1) An n-type sample with a donor doping of \(\;\left ( n_{n}=10^{15}cm^{-3}\right ) \) contains
\(\left ( N_{t}=10^{15}cm^{-3}\right ) \) generation-recombination centers located at the intrinsic
fermi level with \(\sigma _{p}=\sigma _{n}=10^{-15}cm^{2}.\) Assume \(\upsilon _{th}=10^{7}cm/\sec .\)
(a) Calculate the generation rate if the region is depleted of mobile carries
since there are no mobile carries, then \(p_{n}=n_{n}=0\) .
but the recombination rate is\[ U=\frac{\upsilon _{th}\sigma _{n}\sigma _{p}N_{t}\left ( p_{n}n_{n}-n_{i}^{2}\right ) }{\sigma _{p}\left ( p_{n}+n_{i}e^{\left ( \frac{E_{i}-E_{t}}{KT}\right ) }\right ) +\sigma _{n}\left ( n_{n}+n_{i}e^{\left ( \frac{E_{t}-E_{i}}{KT}\right ) }\right ) } \]
but since \(p_{n}=n_{n}=0\) then \(U\) becomes\[ U=\frac{\upsilon _{th}\sigma _{n}\sigma _{p}N_{t}\left ( -n_{i}^{2}\right ) }{\sigma _{p}\left ( n_{i}e^{\left ( \frac{E_{i}-E_{t}}{KT}\right ) }\right ) +\sigma _{n}\left ( n_{i}e^{\left ( \frac{E_{t}-E_{i}}{KT}\right ) }\right ) } \]
substituting in the above equation for the values given and assuming room
tempreture, results in
\[\begin{array} [c]{ccc}U & = & \frac{\left ( 10^{7}\right ) \left ( 10^{-15}\right ) \left ( 10^{-15}\right ) \left ( 10^{15}\right ) \left ( -\left ( 1.45\times 10^{10}\right ) ^{2}\right ) }{10^{-15}\left ( 1.45\times 10^{10}\;e^{\left ( \frac 0{KT}\right ) }\right ) +10^{-15}\left ( 1.45\times 10^{10}\;e^{\left ( \frac 0{KT}\right ) }\right ) }=\frac{-2.1025\times 10^{12}}{2.9\times 10^{-5}}\\ & = & -7.25\times 10^{16}cm^{-3}\sec \end{array} \]
since \(U\) is negative, then this is the generation rate.
(b)Calculate the generation rate in the region where only minority carries
concentration has been reduced appreciably below its equilibrium rate.
since this is an n-type sampe, then the minority carriers are the holes \(p\),
and also we are given that \(p_{n}\) has been reduced below its equilibrium
rate \(p_{no}\).
(side note: in thermal equilibrium, the generation rate must be balanced
by recombination rate).
(side note: if no external source , such as light, \(\Rightarrow p_{n}n_{n}=n_{i}^{2}\) but if
external source exist, then \(p_{n}n_{n}>n_{i}^{2}\))
\(n_{n}\gg p_{n}\) , and since \(E_{t}=E_{i}\) ,then \(n_{n}\gg n_{i}e^{\left ( \frac{E_{t}-E_{i}}{KT}\right ) }\) .i.e. \(n_{n}\gg n_{i}\)
so, from the expression for \(U\)\[\begin{array} [c]{ccc}U & = & \frac{\upsilon _{th}\sigma _{n}\sigma _{p}N_{t}\left ( p_{n}n_{n}-n_{i}^{2}\right ) }{\sigma _{p}\left ( p_{n}+n_{i}e^{\left ( \frac{E_{i}-E_{t}}{KT}\right ) }\right ) +\sigma _{n}\left ( n_{n}+n_{i}e^{\left ( \frac{E_{t}-E_{i}}{KT}\right ) }\right ) }\\ & & \\ & = & \frac{\upsilon _{th}\sigma _{n}\sigma _{p}N_{t}\left ( p_{n}n_{n}-n_{i}^{2}\right ) }{\sigma _{p}\left ( p_{n}+n_{i}\right ) +\sigma _{n}\left ( n_{n}+n_{i}\right ) }\end{array} \]
since only \(p_{n}\) has changed appreciably below its equilibrium, then
\[\begin{array} [c]{ccc}U & = & \frac{\upsilon _{th}\sigma _{n}\sigma _{p}N_{t}\left ( p_{n}n_{no}-p_{no}n_{no}\right ) }{\sigma _{p}\left ( small\right ) +\sigma _{n}\left ( n_{no}+small\right ) }\\ & & \\ & = & \frac{n_{no}\upsilon _{th}\sigma _{n}\sigma _{p}N_{t}\left ( p_{n}-p_{no}\right ) }{\sigma _{n}n_{no}}\\ & & \\ & = & \upsilon _{th}\sigma _{p}N_{t}\left ( p_{n}-p_{no}\right ) \\ & & \\ & = & -\upsilon _{th}\sigma _{p}N_{t}p_{no}\end{array} \]
since \(p_{n}\ll p_{\substack{no. \\\text{ }}}\)
So \(U\) is now given by, after noting that \(p_{no}=\frac{n_{i}^{2}}{n_{n}}\)\[ U=-\left ( 10^{7}\right ) \left ( 10^{-15}\right ) \left ( 10^{15}\right ) \left ( \frac{\left ( 1.45\times 10^{10}\right ) ^{2}}{10^{15}}\right ) =-2.1025\times 10^{12}cm^{-3}\sec \]
since this is negative, then this is the genration rate.
