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Plate Dissipation

When electrons are attracted from the space charge to the plate, they are accelerated by the attraction. Their gain in speed gives them energy that causes them to strike the plate with a considerable force. As the electrons strike the plate, this energy is converted to heat. The plate must be able to withstand the associated increase in temperature. The maximum amount of power (watts) that a given plate can safely dissipate (as heat) is called the PLATE DISSIPATION rating.

To find the amount of plate dissipation for a given tube under a particular set of plate conditions, use the following equation:

0018.GIF (1269 bytes)

This is a relatively small wattage. It's probable that the plate of our example diode is not overheating. A tube manual could tell us for sure.

Plate dissipation is a circuit loss that must be made good by the power source in a circuit. In our example, this is the plate voltage supply.

Peak Current Rating

The maximum instantaneous current that a tube can pass in the normal direction (cathode to plate) without damage is called the PEAK CURRENT RATING. Peak current rating is determined by the amount of electrons available from the cathode and the length of time plate current flows.

Peak Voltage Rating

This is the maximum instantaneous voltage that can be applied to a tube in the normal direction without a breakdown.

Peak Inverse Voltage Rating

This is the maximum voltage that can be applied to a tube in the reverse direction (plate negative with respect to the cathode)-exceeding this will cause arc-over from the plate to the cathode and will damage the tube. PIV, as this is sometimes abbreviated, becomes very important in the rectifier circuit to be discussed as a later major subject.

Transit Time

Things that happen in electricity and electronics are often explained as if they happen instantaneously. As fast as electricity acts, however, the truth is that cause and effect are separated by a certain amount of time.

Each tube has a factor called TRANSIT TIME, which is the time required for an individual electron to move from the cathode to the plate. In certain applications involving high-frequency voltages, transit time places a limitation on tubes. We will explain this limitation when we discuss the circuits it affects.

Summary of Diode Parameters and Limitations

You should now have a basic understanding of diodes, many of their characteristics, and some of their limitations. One of the more important concepts that you should now understand is that most of these characteristics influence each other. For example, practically all plate characteristics are interrelated. Change one and the others change. Another example is heater voltage. Every tube parameter affected by the cathode depends on proper heater voltage. Interrelationships such as these make electronics both fascinating and, at times, frustrating.

Many of the limiting factors that we have discussed are the same ones found in other electrical devices such as motors, stoves, toasters, and so on. Heating and overheating, insulation breakdown, and excessive voltage and current are all limitations that you have noted before.

The point is that you can and should apply just about everything you have learned about electricity to electron tubes. Little is new except the environment.

Q.12 A large negative voltage is applied to the plate of a diode, and a large positive voltage is applied to the cathode. If the tube conducts, what tube parameter has been exceeded? answer.gif (214 bytes)

THE TRIODE

Diode electron tubes can be used as rectifiers, switches, and in many other useful applications. They are still used in Fleming's original application in some radio circuits. You will learn more of these applications in other NEETS modules and later will see the diode in several pieces of electronic equipment.

As with all inventions, Fleming's diode was immediately the subject of much experimentation and many attempts at improvement. An American experimenter, Dr. Lee De Forest, added another active element to the diode in 1906. He was trying to improve the radio application of Fleming's diode. His new tube was eventually called a triode.

DeForest's triode was not very successful as a radio "detector." (Detectors will be studied in a later NEETS module.) However, in 1912, De Forest discovered that his original triode could AMPLIFY or magnify very weak electrical impulses. It is because of the triode's ability to amplify that De Forest is honored as one of the great radio pioneers.

The immediate application of the triode amplifier was in telephone and radio. Both fields were limited because electrical impulses (signals) became weaker and weaker as the distance from the signal source increased. The triode, along with other developments of the time, made long-distance communications possible. Looking back, we can now see that the amplifying tube was the real beginning of modern electronics and influenced everything that followed. Let's find out more about the idea of amplification and how it is done in the triode.

You are already familiar with a type of amplification. In a previous NEETS module, step-up transformers were discussed. You should remember that an input voltage applied to the primary of a step-up transformer is increased in amplitude at the secondary by a factor determined by the step-up turns ratio.

For example, if 5 volts were applied to the primary of a 1:3 step-up transformer, the secondary would produce 15 volts. In other words, the input voltage was amplified by a factor of 3. When applied to electronic circuits, these primary and secondary voltages are more often called signals, or input and output signal, respectively. In electronics, the amplitude of an input signal must sometimes be increased many times-often, hundreds or thousands of times!

Because of size and design limitations, transformers are usually not practical for use in electronics as amplifiers.

DeForest's first experiment with the diode was to place an additional metal plate between the cathode and plate. He then placed an ac signal on the metal plate. When the circuit was energized, De Forest found that the ammeter stayed on zero regardless of the polarity of the input signal.

What was happening was that the new element was blocking (or shadowing) the plate. Any electrons attempting to reach the plate from the cathode would hit the new element instead. As the circuit didn't work, it was back to the drawing board.

