The following is a TCAD simulation of a high speed UTC photodetector. An I-V curve is simulated for the photodetector, forward and reverse. A light beam is simulated to enter the photodetector. The photo-current response to a light impulse is simulated, followed by a frequency response in TCAD.
Beam Simulation Entering Photodetector:
Frequency Response in ATLAS:
The full project (pdf) is here: ece530_final_mbenker
The following problems deal with polarizers, which is a device used to alter the polarization of an optical wave.
Unpolarized light of intensity I is incident on an ideal linear polarizer (no absorption). What is the transmitted intensity?
Unpolarized light contains all possible angles to the linear polarizer. On a two dimensional plane, the linear polarizer will emit only that amount of light intensity that is found in the axis of polarization. Therefore, the Intensity of light emitted from a linear polarizer from incident unpolarized light will be half the intensity of the incident light.
Four ideal linear polarizers are placed in a row with the polarizing axes vertical, 20 degrees to vertical, 55 degrees to vertical, and 90 degrees to vertical. Natural light of intensity I is incident on the first polarizer.
a) Calculate the intensity of light emerging from the last polarizer.
b) Is it possible to reduce the intensity of transmitted light (while maintaining some light transmission) by removing one of the polarizers?
c) Is it possible to reduce the intensity of transmitted light to zero by removing a polarizer(s)?
a) Using Malus’s Law, the intensity of light from a polarizer is equal to the incident intensity multiplied by the cosine squared of the angle between the incident light and the polarizer. This formula is used in subsequent calculations (below). The intensity of light from the last polarizer is 19.8% of the incident light intensity.
b) My removing polarizer three, the total intensity is reduced to 0.0516 times the incident intensity.
c) In order to achieve an intensity of zero on the output of the polarizer, there will need to exist an angle difference of 90 degrees between two of the polarizers. This is not achievable by removing only one of the polarizers, however it would be possible by removing both the second and third polarizer, leaving a difference of 90 degrees between two polarizers.
This post features an LED structure simulated in ATLAS. The goal will be to demonstrate why this structure may be considered an LED. Light Emitting Diodes and Laser Diodes both serve as electronic-to-photonic transducers. Of importance to the operation of LEDs is the radiative recombination rate.
The following LED structure is built using the following layers (top-down):
- GaAs: 0.5 microns, p-type: 1e15
- AlGaAs: 0.5 microns, p-type: 1e15, x=0.35
- GaAs: 0.1 microns, p-type: 1e15, LED
- AlGaAs: 0.5 microns, n-type: 1e18, x=0.35
- GaAs: 2.4 microns, n-type: 1e18
This structure uses alternating GaAs and AlGaAs layers.
For a Diverging lens, derive a formula for the output angle with respect to the refractive indexes and input angle. Assume paraxial approximation and thin lens.
For a Diverging lens, construct a derivation of Newton’s lens equation x_o*x_i = f^2.
Is the focal length of a spherical mirror affected by the medium in which it is immersed? …. of a thin lens? What’s the difference?
A spherical mirror may be either convex or concave. In either case, the focal length for a spherical mirror is one-half the radius of curvature.
The formula for focal length of a mirror is independent of the refractive index of the medium:
The thin lens equation, including the refractive index of the surrounding material (“air”):
The effect of the refractive index of the surrounding material can be summarized as follows:
- The focal length is inversely proportional to the refractive index of the lens minus the refractive index of the surrounding medium.
- As the refractive index of the surrounding medium increases, the focal length also increases.
- If the refractive index of the surrounding medium is larger than the refractive index of the thick lens, the incident ray will diverge upon exiting the lens.
Under what conditions would the lateral magnification (m=-i/o) for lenses and mirrors become infinite? Is there any practical significance to such condition?
Magnification of a lens or mirror is the ratio of projected image distance to object distance. Simply put, how much closer does the object appear as a result of the features of the lens or mirror? The object may seem larger or it may seem smaller as a result of it’s projection through a lens or mirror. Take for instance, positive magnification:
If the virtual image appears further than the real object, there will be negative magnification:
The formula for magnification is the following:
The question then is, how can there be an infinite ratio of image size to object size? Consider the equation for focal length:
For magnification to be infinite, the image distance should be infinite, in which case the object distance is equal to the focal length:
In this case, the magnification is infinite:
The meaning of this case is that the object appears as if it were coming from a distance of infinity, or very far away and is not visible. A negative magnification means that the image is upside-down.
How does the focal length of a glass lens for blue light compare with that for red light? Consider the case of either a diverging lens or a converging lens.
This question really has three parts:
- Focal Length of a lens
- Effect of light frequency (color)
- Diverging and Converging lens
Focal Length of the Converging and Diverging Lens
For the converging and diverging lens, the focal point has a different meaning. First, consider the converging lens. Parallel rays entering a converging lens will be brought to focus at the focal point F of the lens. The distance between the lens and the focal point F is called the focal length, f. The focal length is a function of the radius of curvature of both sides or planes of the lens as well as the refractive index of the lens. The formula for focal length is below,
(1/f) = (n-1)((1/r1)-(1/r2)).
This formula also works for a diverging lens, however the directions of the radius of curvature must be taken into account. If for instance the center of the circle for one side of the lens is to the left of the lens, one may chose that direction to be positive and the other direction to be negative; as long as one maintains the same standard for direction.
If the focal length of a lens is negative, meaning that the focal point is behind the lens, on the side at which the rays entered, this is a diverging lens.
Interaction of Color with Focal Length
The other part of this question dealt with how the focal length would change for one color such as blue versus another color such as red. The key to this relationship is the refractive index of the lens, as the refractive index can change with regards to the color (i.e. frequency).
The material from which the lens is made is not known, however as demonstrated by the following table, the refractive index is consistently higher for smaller wavelength colors.