prove that the recombination rate in the space charge region of the p-n junction
in which \(\sigma _{p}=\sigma _{n}\;\;\)is maximized when \(p=n=n_{i}e^{\left ( \frac{qv}{KT}\right ) }\).
solution
the space charge region of a p-n junction is the depletion region, since outside
the depletion region the space is netural .
from the equation for \(U\)\[ U=\frac{\upsilon _{th}\sigma _{n}\sigma _{p}N_{t}\left ( p_{n}n_{n}-n_{i}^{2}\right ) }{\sigma _{p}\left ( p_{n}+n_{i}e^{\left ( \frac{E_{i}-E_{t}}{KT}\right ) }\right ) +\sigma _{n}\left ( n_{n}+n_{i}e^{\left ( \frac{E_{t}-E_{i}}{KT}\right ) }\right ) } \]
since \(\sigma _{p}=\sigma _{n}=\sigma _{0}\) , then \(U\) become
\[\begin{array} [c]{ccc}U & = & \frac{\upsilon _{th}\sigma _{0}^{2}N_{t}\left ( p_{n}n_{n}-n_{i}^{2}\right ) }{\sigma _{0}\left ( p_{n}+n_{i}e^{\left ( \frac{E_{i}-E_{t}}{KT}\right ) }\right ) +\sigma _{0}\left ( n_{n}+n_{i}e^{\left ( \frac{E_{t}-E_{i}}{KT}\right ) }\right ) }\\ & & \\ & = & \frac{\upsilon _{th}\sigma _{0}^{2}N_{t}\left ( p_{n}n_{n}-n_{i}^{2}\right ) }{\sigma _{0}\left ( p_{n}+n_{n}+n_{i}\left ( e^{\left ( \frac{E_{i}-E_{t}}{KT}\right ) }+e^{\left ( \frac{E_{t}-E_{i}}{KT}\right ) }\right ) \right ) }\\ & & \\ & = & \frac{\upsilon _{th}\sigma _{0}N_{t}\left ( p_{n}n_{n}-n_{i}^{2}\right ) }{p_{n}+n_{n}+2n_{i}\cosh \left ( \frac{E_{t}-E_{i}}{KT}\right ) }\end{array} \]
now, since \(p_{n}=p_{no}e^{\frac{qV}{KT}}\) and \(n_{n}=n_{no}e^{\frac{qV}{KT}}\) then \(U\) becomes
\[ U=\frac{\upsilon _{th}\sigma _{0}N_{t}\left ( p_{no}n_{no}e^{\frac{qV}{KT}}-n_{i}^{2}\right ) }{p_{n}+n_{n}+2n_{i}\cosh \left ( \frac{E_{t}-E_{i}}{KT}\right ) } \]
but \(p_{no}n_{no}=n_{i}^{2}\) so \(U\) becomes
\[ U=\frac{\upsilon _{th}\sigma _{0}N_{t}n_{i}^{2}\left ( e^{\frac{qV}{KT}}-1\right ) }{p_{n}+n_{n}+2n_{i}\cosh \left ( \frac{E_{t}-E_{i}}{KT}\right ) } \]
this \(U\) is max when the denimonator \(p_{n}+n_{n}+2n_{i}\cosh \left ( \frac{E_{t}-E_{i} }{KT}\right ) \) is minimum,
but min value for \(\cosh \left ( *\right ) \) is 1. so min value of denomiator is \(p_{n}+n_{n}+2n_{i}\)
but \(n_{i}\) is a constant, so we need that \(p_{n}+n_{n}\) be min.