In his next attempt, De Forest decided to change the element between the cathode and the plate. Instead of a solid metal plate, he used a wire mesh. This would allow electrons to flow from the cathode, THROUGH THE WIRE MESH, to the plate. This tube circuit is shown in figure 1-15. In view (A) you see De Forest's circuit with 0 volts applied to the third element, (today called a control grid or occasionally just the grid). Under these conditions, assume that the ammeter reads 5 milliamperes. With no voltage applied to the grid, the grid has little effect on the electron stream. For all practical purposes, the control grid is not there. Most electrons flow through the open mesh. The tube functions as a diode.

Figure 1-15. - DeForest's experiment.

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In view (B), you see De Forest's tube with +3 volts applied to the control grid. When De Forest applied this voltage, he found that plate current, I p, increased by a large amount. (We'll say it doubled to simplify the explanation.) You already know that the only way to double the plate current in a diode is to increase the plate voltage by a large amount. Yet, De Forest had doubled plate current by applying only 3 volts positive to the control grid!

The reason for this is fairly easy to understand. It's the old principle of "opposites attract." When the control grid was made positive, electrons surrounding the cathode (negative charges) were attracted to the grid. But remember, the grid is a metal mesh. Most of the electrons, instead of striking the grid wires, were propelled through the holes in the mesh. Once they had passed the grid, they were attracted to the positive charge in the plate.

You might wonder why the grid would make that much difference. After all, the plate has 300 volts on it, while the grid only has 3 volts on it. Surely the plate would have a greater effect on current flow than a grid with only one one-hundredth the attractive potential of the plate. But remember, in your study of capacitors you discovered that opposites attract because of electrostatic lines of force, and that the strength of electrostatic lines of force decreased with distance. In his tube, DeForest had placed the grid very close to the cathode. Therefore, it had a greater effect on current flow from the cathode than did the plate, which was placed at a much greater distance from the cathode. For this reason, De Forest was able to double the current flow through the tube with only +3 volts applied to the grid.

DeForest had certainly hit on something. Now the problem was to find out what would happen when a negative potential was applied to the grid. This is shown in view (C) of figure 1-15. When De Forest applied -3 volts to the grid, he found that plate current decreased to half of what it was when the grid had no voltage applied. The reason for this is found in the principle of "likes repel." The negatively charged grid simply repelled some of the electrons back toward the cathode. In this manner, the attractive effect of the plate was decreased, and less current flowed to the plate.

Now DeForest was getting somewhere. Using his new tube (which he called a triode because it had 3 elements in it), he was able to control relatively large changes of current with very small voltages. But! was it amplification? Remember, amplification is the process of taking a small signal and increasing its amplitude. In De Forest's circuit, the small input signal was 3 volts dc. What De Forest got for an output was a variation in plate current of 7.5 milliamperes. Instead of amplification, De Forest had obtained "conversion," or in other words, converted a signal voltage to a current variation. This wasn't exactly what he had in mind. As it stood, the circuit wasn't very useful. Obviously, something was needed. After examining the circuit, De Forest discovered the answer - Ohm's law. Remember E = I X R? De Forest wanted a voltage change, not a current change. The answer was simple:

0025.GIF (595 bytes)

In other words, run the plate current variation (caused by the voltage on the grid) through a resistor, and cause a varying voltage drop across the resistor. This is shown in figure 1-16.

Figure 1-16. - Operation of the plate load resistor.

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The circuit is identical to the one in figure 1-15 except that now a resistor (called a plate-load resistor, RL) has been added to the plate circuit, and a voltmeter has been added to measure the voltage drop across RL.

In view (A) of figure 1-16, the control grid is at 0 volts. Once again 5 milliamperes flow in the plate circuit. Now, the 5 milliamperes must flow through RL. The voltage drop is equal to:

E = I X R

E = (5 X 10-3 amperes) X (10 X 103 ohms)

E = (5 X 10-3) X (10 X 103)

E = 5 X 10

E = 50 volts

Thus the voltage drop across the plate-load resistor, RL, is 50 volts when no voltage is applied to the grid. In view (B) of the figure, +3 volts is applied to the control grid. Once again plate current increases to 10 milliamperes. The voltage drop across RL is

E = I X R

E = (10 X 10-3 amperes) X (10 X 103 ohms)

E = (10 X 10-3) X (10 X 103)

E = 10 X 10

E = 100 volts

By applying +3 volts to the grid, the voltage drop across RL was increased by 50 volts (from the original 50 volts to 100 volts). In view (C), -3 volts has once again been applied to the control grid. Once again plate current decreases to 2.5 milliamperes, and the voltage drop across RL drops to 25 volts.

We have caused the voltage across RL to vary by varying the grid voltage; but is it amplification? Wen, let's take a look at it. The grid voltage, or input signal, varies from +3 to -3 volts, or 6 volts. The voltage drop across RL varies from 25 volts to 100 volts, or 75 volts. In other words, the triode has caused a 6-volt input signal (varying) to be outputted as a signal that varies by 75 volts. That's amplification!

Q.13 What is the primary difference between a diode and a triode? answer.gif (214 bytes)
Q.14 Why does the grid have a greater effect than the plate on electron flow through a vacuum tube? answer.gif (214 bytes)
Q.15 What component is used in a triode amplifier to convert variation in current flow to voltage variation? answer.gif (214 bytes)







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