Electronics Blog

Lesson 2342B-6 seaming contradiction question
Q. In lesson 2342B-6 I am confused by a seaming contradiction.
Paragraph 2 of Topic 17 describes the leads as shown in fig. 36a & fig 38a. However fig.38d shows what I believe to be a different configuration.
Is the flat sided transistor from fig38b a different configuration that I need to memorize next to that of fig36a or am I missing something? Do I then also need to memorize fig 38 b, c & e?
A. Memorization is such a double edged sword. While it does help in the short term, it defeats the purpose when we are exposed
to the information applied differently than we memorized.
Since it does not fit the same way, we typically cannot see how
to apply it to the new situation. In terms of the different lead arrangements, we have illustrated them using typical pictorial drawings.
If you are actually working with transistors, I would highly recommend that you refer to the specification sheet for the actual lead arrangement.
I tried doing what you are suggesting (memorizing), and blew up several until I realized that what I memorized as connected ECB was actually EBC.
It would be better for you to understand that different types may or may not have the same lead arrangement and to work with them as they come.


How to get the proper voltage reading in Lesson 2324B-8

Q. In Lesson 2324B-8 on page 39 the book shows that connecting a wire between the circuits gives a voltage reading of 9v between points A and B. I’ve been racking my brain trying to figure out how we came up with 9v. Can you please explain?

A. Let's see how the value of 9V was calculated.

The first portion of that figure shows there are two separate circuits involved. The one on the left is a series circuit of two resistors, while the one on the right only has one resistor being used as the one that is not connected on the left hand side cannot and will not have a current flow through it due to the fact there is no complete path through it.

Now if we analyze those two circuits based upon that information, there will be a 15V drop across R2, and a 6V drop across R4. We know those facts from the basis of 20V/20Ω = 1A and 1A x 15Ω = 15V, while the other is just 6V across the only component of 60Ω.

The meter connected between points A and B does not have a potential measurement due to the fact that there is nothing to make a common connection between those two circuits. Granted, you could try to say the meter is the common element, but as it is measuring volts and has a very high resistance when in that mode, it becomes less of a possibility.

Now when we do connect the common line as shown in (b), those two voltages we calculated before are still there, but now we do have the two playing nicely together in the same sandbox.

Therefore, when we measure from point A which is at the 15V potential, and point B which is at the 6V potential, we arrive at the potential difference between them of 9V. That is how we got to the answer.

Hopefully this answers your question, but if not or if there are other questions/concerns that come along as you progress through your lessons, please feel free to contact the instruction staff using the information below.

In addition, please make sure to include your student number in any and all communication with us, as well as putting that information on all exams submitted.

What are the three different transistor configurations?

Q. In my lessons, the three different transistor configurations are described. I am having trouble telling them apart.

A. A transistor has three elements: collector, base, and emitter. The input signal is applied to one element, and the output is taken off a second element. The one left over is the common element.

For example, say the input signal is applied to the emitter and taken off the collector. That leaves the base, so this describes the common base amplifier. The signal is AC. One side of the input signal source must go to AC ground. The other side must go to the input element of the transistor.

The output signal comes off another element. One side of the load to which the output signal is sent must be at AC ground, too. The remaining element of the transistor must also be connected to the AC ground. This is because current must flow from the input source, through the transistor, and back to the input source.

Output current must also from the transistor, through the load, and back to the transistor through the ground. Both of these currents, the input and output, must flow through the common transistor element. In these circuits, the ground acts as a common conducting path for the AC input current, the transistor, and the AC output current.

This is where the “common” comes in. Since the remaining transistor element is connected to AC common ground, we call the configurations common emitter, common base, and common collector.

Students are often puzzled by the common collector configuration. The collector goes to the power source, Vcc, instead of to ground. While Vcc is the source of the DC voltage for the circuit, Vcc is a short to ground for AC signals. In a power supply operated off the AC power lines, there are filter capacitors with high capacitance values. Their reactances are very low at all the signal frequencies.

Even in battery operated amplifiers there are usually electrolytic capacitors connected across the battery, since as the batteries discharge, their internal resistances rise, and can interfere with the operation of the amplifiers they power. The result is that these capacitors act as short circuits to ground for any AC signal current that flows to Vcc. This is why the common collector has its common element (the collector) at AC ground, even though it is connected to Vcc.

