For the following problems, consider a semiconductor for which \(E_g = 1 eV, \quad kT=.025 eV\) and
\(D_c = 12 \times 10^{21} \frac{states}{eV*cm^3}, \quad B_c = 3 eV\)
\(D_v = 12 \times 10^{21} \frac{states}{eV*cm^3}, \quad B_v = 3 eV\).
(Assume that within each band the density of states is independent of E.)
Let's set our zero of energy at the top of the valence band.
Please mention any possible typos, confusing things etc.
It will probably help to discuss a lot of this in the comments here. For example, how do the units work going from charge to electric field to potential? How does one end up with a potential in eV? (You may have to convert from cm to meters when you use surface charge to calculate electric field if epsilon_o is in Farads/meter.)
1. a) If this semiconductor is undoped, then what is the value of n at room temperature? What is the relationship between n and p? What is \(E_f\) for this un-doped case?
b) Suppose that this semiconductor is doped with 10^17 donors/cm^3. In that case we like to assume that each donor contributes one electron to the conduction band. What is \(E_f\) in this case?
(What do you think might be the rationale behind choosing \(12 \times 10^{21} \frac{states}{eV*cm^3}\) for the density of states? Why that value? (post here))
2. Consider a semiconductor as above. Suppose that it is doped with 10^17 donors for the half to the left of the plane x=0, and doped with 10^17 acceptors/cm^3 to the right of x=0. (The plane x=0 defines an interface between to two differently doped regions.) Suppose that within a distance \(x_d\) of the interface all the electrons in the conduction band from the left side cross over to the other side and fill up previously empty valance band states (holes) there.
a) When \(E_f\) is independent of x, that represents an equilibrium state. What value of \(x_d\) would enable to bands to bend and shift by just the right amount to enable \(E_f\) to be independent of x?
b) Plot the conduction band edge and valence band edge as a function of x for this case.
3 (optional extra problem). Suppose the doping on each side is 10^16 cm^-3.
What is the equilibrium value of \(x_d\) for this case?
4. a) For the n-p junctions you solved in problem 2 (doped with \(10^{17} 1/cm^3\)), use your calculated value for \(E_c (x)\) to calculate n(x) as a function of x. You can use the equation \(n(x) = KT D_c e^{-(E_c (x) - E_f)/kT}\)?
b) What is \(n(-x_d)\)? What is \(n(x_d)\)? Sketch a rough graph n(x) as a function of x.
c) Graph the product of the conduction electron density times the electric fields a function of x. At what x does its maximum value occur?
d) Calculate the drift current as a function of x in the junction region. (You can just do this for x less than zero since that's easier and I think the maximum is in that range.)
5. Use your equation for n(x) from the previous problem to calculate the diffusion current in the junction (for x less than zero). This is an important problem. Please spend some time on this one.
6. Graph this two currents. Compare and discuss here.
Notes: \(n(x) e^2 \tau/m^*m\) multiplied time electric field has units of current per unit area. I think that if you use your expression for n(x) (from the previous problem) and multiply it time the electric field in the junction, which varies linearly with x, that correspond to a current associated with electrons accelerated by the electric field.
Special bonus: Calculate the peak value of that current in Coulombs/second (amperes) for a specific junction with an area of 1 cm^2, using m*=0.2 and \(\tau = 10^{-12}\) seconds. (Use the specific 10^17 doping case above. We want an actual number. Is it big, small, negligible, 10^-15 amps, 2 amps or what?
Hints: One can separate out the term \( \mu= e \tau/m = e \tau c^2/(m^*m c^2)\). With mc^2 in eV and c in cm/s that can have units of \(cm^2/(Volt*seconds\). (The e turns eV into Volts...).
Does the maximum in this product occur somewhere between -x_d to zero? Where is it?
PS. Here is how one might approach graphing current density.
Here is a video on the integration to get V(x):

Hi guys,
ReplyDeleteJust commenting on the units question:
How do the units work going from charge to electric field to potential? How does one end up with a potential in eV?
I know that Electric field is measured in Newton/Coulumb which is equivalent to V/meter so by multiplying E*distance we can get voltage i.e Edx=dv. Now that we have Volts we simply multiply by e to get eVolts.
As for charge we know E=F/q where q is the charge. So then q=F/E will give us charge. Is there any other way to do this? How do we get charge to volts? Let me know, thanks guys.
Thanks. good questions.
DeleteStudent: Coming back to this, I'm not sure if I'm using the given values right. Are B_c and B_v the conduction and valence bands? And is E_g is the energy gap between them? Thanks, I'd like to clarify that before I starting posting results.
ReplyDelete\(B_c\) is the conduction band width. So the conduction band goes from 1 to 4 eV. Does that make sense?
DeleteSimilarly, the valence band extends from -3 to 0 eV. So, yes, the gap between them is 1 eV.
