Before starting to build your own digital electronics circuits, you should make sure that you are very familiar with the basic direct current electricity laws that govern how electricity flows through them. Don’t worry if you have not yet been exposed to any direct current electrical theory, it’s actually pretty simple and in the introduction to this chapter, I gave you a quick run down of how direct current circuits operate. I’m sure you were able to get through that without too many problems.
To make sure that you are clear on what direct current (also known as ‘‘DC’’) is, it consists of electricity running in a single direction without any changes. Alternating current (‘‘AC’’) continuously changes from positive to negative (as shown in Fig. 3-2). AC is primarily used for high-power circuitry and not for any kind of digital electronics, except as something that is controlled by it. Digital electronics is powered by direct current, which consists of a fixed voltage which does not change level or polarity, as AC does.
As I indicated in the introduction, there are two components to electricity:
voltageis the ‘‘pressure’’ applied to the electrons andcurrentis the number of electrons that flow past a point or a set amount of time. I use the terms ‘‘pressure’’ and ‘‘flow’’ to help you visualize electricity moving in a wire as being the same as water flowing through a pipe. Using a water/pipe analogy can help you visualize how electricity moves and changes according to the conditions it is subjected to.
It should be obvious that the more pressure you apply to water in a pipe, the more water will pass through it. You can demonstrate this with a garden hose and a tap. By partially closing the tap, you are restricting the flow of the water coming from it, and the stream will not go very far from the end of the hose and very little water will flow out. When you completely open the tap, the water will spray out considerably further and a lot more water will be passing out the end of the hose. Instead of saying that you are closing the tap,
why don’t you think of the closing tap as resisting the flow of water through the pipe and into the hose? This is exactly analogous to theload in a circuit converting electrical energy into something else. Electricity coming out of the load will be at a lower pressure (or voltage) than the electricity going into the load and the amount of current will be reduced as well.
When you visualized the pipe/tap/hose analogy, you probably considered that all the resistance in the circuit was provided by the tap – the pipe and the hose did not impede the water’s flow in any way. This is also how we model how electricity flows in wires; the wires do not cause a drop in voltage and do not restrict the amount of current that is flowing in them. If you think about it for a moment, you will probably realize that this assumption means that the wires are ‘‘superconductors’’; any amount of electricity and at any voltage could be carried in the wires without any loss.
The wires that you use are certainly not superconductors, but the assumption that the wires do not impede the flow of electricity is a good one as their resistance in most circuits is usually negligible. By assuming that the wires are superconductors, you can apply some simple rules to understand the behavior of electricity in a circuit.
Going back to the original schematic diagram in this chapter (see Fig. 3-1), we can relate it to the pipe/tap/hose example of this section. The circuit’s power supply is analogous to the pipe supplying water to the tap (which itself is analogous to the electrical circuit’s load). The hose provides the same function as the wires bringing the electrical current back to the power supply.
In the pipe/tap/hose example, you should be able to visualize that the amount of water coming through the hose is dependent on how much the tap impedes the water flow from the pipe. It should be obvious that the less the tap impedes the water flow, the more water will come out the hose. Exactly the same thing happens in an electrical circuit; the ‘‘load’’ will impede or ‘‘resist’’ the flow of electricity through it and, in the process, take energy from the electricity to do something with it.
The most basic load that can be present in a circuit is known as the ‘‘resistor’’ (Fig. 3-3), which provides a specified amount of resistance,
measured in ‘‘ohms’’, to electricity. The ‘‘schematic symbol’’ is the jagged line you will see in various schematic diagrams in this book and in other sources. The schematic symbol is the graphic representation of the component and can be used along with the graphic symbol for a gate in a schematic diagram. In traditional resistors, the amount of resistance is specified by a number of colored bands that are painted on its sides – the values specified by these bands are calculated using the formula below and the values for each of the colors listed in Table 3-2.