Why is my PC clock not accurate?

Q. I have noticed that the clock in my computer isn't very accurate. It seems to lose time at several minutes a week. Do I have a computer virus?

A. Not likely. The clocks in computers are not very accurate. This function is done by a special clock / calendar chip on the motherboard that runs off a battery. The time base for this chip is a crystal controlled oscillator that would not be affected by a virus. The software merely reads in the output from the chip at regular intervals. The crystals drift a bit, which throws the clocks off.

You can change the date and time by double-clicking on the My Computericon on your desktop. When the window opens, click on Control Panel,” and when the Control Panel window opens, click on Date / Time.” The window for this will then open, and you will see a calendar with the date highlighted and a clock on the right. Below the clock is a box you can use to set the time. You can use your mouse to set the date and the time zone you are in, too.

Q. I have learned about Class A, B and C amplifiers, but have heard that there is also a kind called Class D. What is a Class D amplifier, and how does it differ from the first three classes?

A. In a Class D amplifier, the power transistors are operated as switches. They are driven by pules so that they turn on and offvery rapidly. The duty cycles of their control pulses are made to follow variations in the signal voltage.

In other words, the on” and “off” times are varied, so that the output from the power transistors is a series of pulses of varying widths. The pulses are smoothed into a continuous waveform in the output by LC filter circuits.

The advantage of Class D is that the power transistors are either on or off. When they are on, the current through them is high, but the voltage across them is low.

Since power is the product of current times voltage, the power dissipated in the output transistors is low. The same is true when the transistors are off. Although the voltage across them is high, the current through them is low, so the power they must handle is also low.

The transistors spend very little time in the range where both the current through them and the voltage across them are substantial, and the power is high. Only during the relatively short rise and fall times of the pulses are they in this range.

(One instructor likens this to someone running quickly back and forth over hot coals. He spends most of his time at one end or the other of the coals, where the ground is cool, and little of the time actually over the hot coals.)

Consequently, these amplifiers can produce a powerful output signal without having to dissipate much power. Class D amplifiers are very efficient. Unlike Class C amplifiers, they do not require a tuned load. They are often used in the power output stages of audio amplifiers and radio transmitters.



Q. My lessons say that the current through a capacitor is 90 degrees ahead of the voltage. Why is this?

A. A capacitor draws current whenever the voltage across it changes. The faster the voltage changes, the more current it will draw. Now think about the shape of the sine wave for the voltage across the capacitor. It is steepest at the point where it goes through zero and changes polarity.

In other words the voltage changes fastest in this part of the AC cycle. This occurs at 180° and 360° in the sine wave for the voltage. So the current will peak at these points. The voltage change slows down and stops momentarily when its sine wave reaches 90° and 270°. The current reaches zero at these points. So the sine wave for the current peaks at 180° and 360°. It goes through zero at 90° and 270°. This describes a sine wave that is 90 degrees ahead of the sine wave for the voltage.

Capacitors value

Q. I was looking at the schematic diagram of an amplifier and I noticed that there was a 0.01 µF capacitor connected in parallel with the filter capacitor, which was 5,000µF.

The 0.01 µF capacitors value is so much smaller than the filter capacitors that it doesn't make any difference in the capacitance there.

Do you think this is a mistake?

A. No. The smaller value capacitor is there for a good reason. The 5,000µF capacitor is big, physically. It is basically two foil plates and an insulator rolled up into a tube. As a result, the two plates have some inductance.

At higher audio frequencies, and above the audible range, this inductance has a significant amount of reactance. This counteracts the capacitive reactance of the filter capacitor at these frequencies.

This would lead to oscillations in the amplifier, as some of the output signal would get coupled back to the input stages by way of the common power supply line.

The 0.01 pµF capacitor is much smaller, physically, than the 5,000µF one. It has far less inductance, so it is still effective at the higher frequencies. So it has a low capacitive reactance at the higher frequencies, and short-circuits the filter capacitor for them. This prevents oscillations.

