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EE80T – Modern Electronic Technology and How it Works
Winter 2009
Due in class Tuesday February 10, 2008
Turn this sheet in separately from your other homework problems. Please
be sure to put your name on this topic choice sheet and the other part of
this homework too.
1. You will be writing a term paper later on in this course. The paper will be on a
topic of your choice, subject to approval here. In your paper you will be expected
to investigate some aspect of electrical or electronic technology, its historic
precedent, its invention, the people involved, and how it works. You need to
choose a topic not covered in the class, or if it is, propose specifically how you
will cover it in greater detail than it is covered in the course. It will be around 5
pages, certainly no more than 10, typed and submitted both electronically in Word
format and as a printed copy. Choose three topics that interest you and list them
below in your order of preference. Your paper topic will be selected by me from
this list with consideration paid to your order of preference. For the next
homework set you will be preparing your abstract and outline for your paper. The
purpose of this is to insure that you choose a topic that allows you to write a good
paper. If, when you get this back, you don’t like my choice or can better explain
what you intend to do in person don’t hesitate to see the course instructor.
Proposed topics:
a. __________________________________________________________
b. __________________________________________________________
c. __________________________________________________________
EE80T – Modern Electronic Technology and How it Works
Winter 2009
Due in class Tuesday February 10, 2008
2. Impedance of Capacitors - Capacitors have a different impedance as a function of
frequency and their rated value of “C”. For frequency values listed below, what is
the magnitude of impedance |Z| for a 10 pF capacitor? Picofarad-range capacitors
are often found in Radio Frequency (RF) and microwave circuits and are often
times integrated onto silicon chips.
a. at 60 Hz, The AC power line frequency in North America
b. at 20 KHz, Audio frequency applications, this is around the highest frequency
that we can hear.
c. at 1100 KHz, roughly the carrier frequency of a radio station located near the
center of the AM radio band.
d. at 103 MHz, a carrier frequency in the FM radio band.
e. at 900 MHz, Ultra High Frequency (UHF) band upper edge, used for analog
cable TV channels
f. at 2.4 GHz, A carrier frequency in the 802.11b wireless Local Area Network
(LAN) band, “WiFi”
3. Impedance of Inductors - What is the |Z| of a 20 nH inductor for the frequencies
listed in problem 2 above?
4. Low Pass Filters – The circuit below shows how the voltage divider concept can
be applied to understand how a filter works. A filter passes certain frequencies
and attenuates others between its iput and output terminals. For example, a lowpass filter allows the low frequencies of a signal to pass through from the input to
the output while at the same time, it attenuates the high frequencies in the same
signal. From problem 2 and 3, we saw that the impedances of capacitors and
inductors change depending on the frequency of the signal. Thus, filter circuits
often use capacitors and inductors.
636 ?
1 nF
a. For the circuit above, calculate the ratio |Vo/Vs|, output voltage across the
capacitor over input voltage as a function of ω the frequency.
b. Calculate the frequency ω where the magnitude of the ratio |Vo/Vs| as a
function of ω is equal to 1/√2. In other words find ω so that Vo/Vs equal 0.707.
(this is the point where the power has been attenuated by a factor of 2.)
c. Calculate the value 1/ (RC) and compare it against the answer found in b.
d. On the graph next page, plot the following frequencies ƒ vs. the |Vo/Vs|
magnitude. Remember that ω = 2*π* ƒ. Plot ƒ = 1 Hz, 10 Hz, 100 Hz, 1,000
Hz, 10,000 Hz, 25,000 Hz, 100,000 Hz, and 1,000,000 Hz. The graph template
provided is for what is called a log-log plot. The scales are laid out using the
logarithm of the numbers shown on the axes. This is used so that we can cram
a wide range of frequencies and amplitudes on one piece of paper (or one
computer screen). Very useful for lots of things. In particular quantities that
vary according to a power law show up as straight lines on log-log plots.
