still, close to the depletion depth boundary field
distribution is almost one-dimensional, E vs x dependence is linear, and the slope of this line is exactly the
same as in 1D case with the same substrate doping level. This linear region is very clearly seen in Vyshi’s
plots. There should be another linear region with the same slope (since the doping level is the same) on the
other side of the depleted region - very close to the p − n junction, within a few microns from it 2D effects
should be negligible. In Vyshi’s simulation this region is extremely small, almost unnoticeable. The most
simple way to estimate Ej is to use the same equation,
V0 =
Ej d
2
, which is equivalent to an area of a triangle shown by a dased line. Comparing to Vyshi’s plots it is clear
that it will lead to overestimating of V0 by a factor of 1.5–2. If we nevertherless plug in this value of V0 into
the above expression for Ej , the depletion depth can be written as
s
s
2ǫV0
Adiode
d=
·
qNd
Apix
which is the same as in 1D case, but multiplied by the square root of area ratio. It can be deduced from the
above discussion that this formula probably underestimates the depletion depth d. It is likely to produce
correct functional dependencies, though. For V0 = 10 V and Nd = 1.43 · 1012 cm−3 (substrate doping that
corresponds to 3000 Ohm·cm n-type silicon) depletion depth d = 42 microns. For V0 = 15 V d = 51 microns.
These results are surprizingly close to Vyshi’s simulations. A plot of depletion depth as a function of voltage
is shown in Fig. 3.
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Figure 3: Depletion depth as a function of diode voltage (12 micron diameter circle in a 24 x 24 micron
square pixel.
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