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| ====== Real Analog: Chapter 4 ====== | ====== Real Analog: Chapter 4 ====== | ||
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| + | * Chapter 4 Videos: | ||
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| ====== 4. Introduction and Chapter Objectives ====== | ====== 4. Introduction and Chapter Objectives ====== | ||
| In previous chapters, a number of approaches have been presented for analyzing electrical circuits. In these analysis approaches, we have been provided with a circuit consisting of a number of elements (resistors, power supplies, etc.) and determined some circuit variable of interest (a voltage or current, for example). In the process of determining this variable, we have written equations which allow us to determine __any and all__ variables in the system. For a complex circuit, with many elements, this approach can result in a very large number of equations and a correspondingly large amount of effort expended in the solution of these equations. Unfortunately, | In previous chapters, a number of approaches have been presented for analyzing electrical circuits. In these analysis approaches, we have been provided with a circuit consisting of a number of elements (resistors, power supplies, etc.) and determined some circuit variable of interest (a voltage or current, for example). In the process of determining this variable, we have written equations which allow us to determine __any and all__ variables in the system. For a complex circuit, with many elements, this approach can result in a very large number of equations and a correspondingly large amount of effort expended in the solution of these equations. Unfortunately, | ||
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| A system is commonly represented as shown in the block diagram of Fig. 4.1. The system has some input, //u(t)//, and some output, //y(t)//. In general, both the input and output can be functions of time; the case of constant values is a special case of a time-varying function. The output will be represented as some arbitrary function of the input: | A system is commonly represented as shown in the block diagram of Fig. 4.1. The system has some input, //u(t)//, and some output, //y(t)//. In general, both the input and output can be functions of time; the case of constant values is a special case of a time-varying function. The output will be represented as some arbitrary function of the input: | ||
| - | $$y \left( t \right) = f \{ u \left( t \right) \} (Eq. 4.1)$$ | + | $$y \left( t \right) = f \{ u \left( t \right) \} \qquad |
| Equation (4.1) is said to be the // | Equation (4.1) is said to be the // | ||
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| ==== Example 4.1: Mass Subjected to an External Force ==== | ==== Example 4.1: Mass Subjected to an External Force ==== | ||
| - | Consider the mass-damper system shown in the figure below. The applied force //F(t)// pushes the mass to the right. The mass's velocity resulting from the applied force is //v(t)//. The mass slides on a surface | + | Consider the mass-damper system shown in the figure below. The applied force //F(t)// pushes the mass to the right. The mass's velocity resulting from the applied force is //v(t)//. The mass slides on a surface |
| {{ : | {{ : | ||
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| {{ : | {{ : | ||
| - | We previously wrote mesh equations for this circuit (for a specific value of $V_{in}$) in [[https:// | + | We previously wrote mesh equations for this circuit (for a specific value of $V_{in}$) in [[/ |
| {{ : | {{ : | ||
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| $$V= 2 \left( i_1 - 3V_x \right)$$ | $$V= 2 \left( i_1 - 3V_x \right)$$ | ||
| - | The above four equations provide an inpuot-output description of the circuit. If desired, they can be combined to eliminate all variables except $V_{in}$ and $V$ and re-written in the form $V = f \{ V_{in} \}$ per equation (1). __Note that all information about the original system, except the relationship between the input and output signals, is lost once we do this__. | + | The above four equations provide an inppot-output description of the circuit. If desired, they can be combined to eliminate all variables except $V_{in}$ and $V$ and re-written in the form $V = f \{ V_{in} \}$ per equation (1). __Note that all information about the original system, except the relationship between the input and output signals, is lost once we do this__. |
