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Sirindhorn International Institute of Technology Thammasat University School of Information, Computer and Communication Technology Lecture Notes: ECS 203 Basic Electrical Engineering Semester 1/2013 Asst. Prof. Dr.Prapun Suksompong June 11, 2013 1 1 Special thanks to Dr.Waree Kongprawechnon and Dr.Somsak Kittipiyakul for earlier versions of the notes. Parts of the notes were compiled from C.K. Alexander and M.N.O. Sadiku, Fundamentals of Electric Circuits, 4th ed., McGraw-Hill, International Edition, 2009 and G. Rizzoni, Principles and Applications of Electrical Engineering, 5th ed., Mc-Graw-Hill, International Edition, 2007 CHAPTER 1 Basic Concepts In electrical engineering, we are often interested in communicating or transferring energy from one point to another. To do this requires an interconnection of electrical devices. Such interconnection is referred to as an electric circuit, and each component of the circuit is known as an element. Definition 1.0.1. An electric circuit is an interconnection of electrical elements. 1.1. Systems of Units 1.1.1. As engineers, we deal with measurable quantities. Our measurement must be communicated in standard language that virtually all professionals can understand irrespective of the country. Such an international measurement language is the International System of Units (SI). • In this system, there are six principal units from which the units of all other physical quantities can be derived. 5 6 1. BASIC CONCEPTS Quantity Basic Unit Symbol Length meter m Mass kilogram kg Time second s Electric Current ampere A Temperature kelvin K Luminous Intensity candela cd Charge coulomb C • One great advantage of SI unit is that it uses prefixes based on the power of 10 to relate larger and smaller units to the basic unit. Multiplier 1012 109 106 103 10−2 10−3 10−6 10−9 10−12 Prefix Symbol tera T giga G mega M kilo k centi c milli m micro µ nano n pico p Example 1.1.2. Change of units: 600, 000, 000 mA = 1.2. Circuit Variables 1.2.1. Charge: The concept of electric charge is the underlying principle for all electrical phenomena. Charge is an electrical property of the atomic particles of which matter consists, measured in coulombs (C). The charge of an electron is −1.602 × 10−19 C. • The coulomb is a large unit for charges. In 1 C of charge, there are 1/(1.602 × 10−19 ) = 6.24 × 1018 electrons. Thus realistic or laboratory values of charges are on the order of pC, nC, or µC. • A large power supply capacitor can store up to 0.5 C of charge. 1.2.2. Law of Conservation of Charge: Charge can neither be created nor destroyed, only transferred. 1.2. CIRCUIT VARIABLES 7 Definition 1.2.3. Current: The time rate of change of charge, measured in amperes (A). Mathematically, i(t) = d q(t) dt Note: • 1 ampere (A) = 1 coulomb/second (C/s). • The charge transferred between time t1 and t2 is obtained by Z t2 q= idt t1 1.2.4. Two types of currents: (a) A direct current (dc) is a current that remains constant with time. (b) A time-varying current is a current that varies with time. • An alternating current (ac) is a current that varies sinusoidally with time. 8 1. BASIC CONCEPTS • Such ac current is used in your household, to run the air conditioner, refrigerator, washing machine, and other electric appliances. 1.2.5. By convention the symbol I is used to represent such a constant current. A time-varying current is represented by the symbol i. Definition 1.2.6. Voltage (or potential difference): the energy required to move a unit charge though an element, measured in volts (V). The voltage between two points a and b in a circuit is denoted by vab and can be interpreted in two ways: (a) point a is at a potential of vab volts higher than point b, or (b) the potential at point a with respect to point b is vab . Note: • 1 volt (V) = 1 joule/coulomb = 1 newton-meter/coulomb • vab = −vba • Mathematically, dw vab = dq where w is the energy in joules (J) and q is charge in coulombs (C). + a vab – b The plus (+) and minus (-) signs at the points a and b are used to define reference direction or voltage polarity. + a 9V – – a -9 V b + b 1.2. CIRCUIT VARIABLES 9 1.2.7. Like electric current, a constant voltage is called a dc voltage and is represented by V , whereas a time-varying voltage is called an ac voltage and is represented by v. A dc voltage is commonly produced by a battery; ac voltage is produced by an electric generator. 1.2.8. Current and voltage are the two basic variables in electric circuits. The common term signal is used for an electric quantity such as a current or a voltage (or even electromagnetic wave) when it is used for conveying information. Engineers prefer to call such variables signals rather than mathematical functions of time because of their importance in communications and other disciplines. For practical purposes, we need to be able to find/calculate/measure more than the current and voltage. We all know from experience that a 100-watt bulb gives more light than a 60-watt bulb. We also know that when we pay our bills to the electric utility companies, we are paying for the electric energy consumed over a certain period of time. Thus power and energy calculations are important in circuit analysis. Definition 1.2.9. Power: time rate of expending or absorbing energy, measured in watts (W). Mathematically, the instantaneous power dw dq dw = = vi p= dt dq dt Definition 1.2.10. Sign of power • Plus sign: Power is absorbed by the element. (resistor, inductor) • Minus sign: Power is supplied by the element. (battery, generator) Definition 1.2.11. Passive sign convention: • Passive sign convention is satisfied when the current enters through the positive terminal of an element. • If the current enters through the positive polarity of the voltage, p = vi. • If the current enters through the negative polarity of the voltage, p = −vi. 10 1. BASIC CONCEPTS i i + + v v – – p = +vi p = –vi 3A 3A + – 4V 4V – + 3A 3A + – 4V 4V – + Example 1.2.12. Light bulb or battery 1.2. CIRCUIT VARIABLES 11 1.2.13. Law of Conservation of Energy: Energy can neither be created nor destroyed, only transferred. • For this reason, the algebraic sum of power in a circuit, at any instant of time, must be zero. • The total power supplied to the circuit must balance the total power absorbed. Example 1.2.14. The circuit below has five elements. If p1 = −205 W, p2 = 60 W, p4 = 45 W, p5 = 30 W, calculate the power p3 received or delivered by element 3. 