so
\[\begin{array} [c]{ccc}dp_{n}+dn_{n} & = & 0\\ & & \\ dp_{n} & = & -dn_{n}\end{array} \]
but since \begin{equation} p_{n}n_{n}=n_{i}^{2}e^{\frac{qV}{KT}}=\text{constant}\tag{1} \end{equation}
then, differentiat and get
\[ p_{n}dn_{n}+n_{n}dp_{n}=0 \]
so\[ p_{n}dn_{n}+n_{n}\left ( -dn_{n}\right ) =0 \]
\begin{equation} p_{n}=n_{n}\tag{2} \end{equation}
so from equation (1) and from equation (2) we get
\[ p_{n}^{2}=n_{i}^{2}e^{\frac{qV}{KT}} \]
so \(p_{n}=n_{n}=n_{i}e^{\frac{qV}{2KT}}\) QED.
if recombination of electrons takes place through donorlike recmobination
center, the capture cross section \(\sigma _{n}\) for the site can be crudely estimated
by considering that the free electron enters a region where its thermal
energy \(\frac 32KT\) is less than the energy associated with the columbic
attraction; this means that the carrier is drawn to the center. The raduis
at which this occurse describes an area that may be taken to be equal
to \(\sigma _{n\text{ . }}\)
a)Obtain an expression for \(\sigma _{n}\) based upon this model and evaluate it for
Si at 300K.
the columbic force is given by\[ F=\frac 1{4\pi \epsilon _{0}}\frac{q_{1}q_{2}}{r^{2}} \]
note that this is an attraction force, since the electron is negative charged
and the center is poistive charged relative to the electron (since it is
not occupied). so, the electric field to pull in the elctron is given by\[ V\left ( r\right ) =\int _{0}^{r}F\;dr=\frac 1{4\pi \epsilon _{0}}q^{2}\left ( -1/r\right ) \]
when the electron thermal energy is \(<\) the above potional energy,
the electron is captuered. so , the largest posible cross section raduis
is given by the solution to
\[ \frac 1{4\pi \epsilon _{0}}q^{2}\left ( 1/r\right ) =\frac 32KT \]
\[ r=\frac 1{4\pi \epsilon _{0}}\frac{2\;q^{2}}{3KT}=\frac 1{4\pi \left ( 8.85418\times 10^{-14}\right ) }\frac{2\left ( 1.60218\times 10^{-19}\right ) ^{2} }{3\left ( .0259\right ) \left ( 1.60218\times 10^{-19}\right ) }=3.7064\times 10^{-6}\;cm \]
and \[ \sigma _{n}=\pi r^{2}=4.315\times 10^{-11}cm^{2} \]
b)Considering the dependence on temperature of \(\sigma _{n}\) from this model
togother with other temperature dependent terms, how would the
electron capture cross section depend on termp? (assum the center
exists in extrinsic Si).
as the themral energy of the carrier increases, \(\sigma _{0}\) will decrease,
due to the inverse relationship between \(r\) and thermal energy.so
as temp. increases, the capture cross section will become smaller.
Light in incident on an Si bar doped with \(N_{D}=10^{16}cm^{-3}\) donors, creating
\(G_{L}=10^{21}cm^{-3}\sec \) EH (electron-hole) pairs uniformly throughout the sample.
There are \(N_{t}=10^{15}cm^{-3}\) bulk recombination centers at \(E_{i}\) with electron
and hole concentrations with the light turned on. \(\sigma _{0}=10^{-15}cm^{2}.\)
a)Calculate the steady state hole electron concentration with the light
turned on .
the life time for holes in an n-type semiconductor is given by \begin{equation} \tau _{p}=\frac{1}{v_{th}\sigma _{p}N_{t}}\tag{1} \end{equation}
note that while the light is turned on, \(p_{n}=p_{no}+\delta p\) where the \(\delta p\) are the additional
holes generated by the light effect.
now, \(\delta p=\tau _{p}G_{L}=\frac 1{v_{th}\sigma _{p}N_{t}}G_{L}\) \[ \delta p=\frac 1{\left ( 10^{7}\right ) \left ( 10^{-15}\right ) \left ( 10^{15}\right ) }10^{21}=10^{14}cm^{-3} \]
but also\[ \delta p=\delta n \]
so, the steady state electron concentration is\(\;\)\[ N_{D}+\delta n=10^{16}+10^{14}=1.01\times 10^{16}cm^{-3} \]
the steady state hole concentration is \(\frac{n_{i}^{2}}{N_{D}}+\delta p=\frac{\left ( 1.45\times 10^{10}\right ) ^{2}}{10^{16}}+10^{14}\simeq 10^{14}cm^{-3}\)
b)At time t=0 the light is turned off. Calculate the time response of the total hole density and find the life time. the thermal velosity is \(10^{7}cm/\sec \) and there is no current flowing.