| The system-level block diagram for the circuit might then be drawn as: | The system-level block diagram for the circuit might then be drawn as: | ||
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| All of the above relationships are of the form: | All of the above relationships are of the form: | ||
| - | $$y(t) = Kx(t) (Eq. 4.2)$$ | + | $$y(t) = Kx(t) \qquad |
| Where $x(t)$ and $y(t)$ are voltages or currents in the above examples. More generally, $x(t)$ and $y(t)$ can be considered | Where $x(t)$ and $y(t)$ are voltages or currents in the above examples. More generally, $x(t)$ and $y(t)$ can be considered | ||
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| 1. The 20Ω resistor below obeys Ohm’s law, so that $V=20I$. We will consider the input to be the current through | 1. The 20Ω resistor below obeys Ohm’s law, so that $V=20I$. We will consider the input to be the current through | ||
| the resistor and the output to be the voltage drop across the resistor. Determine: | the resistor and the output to be the voltage drop across the resistor. Determine: | ||
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| - | | + | |
| | | ||
| Do your answers above indicate that the additive property holds for this resistor? Why? | Do your answers above indicate that the additive property holds for this resistor? Why? | ||
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| - | The circuit above can be consider | + | The circuit above can be considered |
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| {{ : | {{ : | ||
| - | Although not the most efficient approach for this problem, we will again use superposition to approach the problem. One source will, of course, be the current source $I_S$. We will assume that our second source is the current $i$ at node a. Killing the current source $I_S$ results in the circuit to the left below; from this figure the voltage $v_1$ can be seen to be $v_1 = i \left( R_1 + R_2 \right). Killing the current source $i$ results in the figure to the right below; from this figure the voltage $v_2$ is seen to be $v_2 = R_1I_S$ (the dead current source results in an open circuit, so no current flows through the resistor $R_2$). The total voltage across the terminals is, therefore, $v= \left( R_1 + R_2 \right) i + R_1I_S$. | + | Although not the most efficient approach for this problem, we will again use superposition to approach the problem. One source will, of course, be the current source $I_S$. We will assume that our second source is the current $i$ at node a. Killing the current source $I_S$ results in the circuit to the left below; from this figure the voltage $v_1$ can be seen to be $v_1 = i \left( R_1 + R_2 \right)$. Killing the current source $i$ results in the figure to the right below; from this figure the voltage $v_2$ is seen to be $v_2 = R_1I_S$ (the dead current source results in an open circuit, so no current flows through the resistor $R_2$). The total voltage across the terminals is, therefore, $v= \left( R_1 + R_2 \right) i + R_1I_S$. |
| {{ : | {{ : | ||
| + | The i-v characteristic for the circuit is, therefore, as shown in the figure below. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | ---- | ||
| + | |||
| + | === Notes on Linear Circuit i-v Characteristics: | ||
| + | - All two-terminal networks which contain only sources and resistors will have i-v relationships of the form shown in examples 1, 2, and 3. That is, they will be straight lines of the form $v=m \cdot i + b$. The y-intercept term, b, is due to sources in the network; if there are no sources in the network, $b = 0$ and the //i-v// characteristic will pass through the origin. | ||
| + | - Due to the form of the i-v characteristic provided in note 1 above, any two-terminal network can be represented as a single source and a single resistor. | ||
| + | - The form of the solution for examples 2 and 3 are the same. Thus, the circuit of example 2 is indistinguishable from a similar circuit with a current source $\frac{V_S}{R_1}$ in parallel with the resistor $R_1$. Likewise, the circuit of example 3 is indistinguishable from a similar circuit with a voltage source $I_SR_1$ in series with the resistor $R_1$. The equivalent circuits are shown below. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Section Summary ==== | ||
| + | * Electrical circuits, sub-circuits, | ||