Resistance¼ ððBand 1 Color Value10Þ þ ðBand 2 Color ValueÞÞ
10Band 3 Color ValueOhms
In the introduction to the chapter, I stated that power supplies provide elec- trons with a specific ‘‘pressure’’ called voltage. Knowing the voltage applied
Table 3-2 Resistor color code values.
Color Band color value Tolerance
Black 0 N/A Brown 1 1% Red 2 2% Orange 3 N/A Yellow 4 N/A Green 5 0.5% Blue 6 0.25% Violet 7 0.1% Gray 8 0.05% White 9 N/A Gold N/A 5% Silver N/A 10%
to a load (or resistor), you can calculate the electrical current using Ohm’s law which states:
The voltage applied to a load is equal to the product of its resistance and the current passing through it.
This can be expressed mathematically as:
V¼iR
where ‘‘V’’ is voltage, ‘‘R’’ is resistance and ‘‘i’’ is current. The letter ‘‘i’’ is used to represent current instead of the more obvious ‘‘C’’ because this char- acter was already for specifying capacitance, as I will explain below. Voltage is measured in ‘‘volts’’, resistance in ‘‘ohms’’ and current in ‘‘amperes’’. For the work done in this book, you can assume that ohms have the units of volts/amperes and is given the symbol; you can look up how these values are derived, but for now just take them for what I’ve presented here. With a bit of basic algebra, once you know two of the values used in Ohm’s law, you can calculate the third.
Voltage, current, resistance, and, indeed, all the electrical values that you will see are part of the ‘‘SI’’ (Syste`me Internationale), and its values are governed by SI standards. Each time a unit deviates by three orders of magnitude from the base value, the units are given a prefix that indicates the magnitude multiplier and these multipliers are listed in Table 3-3. For example, one thousandth of a volt is known as a ‘‘millivolt’’. The actual component values are normally given a single letter symbol that indicates its value. Most electronic devices, like resistors are given a two digit value that is multiplied by the power of ten which the symbol indicates. For example,
Table 3-3 Syste`me Internationale magnitude of prefixes and symbols.
Power multiplier Prefix Symbol Power multiplier Prefix Symbol
103 kilo k 103 milli m
106 mega M 106 micro m 109 Giga G 109 nano n 1012 tera T 1012 pico p 1015 peta P 1015 femto f
thousands of units are given the prefix ‘‘k’’, so a resistor having a value of 10,000 ohms is usually referred to as having a value of ‘‘10 kohms’’, or most popularly ‘‘10 k’’.
Looking at more complex circuits, such as the two resistor ‘‘series’’ circuit shown in Fig. 3-4, you must remember that individual measurements must be taken across each resistor’s two terminals; you doNOTmake measurements relative to a common point. The reason for making this statement is to point out that the voltage across a resistor, which is also known as the ‘‘voltage drop’’, is dependent on the current flowing through it.
Using this knowledge, you can understand how electricity flows through the two series resistors in Fig. 3-4. The voltage applied to the circuit causes current to flow through both of the resistors and the amount of current is equal to the current passing through a single resistor value which is the sum of the two resistors. Knowing this current, and an individual resistor’s value, you can calculate the voltage drop across each one. If you do the calculations, you will discover that the voltage drop across each resistor is equal to the applied voltage.
This may be a bit hard to understand, but go back to the pipe/tap/hose example and think about the situation where you had a pipe/tape/pipe/tap/ hose. In this case, there would be a pressure drop across the first tap and then another pressure drop across the second tap. This is exactly what happens in Fig. 3-4: some voltage ‘‘drops’’ across Resistor 1 and the rest drops across Resistor 2. The amount of the drop across each resistor is proportional to its value relative to the total resistance in the circuit.
To demonstrate this, consider the case where Resistor 1 in Fig. 3-4 is 5 ohms and Resistor 2 is 8 ohms. Current has to flow through Resistor 1 followed by Resistor 2, which means that the total resistance it experiences is equivalent to the sum of the two resistances (13 ohms). The current through the two resistors could be calculated using Ohm’s law, as voltage applied divided by Resistor 1 plus Resistor 2. The general formula for calculating
equivalent the resistance of a series circuit is the sum of the resistances, which is written out as:
Re¼R1þR2þ. . .
