**
****Click number 1 of numbered link array.** There are two quantities that
technicians keep track of in order to monitor circuit performance, for the
purpose of fault isolation. They are current and voltage. The top circuit on
this page shows current as a function of voltage. Or otherwise stated, it
animates the equation current = voltage divided by resistance. This relationship
is known as Ohms Law. Voltage is usually the easiest quantity to measure,
because it can be read simply by probing a single point. You can record the
voltage on both ends of a resistor, and then read the resistor value from the
circuit schematic. With this data Ohms Law can be used to calculate the current
flowing through the resistor. The Animation shows current increasing as the
voltage increases.

**
****Click number 2.** This circuit shows that when a constant current goes
through a variable resistance the voltage increases in proportion to the
resistance. This is in accordance with Ohms Law in the form of Voltage = Current
times Resistance. Power dissipation in a resistor is proportional to the product
of current and Resistance. In this case current is constant; therefore, power
dissipation is directly proportional to the voltage. This relationship is
clearly demonstrated by the animation, when voltage and power are observed. Note
that the power meter is connected in series with and across resistor, so that
meter monitors both current and voltage.

**
****Click number 3.** This animation shows that the current through a fixed
resistor is proportional to the applied voltage. Power is proportional to the
product of voltage and current which are both increasing in step. From the
equation Power = Voltage times current it follows that Power = voltage squared
divided by resistance. Simply substitute voltage divided by resistance into the
equation, if you care to do the algebra. Remember to convert milli-amps, and
milli-volts, to volts when you calculate power in units of watts. It is easy to
do this using scientific notation.

**
****Click number 4.** In a series circuit the current is the same in each
resistor or component in the circuit. Thus, the total voltage across the circuit
equals the current times the total resistance in the circuit. Since measuring
current involves breaking the circuit to insert a ammeter, we usually prefer to
measure voltages with respect to ground using one probe or across components
using two probes. In a series circuit the voltage across any resistor should be
proportional to the resistance of the resistor. A method that always works is
measuring the voltage across each resistor and calculating the current through
each resistor. If one resistor’s calculated current is different from the
other two or more resistors in the series circuit, then that resistor is suspect
of being defective. I use the term "suspect" because until you verify
failure by testing the component or fixing the problem by part replacement,
troubleshooting is not complete.

**
****Click number 5.** In a parallel circuit the voltage across each resistor is
the same. Thus, current is strictly a function of resistance. In the animation
shown on this page resistors are switched in and out of a parallel circuit,
causing total current to the circuit to vary. You are asked to deduce from the
total current variations whether the circuit is defective, and if it is
defective you should determine which resistor is suspect. Previously I stated
current is not easy to measure. However, total current readings are frequently
available. Sometimes they can be read directly off a laboratory power supply.
Sometimes it is easy to connect a current probe around a wire between the power
source and the circuit board.

**
****Click number 6.** Most electronic circuits are a combination of series and
a parallel components. When we analyze parallel, series circuits we reduce the
circuit to an equivalent purely series circuit or equivalent purely parallel
circuit. For example, when the two milli-ampere range is selected, resistor one,
two, and three can be combined into a single resistor, that we can call
"branch one total". Now we have a parallel circuit consisting of
"branch one total" resistance and the meter resistance. We can now
calculate the current that should be flowing through the three resistors and the
resistance of the meter. Troubleshooting is often much easier. If a circuit has
a fault, We can simply measure the voltage drop across each resistor of the
voltage divider. Than using Ohm’s Law calculate the current in each resistor.
If for example, We find that resistor number two’s calculated current is
different than the other two calculated currents, we know that "resistor
number two" is the defective component. Our troubleshooting is complete. If
we find that the voltage divider is not defective, than the Meter must be the
problem.

**Click number 7.** Here you see a Java Script version of the same circuit
that we just saw in GIF format. Observe that the Unit Under Test Meter reading
is not intended to simulate a digital meter, but indicates the position of a
needle on an analog meter movement. The Meter reads pegged when the reading is
more than 105% of full scale. You can study normal operation of the meter or
insert a fault. You must step the voltage up or down in order to activate the
simulation. Then you will have total control of the power supply, and meter
switch position. You can also select which point you wish to probe with your
virtual voltmeter.