the govering equation is \[ p_{n}\left ( t\right ) =p_{no}+\delta p\left ( 0\right ) e^{\frac{-t}{\tau _{p}}} \]
where \(\tau _{p}=\frac 1{v_{th}\sigma _{p}N_{t}}=\frac 1{\left ( 10^{7}\right ) \left ( 10^{-15}\right ) \left ( 10^{15}\right ) }=10^{-7}\sec \) also we know that
\(p_{no}=\) \(\frac{n_{i}^{2}}{N_{D}}=21025\;cm^{\substack{-3 \\}}\)
\[ p_{n}\left ( t\right ) =21025+\left ( 10^{14}\right ) e^{\frac{-t}{10^{-7}}} \]
\[ \frac{\left ( p_{n}\left ( t\right ) -21025\right ) }{10^{14}}=e^{\frac{-t}{10^{-7}}} \]
\[ \ln \left ( p_{n}\left ( t\right ) -21025\right ) -\ln \left ( 10^{14}\right ) =-10^{7}t \]
\[ \ln \left ( p_{n}\left ( t\right ) -21025\right ) -32.236=-10^{7}t \]
the above equation gives \(p_{n}\left ( t\right ) \) at any time after the light is turned off. plug in a value for t to find \(p_{n}\)
Determine the forward current through a \(p^{+}-n\) junction device at forward bias of .675 V, given the following particulars:
\(\sigma _{n}=\sigma _{p}=10^{-14}cm^{2},\;v_{th}=10^{7}cm/\sec ,\;n_{i}=10^{10}cm^{-3},\;N_{A}=10^{18}cm^{-3},\;N_{D}=10^{15}cm^{-3},\;A=10^{-3}cm^{2},\;\tau _{p}=10^{-7}\sec ,\;w=deplectionwidth=1\mu m,\;N_{t}=3\times 10^{13}cm^{-3}\)
in a forward biased junction, the process is recombination U, i.e. \(pn\gg n_{i}^{2}.\)
so the forward current (the current entering the n side) is made of 2 components, the recombination current \(j_{rec}\) and the junction current \(j_{jun}\)
\[ J_{forward}=J_{rec}+J_{jun} \] \[ \left . \begin{array} [c]{c}J_{jun}=J_{saturation}e^{\frac V{V_{T}}}\\ J_{saturation}=q\sqrt{\frac{D_{p}}{\tau _{p}}}\frac{n_{i}^{2}}{N_{D}}\end{array} \right \} \;\;\;\;\;\;\;(1) \]
note, equation (1) above is for low injection, where \(p_{no}\gg n_{po}\) and where \(V\gg 3\frac{KT}q\) this assumption is valid since
\(p_{no}=\frac{n_{i}^{2}}{N_{D}}=\frac{\left ( 1.45\times 10^{10}\right ) ^{2}}{10^{15}}=210250\;cm^{-3}\), and \(n_{po}=\frac{n_{i}^{2}}{N_{A}}=\frac{\left ( 1.45\times 10^{10}\right ) ^{2}}{10^{18}}=210\;cm^{-3}.\) so \(p_{no}\gg n_{po}\)
and \(V=0.675\;V\) while \(3\frac{KT}q\) =\(3\left ( 0.0259\right ) =0.0777V\) so \(V\gg 3\frac{KT}q.\) so this is a reasonable approximation.
if this assumption is not valid, one needs to use this equation instead for finding the saturation current:\[ J_{sat}=q\left ( \frac{D_{p}p_{no}}{L_{p}}+\frac{D_{n}n_{po}}{L_{n}}\right ) \]
\(D_{p}\) is found from table to be \(12\;cm^{2}\sec \) so from equation (1):
\[ J_{jun}=1.60218\times 10^{-19}\sqrt{\frac{12}{10^{-7}}}\frac{\left ( 1.45\times 10^{10}\right ) ^{2}}{10^{15}}\;\;e^{\frac{0.675}{0.0259}}=\fbox{76.82\ \ \ \ Amp\ per\ cm$^2$} \]
\[ J_{rec}=\frac{qWn_{i}}{2\tau _{r}}e^{\frac V{2V_{T}}} \]
but\[ \tau _{r}=\frac 1{\sigma _{o}v_{th}N_{t}}=\frac 1{\left ( 10^{-14}\right ) \left ( 10^{7}\right ) \left ( 3\times 10^{13}\right ) }=3.333\times 10^{-7}\sec \]
so\[ J_{rec}=\frac{1.60218\times 10^{-19}\left ( 10^{-4}\right ) \left ( 1.45\times 10^{10}\right ) }{2\left ( 3.333\times 10^{-7}\right ) }e^{\frac{0.675}{2\left ( 0.0259\right ) }}=\fbox{0.159\ Amp\ per\ cm$^2$} \]
so total forward current density is \(76.82\;+0.159=76.979\;Amp\;per\;cm^{2}\)
but \(J=\frac I{Area}\)
so, \[ \fbox{I=76.979$\times $10$^-3$=0.0769\ Amp} \]
Determine the forward bias required to sustain a forward current of \(10\;mA\) and assume all other particulars are the same as problem 1.