1.2.15. Energy: the energy absorbed or supplied by an element from time 0 to t is Z t Z t w= p dt = vi dt. 0 0 • Integration suggests finding area under the curve. • Need to be careful with negative area. Example 1.2.16. Electricity bills: The electric power utility companies measure energy in kilowatt-hours (kWh), where 1 kWh = 3600 kJ. 12 1. BASIC CONCEPTS 1.3. Circuit Elements Definition 1.3.1. There are 2 types of elements found in electrical circuits. 1) Active elements (is capable of generating energy), e.g., generators, batteries, and operational amplifiers (Op-amp). 2) Passive element, e.g., resistors, capacitors and inductors. Definition 1.3.2. The most important active elements are voltage and current sources: (a) Voltage source provides the circuit with a specified voltage (e.g. a 1.5V battery) (b) Current source provides the circuit with a specified current (e.g. a 1A current source). Definition 1.3.3. In addition, we may characterize the voltage or current sources as: 1) Independent source: An active element that provides a specified voltage or current that is completely independent of other circuit elements. v + V – DC i 1.3. CIRCUIT ELEMENTS 13 2) Dependent source: An active element in which the source quantity is controlled by another voltage or current. v + – i 1.3.4. The key idea to keep in mind is that a voltage source comes with polarities (+ -) in its symbol, while a current source comes with an arrow, irrespective of what it depends on. Example 1.3.5. Current-controlled voltage source A B i + 5V – C + – 10i Example 1.3.6. Draw the general form of a voltage-controlled current cource. 14 1. BASIC CONCEPTS 1.3.7. Remarks: • Dependent sources are useful in modeling elements such as transistors, operational amplifiers and integrated circuits. • Ideal sources – An ideal voltage source (dependent or independent) will produce any current required to ensure that the terminal voltage is as stated. – An ideal current source will produce the necessary voltage to ensure the stated current flow. – Thus an ideal source could in theory supply an infinite amount of energy. • Not only do sources supply power to a circuit, they can absorb power from a circuit too. • For a voltage source, we know the voltage but not the current supplied or drawn by it. By the same token, we know the current supplied by a current source but not the voltage across it. CHAPTER 2 Basic Laws Here we explore two fundamental laws that govern electric circuits (Ohm’s law and Kirchhoff’s laws) and discuss some techniques commonly applied in circuit design and analysis. 2.1. Ohm’s Law Ohm’s law shows a relationship between voltage and current of a resistive element such as conducting wire or light bulb. 2.1.1. Ohm’s Law: The voltage v across a resistor is directly proportional to the current i flowing through the resistor. v = iR, where R = resistance of the resistor, denoting its ability to resist the flow of electric current. The resistance is measured in ohms (Ω). • To apply Ohm’s law, the direction of current i and the polarity of voltage v must conform with the passive sign convention. This implies that current flows from a higher potential to a lower potential 15 16 2. BASIC LAWS in order for v = iR. If current flows from a lower potential to a higher potential, v = −iR. i l Material with Cross-sectional resistivity r area A + v – R 2.1.2. The resistance R of a cylindrical conductor of cross-sectional area A, length L, and conductivity σ is given by L R= . σA Alternatively, L R=ρ A where ρ is known as the resistivity of the material in ohm-meters. Good conductors, such as copper and aluminum, have low resistivities, while insulators, such as mica and paper, have high resistivities. 2.1.3. Remarks: (a) R = v/i (b) Conductance : 1 i = R v 1 The unit of G is the mho (f) or siemens2 (S) G= 1Yes, this is NOT a typo! It was derived from spelling ohm backwards. 2In English, the term siemens is used both for the singular and plural. 2.1. OHM’S LAW 17 (c) The two extreme possible values of R. (i) When R = 0, we have a short circuit and v = iR = 0 showing that v = 0 for any i. + i v=0 R=0 – (ii) When R = ∞, we have an open circuit and v =0 i = lim R→∞ R indicating that i = 0 for any v. + v i=0 R=∞ – 2.1.4. A resistor is either fixed or variable. Most resistors are of the fixed type, meaning their resistance remains constant. 18 2. BASIC LAWS A common variable resistor is known as a potentiometer or pot for short 2.1.5. Not all resistors obey Ohms law. A resistor that obeys Ohms law is known as a linear resistor. • A nonlinear resistor does not obey Ohms law. • Examples of devices with nonlinear resistance are the lightbulb and the diode. • Although all practical resistors may exhibit nonlinear behavior under certain conditions, we will assume in this class that all elements actually designated as resistors are linear. 2.1. OHM’S LAW 19 2.1.6. Using Ohm’s law, the power p dissipated by a resistor R is v2 p = vi = i2 R = . R Example 2.1.7. In the circuit below, calculate the current i, and the power p. i 30 V DC 5 kΩ + v – Definition 2.1.8. The power rating is the maximum allowable power dissipation in the resistor. Exceeding this power rating leads to overheating and can cause the resistor to burn up. Example 2.1.9. Determine the minimum resistor size that can be connected to a 1.5V battery without exceeding the resistor’s 14 -W power rating. more complicated examples, including the one shown in Figure 2.1. 2.1 T E R M I N O L O G Y Lumped circuit elements are the fundamental building blocks of electronic cir2. BASIC LAWS cuits. Virtually all of our analyses will be conducted on circuits containing two-terminal elements; multi-terminal elements will be modeled using combinations of two-terminal elements. We have already seen several two-terminal Node, Branches and Loops elements such 2.2. as resistors, voltage sources, and current sources. Electronic access to an element is made through its terminals. Definition 2.2.1. Since the elements of an electric circuit can be interAn electronic circuit is constructed by connecting together a collection of connected in ways, we need to understand some basic concept of separateseveral elements at their terminals, as shown in Figure 2.2. The junction points networkattopology. which the terminals of two or more elements are connected are referred to as the nodes of=a circuit. Similarly, the connections between the nodes are referred • Network interconnection of elements or devices to as the edges or branches of a circuit. Note that each element in Figure 2.2 • Circuit = a branch. network closedandpaths forms a single Thuswith an element a branch are the same for circuits comprising only two-terminal circuit a loops are defined to besuch as a Definition 2.2.2. Branch: A elements. branch Finally, represents single element closed paths through a circuit along its branches. voltage source or nodes, a resistor. Aand branch represents Several branches, loops are identified inany Figuretwo-terminal 2.2. In the circuitelement. in Figure 2.2, there are 10 branches (and thus, 10 elements) and 6 nodes. Definition 2.2.3. Node: A node is the “point” of connection between As another example, a is a node in the circuit depicted in Figure 2.1 at two or more branches. which three branches meet. Similarly, b is a node at which two branches meet. ab and bc are of branches in thein circuit. The circuit has five branches • It is usuallyexamples indicated by a dot a circuit. nodes. • Ifand a four short circuit (a connecting wire) connects two nodes, the two Since we assume that the interconnections between the elements in a circuit nodes constitute a single node. are perfect (i.e., the wires are ideal), then it is not necessary for a set of elements to be joined at aAsingle in space for their interconnection to beA closed Definition 2.2.4.together Loop: looppoint is any closed path in a circuit. considered a single node. An example of this is shown in Figure 2.3. While path is formed by starting at aare node, passing through a set ofdoes nodes and the four elements in the figure connected together, their connection returning theat starting node without passing throughconnection. any node more notto occur a single point in space. Rather, it is a distributed 20 than once. Nodes Loop Elements E 2.2 An arbitrary circuit. Branch Definition 2.2.5. Series: Two or more elements are in series if they are cascaded or connected sequentially and consequently carry the same current. Definition 2.2.6. Parallel: Two or more elements are in parallel if they are connected to the same two nodes and consequently have the same voltage across them. 2.2.7. Elements may be connected in a way that they are neither in series nor in parallel. 2.2. NODE, BRANCHES AND LOOPS 21 Example 2.2.8. How many branches and nodes does the circuit in the following figure have? Identify the elements that are in series and in parallel. 5Ω 1Ω 2Ω DC 10 V 4Ω 2.2.9. A loop is said to be independent if it contains2.2 a branch which is Kirchhoff’s Laws not in any other loop. Independent loops or paths result in independent sets of equations. A network with b branches, n nodes, and ` independent loops will satisfy the fundamental theorem of network topology: B Elements b = ` + n − 1. B C aA circuit Care Definition 2.2.10. The primary signals within its currents A and voltages, which we denote by the symbols i and Dv, respectively. We Distributed node D define a branch current as the current along a branch of the circuit, and Ideal wires a branch voltage as the potential difference measured across a branch. CHAPTER T F I G U R E 2.3 interconnectio elements that at a single nod i Branch current v + Branch voltage Nonetheless, because the interconnections are perfect, the connection can be considered to be a single node, as indicated in the figure. The primary signals within a circuit are its currents and voltages, which we denote by the symbols i and v, respectively. We define a branch current as the current along a branch of the circuit (see Figure 2.4), and a branch voltage as the potential difference measured across a branch. Since elements and branches are the same for circuits formed of two-terminal elements, the branch voltages and currents are the same as the corresponding terminal variables for the elements forming the branches. Recall, as defined in Chapter 1, the terminal variables for an element are the voltage across and the current through the element. As an example, i4 is a branch current that flows through branch bc in the circuit in Figure 2.1. Similarly, v4 is the branch voltage for the branch bc. F I G U R E 2.4 deﬁnitions illu a circuit. 22 2. BASIC LAWS 2.3. Kirchhoff ’s Laws Ohm’s law coupled with Kirchhoff’s two laws gives a sufficient, powerful set of tools for analyzing a large variety of electric circuits. 2.3.1. Kirchhoff ’s current law (KCL): the algebraic sum of currents departing a node (or a closed boundary) is zero. Mathematically, X in = 0 n KCL is based on the law of conservation of charge. An alternative form of KCL is Sum of currents (or charges) drawn as entering a node = Sum of the currents (charges) drawn as leaving the node. i5 i1 i4 i2 i3 Note that KCL also applies to a closed boundary. This may be regarded as a generalized case, because a node may be regarded as a closed surface shrunk to a point. In two dimensions, a closed boundary is the same as a closed path. The total current entering the closed surface is equal to the total current leaving the surface. Closed boundary 2.3. KIRCHHOFF’S LAWS 23 Example 2.3.2. A simple application of KCL is combining current sources in parallel. IT a I1 b I2 I3 (a) IT a IT = I 1 – I2 + I 3 b (b) A Kirchhoff ’s voltage law (KVL): the algebraic sum of all voltages around a closed path (or loop) is zero. Mathematically, M X vm = 0 m=1 KVL is based on the law of conservation of energy. An alternative form of KVL is Sum of voltage drops = Sum of voltage rises. + v2 – + v1 v3 – v4 – v5 + 24 2. BASIC LAWS Example 2.3.3. When voltage sources are connected in series, KVL can be applied to obtain the total voltage. a + V1 Vab V2 a Vab V3 b – + b VS = V1 + V2 – V3 – Example 2.3.4. Find v1 and v2 in the following circuit. 4Ω + v1 – 10 V 8V + v2 – 2Ω 2.4. SERIES RESISTORS AND VOLTAGE DIVISION 25 2.4. Series Resistors and Voltage Division 2.4.1. When two resistors R1 and R2 ohms are connected in series, they can be replaced by an equivalent resistor Req where Req = R1 + R2 . In particular, the two resistors in series shown in the following circuit i R1 R2 + v1 – + v2 – a v b can be replaced by an equivalent resistor Req where Req = R1 +R2 as shown below. i a Req + v – v b The two circuits above are equivalent in the sense that they exhibit the same voltage-current relationships at the terminals a-b. Voltage Divider: If R1 and R2 are connected in series with a voltage source v volts, the voltage drops across R1 and R2 are R1 R2 v and v2 = v v1 = R1 + R2 R1 + R2 26 2. BASIC LAWS Note: The source voltage v is divided among the resistors in direct proportion to their resistances. 2.4.2. In general, for N resistors whose values are R1 , R2 , . . . , RN ohms connected in series, they can be replaced by an equivalent resistor Req where N X Req = R1 + R2 + · · · + RN = Rj j=1 If a circuit has N resistors in series with a voltage source v, the jth resistor Rj has a voltage drop of Rj v vj = R1 + R2 + · · · + RN 2.5. Parallel Resistors and Current Division When two resistors R1 and R2 ohms are connected in parallel, they can be replaced by an equivalent resistor Req where 1 1 1 = + Req R1 R2 or R1 R2 Req = R1 kR2 = R1 + R2 i Node a i1 v i2 R1 R2 Node b Current Divider: If R1 and R2 are connected in parallel with a current source i, the current passing through R1 and R2 are R2 R1 i1 = i and i2 = i R1 + R2 R1 + R2 Note: The source current i is divided among the resistors in inverse proportion to their resistances. 2.5. PARALLEL RESISTORS AND CURRENT DIVISION 27 Example 2.5.1. Example 2.5.2. 6k3 = Example 2.5.3. (a)k(na) = Example 2.5.4. (ma)k(na) = In general, for N resistors connected in parallel, the equivalent resistor Req = R1 kR2 k · · · kRN is 1 1 1 1 = + + ··· + Req R1 R2 RN Example 2.5.5. Find Req for the following circuit. 4Ω 1Ω 2Ω Req 5Ω 8Ω 6Ω 3Ω Example 2.5.6. Find the equivalent resistance of the following circuit. 28 2. BASIC LAWS i 4Ω i0 a 6Ω 12 V + v0 – 3Ω b Example 2.5.7. Find io , vo , po (power dissipated in the 3Ω resistor). Example 2.5.8. Three light bulbs are connected to a 9V battery as shown below. Calculate: (a) the total current supplied by the battery, (b) the current through each bulb, (c) the resistance of each bulb. I 9V 15 W 20 W 9V 10 W I1 I2 + V2 – R2 + V3 – R3 + V1 – R1 2.7. MEASURING DEVICES 29 2.6. Practical Voltage and Current Sources An ideal voltage source is assumed to supply a constant voltage. This implies that it can supply very large current even when the load resistance is very small. However, a practical voltage source can supply only a finite amount of current. To reflect this limitation, we model a practical voltage source as an ideal voltage source connected in series with an internal resistance rs , as follows: Similarly, a practical current source can be modeled as an ideal current source connected in parallel with an internal resistance rs . 2.7. Measuring Devices Ohmmeter: measures the resistance of the element. Important rule: Measure the resistance only when the element is disconnected from circuits. Ammeter: connected in series with an element to measure current flowing through that element. Since an ideal ammeter should not restrict the flow of current, (i.e., cause a voltage drop), an ideal ammeter has zero internal resistance. Voltmeter:connected in parallel with an element to measure voltage across that element. Since an ideal voltmeter should not draw current away from the element, an ideal voltmeter has infinite internal resistance. Resistor Combination, Voltage Divider, Current Divider Monday, June 10, 2013 8:39 PM Lecture 2013 Page 1 Lecture 2013 Page 2 ECS 203 - Part 1B Asst. Prof. Dr.Prapun Suksompong CHAPTER 3 Methods of Analysis Here we apply the fundamental laws of circuit theory (Ohm’s Law & Kirchhoff’s Laws) to develop two powerful techniques for circuit analysis. 1. Nodal Analysis (based on KCL) 2. Mesh Analysis (based on KVL) This is the most important chapter for our course. 3.1. Nodal Analysis Here, we analyze circuit using node voltages as the circuit variables. Example 3.1.1. Back to Example 2.5.7 i 4Ω i0 a 6Ω 12 V b 31 + v0 – 3Ω 32 3. METHODS OF ANALYSIS 3.1.2. Steps to Determine Node Voltages: Step 0: Determine the number of nodes n. Step 1: Select a node as a reference node (ground node). Assign voltages v1 , v2 , · · · , vn−1 to the remaining n − 1 nodes. • The voltage are now referenced with respect to the reference node. • The ground node is assumed to have 0 potential. • Recall that voltages are measured between two points. In this case, the second point is always the ground node and hence we agree not to talk about it but know that all these node voltages are in fact the values of the voltages with respect to the reference node. • If a voltage source is connected between the reference node and a nonreference node, we simply set the voltage at the nonreference node equal to the voltage of the voltage source. Step 2: Apply KCL to each of the n − 1 nonreference nodes. (a) Use Ohm’s law to express the branch currents in terms of node voltages. (b) Current source automatically gives current value. Caution: Watch out for the direction of the arrow. Step 3: Solve the resulting simultaneous equations to obtain the unknown node voltages. 3.1. NODAL ANALYSIS 33 3.1.3. Remarks: There are multiple methods to solve the simultaneous equations in Step 3. • Method 1: Elimination technique (good for a few variables) • Method 2: Write in term of matrix and vectors (write Ax = b), then use Cramer’s rule. • Method 3: Use – computer (e.g., MATLAB) to find A−1 and then find x = A−1 b – calculator (fx-991MS can solve simultaneous linear equations with two or three unknowns.) Example 3.1.4. Calculate the node voltages in the circuit below. 5A 1 3Ω 4Ω 2 6Ω 10 A 34 3. METHODS OF ANALYSIS Example 3.1.5. Calculate the node voltages in the circuit below. 4Ω ix 1 3A 2Ω 8Ω 2 4Ω 0 3 2ix 3.1. NODAL ANALYSIS 35 3.1.6. Special Case: If there is a voltage source connected between two nonreference nodes, the two nonreference nodes form a supernode. We apply both KCL and KVL to determine the node voltages. Example 3.1.7. Find v and i in the circuit below. 9V 4Ω i 21 V 3Ω + v – 2Ω 6Ω 3.1.8. Note the following properties of a supernode: (a) The voltage source inside the supernode provides a constraint equation needed to solve for the node voltages. (b) A supernode has no voltage of its own. (c) We can have more than two nodes forming a single supernode. (d) The supernodes are treated differently because nodal analysis requires knowing the current through each element. However, there is no way of knowing the current through a voltage source in advance. 36 3. METHODS OF ANALYSIS 3.2. Mesh Analysis Mesh analysis provides another general procedure for analyzing circuits, using mesh currents as the circuit variables. Definition 3.2.1. Mesh is a loop which does not contain any other loop within it. 3.2.2. Steps to Determine Mesh Currents: Step 0: Determine the number of meshes n. Step 1: Assign mesh currents1 i1 , i2 , . . ., in , to the n meshes. • The direction of the mesh current is arbitrary–(clockwise or counterclockwise)–and does not affect the validity of the solution. • For convenience, we define currents flow in the clockwise (CW) direction. Step 2: From the current direction in each mesh, define the voltage drop polarities. Step 3: Apply KVL to each of the n meshes. Use Ohm’s law to express the voltages in terms of the mesh current. • Tip (for combining Step 2 and 3): Go around the loop in the same direction as the mesh current (of that mesh). (Note that according to our agreement above, this is in the CW direction.) When we pass a resistor R, the voltage drops by I × R where I is the branch current (algebraic sum of mesh currents) through that resistor in the CW direction. Step 4: Solve the resulting n simultaneous equations to get the mesh currents. 1Using mesh currents instead of element currents as circuit variables is convenient and reduces the number of equations that must be solved simultaneously. 3.2. MESH ANALYSIS 37 Example 3.2.3. Back to Example 2.5.7 i 4Ω i0 a 6Ω 12 V + v0 – 3Ω b Example 3.2.4. Find the branch currents I1 , I2 , and I3 using mesh analysis. I1 5Ω I2 6Ω I3 10 Ω 15 V i1 i2 10 V 4Ω 38 3. METHODS OF ANALYSIS 3.3. Remarks on Nodal Analysis and Mesh Analysis 3.3.1. Nodal analysis applies KCL to find unknown (node) voltages in a given circuit, while mesh analysis applies KVL to find unknown (mesh) currents. 3.3.2. Mesh analysis is not quite as general as nodal analysis because it is only applicable to a circuit that is planar. • A planar circuit is one that can be drawn in a plane with no branches crossing one another; otherwise it is nonplanar. 3.3.3. Nodal Analysis vs. Mesh Analysis: Given a network to be analyzed, how do we know which method is better or more efficient? Suggestion: You should be familiar with both methods. Choose the method that results in smaller number of variables or equations. • A circuit with fewer nodes than meshes is better analyzed using nodal analysis, while a circuit with fewer meshes than nodes is better analyzed using mesh analysis. You can also use one method to check your results of the other method. 3.3.4. Nodal analysis and mesh analysis can also be used to find equivalent resistance of a part of a circuit. This becomes extremely useful when the techniques that we studied in the previous chapter cannot be directly applied (, e.g., when we can’t find resistors that are in parallel or in series; they are all connected in some “strange” configuration.) The are two approaches to this kind of problems. (a) Apply 1 V voltage source across the terminals, find the corresponding current I through the voltage source. Then, V 1 Req = = . I I (b) Put 1 A current source through the terminals, find the corresponding voltage V across the current source. Then, V V Req = = = V. I 1 3.3. REMARKS ON NODAL ANALYSIS AND MESH ANALYSIS Example 3.3.5. Find the equivalent resistance for the following circuit 1Ω 1Ω Req 1Ω 1Ω 1Ω 39 ECS 203 - Part 1C - For ME2 Asst. Prof. Dr.Prapun Suksompong CHAPTER 4 Circuit Theorems The growth in areas of application of electrical circuits has led to an evolution from simple to complex circuits. To handle such complexity, engineers over the years have developed theorems to simplify circuit analysis. These theorems (Thevenin’s and Norton’s theorems) are applicable to linear circuits which are composed of resistors, voltage and current sources. 4.1. Linearity Property Definition 4.1.1. A linear circuit is a circuit whose output is linearly related (or directly proportional) to its input1. The input and output can be any voltage or current in the circuit. When we says that the input and output are linearly related, we mean they need to satisfies two properties: (a) Homogeneous (Scaling): If the input is multiplied by a constant k, then we should observed that the output is also multiplied by k. (b) Additive: If the inputs are summed then the output are summed. 1The input and output are sometimes referred to as cause and effect, respectively. 41 42 4. CIRCUIT THEOREMS Example 4.1.2. The linear circuit below is excited by a voltage source vs , which serves as the input. i vs Linear circuit R The circuit is terminated by a load R. We take the current i through R as the output. Suppose vs = 10 V gives i = 2 A. According to the linearity principle, vs = 1V will give i = 0.2 A. By the same token, i = 1 mA must be due to vs = 5 mV. Example 4.1.3. A resistor is a linear element when we consider the current i as its input and the voltage v as its output because it has the following properties: • Homogeneous (Scaling): If i is multiplied by a constant k, then the output v is multiplied by k. iR = v ⇒ (ki)R = kv • Additive: If the inputs, i1 and i2 , are summed then the corresponding output are summed. i1 R = v1 , i2 R = v2 ⇒ (i1 + i2 )R = v1 + v2 4.1.4. Because p = i2 R = v 2 /R (making it a quadratic function rather than a linear one), the relationship between power and voltage (or current) is nonlinear. Therefore, the theorems covered in this chapter are not applicable to power. 4.1. LINEARITY PROPERTY 43 Example 4.1.5. For the circuit below, find vo when (a) is = 15 and (b) is = 30. 12 Ω is 4Ω 8Ω + v0 – 44 4. CIRCUIT THEOREMS 4.2. Superposition Example 4.2.1. Find the voltage v in the following circuit. 8Ω 6V 4Ω + v – 3A From the expression of v, observe that there are two contributions. (a) When Is acts alone (set Vs = 0), (b) When Vs acts alone (set Is = 0), Key Idea: Find these contributions from the individual sources and then add them up to get the final answer. Definition 4.2.2. Superposition technique is a way to determine currents and voltages in a circuit that has multiple independent sources by considering the contribution of one source at a time and then add them up. 4.2.3. The superposition principle states that the voltage across (or current through) an element in a linear circuit is the algebraic sum of the voltages across (or currents through) that element due to each independent source acting alone. 4.2. SUPERPOSITION 45 4.2.4. To apply the superposition principle, we must keep two things in mind. (a) We consider one independent source at a time while all other independent source are turned off.2 • Replace other independent voltage sources by 0 V (or short circuits) • Replace other independent current sources by 0 A (or open circuits) This way we obtain a simpler and more manageable circuit. (b) Dependent sources are left intact because they are controlled by circuit variable. 4.2.5. Steps to Apply Superposition Principles: S1: Turn off all independent sources except one source. Find the output due to that active source. (Here, you may use any technique of your choice.) S2: Repeat S1 for each of the other independent sources. S3: Find the total contribution by adding algebraically all the contributions due to the independent sources. Example 4.2.6. Back to Example 4.2.1. 8Ω 6V 4Ω + v – 3A 2Other terms such as killed, made inactive, deadened, or set equal to zero are often used to convey the same idea. 46 4. CIRCUIT THEOREMS Example 4.2.7. Using superposition theorem, find vo in the following circuit. 3Ω + v0 – 2Ω 5Ω 4A 10 V 4.2.8. Remark on linearity: Keep in mind that superposition is based on linearity. Hence, we cannot find the total power from the power due to each source, because the power absorbed by a resistor depends on the square of the voltage or current and hence it is not linear (e.g. because 52 6= 12 + 42 ). 4.2.9. Remark on complexity: Superposition helps reduce a complex circuit to simpler circuits through replacement of voltage sources by short circuits and of current sources by open circuits. However, it may very likely involve more work. For example, if the circuit has three independent sources, we may have to analyze three circuits. The advantage is that each of the three circuits is considerably easier to analyze than the original one. 4.3. SOURCE TRANSFORMATION 47 4.3. Source Transformation We have noticed that series-parallel resistance combination helps simplify circuits. The simplification is done by replacing one part of a circuit by its equivalence.3 Source transformation is another tool for simplifying circuits. 4.3.1. A source transformation is the process of replacing a voltage source in series with a resistor R by a current source in parallel with a resistor R or vice versa. R a vs a is b R b Notice that when terminals a − b are short-circuited, the short-circuit current flowing from a to b is isc = vs /R in the circuit on the left-hand side and isc = is for the circuit on the righthand side. Thus, vs /R = is in order for the two circuits to be equivalent. Hence, source transformation requires that vs (4.1) vs = is R or is = . R 4.3.2. Remarks: (a) The “R” in series with the voltage source and the “R” in parallel with the current source are not the same “resistor” even though they have the same value. In particular, the voltage values across them are generally different and the current values through them are generally different. (b) From (4.1), an ideal voltage source with R = 0 cannot be replaced by a finite current source. Similarly, an ideal current source with R = ∞ cannot be replaced by a finite voltage source. 3Recall that an equivalent circuit is one whose v − i characteristics are identical with the original circuit. 48 4. CIRCUIT THEOREMS Example 4.3.3. Use source transformation to find v0 in the following circuit: 2Ω 4Ω 3A 8Ω 3Ω + v0 – 12 V ECS 203 - Part 1D - For ME2 Asst. Prof. Dr.Prapun Suksompong 4.4. THEVENIN’S THEOREM 49 4.4. Thevenin’s Theorem 4.4.1. It often occurs in practice that a particular element in a circuit or a particular part of a circuit is variable (usually called the load) while other elements are fixed. • As a typical example, a household outlet terminal may be connected to different appliances constituting a variable load. Each time the variable element is changed, the entire circuit has to be analyzed all over again. To avoid this problem, Thevenins theorem provides a technique by which the fixed part of the circuit is replaced by an equivalent circuit. 4.4.2. Thevenin’s Theorem is an important method to simplify a complicated circuit to a very simple circuit. It states that a circuit can be replaced by an equivalent circuit consisting of an independent voltage source VT h in series with a resistor RT h , where VT h : the open circuit voltage at the terminal. RT h : the equivalent resistance at the terminals when the independent sources are turned off. I a + Linear two- terminal circuit V Load – b (a) RTh I a + VTh V Load – b (b) This theorem allows us to convert a complicated network into a simple circuit consisting of a single voltage source and a single resistor connected in series. The circuit is equivalent in the sense that it looks the same from the outside, that is, it behaves the same electrically as seen by an outside observer connected to terminals a and b. 50 4. CIRCUIT THEOREMS 4.4.3. Steps to Apply Thevenin’s theorem. (Case I: No dependent source) S1: Find RT h : Turn off all independent sources. RT h is the input resistance of the network looking between terminals a and b. S2: Find VT h : Open the two terminals (remove the load) which you want to find the Thevenin equivalent circuit. VT h is the open-circuit voltage across the terminals. Linear two- terminal circuit VTh = voc + a voc – b Linear circuit with all independent sources set equal to zero a Rin b RTh = Rin S3: Connect VT h and RT h in series to produce the Thevenin equivalent circuit for the original circuit. Example 4.4.4. Find the Thevenin equivalent circuit of the circuit shown below, to the left of the terminals a-b. 4.4. THEVENIN’S THEOREM 51 Example 4.4.5. Find the Thevenin equivalent circuit of the circuit shown below, to the left of the terminals a-b. Then find the current through RL = 6, 16, and 36 Ω. 4Ω 32 V 1Ω 12 Ω a 2A RL b Solution: 4Ω 1Ω RTh 12 Ω 4Ω a 1Ω VTh + 32 V 12 Ω 2A (a) VTh – b (b) a b We can find vTHA by measuring the open-circuit voltage at the aa port in the network in Figure 3.68b. We find by inspection that vTHA = 1 V 52 Notice that the 1-A current flows each of the 1- resistors in the loop containing 4. through CIRCUIT THEOREMS the current source, and so v1 is 1 V. Since there is no current in the resistor connected to the a terminal, the voltage v2 across that resistor is 0. Thus vTHA = v1 + v2 = 1 V. We find RTHA by measuring thethe resistance looking port in theab network in circuit Example 4.4.6. Determine current I into in the theaabranch in the Figure 3.68c. The current source has been turned into an open circuit for the purpose below. 1A 1A 1Ω 1Ω 1Ω a a′ 1Ω 1Ω 1Ω 1Ω b b′ I 1Ω F I G U R E 3.67 Determ current in the branch ab. Network B Network A Solution: There are many approaches that we can take to obtain the current I. For example, we could apply the node method and determine the node voltages at nodes a and b and thereby determine the current I. However, we will find the Thvenin equivalent network for the subcircuit to the left of the aa0 terminal pair (Network A) and for the subcircuit to the right of the bb0 terminal pair (Network B), and then using these subcircuits solve 164 network theorems for the current I. CHAPTER THREE 1A 1Ω 1A (b) 1 Ω 1Ω F I G U R E 3.68 Finding the Thévenin equivalent for Network A. 1Ω 1Ω 1Ω (a) vTHA - a a′ v1 + a+ v a′ THA - 1Ω 1Ω + v2 - RTHA 1Ω (c) 1Ω 1Ω 1Ω a a′ RTHA 1A 1Ω 1A (b) + vTHB b b′ - vTHB F I G U R E 3.69 Finding the Thévenin equivalent for Network B. 1Ω 1Ω b b′ 1Ω 1Ω 1Ω (a) 1Ω RTHB (c) RTHB b b′ 1Ω 1Ω of measuring RTHA . By inspection, we find that RTHA = 2 . Let us now find the Thévenin equivalent for Network B shown in Figure 3.69a. Let vTHB and RTHB be the Thévenin parameters for this network. vTHB is the open-circuit voltage at the bb port in the network in Figure 3.69b. Using 4.4. THEVENIN’S THEOREM 53 4.4.7. Steps to Apply Thevenin’s theorem.(Case II: with dependent sources) S1: Find RT H : S1.1 Turn off all independent sources (but leave the dependent sources on). S1.2.i Apply a voltage source vo at terminals a and b, determine the resulting current io , then vo RT H = io Note that: We usually set vo = 1 V. Or, equivalently, S1.2.ii Apply a current source io at terminal a and b, find vo , then RT H = vioo S2: Find VT H , as the open-circuit voltage across the terminals. S3: Connect RT H and VT H in series. Remark: It often occurs that RT H takes a negative value. In this case, the negative resistance implies that the circuit is supplying power. This is possible in a circuit with dependent sources. 54 4. CIRCUIT THEOREMS 4.5. Norton’s Theorem Norton’s Theorem gives an alternative equivalent circuit to Thevenin’s Theorem. 4.5.1. Norton’s Theorem: A circuit can be replaced by an equivalent circuit consisting of a current source IN in parallel with a resistor RN , where IN is the short-circuit current through the terminals and RN is the input or equivalent resistance at the terminals when the independent sources are turned off. Note: RN = RT H and IN = RVTTHH . These relations are easily seen via source transformation.4 a Linear two-terminal circuit b (a) a IN RN b (b) 4For this reason, source transformation is often called Thevenin-Norton transformation. 4.5. NORTON’S THEOREM 55 Steps to Apply Norton’s Theorem S1: Find RN (in the same way we find RT H ). S2: Find IN : Short circuit terminals a to b. IN is the current passing through a and b. a Linear two- terminal circuit isc = IN b S3: Connect IN and RN in parallel. Example 4.5.2. Back to the circuit in Example 4.4.5. Find the Norton equivalent circuit of the circuit shown below, to the left of the terminals a-b. 4Ω 32 V 12 Ω 1Ω a 2A RL b 56 4. CIRCUIT THEOREMS Example 4.5.3. Find the Norton equivalent circuit of the circuit in the following figure at terminals a-b. 8Ω a 4Ω 5Ω 2A 12 V b 8Ω 4.6. MAXIMUM POWER TRANSFER 57 4.6. Maximum Power Transfer In many practical situations, a circuit is designed to provide power to a load. In areas such as communications, it is desirable to maximize the power delivered to a load. We now address the problem of delivering the maximum power to a load when given a system with known internal losses. 4.6.1. Questions: (a) How much power can be transferred to the load under the most ideal conditions? (b) What is the value of the load resistance that will absorb the maximum power from the source? 4.6.2. If the entire circuit is replaced by its Thevenin equivalent except for the load, as shown below, the power delivered to the load resistor RL is p = i2 RL where i = RTh Vth Rth + RL a i RL VTh b 58 4. CIRCUIT THEOREMS The derivative of p with respect to RL is given by dp di = 2i RL + i2 dRL dRL 2 Vth Vth Vth − + =2 Rth + RL Rth + RL (Rth + RL )2 2 2RL Vth = − +1 . Rth + RL Rth + RL We then set this derivative equal to zero and get RL = RT H . 4.6.3. The maximum power transfer takes place when the load resistance RL equals the Thevenin resistance RT h . The corresponding maximum power transferred to RL equals to 2 V2 Vth Rth = th . pmax = Rth + Rth 4Rth 4.6. MAXIMUM POWER TRANSFER 59 Example 4.6.4. Connect a load resistor RL across the circuit in Example 4.4.4. Assume that R1 = R2 = 14Ω, Vs = 56V, and Is = 2A. Find the value of RL for maximum power transfer and the corresponding maximum power. Example 4.6.5. Find the value of RL for maximum power transfer in the circuit below. Find the corresponding maximum power. 6Ω 12 V 3Ω 12 Ω 2Ω a 2A RL b ECS 203 - Part 1E - For ME2 Asst. Prof. Dr.Prapun Suksompong CHAPTER 5 Operational Amplifiers In this chapter, we learn how to use a new circuit element called op amp to build circuits that can perform various kinds of mathematical operations. Op amp is a building block of modern electronic instrumentation. Therefore, mastery of operational amplifier fundamentals is paramount to any practical application of electronic circuits. They are popular in practical circuit degigns because they are versatile, inexpensive, easy to use, and fun to work with. 5.1. Introduction to Op Amp 5.1.1. Operational amplifiers (or Op Amp) is an active circuit element that can perform mathematical operations (e.g., amplification, summation, subtraction, multiplication, division, integration, differentiation) between signals . • The ability of the op amp to perform these mathematical operations is the reason it is called an operational amplifier. It is also the reason for the widespread use of op amps in analog design. 5.1.2. An op amp consisting of a complex arrangement of resistors, transistors, capacitors, and diodes. Here, we ignore the details. 61 62 5. OPERATIONAL AMPLIFIERS 5.1.3. The circuit symbol for the op amp is shown below. V+ 7 Inverting input 2 – Noninverting input 3 + 6 Output 4 1 5 V– Offset Null • It has two inputs and one output. • The inputs are marked with minus (-) and plus (+) to specify inverting and noninverting1 inputs, respectively. 5.1.4. As an active element, the op amp must be powered by a voltage supply: i7 i2 – 3 + i+ 7 + VCC – io 6 4 i4 + VCC – Although, in this class, the power supplies are often ignored in op amp circuit diagrams for the sake of simplicity, the power supply currents must not be overlooked: io = i7 + i4 + i + + i− . Caution: Even when pins 7 and 4 are not shown explicitly, they are always there and the corresponding currents are also there. 1An input applied to the noninverting terminal will appear with the same polarity at the output, while an input applied to the inverting terminal will appear inverted at the output. 5.1. INTRODUCTION TO OP AMP 63 5.1.5. The equivalent circuit of non-ideal op amp is shown below. Note that the output section consists of a voltage-controlled source in series with the output resistance Ro . vRo – vd Ri + + – vo Avd v+ Input-output relations: where vo = Avd = A(v+ − v− ) vo = voltage between the output terminal and ground v− = voltage between the inverting terminal and ground v+ = voltage between the noninverting terminal and ground vd = v+ − v− = differential input voltage A = open-loop voltage gain • In words, the op amp senses the difference between the two inputs, multiplies it by the gain A, and causes the resulting voltage to appear at the output. A is called the open-loop voltage gain because it is the gain of the op amp without any external feedback from output to input. 