Thanks for starting things off Ben very nice!
ReplyDeleteEach electron occupies a state. Dc does not change. Only E_f changes. That changing of E_f changes the number of occupied states.
ReplyDeleteThe formula with the logarithm is not used for doped systems. It is only valid for the special case n=p (no doping).
I don't think it's quite that simple. All of the added electrons will easily fill up the valence band and pour into the conduction band. So, we want to find the value of \(E_{f}\) such that
ReplyDelete\begin{equation*}
10^{17} = \int\limits_{1ev}^{4ev}D_{c}\dfrac{1}{e^{(E-E_{f})/kT}+1}dE
\end{equation*}
By assuming that \(E_{f}\) will lie somewhere between 0 and 1 eV, and that the exponent will be pretty big, we can solve this for \(E_{f}\), yielding:
\begin{equation*}
n=kTD_{c}e^{-(1ev-E_{f})/kT}
\end{equation*}
and finally
\begin{equation*}
E_{f}=1eV+kT\ln\left(\dfrac{n}{kTD_{c}}\right) = 0.8eV
\end{equation*}
Basically, we're looking for what \(E_{f}\) would get us \(10^{17}\) occupied states in the conduction band.
Does this make sense?
This is for 1b and our selection of density of states.
ReplyDeleteSo, suppose we have a slab of silicon. It has density 2.390\(\tfrac{g}{cm^{3}}\), and silicon has atomic mass \(28.085\tfrac{\text{u}}{\text{atom}}\). From here we can calculate the number of atoms per \(cm^{3}\):
\begin{equation*}
2.3290\tfrac{\text{cm}^{3}}{\text{g}}\times\tfrac{\text{atom}}{28.085\text{u}} = 2.3290\tfrac{\text{cm}^{3}}{\text{g}}\times\tfrac{\text{atom}}{4.66\times 10^{-23}\text{g}} = 5\times 10^{22}\tfrac{\text{atoms}}{\text{cm}^{3}}
\end{equation*}
This number is close, but not quite right. Each atom will have four valence electrons, and a total of 14 electrons. But, everything I try runs me into a dead end or nonsensical answer.
Does anyone have any ideas?
Woops, I messed up the units when I typed this up. Switch the cm and g in the density term of the \LaTeX\ equation!
DeleteOh, good thinking but actually I picked that just to make things work our so that 10^17 would correspond to 0.8 eV. Just to get round numbers.
DeleteHi all for problem 2 here are the equations we will need:
ReplyDeleten(x) = kTD_c * exp(-(E_c(x)-E_f)/kT)
Jdiffusion = (-e) * (D_n) * (-dn/dx)
Jdrift = n(x) * e * (e*E(x)*tau/m_e)
here E(x) is electric field as a function of x (look at Ben's old post on this).
SO my logic is that
1. we solve for E_c(x)
2. we use that to solve for n(x) since n has E_c in it.
3. We set Jdiff = Jdrift for the equilibrium state
4. We solve for E(x) electric field = somthing that has xd in it
5. Now we solve for xd knowing E(x)_max when x=0, again view Bens post on this.
6. ? are we done? Is this the right logic or is this problem simpler than I think.
Ill learn latex soon! Let me know what you guys are thinking for problem 2.
As for problem 1 a) I agree with Brian and for 1b) I also got the same thing Ben got.
I don't see any problem with those units. You have a charge (coulombs) divided by a length and an epsilon_o. That can have units of volts.
ReplyDeleteIf you multiply by another e, then you can get eV. For example, \(e^2/(\epsilon_o r)\) has units of eV, doesn't it?
I think you may have shifted the location of your place where x=0, or made a mistake in the integral. Otherwise what you have is probably correct. I think V(x) should be zero when x = -xd. Also, which range of x is your V(x) derived for?
ReplyDeleteRiley, what is your equation for Emax?
ReplyDelete"for my Emax I have eNxd/(2ϵϵo). I forgot to mention that I dont think we can get away with only using ϵ of free space when calculating the electric field as crystal are very dense. Its not stated in the problem but I assumed the pn junction consists mostly of silicon (ϵ=11.7)"
ReplyDeleteVery good point Riley. This might explain the differences. Maybe now we can all converge on an xd value for this problem!
That makes sense. Just an area. You can make up an area. Like .01 cm^2 or something.
ReplyDeleteTruth is, it looks like I was careless about the distinction between current density, which is what one can normally calculate, and current, which is current density times some (cross-sectional) area of a hypothetical object. Does that make any sense?
Results in either Amps/cm^2 (current density) or Amps (current) are all fine and very appreciated.
check your signs. electrons diffuse to the right. They carry a negative charge. They drift to the left. There are lots of possibilities for sign errors. we can sort that out.
ReplyDeleteBut anyway, can you calculate the magnitude of either one at its maximum for an area of .01 cm^2?