Notes On Semiconductors: PN junction Practice Problems

Problem 1

An abrupt silicon pn junction has dopant concentrations of Na = 1×10^15 cm^-3 and Nd = 2×10^17cm^-3. (a) Evaluate the built in potential at room temperature. (b) use the depletion approximation to calculate the width of the space-charge layer and the peak electric field for junction voltages Va = 0V, -10V

Turn on voltage

Part b is now asking for the width, W, or referred to as xd. The formula for peak E field and width are below. Note, that epsilon in this equation refers to that of silicon (11.68*8.85*10^-14F/cm). Also note that we must convert this constant to cm in order to use it in the problem

Width of depletion//space charge region
Maximum field based off of V = (1/2)Emax*xd

Solving for both conditions we find xd = 0.97um for 0V and 3.73um for -10V, with E field being 14800V/cm for 0V, and 57500V/cm for -10V.

Problem 2

Find and sketch the built in field and potential for a silicon PIN junction with the doping profile shown. Indicate the length of each depletion region.

To begin solving part a, we must look at the charge densities of each region, and use Poisson’s equation in order to establish a relationship between x, field, and potential. To do this, we will create 4 ticks on the x axis, x1 being somewhere to the left of -1, x2 = -1, x3 = 0.5, and x4 being somewhere beyond x3.

charge density on n-side is equal to qNd
only valid for this boundary condition
Final potential for N-side
Notice due to having no doping in this middle region, we can say that the charge density is 0, thus E through this region has to relate to the field generated by the n-type
Boundary for intermediate region
Note that we are using the potential of where we started when integrating
Final for middle region
Notice that charge density is equal to -qNa, thus the bounds were flipped

Here we see what we are integrating from, and what our bounds will be
Here is final equation
Here we are defining the max fields we need

Relationship of turn on voltage and the 3 zones of potential

Additionally, we know that the vbi is the same for a pn junction in this problem, thus

Note Vbi is the same as phi subscript i
Note here that due to emax, we are able to make a relationship between (x2-x1) and (x4-x3), which our two depletion region zones

To answer this question, we use the equation relating to doping concentrations and turn on voltage to find the turn on voltage, and our relationship between the two depletion regions in order to solve for one. I chose to solve for the (x2-x1) first, using my calculator to solve the quadratic formula. Remember boltzmann’s constant is 8.617*10^-5ev/K, the e and q cancels, and epsilon is equal to 11.68*8.85*10^-14F/cm for silicon, and temperature is in Kelvin. Final answer to (x2-x1) is 2.5*10^-6cm. Using the relationship of the depletion regions, we solve (x4-x3) = 2.5*10^-4cm. Graphing all the fields should be trivial after obtaining these values. The following equation is a visual representation:

Final equation

b. Compare the maximum field to the field in a pn junction that contains no lightly doped intermediate region but has the same dopant concentrations.

Note: for PIN junction only
Maximum field based off of V = (1/2)Emax*xd. This is for abrupt PN junction

Note from above problem that width is equal to ((2eps/q)*(NaNd/(Na+Nd))*(Vi-Va))^1/2

Using the pin junction we solve for a value of 3869 V/cm. Assuming same doping concentrations for pn junction, we solve to find width equal to 9.48*10^-5cm, and Emax = 13354.43V/cm

c. Explain what is happening

Due to the middle region between the p and n type material that holds no doping concentration, voltage is dropped. The reduced voltage induces a smaller depletion region on either material, which is directly related to electric field. Due to the reduced voltage, it makes sense that the electric field would drop in the pin junction.

d. Discuss how the depletion capacitance

To solve for capacitance, we must make a relationship of C = d(Q)/dVa, and Q is equal to (x2-x1). Thus we can use the final equation used to solve part A for this. Taking the derivative with respect to Va will give you the following equation

Taking derivative of x2-x1 in respect to Va.
Final equation

This equation is only differing from a regular p-n junction by one term relating to the lightly doped region. Due to this term, the capacitance is much lower due to the depletion region being wider.