| + | * For linear circuits, the voltage-current relationship across two terminals of the circuit can always be represented as a straight line of the form $v=m \cdot i + b$. If we plot this relationship with voltage on the vertical axis and current on the horizontal axis, the slope of the line corresponds to an equivalent resistance seen across the terminals, while the y-intercept of the line is the voltage across the terminals, if the terminals are open-circuited. We will formalize this important result in section 4.5. | ||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Exercises ==== | ||
| + | - Determine the i-v characteristics of the circuit below, as seen at the terminals a-b. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | ---- | ||
| + | |||
| + | ===== 4.5: Thévenin’s and Norton’s Theorems ===== | ||
| + | In section 4.4, we saw that it is possible to characterize a circuit consisting of sources and resistors by the voltage- current (or //i-v//) characteristic seen at a pair of terminals of the circuit. When we do this, we have essentially simplified our description of the circuit from a detailed model of the internal circuit parameters to a simpler model which describes the overall behavior of the circuit as seen at the terminals of the circuit. This simpler model can then be used to simplify the analysis and/or design of the overall system. | ||
| + | |||
| + | In this section, we will formalize the above result as // | ||
| + | |||
| + | Consider the two interconnected circuits shown in Figure 4.7 below. The circuits are interconnected at the two terminals a and b, as shown. Our goal is to replace circuit A in the system of Figure 4.7 with a simpler circuit which has the same current-voltage characteristic as circuit A. That is, if we replace circuit A with its simpler equivalent circuit, the operation of circuit B will be unaffected. We will make the following assumptions about the overall system: | ||
| + | * Circuit A is linear | ||
| + | * Circuit A has no dependent sources which are controlled by parameters within circuit B | ||
| + | * Circuit B has no dependent sources which are controlled by parameters within circuit A | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | In section 4.3, we determined i-v characteristics for several example two-terminal circuits, using the superposition principle. We will follow the same basic approach here, except for a general linear two-terminal circuit, in order to develop Thévenin’s and Norton’s theorems. | ||
| + | |||
| + | ==== Thévenin’s Theorem ==== | ||
| + | First, we will kill all sources in circuit A and determine the voltage resulting from an applied current, as shown in Figure 4.8 below. With the sources killed, circuit A will look strictly like an equivalent resistance to any external circuitry. This equivalent resistance is designated as $R_{TH}$ in Figure 4.8. The voltage resulting from an applied current, with circuit A dead is: | ||
| + | |||
| + | $$v_1 = R_{TH} \cdot i \qquad | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | Now we will determine the voltage resulting from re-activating circuit A’s sources and open-circuiting terminals a and b. We open-circuit the terminals a-b here since we presented equation (4.3) as resulting from a current source, rather than a voltage source. The circuit being examined is as shown in Figure 4.9. The voltage $v_{OC}$ is the “open-circuit” voltage. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | Superimposing the two voltages above results in: | ||
| + | |||
| + | $$v=v_1 + v_{OC} | ||
| + | |||
| + | Or | ||
| + | |||
| + | $$v= R_{TH} \cdot i + v_{OC} | ||
| + | |||
| + | Equation (4.5) is Thévenin’s theorem. It indicates that the voltage-current characteristic of any linear circuit (with the exception noted below) can be duplicated by an independent voltage source in series with a resistance $R_{TH}$, known as the //Thévenin resistance// | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | === Procedure for Determining Thévenin Equivalent Circuit === | ||
| + | - Identify the circuit and terminals for which the Thévenin equivalent circuit is desired. | ||
| + | - Kill the independent sources (do nothing to any dependent sources) in circuit and determine the equivalent resistance RTH of the circuit. If there are no dependent sources, $R_{TH}$ is simply the equivalent resistance of the resulting resistive network. Otherwise, one can apply an independent current source at the terminals and determine the resulting voltage across the terminals; the voltage-to-current ratio is $R_{TH}$. | ||
| + | - Re-activate the sources and determine the open-circuit voltage $V_{OC}$ across the circuit terminals. Use any analysis approach you choose to determine the open-circuit voltage. | ||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Example 4.9 ==== | ||