Knowing the resistor values, the voltage drop across each resistor can be calculated as its fraction of the total resistance; the voltage across Resistor 1 would be 5/13ths of the applied voltage while the voltage across Resistor 2 would be 8/13ths of the applied voltage. Dividing the resistor values into the individual resistor voltage drops will yield the same current as dividing the applied voltage by the total resistance of the circuit.
Adding the two resistor voltage drops together, you will see that they total the applied voltage. This is a useful test to remember when you are checking your calculations, to make sure they are correct.
The properties of series resistance circuits are summed up quite well as Kirchoff’s voltage law, which states that ‘‘the sum of the voltage drops in a series circuit is equivalent to the applied voltage and current is the same at all points in the circuit.’’
Along with being able to calculate the amount of current passing through a series resistor circuit and the voltage drop across each resistor, you can also calculate the voltage across each resistor in a parallel resistor circuit like Fig. 3-5 as well as the current through all the resistors. To do this, you have to remember Kirchoff’s current law, which states that ‘‘the sum of the currents through each resistance is equivalent to the total current drawn by the circuit and the voltage drops across each resistor is the same as the applied voltage.’’
With each resistor in parallel, it should be fairly obvious that the voltage drop across each one is the same as the applied voltage, and the current flowing through each one can be calculated using Ohm’s law. It should also
be obvious that the current drawn from the power source is equivalent to the sum of the currents passing through each resistor.
If you were to calculate some different current values for different resistances, you would discover that the general formula for the equivalent resistance was:
Re ¼1=ð1=R1Þ þ ð1=R2þ. . .Þ
For the simple case of two resistors in parallel, the equivalent resistance can be expressed using the formula:
Re¼ ðR1R2Þ=ðR1þR2Þ
Complex resistor circuits, made up of resistors wired in both series and parallel, like the one shown in Fig. 3-6, can be simplified to a single equiva- lent resistor by applying the series and resistor formulas that I have presented so far in this section. When doing this, I recommend first finding the equiva- lent to the series resistances and then the equivalent to the parallel resistances until you are left with one single equivalent resistance.
The last piece of basic electrical theory that I would like to leave you with is how to calculate the power dissipated by a resistor. When you took Newtonian physics, you were told that power was the product of the rate at which something was moving and the force applied to it. In electrical circuits, we have both these quantities, voltage being the force applied to the electrons and current being the rate of movement. To find the power being dissipated (in watts), you can use the simple formula:
P¼Vi
or, if you don’t know one of the two input quantities, you can apply Ohm’s law and the formula becomes:
P¼V2=R
¼i2R
I must point out that when you are working with digital electronics, most currents in the circuits are measured somewhere between 100mA to 20 mA. This seemingly small amount of current minimizes the amount of power that is dissipated (or used) in the digital electronic circuits. I’m pointing this out because if you were to get a book on basic electronics you would dis- cover that the examples and questions will usually involve full amperes of current – not thousands or tens of thousands as I have noted here. The reason why basic electronics books work with full amps is because it is easier for students to do the calculations and they don’t have to worry about work- ing with different orders of magnitude.
So far in these few initial pages of this chapter, I have gone through the same amount of material that is presented in multiple courses in electrical theory. Much of the background material has been left out as well as derivations of the various formulas. For the purposes of working with digital electronics, you should be familiar with the following concepts:
1. Electricity flows like water in a closed circuit.
2. The amount of current flow in a circuit is proportional to the amount of resistance it encounters.
3. Voltage across a load or resistance is measured at its two terminals. 4. Voltage is current times resistance (Ohm’s law).
5. Power is simply voltage times current in a DC circuit.
The other rules are derivations of these basic concepts and while I don’t recommend trying to work them out in an exam, what you do remember can be checked against the basic concepts listed above.