since\[ J=\frac I{Area} \]
then\[ I=A\left ( J_{rec}+J_{jun}\right ) \;\;\;\;\;\;\;\;(1) \]
where
\[ \left . \begin{array} [c]{c}J_{jun}=J_{saturation}e^{\frac V{V_{T}}}\\ J_{saturation}=q\sqrt{\frac{D_{p}}{\tau _{p}}}\frac{n_{i}^{2}}{N_{D}}\end{array} \right \} \;\;\;\;\;\;\;(2) \]
\(D_{p}\) is found from table to be \(12\;cm^{2}\sec \) so from equation (2):
\[ J_{jun}=1.60218\times 10^{-19}\sqrt{\frac{12}{10^{-7}}}\frac{\left ( 1.45\times 10^{10}\right ) ^{2}}{10^{15}}\;\;e^{\frac V{0.0259}} \]
\[ J_{rec}=\frac{qWn_{i}}{2\tau _{r}}e^{\frac V{2V_{T}}} \]
but\[ \tau _{r}=\frac 1{\sigma _{o}v_{th}N_{t}}=\frac 1{\left ( 10^{-14}\right ) \left ( 10^{7}\right ) \left ( 3\times 10^{13}\right ) }=3.333\times 10^{-7}\sec \]
so\[ J_{rec}=\frac{1.60218\times 10^{-19}\left ( 10^{-4}\right ) \left ( 1.45\times 10^{10}\right ) }{2\left ( 3.333\times 10^{-7}\right ) }e^{\frac V{2\left ( 0.0259\right ) }} \]
so, equation (1) becomes:
\[\begin{array} [c]{c}I=A\left ( J_{rec}+J_{jun}\right ) \\ \frac{10\times 10^{-3}}A=1.60218\times 10^{-19}\sqrt{\frac{12}{10^{-7}}} \frac{\left ( 1.45\times 10^{10}\right ) ^{2}}{10^{15}}\;\;e^{\frac V{0.0259}}+ \frac{1.60218\times 10^{-19}\left ( 10^{-4}\right ) \left ( 1.45\times 10^{10}\right ) }{2\left ( 3.333\times 10^{-7}\right ) }e^{\frac V{2\left ( 0.0259\right ) }}\\ 10=3.69\times 10^{-10}x+3.485\times 10^{-7}x^{\frac 12}\end{array} \;\;\;\;\;\;\left ( 3\right ) \]
where \(x=\;e^{\frac V{0.0259}}\)
so, from equation (3), solve for \(x\)
let \(y^{2}=x\) so equation (3) becomes\[\begin{array} [c]{c}10=3.69\times 10^{-10}y^{2}+3.485\times 10^{-7}y\\ y=\frac{-4b\pm \sqrt{b^{2}-4ac}}{2a}\end{array} \]
where \(ay^{2}+by+c=0\Rightarrow a=3.69\times 10^{-10},b=3.485\times 10^{-7},c=-10\)\[\begin{array} [c]{c}y= \frac{-4\left ( 3.485\times 10^{-7}\right ) \pm \sqrt{\left ( 3.485\times 10^{-7}\right ) ^{2}\left ( 3.69\times 10^{-10}\right ) 10}}{2\left ( 3.69\times 10^{-10}\right ) }=\frac{-1.394\times 10^{-6}\pm 2.117\times 10^{-11}}{7.38\times 10^{-10}}\\ y=-1888.9 \end{array} \]
but \(y^{2}=x,\) so\[\begin{array} [c]{c}x=3567943\\ \;e^{\frac V{0.0259}}=3567943\\ \frac V{0.0259}=\ln \left ( 3567943\right ) =15.087\\ \fbox{V=0.390\ V}\end{array} \]
Determine the forward current through a \(p^{+}-n\) junction device at forward bias of .675 V, given the following particulars:
\(\sigma _{n}=\sigma _{p}=10^{-14}cm^{2},\;v_{th}=10^{7}cm/\sec ,\;n_{i}=10^{10}cm^{-3},\;N_{A}=10^{18}cm^{-3},\;N_{D}=10^{15}cm^{-3},\;A=10^{-3}cm^{2},\;\tau _{p}=10^{-7}\sec ,\;w=deplectionwidth=1\mu m,\;N_{t}=10^{16}cm^{-3}\)
in a forward biased junction, the process is recombination U, i.e. \(pn\gg n_{i}^{2}.\)
so the forward current (the current entering the n side) is made of 2 components, the recombination current \(j_{rec}\) and the junction current \(j_{jun}\)
\[ J_{forward}=J_{rec}+J_{jun} \] \[ \left . \begin{array} [c]{c}J_{jun}=J_{saturation}e^{\frac V{V_{T}}}\\ J_{saturation}=q\sqrt{\frac{D_{p}}{\tau _{p}}}\frac{n_{i}^{2}}{N_{D}}\end{array} \right \} \;\;\;\;\;\;\;(1) \]
note, equation (1) above is for low injection, where \(p_{no}\gg n_{po}\) and where \(V\gg 3\frac{KT}q\) this assumption is valid since
\(p_{no}=\frac{n_{i}^{2}}{N_{D}}=\frac{\left ( 1.45\times 10^{10}\right ) ^{2}}{10^{15}}=210250\;cm^{-3}\), and \(n_{po}=\frac{n_{i}^{2}}{N_{A}}=\frac{\left ( 1.45\times 10^{10}\right ) ^{2}}{10^{18}}=210\;cm^{-3}.\) so \(p_{no}\gg n_{po}\)
and \(V=0.675\;V\) while \(3\frac{KT}q\) =\(3\left ( 0.0259\right ) =0.0777V\) so \(V\gg 3\frac{KT}q.\) so this is a reasonable approximation.
if this assumption is not valid, one needs to use this equation instead for finding the saturation current:\[ J_{sat}=q\left ( \frac{D_{p}p_{no}}{L_{p}}+\frac{D_{n}n_{po}}{L_{n}}\right ) \]
\(D_{p}\) is found from table to be \(12\;cm^{2}\sec \) so from equation (1):
\[ J_{jun}=1.60218\times 10^{-19}\sqrt{\frac{12}{10^{-7}}}\frac{\left ( 1.45\times 10^{10}\right ) ^{2}}{10^{15}}\;\;e^{\frac{0.675}{0.0259}}=\fbox{76.82\ \ \ \ Amp\ per\ cm$^2$} \]
\[ J_{rec}=\frac{qWn_{i}}{2\tau _{r}}e^{\frac V{2V_{T}}} \]
but\[ \tau _{r}=\frac 1{\sigma _{o}v_{th}N_{t}}=\frac 1{\left ( 10^{-14}\right ) \left ( 10^{7}\right ) \left ( 10^{16}\right ) }=10^{-9}\sec \]
so\[ J_{rec}=\frac{1.60218\times 10^{-19}\left ( 10^{-4}\right ) \left ( 1.45\times 10^{10}\right ) }{2\left ( 10^{-9}\right ) }e^{\frac{0.675}{2\left ( 0.0259\right ) }}=\fbox{53\ Amp\ per\ cm$^2$} \]
so total forward current density is \(76.82\;+53=129.8\;Amp\;per\;cm^{2}\)
but \(J=\frac I{Area}\)
so, \[ \fbox{I=129.8$\times $10$^-3$=0.1298\ Amp} \]
notice that when the recombination centers numbers increased, the recombination current component increased, about 1000 times increase in the number of centers caused an increase from 0.159 to 53 \(Amp/\sec \,\)increase.
