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This site provides the free electronic tutorials that teach the basic electronics principles needed to perform the interactive troubleshooting that we specialize in the Basic Electronics interactive Web..  



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Lecture for those who prefer to read rather than listen to lecture


Analog Electronics Lecture.


Ohms Law Lecture.

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 resistorís 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 Ohmís Law calculate the current in each resistor. If for example, We find that resistor number twoís 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 itís 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 resistorís 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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