M agnitud e of |V o/V s |
Frequency of Vs [Hz]
5. High Pass Filters – Similar to problem 4 where we use a capacitor to filter out high
frequencies from the source voltage, the circuit below uses an inductor to filter out lower
frequencies from the source voltage. The schematic below shows the high pass filter
circuit. Again, a voltage divider circuit is used and the output is the voltage across the
inductor. Notice that in this circuit, the output is taken across the inductor, compare this
to the circuit in problem 4 where the output is taken across the capacitor.
10 ?
1.6 mH
a. From the circuit above, calculate the ratio |Vo/Vs|; the output voltage across the
inductor over input voltage as a function of ω the frequency.
b. Calculate the frequency ω where the magnitude of the ratio |Vo/Vs| as a function
of ω is equal to 1/√2. In other words find ω so that Vo/Vs equal 0.707. Again
this is the half power point, where only half the available power is showing up
across the output.
c. Calculate the value R/L and compare it against the answer found in c.
d. On the graph next page, plot the following frequencies ƒ vs. the |Vo/Vs|
magnitude. Remember that ω = 2*π* ƒ. Plot ƒ = 1 Hz, 10 Hz, 100 Hz, 1,000 Hz,
10,000 Hz, 100,000 Hz, and 1,000,000 Hz. The scales on the following graph
papers are different than the last, be sure to use the one that follows.
M agnitude of |V o/V s |
Frequency of Vs [Hz]
6. Narrow Band Frequency Selection Using LC resonant problem – LC resonant
circuits are used to tune radios. Their resonance frequency can be tuned to the
carrier frequency of the station that you want to listen to. To see how this works
consider the schematic below that shows the antenna wired across a resonant
circuit. The antenna intercepts electromagnetic waves from the transmitter and in
so doing a voltage appears between its output and ground. Real antennas also
have some resistance associated with them. The schematic below is a simplified
model of the circuit in which the antenna is modeled as a voltage source with a
series resistance RS. The circuit uses an LC resonator immediately after the
reception antenna to tune to the desired frequency. The operation of this circuit
can be understood as a type of voltage divider circuit in which the input voltage is
divided between the antenna resistance and the equivalent impedance of the LC
parallel combination in proportion to their impedance. This is a “bandpass” filter
and works pretty much like the high pass and low pass filters we considered
a. From the circuit above, calculate the resonant frequency of the LC circuit
where ƒ = 1 / (2 * π * √(LC) ).
b. Although we haven’t proved it, at this frequency the equivalent impedance
of the L and C in parallel gets very very large (approaches infinity, an
open circuit if there are no losses). If this is the case what proportion of
the input voltage will appear across the indicated output terminals?
c. Calculate the impedance of the inductor and capacitor at ƒ = 100 Hz. Next,
use the calculated impedances of the capacitor and inductor to find the
combined equivalent parallel impedance of the capacitor and inductor at ƒ
= 100 Hz. Assume that the impedances combine like resistances in
parallel (although we haven’t proved it they do). The combined
impedance is close to the impedance of which component alone? Using
the voltage divider concept, calculate the ratio Vo/Vs at ƒ = 100 Hz.
d. Now repeat part b except that ƒ = 1MHz. (note: this gives an incorrect
result for the combined impedance, in reality the combined impedance is
much bigger, this has to do with the slightly mysterious “phase” which we
are ignoring in this class, but if you are going on in this field you will need
eventually to understand, come to office hours if you want to know more
about this...)
e. Now repeat part c again except that ƒ = 100MHz.
Now by comparing the answers to part b, c, and d, we can see how LC tank
circuits are used as frequency selection circuits. This is an example of tuning
a radio is something that is relatively easy to understand in the “frequency
domain” but seems truly miraculous in the “time domain”. If this doesn’t
surprise you then perhaps you haven’t been paying attention…
7. Antenna Lengths – For all of the frequencies listed in problem 2, please
determine the lengths of the antennas that can be used for each one of those
frequencies. Remember, we would like to have the antenna to be half the
wavelength at the frequency of operation. To find the wavelength, we use the
equation c = ƒ*λ, where c is the speed-of-light, ƒ is the frequency of the wave,
and λ is the wavelength of the wave. From this you can see why high frequencies
are used for cellphones and truly giant antennas are used for AM radios. A
different (and less efficient) sort of antenna choice is made for compact AM radio
8. Series and Parallel Capacitors – Much like resistors, capacitors connected in
series and parallel configurations also can be combined into effectively one
capacitor mathematically. Please find the equivalent capacitance of the capacitors
connected in the configurations as shown. Assume each capacitor is 10 nF. The
formulas for capacitors and inductors in series and parallel come directly from a
generalization of the series and parallel resistor formulae to apply to impedances
combined with a bit of mathematical elbow grease.