Q. How does AC flow through capacitors? I thought electrons couldn’t flow through the dielectric in a capacitor.

A. You’re right, Electrons can’t flow through the dielectric in a capacitor because it is an insulator. But the electric fields from the charged plates on either side of the dielectric do pass through the dielectric.

On one half-cycle of the AC, the AC source puts a positive voltage on one plate and a negative voltage on the other plate. Electrons are drawn off the positive plate and an equal number are pushed onto the negative plate. No electrons pass through the dielectric -only the electric fields.

Electrons move through the rest of the circuit, though, as the capacitor starts to charge. But before it does, the polarity of the AC reverses. Now electrons are moved in the opposite direction. They are pushed onto the plate that had been positive and pulled off the plate that had been negative. Again, current flows through the rest of the circuit, but not between the plates.

In circuits like RC coupled amplifiers there are resistors in the current path. As a result, most of the AC voltage appears across these resistors. The voltage across the capacitor doesn’t have time to change because the AC changes polarity too fast. This is another way of saying that the reactance of the capacitor is low at the frequency ofthe AC.

DC, on the other hand, can’t get through a capacitor. This is because the dielectric is an insulator. Current will flow when the capacitor is first charged by the DC, with electrons flowing onto the negative plate and an equal number flowing off the positive plate. But once the capacitor has charged, no further current will flow.

Boolean equation truth table

Q. I am studying Lesson 5516. In Appendix A-5 in the back of the lesson there is a sample problem. It starts with a Boolean equation you are supposed to use to fill out a truth table. I don’t know how to do this.

A. Each term in the Boolean equation describes where ones are to be put in the output column of the truth table. For example, the first term in the equation is AB. This means that Whenever A is zero and B is one, there is a 1 in the output column, regardless of the states of C and D. So you put a 1 in the output columns where A, B, C, and D are 0100, 0101, 0110, and 0111. The next term is ABCD, with all the terms NOT-ed. This means that where A, B, C, and D are all zeros, we have a 1 in the output column. Only one line of the truth table meets this condition: the first one, where ABCD are 0000.

The next term is ABCD. So We put a 1 in the output column for 1111. In the last term we have A,B, C, D. This means that we put a 1 in the output column for 1011. Note that where there are four letters in a term, it refers to only one line in the truth table. Where there are three letters, two lines of the table will be involved. Where there are two letter, it refers to four lines of the truth table. If there is a one-letter term, eight lines will have a l in their output column.

Regulator IC voltages


Q. I am an electronics hobbyist. I like to build circuits for different applications. I recently built a regulated 5V power supply for experimental use.

I used a 7805 regulator chip, which is supposed to supply 1A of output current. I tested it by putting different power resistors across the output to get different amounts of load current. I then measured the output voltage.

I found that the output voltage started dropping off at much less load current than 1A. Apparently the regulator went into shutdown, because when I disconnected the load, the output voltage slowly came back to 5V. What is going on?


A. Most regulator ICs contain an additional circuit that shuts them down when the IC gets too hot.

The 1A current rating applies only when the IC is at 25°C (77°F). To keep the IC cool, you need to attach it to a heat sink. This is a chunk of aluminum with cooling fins. Heat sinks are rated in degrees Celsius per Watt. In other words, this is the temperature rise per Watt above the temperature of the air in the device’s environment. You also have to take into account the degrees per Watt of the interface between the device and the heat sink, and add this to the heat sink rating.

Most regulator ICs go into thermal limiting at about 1250€ (257°F). The ratings for the IC you are using are listed in its spec sheets.

It also helps to use the minimum possible voltage for the unregulated supply. The regulator IC is basically connected in series with the load, so all the load current, plus a small amount for the regulator itself, flows through the IC. The Voltage across the IC is, from Kirchhoff’ s voltage law, the difference between the unregulated voltage and the regulated voltage.

The power the chip must handle is thus about equal to this voltage times the load current. Keeping the drop across the regulator chip low helps minimize the power it must handle.

Regulator ICs sometimes fail, and some types do so by shorting between the input and output. This places the unregulated voltage across the load. Since you are building a 5V supply, your load is probably TTL chips. If the unregulated voltage is 7V or more, if the regulator fails, it will destroy the TTL chips, too. There are ways to prevent this, such as using a “crowbar” circuit across the output, which you may want to investigate.