3/1/1993
Determine \(N_{t}\;\)through a \(p^{+}-n\) junction device at forward bias of .64 V, and diod current is 35 mA given the following particulars:
\(\sigma _{n}=\sigma _{p}=10^{-14}cm^{2},\;v_{th}=10^{7}cm/\sec ,\;n_{i}=10^{10}cm^{-3},\;N_{A}=10^{18}cm^{-3},\;N_{D}=10^{15}cm^{-3},\;A=10^{-3}cm^{2},\;\tau _{p}=10^{-7}\sec ,\;w=deplectionwidth=1\mu m\)
in a forward biased junction, the process is recombination U, i.e. \(pn\gg n_{i}^{2}.\)
so the forward current (the current entering the n side) is made of 2 components, the recombination current \(j_{rec}\) and the junction current \(j_{jun}\)
\[ J_{forward}=J_{rec}+J_{jun} \] \[ \left . \begin{array} [c]{c}J_{jun}=J_{saturation}e^{\frac V{V_{T}}}\\ J_{saturation}=q\sqrt{\frac{D_{p}}{\tau _{p}}}\frac{n_{i}^{2}}{N_{D}}\end{array} \right \} \;\;\;\;\;\;\;(1) \]
\(D_{p}\) is found from table to be \(12\;cm^{2}\sec \) so from equation (1):
\[ J_{jun}=1.60218\times 10^{-19}\sqrt{\frac{12}{10^{-7}}}\frac{\left ( 1.45\times 10^{10}\right ) ^{2}}{10^{15}}\;\;e^{\frac{0.64}{0.0259}}=\fbox{19\ \ \ \ Amp\ per\ cm$^2$} \]
now,
\[ J_{rec}=\frac{qWn_{i}}{2\tau _{r}}e^{\frac V{2V_{T}}} \]
but\[ \tau _{r}=\frac 1{\sigma _{o}v_{th}N_{t}}=\frac 1{\left ( 10^{-14}\right ) \left ( 10^{7}\right ) N_{t}}=\frac{10^{7}}{N_{t}}\sec \]
so\[ J_{rec}=\frac{N_{t\;}\times 1.60218\times 10^{-19}\left ( 10^{-4}\right ) \left ( 1.45\times 10^{10}\right ) }{2\left ( 10^{7}\right ) }e^{\frac{0.675}{2\left ( 0.0259\right ) }}=\fbox{$5.3\times 10^-15\times N_t$\ Amp\ per\ cm$^2$} \]
so total forward current density is
\(19+5.3\times 10^{-15}\times N_{t}=35\;\)Amp per cm\(^{\text{2}}\)
so, \[ N_{t}=\frac{35-19}{5.3\times 10^{-15}}=\fbox{$3.0188\times 10^15$\ cm$^-3 $} \]
3/10/93
\[ V_{bi}=V_{T}\ln \left ( \frac{N_{A}N_{D}}{n_{i}^{2}}\right ) =0.0259\ln \left ( \frac{10^{18}10^{16}}{\left ( 1.45\times 10^{10}\right ) ^{2}}\right ) =0.8156\;\text{V} \]
depletion junction width\[ W=\sqrt{\frac{2\epsilon _{s}}q}\sqrt{\left ( \frac 1{N_{D}}+\frac 1{N_{A}}\right ) \left ( V_{bi}-V_{bias}\right ) }=3604\sqrt{\left ( \frac 1{10^{18}}+\frac 1{10^{16}}\right ) \left ( 0.8156-V_{bias}\right ) } \]
but for p-n abrupt junction \[ E_{\max }=\frac{2\left ( V_{bi}-V_{bias}\right ) }W \]
since break down will occur when \(E_{\max }\;\)reaches \(E_{crtitcal}\)
so from above equation\[ 2\times 10^{5}=\frac{2\left ( V_{bi}-V_{bias}\right ) }W=\frac{2\left ( 0.8156-V_{bias}\right ) }{3604\sqrt{\left ( \frac 1{10^{18}}+\frac 1{10^{16}}\right ) \left ( 0.8156-V_{bias}\right ) }} \]
solve the above for \(V_{bias}\)\[ V_{bias}=0.8156-13.11=-12.3\text{ V } \]
to the break down voltage is \(0.8156-12.3=13.11\) V
depletion width at \(V_{br}\)\[ W=3604\sqrt{\left ( \frac 1{10^{18}}+\frac 1{10^{16}}\right ) \left ( 13.11\right ) }=1.31\times 10^{-4}cm \]
\[ V_{bi}=V_{T}\ln \left ( \frac{N_{A}N_{D}}{n_{i}^{2}}\right ) =0.0259\ln \left ( \frac{10^{18}10^{18}}{\left ( 1.45\times 10^{10}\right ) ^{2}}\right ) =0.9349\;\text{V} \]
depletion junction width\[ W=\sqrt{\frac{2\epsilon _{s}}q}\sqrt{\left ( \frac 1{N_{D}}+\frac 1{N_{A}}\right ) \left ( V_{bi}-V_{bias}\right ) }=3604\sqrt{\left ( \frac 1{10^{18}}+\frac 1{10^{18}}\right ) \left ( 0.9349-V_{bias}\right ) } \]
but for p-n abrupt junction\[ E_{\max }=\frac{2\left ( V_{bi}-V_{bias}\right ) }W \]
since we assume that break down will occur when \(E_{\max }=E_{crtitcal}\)
so\[ 10^{6}=\frac{2\left ( V_{bi}-V_{bias}\right ) }W=\frac{2\left ( 0.9349-V_{bias}\right ) }{3604\sqrt{\left ( \frac 1{10^{18}}+\frac 1{10^{18}}\right ) \left ( 0.9349-V_{bias}\right ) }} \]
solve the above for \(V_{bias}\)\[ V_{bias}=0.9349-6.49=-5.559\text{ V } \]
to the break down voltage is \(6.49\) V
depletion width at \(V_{br}\)\[ W=3604\sqrt{\left ( \frac 1{10^{18}}+\frac 1{10^{18}}\right ) \left ( 6.49\right ) }=12.9\times 10^{-4}cm \]
in BJT punch through occurs when the collector depletion region extends too far into the base that it reaches the emitter depletion region, i.e. making the active base width to be zero.