You might actually want to do this to increase the maximum voltage that
you can put across this effective capacitance.
This type of connection is used to make a bigger capacitor out of smaller
This is probably something that you would never want to nor try to do, it
is however a good learning exercise to figure out how to apply the series
and parallel capacitor formulas. It is good for you, like broccoli. Do this
like you did for similar problems using resistors. This approach is much
simpler than using the fancy math ways of doing this in our advanced
9. Series and Parallel Inductors – Much like resistors and capacitors, inductors
connected in series and parallel configurations also can be combined into
effectively one inductor mathematically. Please find the equivalent inductance of
the inductors connected in the configurations as shown below. Assume each
capacitor is 2 μH.
10. Series Capacitors and Voltage Handling – Capacitors are made by having an
insulating dielectric sandwiched between two parallel conducting plates. The
electric field points from the positive plate to the negative plate through the
dielectric material in between. The energy is stored in this electric field.
However, the insulating dielectric material can only handle a maximum electric
field before the electric force pulling the negative electrons away from the
positive nuclei becomes so big that the material breaks down and current flows,
causing permanent damage and usually a conducting path through the former
insulator; the capacitor is said to have shorted. In essence, the voltage between
the two plates of the capacitor has a maximum limit. However, to get around this
limitation, capacitors can be placed in series. By doing so the voltage is divided
across the two capacitors. That is, is V is placed across the pair in series then
(assuming that they have equal capacitance) V/2 will appear across each one.
Please answer the following questions based on your new found knowledge of
a. In order to build a Tesla coil, we need to be able to have a capacitor that
can operate at 10 kilovolts. However, we only have capacitors that can
handle 2.5 kilovolts with a capacitance of 1 uF. How should we connect
these capacitors and how many do we need in order to create one
equivalent capacitor that can handle 10 kilovolts?
b. Find the effective capacitance of the equivalent capacitor created in part a?
c. What is the energy stored in the “large” capacitor created in part a when it
is fully charged to 10 kV? This would easily electrocute a turkey if
delivered in one shot…
11. Frequency Conversion in Radios – In FM radio receivers, the modulated carrier
wave operating in the 87.5 MHz to 107.9 MHz range is converted into a
modulated signal at a different frequency called the IF (Intermediate Frequency)
before going through tuned amplifiers that only amplify the signal over a narrow
band of frequencies around the IF. The IF frequency for FM receivers is at 10.7
MHz. A device called a mixer shifts the incoming FM radio signal from the
antenna to 10.7 MHz IF. The mixer multiplies the signal of the local oscillator
with the received FM signal and in so doing down converts the carrier fo the
modulated signal to the 10.7 MHz IF. Essentially, the mixer takes the frequency
of the received FM signal and the frequency of the local oscillator and produces
the sum and difference frequencies at its output. For down conversion we are only
interested in the difference frequency and the high frequency companion is
filtered out using a low pass filter. The following diagram illustrates the mixer.
a. If we desire to listen to a station at 103.7 MHz and the frequency we want
to convert to is the IF of 10.7 MHz, what frequency should you tune the
local oscillator (LO) to so that modulated 103.7 MHz signal is converted
to a modulated signal at 10.7 MHz?
Questions from Reading
12. Name the German scientist that experimentally verified Maxwell’s theories on
13. What is the device that Edison unwittingly invented when he observed the process
known as the “Edison Effect”?
14. Name the scientist who first grasped the concept of the television system by
sending a series of rapid images, which he called “electric telescope”.