Solved Assignment 3 - Microelectronic Circuits | ECE 3040, Assignments of Electrical and Electronics Engineering

Material Type: Assignment; Professor: Doolittle; Class: Microelectronic Circuits; Subject: Electrical & Computer Engr; University: Georgia Institute of Technology-Main Campus; Term: Unknown 1989;

Typology: Assignments

Pre 2010

Uploaded on 08/05/2009

koofers-user-s3b-1
koofers-user-s3b-1 🇺🇸

10 documents

1 / 2

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
ECE 3040 Homework 3
The goal of this homework is to design a photocell (also known as a photoelectric cell or photoelectric resistor)
that will respond to light and change it’s resistance. This device may be used in a later homework problem.
The photocell is basically nothing more than a resistor made out of semiconductor material and is typically
arranged in a package as a long skinny slither of very thin semiconductor, most often CdS due to it’s near
optimal bandgap for adsorbing the solar spectrum and relative inexpensive materials. While a real device is has
a long serpentine shape (see below) we will simply use a rectangular block as was done in class.
The working specifications of your design are for a resistance of 1 Giga-ohm in dark and 3Mega-ohms in bright
light (Intensity, Io=50 mW/cm2 light intensity).
Given:
The mobilities of un=100 cm2/V-sec and up=500 cm2/V-sec
The CdS material is p-type with Na=1e14 cm-3 and ni=2e6 cm-3.
Note: For this problem it is okay to assume low level injection and thus it is okay to use the formula:
BUT if this were a real world problem, the type of variations we want to achieve in resistance are so large that
High level Injection conditions apply and strictly speaking we should use (do not do this in this homework) the
formula:
1) It is determined that the resistor “trace” (width viewed from the light’s perspective as shown in figures above)
will have a width of 0.2 mm and a thickness (same as depth “L” in our light absorption lecture 9) of 4 um.
What is the needed length to achieve 1 Giga-ohm resistance in the dark?
2) Assuming that the bandgap energy is 1.4 eV and that the light has the same photon energy (1.4 eV) and is
completely absorbed uniformly in the semiconductor, what is the generation rate, GL in the semiconductor?
Hint: calculate the flux of photons in 1 cm2 from the power density (intensity) and the amount of energy needed
to generate each electron-hole pair, then find how much of this flux is incident on the photocell given your
results from part 1, then convert this to a generation rate using the assumption of uniform absorption/generation
throughout the thickness of the semiconductor film.
n
GRthermal
n
t
n
τ
=
)()( 11
2
ppnn
npn
t
p
t
n
np
i
GRthermalGRthermal +++
=
=
ττ
pf2

Partial preview of the text

Download Solved Assignment 3 - Microelectronic Circuits | ECE 3040 and more Assignments Electrical and Electronics Engineering in PDF only on Docsity!

ECE 3040 Homework 3

The goal of this homework is to design a photocell (also known as a photoelectric cell or photoelectric resistor) that will respond to light and change it’s resistance. This device may be used in a later homework problem. The photocell is basically nothing more than a resistor made out of semiconductor material and is typically arranged in a package as a long skinny slither of very thin semiconductor, most often CdS due to it’s near optimal bandgap for adsorbing the solar spectrum and relative inexpensive materials. While a real device is has a long serpentine shape (see below) we will simply use a rectangular block as was done in class.

The working specifications of your design are for a resistance of 1 Giga-ohm in dark and 3Mega-ohms in bright light (Intensity, Io =50 mW/cm^2 light intensity). Given: The mobilities of u (^) n =100 cm^2 /V-sec and u (^) p =500 cm^2 /V-sec The CdS material is p-type with Na=1e14 cm-3^ and n (^) i=2e6 cm-^.

Note: For this problem it is okay to assume low level injection and thus it is okay to use the formula:

BUT if this were a real world problem, the type of variations we want to achieve in resistance are so large that High level Injection conditions apply and strictly speaking we should use (do not do this in this homework) the formula:

  1. It is determined that the resistor “trace” (width viewed from the light’s perspective as shown in figures above) will have a width of 0.2 mm and a thickness (same as depth “L” in our light absorption lecture 9) of 4 um. What is the needed length to achieve 1 Giga-ohm resistance in the dark?

  2. Assuming that the bandgap energy is 1.4 eV and that the light has the same photon energy (1.4 eV) and is completely absorbed uniformly in the semiconductor, what is the generation rate, GL in the semiconductor? Hint: calculate the flux of photons in 1 cm^2 from the power density (intensity) and the amount of energy needed to generate each electron-hole pair, then find how much of this flux is incident on the photocell given your results from part 1, then convert this to a generation rate using the assumption of uniform absorption/generation throughout the thickness of the semiconductor film.

n

thermal RG

n

t

n

τ

2

n n p p

n np

t

p

t

n

p n

i

thermal RG thermal RG + + +

− − τ τ

  1. If the semiconductor has dimensions as defined by you in part 1), what minority carrier lifetime is needed to have the desired “Light on resistance” of 3 Mega-ohms?

  2. If 120 volts DC is applied to the resistor (i.e. applied along the longest dimension) and if the semiconductor is assumed to be a single rectangular resistor (not serpentine as shown in the figures above but simply a block), a. What is the electric field on the resistor? b. What is the electron current DENSITY in dark? c. What is the electron current flow in dark? d. What is the hole current DENSITY in dark? e. What is the hole current flow in dark? f. What are the answers to b, c, d, and e in the light?

  3. If the light is instantly turned off, sketch and label the electron concentration as a function of time clearly marking the equilibrium value, no, the total concentration, n, and the excess electron concentration, ∆n.

  4. Purpose: Demonstrate the relationship between diffusion current and excess minority carrier concentration. For the semiconductor above (photocell) a non-uniform light source is exposed on the cell such that the excess electron concentration is described by the following equation:

1 13 114 sin ⎟ −

∆ = + cm

S

x

nx e e

Where S is the length you determined in problem 1. What is the steady state minority carrier current density (no electric field in this case) at all positions in the photocell?

  1. Strong suggestion: Review the minority carrier diffusion equation problems on the web test solutions page.