for the emitter-base depletion region
\[ V_{bi}=0.0259\ln \left ( \frac{5\times 10^{18}\times 10^{15}}{\left ( 2.1\times 10^{10}\right ) ^{2}}\right ) =0.778\text{ v} \]
the emitter -base width
\[ W=\sqrt{\frac{2\epsilon _{s}}q}\sqrt{\left ( \frac 1{N_{E}}+\frac 1{N_{B}}\right ) \left ( V_{bi}-V_{bias}\right ) } \] \[ W=\sqrt{\frac{2\epsilon _{s}}q\left ( \frac 1{5\times 10^{18}}+\frac 1{10^{15}}\right ) \left ( 0.778-0.5\right ) }=6\times 10^{-5}cm \]
emitter depletion region part into the base \[ W_{EB}^{^{\prime }}=W\left ( \frac{N_{E}}{N_{E}+N_{B}}\right ) =6\times 10^{-5}\left ( \frac{5\times 10^{18}}{5\times 10^{18}+10^{15}}\right ) =5.99\times 10^{-5}cm \]
for the collector-base
\[ V_{bi}=0.0259\ln \left ( \frac{10^{15}\times 10^{15}}{\left ( 2.1\times 10^{10}\right ) ^{2}}\right ) =0.558\text{ v} \]
collector-base depletion width \begin{equation} \label{1}W=\sqrt{\frac{2\epsilon _{s}}q\left ( \frac 1{10^{15}}+\frac 1{10^{15}}\right ) \left ( 0.558-V_{CB}\right ) } \end{equation}
collector depeltion region into the base \[\begin{array} [c]{c}W_{CB}^{^{\prime }}= \sqrt{\frac{2\epsilon _{s}}q\left ( \frac 1{10^{15}}+\frac 1{10^{15}}\right ) \left ( 0.558-V_{CB}\right ) }\left ( \frac{N_{c}}{N_{c}+N_{B}}\right ) \\ =\frac 12 \sqrt{\frac{2\epsilon _{s}}q\left ( \frac 1{10^{15}}+\frac 1{10^{15}}\right ) \left ( 0.558-V_{CB}\right ) }\end{array} \]
but total W is given as \(5\mu m=5\times 10^{-4}cm\)
so\[\begin{array} [c]{c}W_{EB}^{^{\prime }}+W_{CB}^{^{\prime }}=5\times 10^{-4}cm\\ 5.99\times 10^{-5}+\frac 12\sqrt{\frac{2\epsilon _{s}}q\left ( \frac 1{10^{15}}+\frac 1{10^{15}}\right ) \left ( 0.558-V_{CB}\right ) }=5\times 10^{-4}cm \end{array} \]
so\[ V_{CB_{critical}}=-29.8\text{ V} \]
so\[ \fbox{V$_br$=30.35 V} \]
for abrupt junction, \[ 30.35=\frac 12E_{critical}W \]
but from (1)\[ W=\sqrt{\frac{2\epsilon _{s}}q\left ( \frac 1{10^{15}}+\frac 1{10^{15}}\right ) \left ( 30.35\right ) }=\sqrt{\frac{2\epsilon _{s}}q\left ( \frac 1{10^{15}}+\frac 1{10^{15}}\right ) \left ( 30.35\right ) }=8.88\times 10^{-4}cm \]
so\[ E_{critical}=68370\text{ V/cm} \]
now, for avalanche to happen\[ \int _{0}^{w}\alpha dx=1 \]
where \(\alpha \) is from figure 27 page 67 for the \(E_{cirtical}\) we just found. so find \(\alpha \) and use \(W\) \(=\) the collector-base depletion width found above \(W=8.88\times 10^{-4}cm,\) and see if the integral goes to 1 or not, if it is, then avalanche will occur first else punchthrough.
also we see that V\(_{br}>\frac{6E_{g}}q=\frac{6\left ( 1.12\right ) }q,\) so avalanche will occure first.