Reviewing the focal length formula, it is understood from the inverse proportionality of the equation that as the refractive index increases, the focal length will decrease. Blue has a higher refractive index than red. Therefore, blue will have a smaller focal length than red.
The following project uses Silvaco TCAD semiconductor software to build and plot the I-V curve of a waveguide UTC photodetector. The design specifications including material layers are outlined below.
The structure is shown below:
Forward Bias Curve:
Negative Bias Curve:
Current Density Plot:
Acceptor and Donor Concentration Plot:
Bandgap, Conduction Band and Valence Band Plots:
Construct an Atlas model for a waveguide UTC photodetector. The P contact is on top of layer R5, and N contact is on layer 16. The PIN diode’s ridge width is 3 microns. Please find: The IV curve of the photodetector (both reverse biased and forward bias).
The material layers and ATLAS code is shown in the following PDF: ece530proj1_mbenker
Introduction to ATLAS
ATLAS by Silvaco is a powerful tool for modeling for simulating a great number of electronic and optoelectronic components, particularly related to semiconductors. Electrical structures are developed using scripts, which are simulated to display a wide range of parameters, including solutions to equations otherwise requiring extensive calculation.
The function of the PN junction diode typically fall off at higher frequencies (~3GHz), where the depletion layer begins to be very small. Beyond that point, an intrinsic semiconductor is typically added between the p-doped and n-doped semiconductors to extend the depletion layer, allowing for a working PN junction structure in the RF domain and to the optical domain. The following file, a P-I-N junction diode is an example provided with ATLAS by Silvaco. The net doping regions are, as expected at either end of the PIN diode. This structure is 10 microns by 10 microns.
The code used to create this structure is depicted below.
The cutline tool is used through the center of the PIN diode after simulating the code. The Tonyplot tool allows for the plotting of a variety of parameters, such as electric field, electron fermi level, net doping, voltage potential, electron and hole concentration and more.
Previously featured was an article that derived a matrix formation of an equation for a thick lens. This matrix equation, it was said can be used to build a variety of optical systems. This will be undertaken using MATLAB. One of the great parts of using a matrix formula in MATLAB is that essentially any known parameter in the optical system can not only be altered directly, but a parameter sweep can be used to see how the parameter will effect the system. Parameters that can be altered include radius of curvature in the lens, thickness of the lens or distance between two lenses, wavelength, incidence angle, refractive indexes and more. You could also have MATLAB solve for a parameter such as the radius of curvature, given a desired angle. All of these parameters can be varied and the results can be plotted.
Matrix Formation for Thick Lens Equation
The matrix equation for the thick lens is modeled below:
- nt2 is the refractive index beyond surface 2
- αt2 is the angle of the exiting or transmitted ray
- Yt2 is the height of the transmitted ray
- D2 is the power of curvature of surface 2
- D1 is the power of curvature of surface 1
- R1 is the radius of curvature of surface 1
- R2 is the radius of curvature of surface 2
- d1 is the thickness of the lens or distance between surface 1 and 2
- ni is the refractive index before surface 1
- αi is the angle of the incident ray
- Yi1 is the height of the incident ray
The following plots show a parameter sweep on an number of these variables. The following attachment includes the code that was used for these calculations and plots: optics1hw
The following set of notes presents first a trigonometric derivation of the thick lens equation using principles such as Snell’s law and the paraxial approximation. A final formula for the thick lens equation is rather unwieldy. A matrix form is much more usable, we will find. Moreover, a matrix form allows for one to add a number of lenses together in series with ease. Parameters of the lenses can be altered as well. Soon, the matrix formation of these equations will be used in MATLAB to demonstrate the ease at which an optical system can be built using matrix formations. The matrix formation of the thick lens equation can be summarized as three matrices multiplied, for the first curved surface, the separation between the next curved surface and the final curved surface. By altering the radius of curvature, the refractive indexes at each position, distances between them using these matrices, a new lens can also be made, such as a convex thin lens by inverting the curvature of the lens and reducing the thickness on the lens. A second lens can be added in series. Once a matrix formation is made handy, there are numerous applications that then become simple.
The following ray tracing examples all utilize Fermat’s principle in examining ray traces incident at a mirror.
Example 1. Draw a ray trace for a ray angled at a convex mirror.
The ray makes a 40 degree angle with the normal of the mirror at the point of incidence. In accordance with the law of reflection (Fermat’s Principle), the ray will exit at 40 degrees on the other side of the normal.
The above example shows a single ray at an angle. Often, rays are drawn together in a group of parrallel rays. This example shows how an incident set of parallel rays will no longer be parallel when reflected by a non-uniform (not flat) mirror surface.
This example brings up an important concept that happens especially with concave mirrors. Two rays drawn seem to be directed towards the same point, known as the focal point. A focal point however is only consistent for smaller angles. The third ray at the bottom makes a 55 degree incident angle with the normal of the surface. The reflected ray is also 55 degrees separated from the normal but is directed to the other side of the normal. The ray does not converge at the focal point as the others do. This effect is known as an aberration and may be discussed further at length in a later article.
This example makes use of the above concept of focal point. An object placed at the focal point will not make an image at the focal point. This is useful if for instance, some type of lense or collecter should be placed at the focus of the mirror. This can be done without worry for it causing disturbances to the image that is formed at the focal point by the reflected rays.
Assignment Sheet MIT OpenCourseWare – Quantum Physics I
PDF of solutions
Barton Zwiebach. 8.04 Quantum Physics I. Spring 2016. Massachusetts Institute of Technology: MIT OpenCourseWare, https://ocw.mit.edu. License: Creative Commons BY-NC-SA.
Of the four ways of manipulating light, these examples employ shaping of a lens and the refractive index to change the path of a ray.