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.

Notes on Semiconductors: P-N junctions: Linearly Graded Regions

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:

Poisson 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.

Relating holes with the linear graded limit
Rearranging Equation
Relating linear with electrons
rearranging eqn
Solving for Vbi

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.

Finding charge density
Setting up Poisson
Integrating Poisson
Solving for C
Finished Linear Electric Field

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

Potential relating to integral of E field

Finished Equation for Potential

Width of linear junction

For capacitance, it is the same as abrupt junctions:

Notes on Semiconductors: P-N 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.

This is what a schematic for a diode looks like

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).

Representation of PN junction fields from Wikipedia

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:

Generic Equation to base Vbi
Equation to find holes based of Fermi level
Rearranged Equation
Relationship between Fermi level and Voltage
Donor equation and Fermi level
Voltage at n side in respect to fermi level
Finished Equation

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)

Poisson Equation
Gradient of Electric field for n-type
Gradient of Electric field for p-type

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).

Representation of PN from Solid State Electronic Devices 6th edition
The reason for the bounds is due to the fact we are integrating from Electric field max to the electric field where it is 0 (boundary)
Integrating on p-type
Graph representing where Emax (Eo) from Solid State Electronic Devices 6th edition. At boundary xp, or xn, electric field must equal zero
Finding maximum electric field
Final relationship between xn and xp

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.

Relationship between E and V on p-side
Relationship between electric field and potential on p-side
Final for p-side in an abrupt junction

Integrating the N side we will find:

Relationship Between E field and Potential
Final Relationship for an abrupt junction

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:

Setting eqn to Zero
Relationship between turn on voltage and Electric field max using relationship of left equation and xn and xp
Definition of Width
Final equation to determine with of space charge region


Notes on Semiconductors: An Introduction

Preparing for a new semester, I’ve decided the best way to keep all my notes intact is to add them here to a website. My notes for semiconductors for example, are scattered upon 3 different textbooks and I’m personally annoyed to carry so much useless weight.

And so it begins:

Semiconductors are initially created from a metal, and is usually doped in order to allow a gradient that current can flow through. An INTRINSIC semiconductor is an un-doped material at which the statement n = p = ni is true.

Further explanation of n, p and ni: n stands for electrons that can be broken from their molecule and moved. p stands for holes which can be filled by electrons. ni is the material specific intrinsic carrier concentration. In a doped semiconductor, the equation n*p = ni^2
If doped with DONORS, the concentration Nd = n, if doped with ACCEPTORS, the concentration Na = p. If both were used the concentration depends on type. n = Nd-Na, p = Na-Nd.

With the general concept of holes, electron concentration, and doping covered, next is deriving a relationship between the energy band spectrum of the semiconductor and inferences to its type and concentrations. Normally, the intrinsic carrier concentration is known for the element, so solving for the type, and concentration of doping element is usually the problem structure.

Example of a band diagram

Note only the conduction band (Ec), Fermi level (Ef), the intrinsic level (Ei, which is in middle of band gap), and Valence band (Ev). In these semiconductor problems, you can identify the type off semiconductor by where a band labeled Ef is placed. If Ef is ABOVE Ei, then the semiconductor is N-type, if it is below Ei, it is P-type.

Now for the equations that relate to this->

Slides taken from: A.R. Hambley, Electronics, © Prentice Hall, 2/e, 2000 A. Sedra and K.C. Smith, Microelectronic Circuits, © Oxford University Press, 5/e, 2004


Note: K is the boltzmann constant (𝑘=8.62×10−5 eV/K) and T is temperature in Kelvin