# 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

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

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. 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

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

### 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 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

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.

## Problem 3

### 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,

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.

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

## 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:

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:

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.