AC Electricity Lecture.

**
****Click number 8.** This page has three animations demonstrating the nature
of alternating current. The first animation demonstrates that current is a
function of voltage and frequency in a purely capacitive circuit. Note that the
current and voltage are ninety degrees out of phase at all frequencies in this
purely capacitive circuit. The second animation on this page shows effect of a
capacitor on a square wave. The time it takes a capacitor to discharge to 37% of
it’s initial voltage through a resistor is called the R, C time constant.
Measure the time constants for .01, .02, .05 and the .1 micro-farad capacitor by
observing their respective waveforms. Next, calculate the time constant for each
case. Try to do this calculation in your head. Note that the product of 1 kilo
ohm and one micro-farad equals one millisecond. Compare the calculated values
with those acquired by observation of the waveform. The advantage of being able
to make calculations in your head is that you do not have to put down the probe
and pick up a pencil or calculator while troubleshooting. This can greatly speed
up the troubleshooting process. The third animation at bottom of page shows the
effect of a resister and capacitor on a sine-wave signal. Circuits like this
must be analyzed using vector geometry or complex algebra. However, at
frequencies of 5 kilohertz and above, or when capacitance reaction is small with
respect to the resistor’s resistance, one can roughly approximate the portion
of the applied voltage across the capacitor as the ratio of capacitive reactance
over the total impedance. In this case, you can simply add the total capacitive
reactance to the resistance algebraically. No Vector summing required. However,
you can also make vector summation easier by using online calculators that are
available at links near bottom of page. You can also build your own calculators
using Java Script. Go to my Java Script site to learn how. I emphasize
estimation techniques because, I started troubleshooting when calculators were
called slide rules. However, until voice activated technical programs are
available for your laptop, calculations not done in your head will slow down
troubleshooting. Voice activated programs are already in use on personal
computers, but technicians represent only a small market for such a specialized
program. Therefore, such a program might be a long time in coming.

**
****Click number 9.** You are now looking at a 5 element Chepyshev High Pass
Filter. The complex circuit response curve voltage amplitude fall off rate is
far greater than the 3 decibels per octave fall off rate of a simple resistor
capacitor filter. This filter is designed for a fifty-ohm signal source and a
fifty-ohm load. Note that at high frequencies the filter acts almost like a
short. If you think of the filter as a short, the circuit reduces to a voltage
divider consisting of two fifty-ohm resistors. Thus, at all frequencies the
resistors divide the voltage by a factor of two, which is a three decibel
reduction in voltage, and a six-decibel reduction in power. Note that at some
frequencies the output is greater than the input at test point one. This makes
sense if you consider that the capacitors and inductors form series resonant
circuits.

**
****Click number 10.** The Chepyshev Low Pass Filter is similar to the
Chepyshev High Pass Filter with respect to fall off rate and number of
components. However, the coils pass low frequencies, while the capacitors
shorted very high frequencies to ground. In the High Pass filter the capacitors
passed high frequencies and the coils shorted low frequencies to ground. The
Radio Frequency and Microwave filters that you are likely to encounter in
industry will be inside of metal containers. This type of construction prevents
radiation output and external radiation. It also provides a defined ground plane
for the components. The metal container does not eliminate stray capacitance,
but a design engineer can consider it a constant rather than a random factor.
This also simplifies fault isolation, because repair is accomplished simply by
replacing the entire filter.

End of A-C Electricity Lecture.

Semiconductor Circuits Lecture.

**
****Click number 12.** Here, we see a GIF animation of a full wave rectifier.
On the positive half cycle of the input voltage the capacitor is charged through
diode number one, and on the negative half cycle the capacitor charges through
diode number two. Observe, that when the capacitor is only one micro-farad, the
output is not significantly filtered, and we see a rectified waveform. As the
size of the filter capacitor increases so does the degree of filtration. At a
thousand micro-farads the ripple voltage is almost totally eliminated; however,
the capacitor is too large and expensive. A voltage regulator is a far more
practical way of obtaining very low ripple voltage. The 47 micro-farad capacitor
reduces ripple voltage to a level sufficient for most voltage regulators.