| + | Determine the Thévenin equivalent of the circuit below, as seen by the load, $R_L$. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | We want to create a Thévenin equivalent circuit of the circuit to the left of the terminals a-b. The load resistor, $R_L$, takes the place of “circuit B” in Fig. 1. The circuit has no dependent sources, so we kill the independent sources and determine the equivalent resistance seen by the load. The resulting circuit is shown below. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | From the above figure, it can be seen that the Thévenin resistance $R_{TH}$ is a parallel combination of a $3\Omega$ resistor and a $6\Omega$ resistor, in series with a $2\Omega$ resistor. Thus, $R_{TH} = \frac{\left( 6\Omega \right) \left( 3\Omega \right)}{6\Omega + 3\Omega} + 2\Omega = 4\Omega$. | ||
| + | |||
| + | The open-circuit voltage $v_{OC}$ is determined from the circuit below. We (arbitrarily) choose nodal analysis to determine the open-circuit voltage. There is one independent voltage in the circuit; it is labeled as $v_0$ in the circuit below. Since there is no current through the $2\Omega$ resistor, $v_{OC} = v_0$. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | Applying KCL at $v_0$, we obtain: $-2A + \frac{v_0 - 6V}{6\Omega} + \frac{v_0}{3\Omega} = 0 \Rightarrow v_0 = v_{OC} = 6V$. Thus, the Thévenin equivalent circuit is on the left below. Re-introducing the load resistance, as shown on the right below, allows us to easily analyze the overall circuit. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Norton’s Theorem ==== | ||
| + | The approach toward generating Norton’s theorem is almost identical to the development of Thévenin’s theorem, except that we apply superposition slightly differently. In Thévenin’s theorem, we looked at the voltage response to an input current; to develop Norton’s theorem, we look at the current response to an applied voltage. The procedure is provided below. | ||
| + | |||
| + | Once again, we kill all sources in circuit A, as shown in Figure 4.8 above but this time we determine the current resulting from an applied voltage. With the sources killed, circuit A still looks like an equivalent resistance to any external circuitry. This equivalent resistance is designated as $R_{TH}$ in Figure 4.8. The current resulting from an applied voltage, with circuit A dead is: | ||
| + | |||
| + | $$i_1 = \frac{v}{R_{TH}} | ||
| + | |||
| + | Notice that equation (4.6) can be obtained by rearranging equation (4.3). | ||
| + | |||
| + | Now we will determine the current resulting from re-activating circuit A’s sources and short-circuiting terminals a and b. We short-circuit the terminals a-b here since we presented equation (4.4) as resulting from a voltage source. The circuit being examined is as shown in Figure 4.11. The current $i_{SC}$ is the “short-circuit” current. It is typical to assume that under short-circuit conditions the short-circuit current enters the node at a; this is consistent with an assumption that circuit A is generating power under short-circuit conditions. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | Employing superposition, | ||
| + | |||
| + | $$i = i_1 - i_{SC} | ||
| + | |||
| + | So | ||
| + | |||
| + | $$i = \frac{v}{R_{TH}} - i_{SC} | ||
| + | |||
| + | Equation (4.8) is Norton’s theorem. It indicates that the voltage-current characteristic of any linear circuit (with the exception noted below) can be duplicated by an independent current source in parallel with a resistance. The current source has the magnitude $i_{SC}$ and the resistance is $R_{TH}$, where $i_{SC}$ is the current seen at the circuit’s terminals if the terminals are short-circuited and $R_{TH}$ is the equivalent resistance of the circuit seen from the two terminals, with all independent sources in the circuit killed. The equivalent Norton circuit is shown in Fig. 4.12. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | === Procedure for Determining Norton Equivalent Circuit: === | ||
| + | - Identify the circuit and terminals for which the Norton equivalent circuit is desired. | ||
| + | - Determine the equivalent resistance $R_{TH}$ of the circuit. The approach for determining $R_{TH}$ is the same for Norton circuits as Thévenin circuits. | ||
| + | - Re-activate the sources and determine the short-circuit current $i_{SC}$ across the circuit terminals. Use any analysis approach you choose to determine the short-circuit current. | ||
| + | |||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Example 4.10 ==== | ||
