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.

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

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

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,

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

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

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.

After taking a look at abrupt junctions, we can now move onto linear graded junctions. To solve for the equations relating positioning, doping, field, and potential, we shall be using Poisson’s Equation:

For linearly graded junctions, the graphs look like this:

Due to the symmetry of linearly graded junctions, we can equate xn = xp = xd/2. Based on this assumption, we can begin to solve for the turn on voltage, the electric field based on position x, and the potential based on x. First we will find the turn on voltage.

Once we have derived the turn on voltage, now we must relate the electric field based on what point you are in the linearly graded region.

Now with the electric field found, we must now solve for potential. Note, that potential is equal to the negative integral of the electric field

For capacitance, it is the same as abrupt junctions:

Familiar with P-type semiconductors and N-type semiconductors, coupled with the basic ways to calculate their values for holes and electrons, we can now move onto the next subject: P-N junction diodes. Diodes are shown by the following figure, and are characterized by allowing current to flow through them only if the voltage drop is greater than their turn on voltage. The diode will resist all current flow when the voltage dropped across its terminals are below the turn on voltage. The voltage dropped by the diode can usually be characterized by approximately 0.7 V, i.e., Vanode-Vcathode = 0.7V. The 0.7V is an example of a turn on voltage. If the voltage was less than this value, there would be no current flow.

Now, we must take a look under the hood of the diode, with the equations we currently know, to understand what physics determine our specified behavior. When a PN junction diode is created, there is a point where the two semiconductor material meets, and allows for a flow of electrons. Eventually, a equilibrium state is reached, at which point there is a space in the P side that is negatively charged (due to its doping atoms accepting electrons from the N type semiconductor’s donors), and conversely a space in the N side is positively charged. Remembering basic physics, we know that a separation of charges will create an electric field pointing from the positive charges to the negative charges. Recall that E = -grad(V).

Noticing this orientation, and direction of the electric field, one could wonder, what side of the PN junction is the cathode and anode? (Answer, the voltage drop has to overcome the electric field direction, thus the P side is the anode, and the N side is the cathode)

Now, it is important to create a relationship between the concentration of the doped atoms in either semiconductor, and the voltage needed to overcome the electric field to turn the device on. All of these variables will also be used to determine the length of the space charged region in either semiconductor.

Note the direction of the electric field in the diagram of a PN junction, and recall E = -grad(V), thus as the electric field increases, the voltage will decrease. Using this principle we can estimate that:

Note: The equations used in this post are for ABRUPT junctions. The graphs for abrupt junctions look like this:

Using the equations relating to holes and electron density in the semiconductor materials, we can derive the final equation to find the amount of voltage needed to turn the diode on. NOTE: the holes concentration and electron concentration are different in the two semiconductors, meaning it does not hold true that the doping concentration of the p-type, multiplied by the doping concentration of the n-type will equal the intrinsic concentration squared. That is, the two semiconductors have two separate equations to find the concentrations of holes and electrons.

With a relationship describing the turn on voltage, now we must create a relationship with the doping concentrations, applied voltage, turn on voltage, and depletion region width.

Using the Poisson Equation, we can make a relationship between the electric field (V/cm), the charge density (C/cm^3) and the permittivity (F/cm)

Now we have established what the gradient of the electric field for both sides of the PN junction in the depletion region. The xn and xp are the boundaries of the depletion (or charged) region. To make a relationship between this equation and electric field, we must integrate this equation. Due to the fact that we are only concerned with the gradient in the x direction, we are only mentioning the x direction in respect to voltage. Recall a gradient is equal to partial derivative x (i) + partial derivative y (j)+ partial derivative z (k).

With the relationship with electric field found, we must integrate again to find the relationship of the doping concentrations and voltage. Due to electric field being equal to the -grad(V), we know that the turn on voltage has to be equal to the electric field integrated from xp to xn. To find this value, we should take our previously derived equations for electric field along the p and n ranges, integrating them to find their associated voltage value.

Integrating the N side we will find:

To solve for the charged region aka depletion region, all one must do is set the equation to x = 0, where both p-type and n-type will equal each other.  Thus: