Chapter 6: Circuits

Chapter 6 Circuits 6.01— Spring 2011— April 25, 2011 233 Chapter 6 Circuits 6.1 What is a circuit? In conventional English, a circuit is a path or route that starts at one place and ultimately returns to that same place. The engineering sense of the word is similar. In electronics, a circuit is a closed path through which electrical currents can flow. The flow of electrical current through a flashlight illustrates the idea. Three parts of the flashlight comprise the circuit: the battery, which supplies power, a bulb, which produces light when connected to the battery, and a switch, which can either be on (connected) or off (disconnected). The important feature of this circuit is that electrical currents flow through it in a loop, as illustrated in the right part of the figure above. The rules that govern the flow of electrical current are similar to the rules that govern the flow of certain types of fluids; and it is often helpful to draw analogies between these flows. The following figure illustrates the flow of blood through the human circulatory system. head and arms lungs heart liver intestines kidneys trunk and legs Chapter 6 Circuits 6.01— Spring 2011— April 25, 2011 234 The right side (from the person’s viewpoint) of the heart pumps blood through the lungs and into the left side of the heart, where it is then pumped throughout the body. The heart and associated network of arteries, capillaries, and veins can be thought of and analyzed as a circuit. As with the flow of electrical current in the flashlight example, blood flows through the circulatory system in loops: it starts at one place and ultimately returns to that same place. As we learn about circuits, we will point out analogies to fluid dynamics for two reasons. First, we all have developed some intuition for the flow of fluid as a result of our interactions with fluids as a part of everyday experience. We can leverage the similarity between fluid flow and the flow of electrical current to increase our intuition for electrical circuits. Second, the analogy between electrical circuits and fluid dynamics is a simple example of the use of circuits to make models of other phenomena. Such models are widely used in acoustics, hydrodynamics, cellular biophysics, systems biology, neurophysiology, and many other fields. Electrical circuits are made up of components, such as resistors, capacitors, inductors, and tran­ sistors, connected together by wires. You can make arbitrarily amazing, complicated devices by hooking these things up in different ways, but in order to help with analysis and design of circuits, we need a systematic way of understanding how they work. As usual, we can’t comprehend the whole thing at once: it’s too hard to analyze the system at the level of individual components, so, again, we’re going to build a model in terms of primitives, means of combination, and means of abstraction. The primitives will be the basic components, such as resistors and op-amps; the means of combination is wiring the primitives together into circuits. We’ll find that abstraction in circuits is different than in software or LTI systems. You can’t think of a circuit as “computing” the voltages on its wires: if you connect it to another circuit, then the voltages are very likely to be different. However, you can think of a circuit as enforcing a constraint on the voltages and currents that enter and exit it; this constraint will remain true, no matter what else you connect to the circuit. 6.1.1 Electrical circuits Circuits are made up of elements connected by nodes. The following circuit has three elements, each represented with a box. i1 i2 i3 There are two nodes, each indicated by a dot. In an electrical circuit nodes can be thought of as wires 39 that connect a component. Voltage is a difference in electrical potential between two different points in a circuit. We will often pick some point in a circuit and say that it is “ground” or has voltage 0. Now, every other 39 Wire is typically made from metals through which electrons can flow to produce electrical currents. Generally, the strong forces of repulsion between electrons prevent their accumulation in wires. However, thinking about nodes as wires is only an approximation, because physical wires can have more complicated electrical properties. Chapter 6 Circuits 6.01— Spring 2011— April 25, 2011 235 point has a voltage defined with respect to ground. Because voltage is a relative concept, we could pick any point in the circuit and call it ground, and we would still get the same results. Current is a flow of electrical charge through a path in the circuit. A positive current in a direction is generated by negative charges (electrons) moving in the opposite direction. We’ll start by considering a simple set of components that have two terminals (connection points into the circuit). Each component has a current flowing through it, and a voltage difference across its terminals. Each type of component has some special characteristics that govern the relationship between its voltage and current. We will restrict our attention, in this course, to components that exert a linear constraint on their current and voltage. One way to model circuits is in terms of their dynamics. That is, to think of the currents and voltages in the system and how they change over time. Such systems are appropriately modeled, in fact, using differential or difference equations, connected together into complex systems, as we saw in the last couple of weeks. But for many purposes, the dynamic properties of a circuit converge quickly, and we can directly model the equilibrium state that they will converge to. The combination of the behavior of the components and the structure in which they’re connected provides a set of constraints on the equilibrium state of the circuit. We’ll work through this view by starting with the constraints that come from the structure, and then examining constraints for three simple types of components. 6.2 Conservation laws Conservation laws describe properties of a circuit that must hold, no matter what the particular elements are. There are two fundamental conservation laws, one that relates to currents and one that relates to voltages. 6.2.1 Kirchoff’s current law Fluids can be roughly classified as compressible or incompressible. Air is compressible. Doubling the pressure on a volume of air will decrease the volume by a factor of two. By contrast, water is nearly incompressible. Doubling the pressure on a volume of water (from 1 atmosphere to 2) will decrease the volume by less than one part in a billion. The laws that govern incompressible fluid flows are relatively simple. The net flow of incom­ pressible fluid into a region of fixed volume must be zero. It follows that if there is flow into one part of a fixed volume, then there most be an equal flow out of the other part. As an example, consider the flow of water through a branching point, as shown below. → i2 i1 → →i 3 Let i1 represent the flow into the branching point and i2 and i3 represent the flows out of it. Assume that i1 , i2 , and i3 have the units of volume per unit time (e.g., m3 /s). Chapter 6 Circuits 6.01— Spring 2011— April 25, 2011 236 What is the relation between i1 , i2 , and i3 ? Example 1. In this example, i1 = i2 + i3 . What if i2 represented flow in the opposite direction, as shown below? Example 2. ← i2 i1 → →i 3 Now i1 + i2 = i3 or equivalently i1 = −i2 + i3 . Changing the direction of a flow variable is equivalent to changing its sign. For that reason, we think about flow variables as signed variables. The direction associated with the variable represents the direction that results when the variable is positive. The flow is in the opposite direction if the variable is negative. Thus, the direction that we associate with a flow is arbitrary. If we happen to associate the one direction, the flow variable will be positive. If we happen to associate the opposite direction, the flow variable will be negative. The laws for the flow of electrical current are similar to those for the flow of an incompressible fluid. The net flow of electrical current into a node must be zero. The following circuit has three elements, each represented with a box. i1 i2 i3 There are two nodes, each indicated by a dot. The net current into or out of each of these nodes is zero. Therefore i1 + i2 + i3 = 0. Similar current relations apply for all nodes in a circuit. Such relations are called Kirchoff’s Cur­ rent Laws (KCL) and represent the first of two fundamental conservation laws for circuits. Kirchoff’s Current Law: the sum of the currents that flow into a node is zero. Electrical currents cannot accumulate in elements, so the current that flows into a circuit element must equal to the current that flows out. Chapter 6 Circuits 6.01— Spring 2011— April 25, 2011 i1 i2 i3 i4 i5 i6 237 Therefore i1 = i4 , i2 = i5 , and i3 = i6 . Teach Yourself 1. Part a. A circuit is divided into the part that is inside the dashed red box (below) and the part that is outside the box (not shown). Find an equation that relates the currents i1 and i2 that flow through the dashed red box. i1 i2 Part b. A more complicated circuit is shown inside the red box below. Find an equation that relates the currents i1 , i2 , and i3 that flow through the dashed red box. i1 i2 i3 Part c. Generalize your results to an arbitrary closed surface. Explain how to prove the generalization. Chapter 6 Circuits 6.01— Spring 2011— April 25, 2011 238 Teach Yourself 2. Part a. How many linearly independent 40 KCL equations can be written for the following circuit? Part b. Generalize your result to determine the number of linearly inde­ pendent KCL equations in an arbitrary circuit with n nodes. Prove your generalization. 6.2.2 Kirchoff’s Voltage Law What physical mechanisms cause flow? Blood circulates due to pressures generated in the heart. Hydraulic pressure also moves water through the pipes of your house. In similar fashion, voltages propel electrical currents. A voltage can be associated with every node in a circuit; and it is the difference between the voltages at two nodes that excites electrical currents to pass between those nodes. The flow of water between water storage tanks provides a useful analogy for the flow of electrical currents. Consider three tanks connected as shown below. h1 h0 h2 i1 40 i2 None of the equations in a linearly independent set can be written as a linear combination of the other equations. For example, x+y=1 x + 2y = 1 are linearly independent equations. However, the equations x+y+z=1 x+y=1 z=0 are not, since the last equation is the difference between the first two. Chapter 6 Circuits 6.01— Spring 2011— April 25, 2011 239 What will be the signs of i1 and i2 ? Water will flow from the middle tank toward the left tank because h1 > h0 . Thus i1 < 0. Similarly, water will flow from the middle tank toward the right tank because h1 > h2 . Thus i2 > 0. Example 3. Voltages work similarly for driving electrical currents. It is the difference in voltage between two nodes that drives electrical current between the nodes. Absolute voltage is not important (or even well defined). Voltage differences are all that matter. Teach Yourself 3. How would the flows in the tank system above change if each height of water in each tank were increased by 1 unit? ... by 1000 units? Another similarity between the hydraulic pressures in the tank system and voltages in a circuit is that the voltage differences that accumulate along any closed path is zero. This is obvious for an array of tanks. Consider the array illustrated below in a top down view. 3 1 3 7 2 2 1 Assume that the lines between circles represent pipes and that the numbers indicate the height (in meters) of the water in each tank. Consider the clockwise path around the perimeter of the network starting and ending with the leftmost tank. The path visits heights of 1, 3, 3, 2, 1, 2, and 1. These heights increase by 2, 0, −1, −1, 1, and −1, which sum to 0. It’s easy to see that the sum of differences in heights around any closed loop will be zero. If it were not, then the height of the first tank would not end up at the same value that it began. This same rule applies to voltage differences, where the rule is called Kirchoff’s Voltage Law (KVL). Kirchoff’s Voltage Law: the sum of voltage differences along a closed path is zero. Chapter 6 Circuits Example 4. 6.01— Spring 2011— April 25, 2011 240 How many KVL equations can be written for the following circuit. + v1 − + v2 − + v3 − How many of the KVL equations are linearly independent? There are three loops through this circuit. One through the left two ele­ ments yields the KVL equation −v1 + v2 = 0 or v1 = v2 . One through the right two elements yields the KVL equation −v2 + v3 = 0 or v2 = v3 . One through the outer two elements yields the KVL equation −v1 + v3 = 0 or v 1 = v3 . Only two of the three KVL equations are linearly independent, since each equation can be derived from the other two. The particularly simple solution to this circuit is that v1 = v2 = v3 . This solution could also have been derived directly from the fact that there are just two nodes, and therefore only one possible potential difference. Chapter 6 Circuits Example 5. 6.01— Spring 2011— April 25, 2011 241 How many different KVL equations can be written for this circuit? + v1 + − V0 + v2 − + v4 − − + v3 − + v6 − + v5 − How many of the KCL equations are linearly independent? One KVL equation can be written for every possible closed loop through the circuit. The most obvious loops are A, B, and C shown below. + v1 + − V0 − A + v2 − + v3 − B + v6 − + v4 − C A: −v1 + v2 + v4 = 0 B: −v2 + v3 − v6 = 0 + v5 − C: −v4 + v6 + v5 = 0 But there are several more, shown below. + v1 + − V0 − D + v2 − + v4 − + v3 − + v6 − + v5 − D=A+B: −v1 + v3 − v6 + v4 = 0 + v1 + − V0 − E + v2 − + v4 − + v3 − + v6 − + v5 − E=A+C: −v1 + v2 + v6 + v5 = 0 Chapter 6 Circuits 6.01— Spring 2011— April 25, 2011 + v1 + − V0 + v2 − + v4 − − + v3 − + v6 − F + v1 + v5 − + − V0 − G F=B+C: −v4 − v2 + v3 + v5 = 0 + v2 − + v4 − 242 + v3 − + v6 − + v5 − G=A+B+C: −v1 + v3 + v5 = 0 These seven equations are not linearly independent. Equation D is the sum of Equations A and B (notice that the v2 terms in Equations A and B cancel). Similarly, equation E is the sum of Equations A and C, equation F is the sum of Equations B and C, and equation G is the sum of Equations A, B, and C. Thus this circuit has three independent KVL equations and four additional dependent KVL equations. Teach Yourself 4. The following set of voltages are not consistent with Kirchoff’s voltage laws but can be made consistent by changing just one. Which one? 2 3 1 3 2 4 1 2 2 3 2 0 Prove that the answer is unique. 6.3 Circuit elements Kirchoff’s Current and Voltage Laws do not alone determine the behavior of a circuit. Consider the following circuit with just two elements. There are two elements, so there are two element currents i1 and i2 as well as two element voltages v1 and v2 , all of which are unknown. + v1 − i1 i2 + v2 − Chapter 6 Circuits Example 6. 6.01— Spring 2011— April 25, 2011 243 List all of the KCL and KVL equations for this circuit. There are two nodes in this circuit. As we saw in TeachYourself 2, this implies that there is a single independent KCL equation, which is i1 + i2 = 0. There is a single loop and therefore a single KVL equation: v1 − v2 = 0. Thus we have four unknowns and just two equations. More information is needed to solve for the currents and voltages. In general, every electrical element imposes a relation between the voltage across the element and the current through the element. This relation is called a constitutive or element relation. The simplest relations are those for sources. A voltage source is an element whose voltage is a constant (e.g., v = V0 ) independent of the current through the element. We will denote a voltage source by a circle enclosing plus and minus symbols to denote voltage. A number beside the source will indicate the constant voltage generated by the source, as illustrated below. Voltages are specified with units called volts which are abbreviated as V. V0 + − A current source is an element whose current is a constant (e.g., i = I0 ) independent of the voltage across the element. We will denote a current source by a circle enclosing an arrow to denote current. A number beside the source will indicate the constant current generated by the source, as illustrated below. Currents are specified with units called amperes or simply amps which are abbreviated as A. I0 Our third simple element is a resistor, in which the voltage is proportional to the current and the proportionality constant is the resistance R, so that vR = RiR . We will denote a resistor as follows. Resistances have the dimensions of ohms (abbreviated as Ω), which are equivalent to volts divided by amps. R A circuit consisting of a 10V voltage source and 2Ω resistor can easily be solved. Chapter 6 Circuits 6.01— Spring 2011— April 25, 2011 i1 + v1 − + − 244 i2 + 10V 2Ω v2 − Example 7. Find v1 , v2 , i1 , and i2 . As before, i1 + i2 = 0 and v1 = v2 . However, now we also know two consitutive relations: v1 = 10V and v2 = 2Ω × i2 . Solving, we find that v1 = v2 = 10V, i2 = v2 /2Ω = 5A, and i1 = −i2 = −5A. Example 8. Determine the currents and voltages in the following circuit. i1 i2 + + v1 − 14A 3Ω v2 − i3 + 4Ω v3 − There is a single independent KCL equation: i1 + i2 + i3 = 0. There are two independent KVL equations that together yield v1 = v2 = v3 . Thus, KCL and KVL together yield three equations in six unknowns. We need three more equations — one from each element. From the current source constitutive relation we have i1 = −14A. From the resistor constitutive relations we have v2 = 3Ω × i2 and v3 = 4Ω × i3 . Solving v1 = v2 = 3Ω×i2 = v3 = 4Ω×i3 = 4Ω×(−i1 −i2 ) = 4Ω×(14−i2 ) so that 3Ω × i2 = 4Ω × (14 − i2 ) so that i2 = 8A. Therefore i1 = −14A, i2 = 8A, and i3 = 6A. Also, v1 = v2 = 3Ω × 8A= v3 = 4Ω × 6A= 24V. Chapter 6 Circuits Example 9. 6.01— Spring 2011— April 25, 2011 245 Solve the following circuit. + v2 − i 2 i1 + v1 − i3 + 6Ω + − 36V 3Ω v3 i4 + 2Ω − v4 − Answer: i1 = −5A ; i2 = 5A ; i3 = 2A ; i4 = 3A ; v1 = 36V ; v2 = 30V ; v3 = 6V ; v4 = 6V 6.4 Solving circuits Given a description of the circuit’s wiring and a specification of the individual components, we can solve the circuit to answer questions about particular voltages or currents in the circuit. There are several ways to approach this problem: we will start with a completely systematic method that is particularly useful if we want to solve circuits using a computer; then, we’ll see some simpler method and patterns that are easier for humans to use. The most obvious strategy would be to write down all the KVL and KCL equations, add the con­ stituent equations for each element, and then solve for the voltage and current at each component. But we have a problem with redundant KVL equations. The voltages across circuit elements are constrained by KVL so that the sum of the voltages around any closed loop is zero. However, there are many possible closed loops through even simple circuits (see section 6.2.2). If we are dealing with a planar circuit, 41 then we know that all loops can be written as combinations of primitive loops with non-overlapping interiors (see example 5). Teach Yourself 5. Each of four nodes are connected to each of the others by a 1Ω resistor. How many resistors are there? Is the network planar? Teach Yourself 6. Each of five nodes are connected to each of the others by a 1Ω resistor. How many resistors are there? Is the network planar? So far, the only general way that we have seen to deal with this redundancy is to (1) find all of the possible KVL equations, and then (2) eliminate the redundant ones. This process seems needlessly complicated. 41 A circuit is “planar” if it is possible to draw all of the elements and all of the lines that connect the elements without any of the lines or elements crossing over one another. Chapter 6 Circuits 6.01— Spring 2011— April 25, 2011 246 One good way to eliminate KVL redundancy is to use node voltages. Node voltages are voltages associated with circuit nodes rather than with circuit elements. Given a complete set of node volt­ ages, it is easy to determine the element voltages: each element voltage is the difference between the voltage of the node at the positive end of the element minus that at its negative end. Example 10. Determine expressions for each of the element voltages from the node volt­ ages e0 , e1 , e2 , and e3 . e3 + v1 − + − + v2 − V0 e1 + v4 − + v3 − + v6 − e2 + v5 − e0 Answers: v1 = e3 − e0 ; v2 = e3 − e1 ; v3 = e3 − e2 ; v4 = e1 − e0 ; v5 = e2 − e0 ; v6 = e1 − e2 ; Example 11. How would the element voltages in the previous example change if 1V were added to each of the node voltages? Adding a constant voltage to each of the node voltages of a circuit has no effect on any of the element voltages. Thus, substituting node voltages for element voltages removes redundancies that are implicit in KCL equations, but also introduces a new redundancy, which we can think of as a global offset voltage. If we are only interested in element voltages (which is typically the case), we can exclude effects of a global offset voltage by arbitrarily assigning the potential of one node to zero. We refer to that special node with zero potential as the “ground.” We will explore two methods that use node voltages: the node-voltage-and-component-current (NVCC) method, which has variables for voltages at all of the nodes and currents through all of the components, and the node method, which only has variables for voltages at the nodes. The NVCC method is completely general, but rather long-winded (and is perhaps better for comput­ ers to use than humans). The node method is terser, with potentially many fewer equations, but can become complicated when there are multiple voltage sources. 6.4.1 NVCC Method In the NVCC method, we take the following steps: Chapter 6 Circuits 6.01— Spring 2011— April 25, 2011 247 1. Label all of the nodes (places where there is a wire between two or more components) with names n1 , . . . , nn , and make variables v1 , . . . , vn , one for the voltage at each node. 2. Declare one of them, ng , to be the ground node (this can be any node; the voltages will be computed relative to the voltage at that node) and set vg = 0. 3. Make current variables i1 , . . . , im for each component (resistor or source) in the network. La­ bel a direction for each current in your network (it doesn’t matter what directions you pick as long as you handle them consistently from here on out). 4. Write down n − 1 KCL equations, one for each node except for ng . These equations assert that the sum of currents entering each node is 0. 5. Write down m constitutive equations, one for each component, describing that component’s linear relationship between its current ik and the voltage difference across the component. The voltage across the component is vk+ − vk− , where vk+ is the node voltage at the positive terminal of the component and vk− is the node voltage at its negative terminal voltage. The direction of the current defines what constitutes the ’positive’ and ’negative’ terminals of the component: the current runs from positive to negative. For a resistor with resistance R, the equation is vk+ − vk− = ik R; for a voltage source with voltage Vs , the equation is vk+ − vk− = Vs ; for a current source with current Cs , the equation is ik = Cs . 6. Solve these equations to determine the node voltage and component currents. So, for this circuit, n1 iA RA n2 iC RB Vc iB iD n3 RD using the current directions for each component as drawn, we have the following equations: • Ground v3 = 0 . • KCL iA − iB − iD − iC = 0 iB − iA = 0 . • Constitutive equations Chapter 6 Circuits 6.01— Spring 2011— April 25, 2011 248 (v1 − v2 ) = iA · RA (v2 − v1 ) = iB · RB (v2 − v3 ) = iD · RD v2 − v3 = Vc . So, now, if we know RA , RB , RD , and VC , which are the specifications of our components, we have 7 linear equations in 7 unknowns (v1 , v2 , v3 , iA , iB , iC , and iD ). Just a small (though possibly tedious) matter of algebra, and we’re done. As an example, let RA = 100Ω, RB = 200Ω, RD = 100Ω, and VC = 10V . Then, we get v2 = 10V ; iA = iB = 0A (that’s reasonable: why would any current bother going that way, when it can just run through the diagonal wire?); and iD = 0.1A, which is pretty straightforward. 6.4.2 Solution strategy You can see that the KCL equations are all in terms of currents; and the constitutive equations give us direct expressions for almost all of the currents. So, a good solution strategy is to work primarily with the KCL equations, trying to eliminate as many current variables as possible, by substituting in expressions derived from the constitutive equations. In the example above, here are the KCL equations: iA − i B − i D − i C = 0 iB − i A = 0 . Now, substituting in expressions from the constitutive equations, we can rewrite them as: Vc v1 − v 2 v2 − v 1 − − − iC = 0 RB RD RA v2 − v 1 v1 − v 2 − =0 . RA RB We also know, from the ground equation v3 = 0, that v2 = VC , so we can rewrite our equations as: v1 − Vc Vc − v1 Vc − − − iC = 0 RB RD RA Vc − v1 v1 − Vc − =0 . RA RB In most cases, you can write equations of this form down directly, without bothering to write the KCL and the constituent equations separately, and then substituting in for the currents. Now we only have two unknowns: v1 and iC , and two equations. The second equation tells us that Vc = V1 , and, substituting that into the first equation, we have iC = Vc RD Chapter 6 Circuits Exercise 6.1. 6.01— Spring 2011— April 25, 2011 249 What happens when we take out the diagonal wire? We have to add a new node, v4 , and then proceed as we did above. RA i A n1 n2 iC RB Vc iB n4 iD n3 RD Ultimately, we can find that iA = iB = iC = −0.025A, iD = 0.025A, v1 = 7.5V , v2 = 10V , and v4 = 2.5V . Verify this solution by writing out the equations and solving them using a process similar to the one outlined above. 6.4.3 An example Let’s try to find the node voltages and constituent currents for this circuit: n1 i4 i1 +15 V n2 3Ω i2 2Ω +10 A i3 n3 NVCC Here are the steps of the NVCC method: 1. We have labeled the nodes n1 , n2 , and n3 . Remember that we don’t need a node for every corner or join in the circuit: a single ’node’ can cover a stretch of wire with several connections, because it will all have the same voltage as long as no components intervene. 2. Declare one of them to be the ground node. When there’s a single voltage source, it is con­ ventional (but not necessary!) to set its negative terminal to be the ground node; so we’ll set v 3 = 0. 3. Make current variables for each component in the network and label them with arrows. We did this already in the figure. 4. Write down n − 1 KCL equations, one for each node except for the ground node. − i4 − i1 = 0 i1 − i2 + i3 = 0 Chapter 6 Circuits 6.01— Spring 2011— April 25, 2011 250 5. Write down m constitutive equations, one for each component, describing that component’s linear relationship between its current ik and the voltage difference across the component. v1 = 15 v1 − v2 = 3i1 v2 − v3 = 2i2 i3 = 10 6. Solve these equations to determine the node voltage and component currents. There are lots of ways to work through this algebra. Here’s one meandering path to the right answer. We can do some plugging in of v1 , v3 , and i3 to simplify matters. 15 − v2 = 3i1 v2 = 2i2 − i4 − i1 = 0 i1 − i2 + 10 = 0 Plugging expressions for i1 and i2 derived from the first two equations into the last one, we get 15 − v2 v2 − = −10 3 2 so v2 = 18 And now it’s easy to see that i1 = −1, i2 = 9, and i4 = 1. NVCC Shortcut If we want to shortcut the NVCC method, we can follow the first three steps above, but then try to go straight to versions of the KCL equations that substitute in expressions for current that are derived from the constituent constraints. So, we could observe that if v3 = 0 then v1 = 15. And then we could write the KCL equations as: 15 − v2 =0 3 v2 15 − v2 + 10 − =0 3 2 − i4 − Solving the second equation we find that v2 = 18; from this we can easily determine the other relevant quantities. 6.4.4 Node method (optional) Node analysis is similar to NVCC, but with the goal of writing down one equation per node. We can further simplify the node representation of circuits that contain voltage sources. When a voltage source connects two nodes, it constrains the two node voltages so that one can be deter­ mined from the other, as shown in the following figure, where e0 in the previous figure is taken as ground. Chapter 6 Circuits 6.01— Spring 2011— April 25, 2011 251 V0 + + − v1 + v2 − V0 e1 + v4 − − + v3 − + v6 − e2 + v5 − gnd So far, we have seen that node voltages are a concise representation of all of the information that is necessary to compute the element voltages of a circuit. Furthermore, node voltages automatically guarantee that KVL is satisfied by all of the element voltages. To complete the analysis of a circuit using node voltages, we must next consider the element currents and KCL. In general, one can write KCL at each of the nodes of a circuit. However, as we saw in TeachYourself 2, one of those equations will always be linearly dependent on the others. Thus, if the circuit has n nodes, we can write n − 1 independent KCL equations. If there are no voltage sources in the circuit, then we can express each KCL equation as a sum of element currents that can each be determined from the node voltages by using the element’s constitutive equation. The result is a system of n−1 KCL equations that can be expressed in terms of n − 1 node voltage variables. If there are voltage sources in the circuit, then two additional issues arise. First, each voltage source connects two nodes, but is represented by just one node variable (since the voltage of the source determines the voltage at the other node). Second, the current through a voltage source cannot be determined from its constitutive equation. The constitutive equation for a voltage source is vs = V0 , which must be true for all currents. Thus, knowing that vs = V0 provides no information about the current through the source. These constraints are illustrated below. i1 V0 i2 i3 e k + V0 + − ek i4 i5 i6 A consistent set of equations can be derived by treating the entire voltage source as a single node, represented by the dashed green box above. The KCL equation for that node, i1 + i 2 + i 3 + i 4 + i 5 + i 6 = 0 Chapter 6 Circuits 6.01— Spring 2011— April 25, 2011 252 involves element currents that can each be written (using their constitutive equations) in terms of node voltages outside the box plus ek and ek + V0 . Like the KCL equations outside the box, this KCL equation adds a single new unknown, which is ek . Example 12. Does the above analysis imply that the current through the voltage source is zero? No. Let iv represent the current through the voltage source. KCL at the top node requires that i1 + i 2 + i 3 + i v = 0 KCL at the bottom node requires that i4 + i 5 + i 6 − i v = 0 The sum of these equations is the KCL equation given above for the dashed green box. It holds reguardless of the value of the current through the voltage source. To find the value of the current through the current source, first solve for i1 , i2 , i3 , i4 , i5 , i6 , and ek , ignoring iv . Then iv = −(i1 + i2 + i3 ), which will be the same as i4 + i5 + i6 . Teach Yourself 7. How would you handle a circuit in which i1 (above) connects directly to another voltage source outside the dashed green box? Node Analysis: Summary • Assign a voltage expression to each node in the circuit, as follows: − Assign a voltage of zero to one node, which is called the “ground.” − Assign voltages to nodes that are connected to voltage sources so that the voltage source relations are satisfied: ? If the voltage on one terminal of a voltage source is known (e.g., ground), then use the voltage source relation to determine the node voltage on the other terminal. ? If neither terminal of the voltage source connects to a known voltage, then assign a variable to represent the node voltage at one terminal and use the voltage source relation to determine the node voltage at the other terminal. − Assign variables to represent the voltages at all remaining nodes. • For each element in the circuit that is not a voltage source, use the associated constitu­ tive law to express the current through the element in terms of the node voltages at its terminals. • Write one KCL equation for each unknown node voltage, as follows: − Express KCL for each node that is not connected to a voltage source as a sum of ele­ ment currents. − For each voltage source, write a single KCL to express the sum of currents entering the nodes connected to both its positive and negative terminals. Chapter 6 Circuits 6.01— Spring 2011— April 25, 2011 253 6.4.5 Common Patterns There are some common patterns of resistors that are important to understand and that can be used over and over again as design elements. In this section, we will lead into a systematic approach to solving circuits. The first step in that approach is to associate voltages with nodes (points at which components connect) in the circuit rather than with the components themselves. This only makes sense if we establish a fixed reference voltage. We pick a particular node in the circuit (it won’t matter which one, but it is, for example, conventional to pick the negative terminal of a voltage source) and call it ground and assert that it has voltage 0. Now, when we speak of a node in the circuit having voltage v, what we mean is that the voltage difference between that node and the ground node is v. Ground is indicated in a circuit diagram by a special single-terminal component made of three parallel lines. 6.4.5.1 Resistors in series The figure below shows two resistors connected together in a circuit with a voltage source, the negative terminal of which is connected to ground. n1 iA RA n2 iC RB Vc iB n3 It induces a simple set of constraints (remember that we don’t need a KVL equation for the ground node): iA − i C = 0 i B − iA = 0 v3 = 0 v1 − v2 = iA · RA v3 − v1 = iB · RB v2 − v3 = Vc What happens when we solve? First, it’s easy to see that because there’s a single loop, KCL implies that the current across each of the nodes is the same. Let’s call it i. Now, we can add together the fourth and fifth equations, and then use the last equation to get Chapter 6 Circuits 6.01— Spring 2011— April 25, 2011 254 v3 − v2 = iA RA + iB RB v3 − v2 = i(RA + RB ) − Vc = i(RA + RB ) −i= Vc RA + RB The interesting thing to see here is that we get exactly the same result as we would have had if there were a single resistor R, with resistance RA + RB . So, if you ever see two or more resistors in series in a circuit, with no other connections from the point between them to other components, you can treat them as if it were one resistor with the sum of the resistance values. This is a nice small piece of abstraction. It might bother you that we got something that looks like v = −iR instead of v = iR. Did we do something wrong? Not really. The reason that it seems funny is that the directions we picked for the currents iA and iB turn out to be “backwards”, in the sense that, in fact, the current is running in the other direction, given the way we hooked them up to the voltage source. But the answer is still correct. Exercise 6.2. Consider the circuit we just analayzed. You should be able to construct an equivalent circuit with only one resistor. What is its resistance value? 6.4.5.2 Resistors in parallel Now, in this figure, n1 iC RB iB RA Vc iA n2 we have a simple circuit with two resistors in parallel. Even though there are a lot of wires being connected together, there are really only two nodes: places where multiple components are connected. Let’s write down the equations governing this system. First, applying KCL to n1 , iA + iB − iC = 0 . Now, setting v2 to ground, and describing the individual components, we have: Chapter 6 Circuits 6.01— Spring 2011— April 25, 2011 255 v2 = 0 v2 − v1 = iA · RA v2 − v1 = iB · RB v1 − v2 = Vc We can simplify this last set of constraints to − VC = iA · RA − VC = iB · RB so Vc RA Vc iB = − RB iA = − Plugging these into the KCL equation, we get: iA + i B − i C = 0 Vc Vc − = iC RA RB RA + RB − Vc = iC RA RB − − Vc = iC RA RB RA + RB What we can see from this is that two resistances, RA and RB , wired up in parallel, act like a single resistor with resistance RA RB . RA + RB This is another common pattern for both analysis and design. If you see a circuit with parallel resistors connected at nodes n1 and n2 , you can simplify it to a circuit that replaces those two paths between n1 and n2 with a single path with a single resistor. To get some intuition, think about what happens when RA is much bigger than RB . Because the voltage across both resistors is the same, it must be that the current through RA is much smaller than the current through RB . The appropriate intuition here is that current goes through the resistor that’s easier to pass through. Exercise 6.3. If RA = 10000Ω and RB = 10Ω, what is the effective resistance of RA and RB in parallel? 6.4.5.3 Voltage divider The figure below shows part of a circuit, in a configuration known as a voltage divider. Chapter 6 Circuits 6.01— Spring 2011— April 25, 2011 256 nin iA RA nout iB RB n0 Using what we know about circuit constraints, we can determine the following relationship be­ tween Vout and Vin : RB Vout = Vin . RA + RB Let’s go step by step. Here are the basic equations: v0 = 0 i A − iB = 0 Vin − Vout = iA RA Vout − v0 = iB RB We can start by seeing that iA = iB ; let’s just call it i. Now, we add the last two equations to each other, and do some algebra: Vin − v0 = iRA + iRB Vin = i(RA + RB ) i= Vin RA + RB Vin − Vout = iRA Vin − Vout = Vin RA RA + RB Vin (RA + RB ) − Vout (RA + RB ) = Vin RA Vin RB = Vout (RA + RB ) Vout = Vin RB RA + RB So, for example, if RA = RB , then Vout = Vin /2. This is a very handy thing: if you need a voltage in your circuit that is between two values that you already have available, you can choose an appropriate RA and RB to create that voltage. Chapter 6 Circuits 6.01— Spring 2011— April 25, 2011 257 Well, almost. When we wrote iA − iB = 0, we were assuming that there was no current flowing out Vout . But, of course, in general, that won’t be true. Consider the following figure: nin iA RA nout iB iL RB RL n0 We’ve shown an additional “load” on the circuit at Vout with a resistor RL standing for whatever resistance that additional load might offer to the ground node n0 . This changes matters considerably. To see what is going to happen, we could solve the whole circuit again. Or, we could observe that, between the node labeled Vout and n0 , we have two resistances, RB and RL in parallel. And we’ve already see that resistances in parallel behave as if they are a single resistor with value RB RL /(RB + RL ). So, (you do the algebra), our result will be that RB Vout = Vin . B RA + RB + RARR L The lesson here is that the modularity in circuits is not as strong as that in programs or our dif­ ference equation models of linear systems. How a circuit will behave can be highly dependent on how it is connected to other components. Still, the constraints that it exerts on the overall system remain the same. 6.5 Circuit Equivalents We just saw that pieces of circuits cannot be abstracted as input/output elements; the actual volt­ ages and currents in them will depend on how they are connected to the rest of a larger circuit. However, we can still abstract them as sets of constraints on the values involved. In fact, when a circuit includes only resistors, current sources, and voltage sources, we can derive a much simpler circuit that induces the same constraints on currents and voltages as the original one. This is a kind of abstraction that’s similar to the abstraction that we saw in linear systems: we can take a complex circuit and treat it as if it were a much simpler circuit. If somebody gave you a circuit and put it in a black box with two wires coming out, labeled + and -, what could you do with it? You could try to figure out what constraints that box puts on the voltage between and current through the wires coming out of the box. Chapter 6 Circuits 6.01— Spring 2011— April 25, 2011 258 We can start by figuring out the open-circuit voltage across the two terminals. That is the voltage drop we’d see across the two wires if nothing were connected to them. We’ll call that Voc . Another thing we could do is connect the two wires together, and see how much current runs through them; this is called the short-circuit current. We’ll call that isc . Because all of the relationships in our circuits are linear, these two values are sufficient to char­ acterize the constraint that this whole box will exert on a circuit connected to it. The constraint will be a relationship between the voltage across its terminals and the current flowing through the box. We can derive it by using Thévenin’s theorem: Any combination of current sources, voltage sources, and resistances with two terminals can be replaced by a single voltage source Vth and a single series resistor Rth . The value of Vth is the open circuit voltage at the terminals Voc , and the value of Rth is Vth divided by the current with the terminals short circuited (−isc ). Let’s look at a picture, then an example. iTH RTH i n+ + VTH n- Here, we show a picture of a black (well, gray) box, abstracted as being made up of a circuit with a single voltage source Vth and a single resistor Rth in series. The open-circuit voltage from n+ to n− is clearly Vth . The short-circuit current isc (in the direction of the arrow) is −Vth /Rth . So, this circuit would have the desired measured properties. 42 Now, here is an actual circuit. We’ll compute its associated open-circuit voltage and short-circuit current, construct the associated Thévenin equivalent circuit, and be sure it has the same proper­ ties. 42 The minus sign here can be kind of confusing. The issue is this: when we are treating this circuit as a black box with terminals n+ and n− , we think of the current flowing out of n+ and in to n− , which is consistent with the voltage difference Vth = V+ − V− . But when we compute the short-circuit current by wiring n+ and n− together, we are continuing to think of isc as flowing out of n+ , but now it is coming out of n− and in to n+ , which is the opposite direction. So, we have to change its sign to compute Rth . Chapter 6 Circuits n1 6.01— Spring 2011— April 25, 2011 RA 259 i n+ + VS - RB nThe first step is to compute the open-circuit voltage. This just means figuring out the difference between the voltage at nodes n+ and n− , under the assumption that the current i = 0. An easy way to do this is to set n− as ground and then find the node voltage at n+ . Let’s write down the equations: v+ − v1 = iA RA v1 − v− = Vs v+ − v− = iB RB − iA − iB = 0 iA − iS = 0 v− = 0 We can solve these pretty straightforwardly to find that RB v+ = Vs . RA + RB So, we know that, for this circuit, Vth = Vs RAR+BRB . Now, we need the short-circuit current, isc . To find this, imagine a wire connecting n+ to n− ; we want to solve for the current passing through this wire. We can use the equations we had before, but adding equation 6.1 wiring n+ to n− , and adding the current isc to the KCL equation 6.2. v+ − v1 = iA RA v1 − v− = Vs v+ − v− = iB RB v + = v− isc − iA − iB = 0 iA − iS = 0 v− = 0 We can solve this system to find that (6.1) (6.2) Chapter 6 Circuits 6.01— Spring 2011— April 25, 2011 isc = − and therefore that Rth = − 260 Vs , RA Vth isc RB Vs RA + RB RA RA RB = RA + RB = Vs What can we do with this information? We could use it during circuit analysis to simplify parts of a circuit model, individually, making it easier to solve the whole system. We could also use it in design, to construct a simpler implementation of a more complex network design. One important point is that the Thévenin equivalent circuit is not exactly the same as the original one. It will exert the same constraints on the voltages and currents of a circuit that it is connected to, but will, for example, have different heat dissipation properties. 6.5.1 Example Here’s another example. n1 + RA RB i n+ RC 2KΩ i n+ + 7.5V VS - - RD n- n- Another example circuit Its Thévenin equivalent It’s a bit more hassle than the previous one, but you can write down the equations to describe the constituents and KCL constraints, as before. If we let RA = 2KΩ, RB = RC = RD = 1KΩ, and VS = 15V , then we can solve for Vth = 7.5V and Rth = 2KΩ. So, it is indistinguishable by current and voltage from the circuit shown in the equivalent circuit. Here is the same circuit, but with the connections that run outside the box made to different nodes in the circuit. Chapter 6 Circuits n1 RA + 6.01— Spring 2011— April 25, 2011 n- RB RC VS - 1750Ω 261 i n+ 3.75V - i n+ + RD nSame circuit, different ports Its Thévenin equivalent Note also that the top lead is marked n− and the bottom one n+ . If we solve, using the same values for the resistors and voltage source as before, we find that Vth = −3.75V and Rth = 1750Ω. We show the Thévenin equivalent circuit next to the original. We’ve changed the polarity of the voltage source and made it 3.75V (instead of having the + terminal at the top and a voltage of -3.75), but that’s just a matter of drawing. These results are quite different: so, the moral is, it matters which wires you connect up to what! 6.5.2 Norton Equivalents It is also possible to derive an equivalent circuit consisting of a single resistor in parallel with a single current source. Such a circuit is called a Norton equivalent circuit. The process for deriving Norton equivalents is similar in spirit to the process outlined above. 6.6 Op Amps So far, we have considered circuits with resistors, voltage sources, and current sources. Now we are going introduce a new component, called an operational amplifier or op-amp, for short. Op-amps are a crucial tool for decoupling some aspects of complex circuits, so that they can be designed independently and then connected. They are also important for amplifying small voltage differences into larger ones, and can be used to scale, invert, and add voltages. 6.6.1 Voltage-controlled voltage sources In fact, op-amps are complex assemblages of transistors and other circuit components. We’ll need a simpler model than that of what they do, so that we can analyze circuits that contain them, and design new circuits that use them. We will think about an op-amp as a voltage-controlled voltage source (VCVS). Op-amps have four terminals, two of which are often called the input terminals and two, the output terminals. We can think of the op-amp as a voltage source, in which the voltage difference across the output terminals is controlled by the voltage difference between the input terminals. The crucial thing about a voltage source is that it maintains the voltage difference that it is trying to source, no matter what it is connected to. Chapter 6 Circuits 6.01— Spring 2011— April 25, 2011 n+ i 262 nout K(v+ − v−) ngnd n- A model of the constraint that an op-amp exerts on the circuit it is connected to relates the voltages at its four terminals in this way: vout − vgnd = K(v+ − v− ) , where K is a very large gain, on the order of 10,000, and asserts that no current flows into the op amp. i+ = i− = 0 . So, we can think of nout and ngnd as constituting a voltage source, whose voltage is defined to be K(v+ − v− ). We can see it as amplifying the voltage difference v+ − v− . It’s hard to really understand how this model works without seeing it in context. So, we’ll study an op-amp that is wired up in a configuration called a voltage follower: n+ + nout K(v+ − v−) VC - i n- Note that we have connected a wire from the output, nout , to the negative input, n− , and so those constitute one node with a single voltage, which we call vout . We can write down the equations: v+ = Vc vout − vgnd = K(v+ − vout ) vgnd = 0 Solving this system, we find that vout = Vc K . K+1 So, for large K, the voltage difference between the output terminals is very close to the voltage difference between the input terminals. In this circuit we have set vgnd = 0; in practice this doesn’t matter. If we had left vgnd in our equations, we would have found that: vout = Vc K 1 + vgnd . K+1 K+1 Chapter 6 Circuits 6.01— Spring 2011— April 25, 2011 263 Since K is very large, that second term can be ignored. In fact, if you look at an actual op-amp, you won’t even find an output corresponding to ngnd . Therefore, we will always treat it as being zero. The crucial additional fact about an op-amp is that no current flows between the input and output terminals, so the op-amp acts as a buffer between the input and output signal. It provides a disconnect between the currents on either side of it. This is a big deal: it gives us a kind of modularity in our circuits that we haven’t had before, by limiting the kinds of influence of one part of the circuit on the other. The ability to partially disconnect sub-parts of our circuits will make it easier to do complex designs. We can appreciate the value of a buffer in the context of using a variable voltage to control a motor. If we have a 15V supply, but only want to put 7.5V across the motor terminals, what should we do? A voltage divider seems like a good strategy: we can use one with two equal resistances, to make 7.5V , and then connect it to the motor as shown here +15V 10K nmotor 10K Motor Rmotor But what will the voltage vmotor end up being? It all depends on the resistance of the motor. If the motor is offering little resistance, say 100Ω, then the voltage vmotor will be very close to 0. 43 So, this is not an effective solution to the problem of supplying 7.5V to the motor. So, now, back to the motor. In this figure, 43 Go back and review the discussion of adding a load to a voltage divider, if this doesn’t seem clear. Chapter 6 Circuits 6.01— Spring 2011— April 25, 2011 264 +15V 10K n+ n10K i K(v+ − v−) nout Rmotor Motor we have used a voltage follower to connect the voltage divider to the motor. Based on our previ­ ous analysis of the follower, we expect the voltage at nout to be 7.5V , at least before we connect it up to the motor. Let’s see how this all works out by writing the equations out for the whole circuit: Letting the current through the top resistor be i1 , the current through the second resistor be i2 , the current through the motor be im , and recalling that vout = v− (since they are the same node), we have i1 − i2 − i+ = 0 − iout − i− − im = 0 i+ = 0 i− = 0 (15 − v+ ) = 10000i1 (v+ − 0) = 10000i2 (v− − 0) = K(v+ − v− ) (v− − 0) = im Rm Solving these equations, we find that v+ = 7.5 K 7.5 K+1 K 7.5 = K + 1 Rm v− = im Chapter 6 Circuits 6.01— Spring 2011— April 25, 2011 265 So, now, the voltage across the motor stays near 7.5V , and because of the isolation provided by the op-amp, it will remain that way, no matter what the resistance of the motor. Further, if in fact the motor has resistance of 5Ω, then we can find that the (maximum) current through the motor, im , is approximately 1.5A. 6.6.2 Simplified model The voltage-controlled voltage-source model allows us to use our standard linear-equation form of reasoning about circuit values, but it can become cumbersome to deal with the K’s all the time. In most of the usual applications of an op-amp, we will find that we can use a simpler model to reason about what is happening. That simpler model also comes with a simpler picture: + n+ nout - n- The simplified behavioral model is that the op-amp adjusts vout in order to try to maintain the constraint that v+ ≈ v− and that no current flows in to n+ or n− . In what follows we will assume that the op-amp enforces v+ = v− . In this section, we will explore several standard applications of op-amps, using this simplified model. If it becomes confusing, you can always go back and apply the VCVS model from the previous section. Non-inverting amplifier Not surprisingly, a primary use of an op-amp is as an amplifier. Here is an amplifier configuration, like this. nin + nout - nRF RI Chapter 6 Circuits 6.01— Spring 2011— April 25, 2011 266 Let’s see if we can figure out the relationship between vin and vout . The circuit constraints tell us that v− = iI RI v− − vout = iF RF − iI − iF = 0 (6.3) vin = v− (6.4) The KCL equation (6.3) has no term for the current into the op-amp, because we assume it is zero. Equation 6.4 is the op-amp constraint. So, we find that vout = vin RF + RI . RI This is cool. We’ve arranged for the output voltage to be greater than the input voltage. We can get any gain (greater than one) by choosing the values of RF and RI . We can think intuitively about how it works by examining some cases. First, if RF = 0, then we’ll have vout = vin , so there’s not a particularly interesting change in the voltages. This is still a useful device, essentially the voltage follower, which we have already seen, with an additional resistor. This extra resistor does not change the voltage relationsips but does increase the current flowing from the op-amp. Note also that choosing RI very much larger than RF also gives us a follower. Now let’s think about a more interesting case, but simplify matters by setting RF = RI . We can look at the part of the circuit running from Vout through RF and RI to ground. This looks a lot like a voltage divider, with v− coming out of the middle of it. Because v− needs to be the same as vin , and it is vout being divided in half, then vout clearly has to be 2vin . Inverting amplifier Here is a very similar configuration, called an inverting amplifier. n+ + nout - nnin RI RF The difference is that the + terminal of the op-amp is connected to ground, and the we’re think­ ing of the path through the resistors as the terminal of the resulting circuit. Let’s figure out the relationship between vin and vout for this one. The circuit constraints tell us that Chapter 6 Circuits 6.01— Spring 2011— April 25, 2011 267 vin − v− = iI RI v− − vout = iF RF iI − iF = 0 v + = v− v+ = 0 Solving, we discover that RF . RI If RF = RI , then this circuit simply inverts the incoming voltage. So, for example, if vin is +10V with respect to ground, then vout will be −10V . Again, we can see the path from nin through the resistors, to nout , as a voltage divider. Knowing that v− has to be 0, we can see that vout has to be equal to −vin . If we want to scale the voltage, as well as invert it, we can do that by selecting appropriate values of RF and RI . vout = −vin Voltage summer A voltage summer 44 circuit, shown here, can be thought of as having three terminals, with the voltage at nout constrained to be a scaled, inverted, sum of the voltages at n1 and n2 . n+ + nout - nn1 n2 RI RF RI You should be able to write down the equations for this circuit, which is very similar to the in­ verting amplifier, and derive the relationship: RF vout = − (v1 + v2 ) . RI 6.6.3 Op-amps and feedback We have now seen two different models of op-amps. Each of those models can be useful in some circumstances, but now we’ll explore some situations that require us to use an even more detailed and faithful model of how an op-amp really behaves. 44 As in thing that sums, not as in endless summer. Chapter 6 Circuits 6.01— Spring 2011— April 25, 2011 268 Here are two voltage followers, a “good” one and a “bad” one. n+ + n+ nout - VC - + + nout - n- + n- VC - “Good” voltage follower “Bad” voltage follower If we use our simplest model to predict their behavior, in both cases, we’ll predict that vout = Vc . K If we use the VCVS model, we’ll predict that, for the good follower vout = Vc K+1 . For the bad follower, things are connected a little bit differently: v− = Vc vout − vgnd = K(v+ − v− ) v+ = vout vgnd = 0 Solving this system, we find that K . K−1 Those two predictions are basically the same for large K. But, in fact, the prediction about the behavior of the bad follower is completely bogus! In fact, if you connect an op-amp this way, it is likely to burn out, possibly generating smoke and noise in the process. vout = Vc To see why this happens, we’ll actually need to make a more detailed model. Dynamics model Our new model is going to take the actual dynamics of the circuit into account. We can model what is going on in an op-amp by using a difference equation to describe the value of the output at a time n as a linear function of its values at previous times and of the values of v+ and v− . Note that we’re using vo = vout in the following. You might imagine that the way the op-amp ’tries’ to make Vo equal K(V+ − V− ) is by gradually adjusting Vo to decrease the error, E, making adjustments that are proportional, with a positive ’gain’ g, to the error. That is, that vo [t] = vo [t − 1] − g · e[t − 1] = vo [t − 1] − g · (vo [t − 1] − K(v+ [t − 1] − v− [t − 1])) (6.5) We would expect this system to adjust vo to be closer to K(v+ − v− ) on every step. The gain g is a property of the design of the op-amp. Chapter 6 Circuits 6.01— Spring 2011— April 25, 2011 269 Now, what happens when we use this model to predict the temporal behavior of the voltage followers? Let’s start with the good one. Good follower In this case, we have v− [t] = vo [t] for all t. So, equation 6.5 becomes vo [t] = vo [t − 1] − g · (vo [t − 1] − K(v+ [t − 1] − vo [t − 1])) = (1 − g − gK)vo [t − 1] + gKv+ [t − 1] Writing this as an operator equation, and then a system function, we get Vo = (1 − g − gK)RVo + gKRV+ Vo (1 − R(1 − g − gK)) = gKRV+ Vo gKR = V+ 1 − R(1 − g − gK) This system has a pole at 1 − g − gK, so if we let g = 1/(1 + K), then we have a pole at 0, which means the system will, ideally, converge in a single step, with Vo = K RV+ , K+1 which agrees with our previous model. Bad follower In this case, we have v+ [t] = vo [t] for all t. This time, equation 6.5 becomes vo [t] = vo [t − 1] − g · (vo [t − 1] − K(vo [t − 1] − v− [t − 1])) = (1 − g + gK)vo [t − 1] − gKv− [t − 1] Writing this as an operator equation, and finding the system function, we get Vo = (1 − g + gK)RVo − gKRV− Vo (1 − R(1 − g + gK)) = −gKRV− Vo −gKR = V− 1 − R(1 − g + gK) This system has a pole at 1 − g + gK, so with g = 1/(1 + K) as before, we have a pole at 2K/(1 + K), which means the system will diverge for large positive values of K. There is no positive value of g that will make this system stable (and, if we were to pick a negative value, it would make the good follower become a bad follower). The prediction we get for g = 1/(1 + K) agrees with the empirical behavior of op-amps, at least in so far as op-amps wired up in feedback to the postive input terminal do not converge to a stable output voltage, but typically burn themselves out, or behave otherwise unpredictably. Chapter 6 Circuits 6.01— Spring 2011— April 25, 2011 270 6.6.4 Where does the current come from? In all of the models we have seen so far, we’ve left out an important point. As you know from lab, op-amps, in fact, need two more connections, which is to VCC , the main positive voltage supply, and to VEE/GND , which is the negative power supply (which can be ground). Why? If no current flows “through” the op-amp, from n+ or n− to the other side, then how can it maintain the necessary voltage difference on the output side, and supply current to, e.g., drive the motor? The answer is that it uses current from the power supplies to regulate the output voltage and to supply the necessary output current. One metaphorical way of thinking about this is that VCC is a big main water-supply pipe and that v+ − v− is the controlling the handle of a faucet. As we change the size of the difference v+ − v− , we open or close the faucet; but the water is being supplied from somewhere else entirely. This metaphor is appropriate in another way: the total voltage on the output of the op-amp is limited by VCC . So, if the VCC = 10, then the output of the op-amp is limited to be between 0 and 10. If we connect the ground connection of the op-amp to a negative supply voltage, then the output of the op-amp can be negative as well as positive; this has many applications.