Capacitance of a pn junction

Problem 3

Problem 3

Assume that the dopant distributions in a piece of silicon are indicated above. (Na0 = 10^18cm^-3, x0 = 10^-4cm unless otherwise indicated, lambda a = 10^-4cm, and lambda d = 2*10^-4 cm)

a. If the pn junction is desired at x0 = 1um, what should be the value of Nd0?

We know that at x0, Na = Nd,  thus,


Equation to Solve part A

Using the second equation, we find that Ndo = 6.1×10^17cm^-3

b. Assume the depletion approximation and make a sketch of space charge near the junction. Approximate this space charge as if this were a linearly graded junction.

Now with part b, we are using x0 = 10^-4cm. Due to the fact that they said linearly graded junction, we know immediately that the charge density will be equal to q(Nd-Na) = qa(x-x0)

Noting this relationship, we are tasked to approximate the linearly graded junction at the intersect. Therefore, we must take the derivative of (Nd-Na) as x approaches x0.

Answer to problem

Plugging in the found value for Ndo, and x0 = 10^-4cm, we find that a = 1.83×10^21cm^-4

c. Under the approximation of part B, take turn on voltage = 0.7V to calculate Emax at thermal equilibrium. Using:

Width of linear junction
Emax related to voltage in linearly graded junction

Problem 4

The small signal capacitance Cd of a pn junciton diode with an area of 10^-5cm^2 is measured. A plot of 1/Cd^2 vs Va is shown below

a. If the diode is considered as a one sided step junction find the indicated oping level on the lower conductivity side (use the slope)

So here, we know we can relate the doping concentrations with capacitance. We know that Cd = Aeps/xd, and we know that xd is related to both doping concentrations. Assuming a one sided step junction means that one of the concentrations will be considered negligible, thus, our equation will simplify into:

Equation relating doping concentration with capacitance
This equation relates the slope that we will find from the graph to doping concentration

Finding the slope is simply (1.5*10^26-0)/(-1-0.8) which gives a negative slope who’s magnitude is 8.33*10^25F^-2V-1. Using this slope with the other values given (recall that eps in silicon is 11.68*8.85*10^-14F/cm. Once solved, we find that N = 1.45*10^15cm^-3

b. Sketch the doping density on the low conductivity side of the junction. Calculate the location of any point at which the dopant density changes.

Here we use the same equation of Cd = Aeps/xd, however, we rearrange it to xd = Aeps/Cd. We know that at the lowest conductivity, Cd will be at its highest on the graph. Thus, we know Cd = (1.5*10^26)^-1/2, and we solve for xd, which equals to 1.27um.

c. Use the intercept on the (1/Cd^2) plot to find the doping density on the highly doped side.

Here we must use a specialized equation relating doping concentration and built in turn on voltage. The x intercept is of interest here, as it is the turn on voltage. Vbi = 0.8V. Using this eqn:

Note, the ni^2/N is relating to the eqn n*p = ni^2

Solving this equation we find Nother = 3.34*10^18

Problem 5

To treat avalanche from first principles, consider than an incident electron collides with the lattice and frees a hole-electron pair. Assume after the interaction, the three particles share the same kinetic energies. Also assume all have the same mass. Use conservation of energy and momentum principles to find that the threshold occurs at (3/2)Eg units of kinetic energy.

So this problem is simply physics. We know that initially we have 1 electron, and at the end we have 2 electrons and one hole. Thus, Ei = Eg + 3Ef, and mvo = 3mvf. Using the momentum conservation equation, we find that vf = 1/3v0. Knowing this is kinetic, know that Ei = 1/2m(vo)^2, and Ef = 1/2m(vf)^2. Substituting we find that

1/2m(vo)^2 = Eg+1/6m(vo)^2; moving to like sides we find -> Eg = 1/3m(vo)^2, thus m(vo)^2 = 3Eg. Finally relating this to Ei, divide both sides by two to find

Ei = 1/2m(vo)^2 = (3/2)Eg

Problem 6

Is Zener breakdown more likely to occur in a reverse-biased silicon or germanium pn junction diode if the peak electric field is the same in both diodes? Discuss. (Consider Band Gap)

Zener breakdown is attributed to the tunneling effect, therefore the smaller band gap in germanium would allow for a zener breakdown at a certain electric field over the silicon junction.

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