**
****Click number 13.** In this circuit all resistors are in ohms. Unregulated
power source is 15.0 dc. 2.0 ohms represent the internal resistance of the power
source. The Zener diode provides a reference voltage of 7.4 volts. The load will
be varied during testing from 0 to 100 ohms. Regulation is specified as
indicated by curve below. Zener Diode maintains transistor base voltage at 7.4
volts. When the load increases, the voltage output drops. When the voltage
output decreases, the base emitter forward bias increases causing transistor
current to increase. In turn, voltage output increases. The reverse occurs when
voltage output drifts upward. I describe the behavior of this circuit using a
graph of the voltage output verses load. Voltage is monitored at several test
points and plotted on graph. The good unit test results are the specification
for the circuit. The other graphs show output of defective circuits. The graphs
for the defective circuits also plot several test points that make it possible
to fault isolate to the component level.

**
****Click number 14.** Below we see signal generator providing a one-milli-volt
signal to an amplifier circuit. The A-C voltage verses frequency is plotted for
four points. They are as follows: the Signal Generator output, the transistor
base, the transistor collector, and the amplifier output. Zero decibels
correspond to the one-milli volt generator output. The Base Line curves are for
the good unit. You may troubleshoot using the plots for the three faulty units.

**
****Click number 15.** The lower partial view of the complete circuit shows the
1 kilo-hertz signal levels at the Base, Emitter and Collector of the transistor.
Note the effect of the capacitors being switched in and out with the relays. The
capacitors bypass the 1kHz signal to ground. This maintains the emitter voltage
at a steady D-C level. This increases the A-C signal seen across the
Base/Emitter junction, which increases the A-C signal gain of the circuit.

**
****Click number 16. **Operational amplifiers have extremely high gain and draw
almost no current at their inputs. A cheap operational amplifier will have a
gain of over a hundred thousand. Thus, even a milli volt of input will produce
an output of over a hundred volts. Actually the outputs max out or saturate near
the supply voltages. Plus and minus 15 volts are typical supply voltages.
Feedback resistors are always used with operation amplifiers in order to keep
them from saturating. The feedback resistors always keep the voltage difference
at its inputs near zero. In the inverted amplifier animation, one of the inputs
is tied to ground; therefore, the other input is held to zero by virtue of
feedback. A few milli-volts might exist between the inputs of an operation
amplifier due to the offset voltage of the op amp. You can simply calculate the
current going through the input resistor as the input voltage divided by the
value of the input resistor. That is also the value of the current going through
the feedback resistor. Thus, the output voltage is simply the product of this
current and the feedback resistance. Note that the polarity of the output must
be opposite to the input voltage if zero voltage is to be maintained at the
junction of the two resistors. A simple method of troubleshooting an op amp is
as follows: First check that the voltage between the inputs is near zero. If the
voltage is not near zero, the output should be in saturation. If it is not in
saturation, the amplifier is defective. If it is in saturation the operational
amplifier or the resistors could be at fault. If the op amp inputs are near zero
and the amplifier is not saturated, than incorrect or operation of circuit is
due to resistor failure. Resistor value changes will result in a linear failure.
That is the gain of the circuit will change even at low frequencies. The gain of
the circuit falling at to fast a rate, causing the band-pass of the circuit to
be out of specification indicates a defective operational amplifier. Frequency
response is a property of the amplifier not the resistors.

**
****Click number 24. **We are now looking at a transistor regulated power
supply. Observe Baseline operation of regulated power supply below. Then
troubleshoot fault number one to fault number four below. This voltage regulator
simulation uses a battery rather than a rectifier as a power source. Since
batteries do not generate ripple voltage, I provided a dynamic load to
demonstrate that voltage regulators not only adjust for slow changes in current
but also can attenuate ripple voltages. They also adjust for rapid changes in
load current demands. The frequencies I used were 50 hertz, 60, 400, and 1200
hertz corresponding to European, American, single-phase 400 hertz and three
phase 400 hertz ripple frequencies respectively. Thus, it illustrates the effect
of the regulator on ripple voltage or rapid load changes.

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