| + | Determine the Norton equivalent of the circuit seen by the load, $R_L$, in the circuit below. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | This is the same circuit as our previous example. The Thévenin resistance, $R_{TH}$, is thus the same as calculated previously: $R_{TH} = 4\Omega$. Removing the load resistance and placing a short-circuit between the nodes a and b, as shown below, allows us to calculate the short-circuit current, $i_{SC}$. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | Performing KCL at the node $v_0$, results in: | ||
| + | |||
| + | $$\frac{v_0}{2\Omega} + \frac{v_0 - 6V}{6\Omega} + \frac{v_0}{3\Omega} = 2A$$ | ||
| + | |||
| + | So | ||
| + | |||
| + | $$v_0 = 3V$$ | ||
| + | |||
| + | Ohm's law can then be used to determine $i_{SC}$: | ||
| + | |||
| + | $$i_{SC} = \frac{3V}{2\Omega} = 1.5A$$ | ||
| + | |||
| + | And the Norton equivalent circuit is shown on the left below. Replacing the load resistance in the equivalent overall circuit is shown to the right below. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | === Exceptions: === | ||
| + | Not all circuits have Thévenin and Norton equivalent circuits. Exceptions are: | ||
| + | - An ideal current source does not have a Thévenin equivalent circuit. (It cannot be represented as a voltage source in series with a resistance.) It is, however, its own Norton equivalent circuit. | ||
| + | - An ideal voltage source does not have a Norton equivalent circuit. (It cannot be represented as a current source in parallel with a resistance.) It is, however, its own Thévenin equivalent circuit. | ||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Source Transformations ==== | ||
| + | Circuit analysis can sometimes be simplified by the use of //source transformations// | ||
| + | |||
| + | Equations (4.5) and (4.8) are both representations of the //i-v// characteristic of the same circuit. Rearranging equation (4.5) to solve for the current //i// results in: | ||
| + | |||
| + | $$i = \frac{v}{R_{TH}} - \frac{v_{OC}}{R_{TH}} | ||
| + | |||
| + | Equating equations (4.8) and (4.9) leads to the conclusion that: | ||
| + | |||
| + | $$i_{SC} = \frac{v_{OC}}{R_{TH}} | ||
| + | |||
| + | Likewise, rearranging equation (4.8) to obtain an expression for v gives: | ||
| + | |||
| + | $$v=i \cdot R_{TH} + i_{SC} \cdot R_{TH} | ||
| + | |||
| + | Equating equations (4.11) and (4.5) results in: | ||
| + | |||
| + | $$V_{OC} = i_{SC} \cdot R_{TH} | ||
| + | |||
| + | Which is the same result as equation (4.10). | ||
| + | |||
| + | Equations (4.10) and (4.12) lead us to the conclusion that any circuit consisting of a voltage source in series with a resistor can be transformed into a current source in parallel with the same resistance. Likewise, a current source in parallel with a resistance can be transformed into a voltage source in series with the same resistance. The values of the transformed sources must be scaled by the resistance value according to equations (4.10) and (4.12). The transformations are depicted in Figure 4.13. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | Source transformations can simplify the analysis of some circuits significantly, | ||
| + | |||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Example 4.11 ===== | ||
| + | Determine the current //i// in the circuit shown below. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | We can use a source transformation to replace the 9V source and $3\Omega$ resistor series combination with a 3A source in parallel with a $3\Omega$ resistor. Likewise, the 2A source and $2\Omega$ resistor parallel combination can be replaced with a 4V source in series with a $2\Omega$ resistor. After these transformations have been made, the parallel resistors can be combined as shown in the figure below. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | The 3A source and $2\Omega$ resistor parallel combination can be combined to a 6V source in series with a $2\Omega$ resistor, as shown below. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | The current //i// can now be determined by direct application of Ohm’s law to the three series resistors, so that $i= \frac{6V - 4V}{2\Omega + 4\Omega + 2\Omega} = 0.25A$. | ||
| + | |||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Voltage-current Characteristics of Thévenin and Norton Circuits ==== | ||
| + | Previously, in section 4.4, we noted that the i-v characteristics of linear two-terminal networks containing only sources and resistors are straight lines. We now look at the voltage-current characteristics in terms of Thévenin and Norton equivalent circuits. | ||
