Analog Building Blocks Electrical Assignment Help

Electronic systems normally process information in either analog form or digital form.’ The two kinds of signals require two different kinds of circuit technology, which are naturally called analog circuits and digital circuits. Until around 1960 almost all technology was of the analog type. Owing to the development of integrated circuits, digital technology has grown tremendously since then, and by now has reached a position of roughly equal importance.

In analog systems, a signal voltage or current is made proportional to some physical quantity. (To the extent that the rather vague word “analog” means anything in this context, it means “proportional.”) For example, the time-varying voltage produced by a microphone is proportional to the acoustic pressure; thus it is an analog signal. Since a physical quantity, such as acoustic pressure, can take on a continuous range of values, the analog signal also must vary continuously. In other words, if a certain voltage is an analog signal, it must be allowed to take on any value between some minimum and some maximum. For this reason analog circuits are sometimes called continuous-state circuits. This distinguishes them from the discrete-state circuits used in digital technology, where only certain values of voltage are allowed.

A circuit in which one voltage is meant to be linearly proportional to another can be called a linear circuit. Although there are exceptions. most circuits found in analog systems are linear circuits. Linear active circuits are also known as amplifiers. Amplifiers are the building blocks of linear systems, and the principal building blocks of analog technology.

This will allow us to discuss important aspects of analog systems without being burdened with details of device physics. Moreover, the buildingblock point of view is consistent with the modern trend toward integrated circuits. Today only special purpose amplifiers (such as high-power or highfrequency amplifiers) are built with individual transistors; for most purposes, assembling separate transistors, resistors, and capacitors on a circuit board is an expensive and wasteful procedure. Modern integrated-circuit (IC) amplifiers are versatile, compact, and remarkably inexpensive. No matter how complicated it may be inside, from the user’s point of view an IC functions as a unit according to certain rules. Thus the building-block point of view is appropriate.


One of the most significant concepts in electronics is the idea of modeling. This subject arises because we wish to predict the performance of electronic systems through the use of equations. since each part of the circuit could be described mathematically: resistors by V = IR, capacitors by i = C dv/dt, and so forth. We must remember, however, that even Ohm’s law is really an idealization of a resistor’s actual performance. In reality, when current passes through a resistor, the resistor gets warmer, which will cause the value of R to change. When we say that the resistor obeys V = IR we are treating R as though it were really constant; we can do this only because the change in R is small and usually has little effect on the circuit. V = IR is thus actually a model that provides a useful approximate description of resistor performance.

Since copper, though a good conductor, is not a perfect one, a coil of wire often has a sizable resistance. To take this into account, an inductor can be represented in a circuit diagram by an inductor symbol and a resistor symbol in series. We would then consider the combination of R and L to be a model which imitates the behavior of the actual coil of wire.

When an amplifier block appears in a system, we likewise have to describe its performance by circuit equations. But if the amplifier is simply a mysterious box with several wires coming out of it, there is no way of saying what it will do. We need to represent the amplifier block by a simple collection of circuit elements (just as the coil of wire is represented by a series connection of R and L.) Once this has been done, we can easily analyze the system containing the block by writing conventional circuit equations.

Dependent Sources

In order to develop the amplifier model, we shall first introduce a new family of ideal circuit elements known as dependent voltage and current sources. Since dependent sources are quite important, it is worthwhile to discuss the concept in detail.

The sources encountered until now in this book have been independent sources. An independent source is one whose value can be independently specified. In contrast.:a dependent source obeys the rule that its value depends on a voltage or current somewhere else in the-circuit. To be sure, such devices are imaginary; we introduce them in order to use them in models for things that actually exist. From the student’s point of view, the vital thing is to understand the rules that these circuit elements always obey. The fact that they do obey simple rules means that circuit analysis can proceed in an orderly way.

Let us begin with the dependent current source shown in Fig. 7.1. In this figure all the components are part of some large circuit, only a piece of which is shown. We see that the value of the current source is βi1. The number β is a given constant. In order to find out the value of the current source we have to first locate the current ij, which is the reference current for the dependent source. The position of ij is indicated on the diagram. We then calculate the value of iI, and from that we can find the value of the dependent source.

Note the following points:

1. Except that its value depends on its reference, a dependent source functions like any other ideal source. For instance, in Fig. 7.1, the current through R must equal βi1. The voltage across R will be βi1R.

2. The relationship between the ideal dependent source and its reference cannot be broken. No matter what the value of t.. the value of the dependent source is β times as large

3. In the case of independent sources, we were free to turn sources off for various purposes (for instance, in finding Thevenin resistance by the alternate method, in using superposition, or in finding the roots of the characteristic equation by the impedance method). However, a dependent source cannot be turned off. Its value must depend on its reference. In Fig. 7.1,. if the value of i1 is not zero, it would violate the rules to set the dependent source to zero.

Dependent Current Source with Current Reference. (Figure 7.1)

Dependent Current Source with Current Reference. (Figure 7.1)

4. The reference current t. in Fig. 7.1 is not a given quantity. In general its value must be found by calculation, and will depend on the inputs applied to the circuit. Thus the reference current should not appear in the result of a circuit calculation. If a reference current should appear in an answer, it must be eliminated by solving for-it in terms of given quantities. The student should watch out for this; it is often a troublesome point.


Find the voltage at A in Fig. 7.2.

Figure 7.2

Figure 7.2


From Ohm’s law it is clear that VA βi1 R2. However, we are not yet finished, because i1 is not a given quantity. To find its value we must refer to the rest of the circuit. In this case we see that i1 = – V0/R1. (Note the minus sign, which comes from the given sign convention for il’) The voltage source Vo is independent, since its value does not depend on something else; V0 is a given quantity and can appear in the answer. Substituting i1 = – V0/R1 into vA = βi1 R2′ we have VA = – βV0R2/R1.

A good way to solve circuit problems when dependent sources are present is to write an equation for the unknown to be calculated, and then see if this equation also contains an unknown reference current. If it does, we write another equation to obtain the reference current. If the reference current itself depends on something else, still another equation is needed; we work backward through the problem until we arrive at the input, which is usually given. In Example 7.1, for instance, we first noted that we could find VA if i1 were known; then we found il from the equation i1 = – V0/R and eliminated it from the expression for vA’ In less straightforward problems one can usually write enough node (or loop) equations to eliminate the reference currents.


Find VOUT in the circuit shown in Fig. 7.3.

Figure 7.3

Figure 7.3


Clearly, VOUT = βi/R3′ but this is not sufficient; the unknown reference current must be eliminated. In order to find i1 we write a node equation for node A:

V0 – VA/R1 – VA/ R2 – βi1 = 0

In writing this equation, however, we have introduced a new unknown, vA- It can be eliminated by means of the Ohm’s law equation VA = – i1R2• Substituting this expression for VA’ we can solve for i1• Thus we obtain

The reader should remember that dependent sources cannot arbitrarily be turned off. This is illustrated by the following example.

Posted on April 29, 2016 in Analog Building Blocks

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