5.1.6. The concept of feedback is crucial to our understanding of op amp circuits. A negative feedback is achieved when the output is fed back to the inverting terminal of the op amp. 64 5. OPERATIONAL AMPLIFIERS Example 5.1.7. Consider the circuit below. There is a feedback path from output to input. The ratio of the output voltage to the input voltage is called the closed-loop gain. Rf R1 Rf R1 – Ro – vi + + vo vd Ri vi + + – Avd – + vo – 5.1.8. Typical ranges for op amp parameters are shown in the following table Working with a nonideal op amp is tedious because it involves dealing Parameter Typical range Ideal values Open-loop gain, A 105 to 108 ∞ 5 13 Input resistance, Ri 10 to 10 Ω ∞Ω Output resistance, Ro 10 to 100 Ω 0Ω Supply voltage, VCC 5 to 24 V with very large numbers. Example 5.1.9. Consider, again, the circuit in Example 5.1.7. Suppose R1 = 10 kΩ and Rf = 20 kΩ. Assume that the op amp has an open-loop voltage gain of 2 × 105 , input resistance Ri of 2 MΩ, and output resistance Ro of 50 Ω. Find the closed-loop gain vo /vi . It can be shown that the closed-loop gain is almost insensitive to the open-loop gain A of the op amp. For this reason, op amps are used in circuits with feedback paths. 5.2. IDEAL OP-AMP 65 5.2. Ideal Op-Amp To facilitate understanding, we assume ideal op amps with the ideal values above. Definition 5.2.1. An ideal op amp is an amplifier with infinite openloop gain, infinite input resistance, and zero output resistance. Unless stated otherwise, we will assume from now on that every op amp is ideal. 5.2.2. Two important characteristics of the ideal op-amp: (a) The current into both input terminals are zero. i− = 0 and i+ = 0. (b) The voltage across the input terminals is negligibly small. vd = v+ − v− ≈ 0 or v+ ≈ v− . i- = 0 – + i+ = 0 v- + v+ = v– 5.2.3. Caution: – – vd + io + + vo – 66 5. OPERATIONAL AMPLIFIERS Example 5.2.4. An ideal op amp is used in the circuit below. Find the closed-loop gain2 vo /vs . Determine current io when vs = 1 V. i0 + – 40 kΩ vs 5 kΩ 20 kΩ 2closed-loop gain = ratio of the output voltage to the input voltage. + vo – 5.3. INVERTING AMPLIFIER 67 5.3. Inverting Amplifier Op amp can be used in circuits as modules for creating more complex circuits. The first of such op-amp circuits is the inverting amplifier which reverses the polarity of the input signal while amplifying it. A key feature of the inverting amplifier is that both the input signal and the feedback are applied at the inverting terminal of the op amp. if i1 vi R1 Rf – + + vo – 68 5. OPERATIONAL AMPLIFIERS The equivalent circuit for the inverting amplifier is: + + vi Rf vi R1 – + R1 – vo – The voltage gain is Av = vo /vi = −Rf /R1 . 5.4. Noninverting Amplifier A noninverting amplifier amplifies a signal by a constant positive gain (no inversion of polarity). The circuit for a noninverting amplifier is if R1 Rf i1 – + vi + vo – 5.4. NONINVERTING AMPLIFIER 69 R The voltage gain is Av = vvoi = 1 + Rf1 , which does not have a negative sign. Thus the output has the same polarity as the input. Special case: If Rf = 0 or R1 = ∞, or both, the gain becomes 1. Under this conditions, the circuit becomes a voltage follower (The output follows the input). – + + vo = vi vi – A voltage follower is used to isolate two cascaded stages of a circuit. First stage + – + + vi vo – – Second stage 70 5. OPERATIONAL AMPLIFIERS Example 5.4.1. Calculate the output voltage vo for the op amp circuit below. 10 kΩ 4 kΩ 6V – + 4V + vo – 5.5. SUMMING AMPLIFIER 71 5.5. Summing Amplifier A summing amplifier is an op-amp circuit that combines several inputs and produces an output that is the weighted sum of the inputs. For this reason, the circuit is called a summer. v1 v2 v3 R1 i1 R2 i2 Rf i a R3 i3 if – + + vo – Rf Rf Rf vo = − v1 + v2 + v3 . R1 R2 R3 Needless to say, the summer can have more that three inputs. 72 5. OPERATIONAL AMPLIFIERS 5.6. Difference Amplifier A difference amplifier is a device that amplifies the difference between two inputs but rejects any signals common to the two inputs. R2 R1 R3 v1 v2 0 va vb 0 – + R4 + vo – (1 + R2 /R1 ) R2 R2 (1 + R1 /R2 ) R2 v2 − v1 = v2 − v1 (1 + R3 /R4 ) R1 R1 (1 + R3 /R4 ) R1 Since a difference amplifier must reject a signal common to the two inputs, the amplifier must have the property that vo = 0 when v1 = v2 . This property exists when vo = R1 R3 = . R2 R4 5.6. DIFFERENCE AMPLIFIER 73 Thus, R2 (v2 − v1 ) . R1 If R2 = R1 and R3 = R4 , the difference amplifier becomes a subtractor, with the output vo = vo = v2 − v1 . Example 5.6.1. Design an op amp circuit with inputs v1 and v2 such that vo = −5v1 + 3v2 . 74 5. OPERATIONAL AMPLIFIERS 5.7. Cascaded of Op Amp Circuits In practice, we can connect op amp circuits in cascade (i.e., head to tail) to achieve a large overall gain. Each circuit in the cascade is called stage. The output of one stage is the input to the next stage. Op amp circuits have the advantage that they can be cascaded without changing their input-output relationships. This is due to the fact that each (ideal) op amp circuit has infinite input resistance and zero output resistance. + v1 – + Stage 1 A1 + Stage 2 A2 v2 = A1v1 – v3 = A2v2 – Stage 3 A3 + vo = A3v3 – 5.8. Application: Digital-to-Analog Converter (DAC) The digital-to-analog converter (DAC) transforms digital signals into analog form. A typical example of a four-bit DAC is shown in (a) below. Digital Input (0000–1111) Analog output Four-bit DAC (a) V2 V1 R1 MSB V4 V3 R2 R3 R4 LSB (b) Rf – + Vo 5.9. APPLICATION: INSTRUMENTATION(AL) AMPLIFIERS (IA) 75 5.9. Application: Instrumentation(al) Amplifiers (IA) One of the most useful and versatile op amp circuits for precision measurement and process control. IA amplifies the difference between the input signals. Inverting input v1 Gain set R – +1 R R RG R Gain set Noninverting input – +3 v2 vo Output R – +2 – R + (a) 2R vo = 1 + (v2 − v1 ) . RG (b) 76 5. OPERATIONAL AMPLIFIERS The instrumentation amplifier amplifies small differential signal voltages superimposed on larger common-mode voltages. Since the common-mode voltages are equal, they cancel each other. The IA has three major characteristics: (a) The voltage gain is adjusted by one external resistor RG . (b) The input impedance of both inputs is very high and does not vary as the gain is adjusted. (c) The output vo depends on the difference between the inputs, not on the voltage common to them. Typical example of IA has gain from 1 to 1000.