| + | |||
| + | Equations (4.5) and (4.8) both provide a linear voltage-current characteristic as shown in Figure 4.14. When the current into the circuit is zero (open-circuited conditions), | ||
| + | |||
| + | $$v= R_{TH} \cdot i_{OC} + v_{OC} = R_{TH} \cdot 0 + v_{OC} = v_{OC}$$ | ||
| + | |||
| + | Likewise, under short-circuited conditions, the voltage differential across the terminals is zero and equation (4.8) readily provides: | ||
| + | |||
| + | $$i= \frac{v_{SC}}{R_{TH}} - i_{sc} = \frac{0}{R_{TH}} - i_{sc} = - i_{sc}$$ | ||
| + | |||
| + | Which is consistent with Fig. 4.14. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | Figure 4.14 is also consistent with equations (4.10) and (4.12) above, since graphically the slope of the line is obviously $R_{TH} = \frac{v_{OC}}{i_{SC}}$. | ||
| + | |||
| + | Figure 4.14 also indicates that there are three simple ways to create Thévenin and Norton equivalent circuits: | ||
| + | - Determine $R_{TH}$ and $v_{OC}$. This provides the slope and y-intercept of the //i-v// characteristic. This approach is outlined above as the method for creating a Thévenin equivalent circuit. | ||
| + | - Determine $R_{TH}$ and $i_{SC}$. This provides the slope and x-intercept of the //i-v// characteristic. This approach is outlined above as the method for creating a Norton equivalent circuit. | ||
| + | - Determine $v_{OC}$ and $i_{SC}$. The equivalent resistance $R_{TH}$ can then be calculated from $R_{TH} = \frac{v_{OC}}{i_{SC}}$ to determine the slope of the //i-v// characteristic. Either a Thévenin or Norton equivalent circuit can then be created. This approach is not commonly used, since determining $R_{TH}$ – the equivalent resistance of the circuit – is usually easier than determining either $v_{OC}$ or $i_{SC}$. | ||
| + | |||
| + | === Note: === | ||
| + | It should be emphasized that the Thévenin and Norton circuits are not independent entities. One can always be determined from the other via a source transformation. Thévenin and Norton circuits are simply two different ways of expressing the same voltage-current characteristic. | ||
| + | |||
| + | ==== Section Summary ==== | ||
| + | * Thévenin’s theorem allows us to replace any linear portion of a circuit with equivalent circuit consisting of a __voltage source in series with a resistance__. This circuit is called the Thévenin equivalent, and provides the same voltage-current relationship at the terminals as the original circuit. The voltage source in the equivalent circuit is the same as the voltage which would be measured across the terminals of the original circuit, if those terminals were open-circuited. The resistance in the equivalent circuit is called the Thévenin resistance, it is the resistance that would be seen across the terminals of the original circuit, if all sources in the circuit were killed. | ||
| + | * Norton’s theorem allows us to replace any linear portion of a circuit with equivalent circuit consisting of a __current source in parallel with a resistance__. This circuit is called the Norton equivalent, and provides the same voltage-current relationship at the terminals as the original circuit. The current source in the equivalent circuit is the same as the current which would be measured across the terminals of the original circuit, if those terminals were short-circuited. The resistance in the equivalent circuit is the resistance that would be seen across the terminals of the original circuit, if all sources in the circuit were killed; it is the same as the Thévenin resistance. | ||
| + | * Thevenin and Norton’s theorems allow us to perform //source transformations// | ||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Exercises ==== | ||
| + | 1. Replace everything except the 1A current source with its Thévenin equivalent circuit and use the result to find $V_1$. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | 2. Replace everything except the 1A current source with its Norton equivalent circuit and use the result to find $V_1$. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | 3. Determine a Norton equivalent circuit for the circuit below. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | ---- | ||
| + | |||
| + | ===== 4.6: Maximum Power Transfer ===== | ||
| + | It is often important for our electrical system to transfer as much power as possible to some related system. For example, in an audio system it is important that the amplifier transfer as much power as possible to the loudspeakers. Otherwise, the amplifier generates power which is not used for any productive purpose((Other than, perhaps, heating the room it is in.)) and the efficiency of the overall system suffers. | ||
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| + | In this section, we will develop design guidelines which will ensure that the maximum possible amount of power is transferred from our electrical circuit to the load that the circuit is driving. These guidelines will be based on Thevenin’s theorem. | ||
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| + | Consider the system shown in Figure 4.15. The overall system consists of an electrical circuit which is being used to drive a load. Physically, the load can be either another electrical system or some electromechanical system such as an electric motor or a loudspeaker. We will model the load as an electrical resistance, $R_L$, though the principles presented here are applicable to more general loading conditions. | ||
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| + | We will replace our electrical system with its Thevenin equivalent in order to analyze the power delivered by the circuit to the load. The overall circuit that we are analyzing is now modeled as shown in Figure 4.16. | ||
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| + | From Figure 4.16, the voltage delivered to the load can be readily determined from a voltage divider relation: | ||
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| + | $$V_L = V_{OC} \frac{R_L}{R_L + R_{TH}} | ||
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| + | Thus, the power delivered to the load is: | ||
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| + | $$P_L = \frac{V_L^2}{R_L} = \frac{V^2_{OC}}{R_L} \left( \frac{R_L}{R_L + R_{TH}} \right) ^2 (Eq. 4.14)$$ | ||
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| + | Figure 4.17 shows a plot of the power delivered to the load, as a function of the load resistance. The power delivered to the load is zero when the load resistance is zero (since there is no voltage drop across the load under this condition) and goes to zero as the load resistance approaches infinity (since there is no current provided to the load under this condition). At some value of $R_L$ the power transfer will be maximized -- our goal will be to determine the value for $R_L$ which maximizes the power delivered to the load. | ||
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| + | The maximum value of power on the curve shown in Figure 4.17 can be determined by differentiating equation (4.14) with respect to the load resistance $R_L$ and setting the result to zero. This leads to: | ||
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| + | $$\frac{\partial P_L}{\partial R_L} = V^2_{OC} \left[ \frac{\left( R_L + R_{TH} \right) ^2 - 2R_L \left( R_{TH} + R_L \right) }{\left( R_L + R_{TH} \right) ^4} \right] = 0 (Eq. 4.15)$$ | ||
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| + | The above condition is satisfied if the numerator of equation (4.15) is zero, so our condition becomes: | ||
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| + | $$\left( R_L + R_{TH} \right) ^2 = 2R_L \left( R_{TH} + R_L \right) $$ | ||
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| + | Dividing both sides by $R_{TH} + R_L$ results in: | ||
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| + | $$R_L + R_{TH} = 2R_L$$ | ||
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| + | Or | ||
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| + | $$R_L = R_{TH} | ||
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| + | Thus, maximum power transfer takes place when the load resistance and the Thevenin resistance of the circuit supplying the power are equal. The above result is sometimes called the //maximum power transfer theorem//. When the conditions of the maximum power transfer theorem are met, the total power delivered to the load is: | ||
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| + | $$P_L = \frac{(\frac{V_{OC}}{2})^2}{R_{TH}} = \frac{V^2_{OC}}{4R_{TH}} (Eq. 4.17)$$ | ||
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| + | This is one half of the total power generated by the circuit, half the power is absorbed in the resistance $R_{TH}$. | ||
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| + | === Conclusion === | ||
| + | The power delivered to the load is maximized if the load resistance is equal to the Thevenin resistance of the circuit supplying the power. When this condition is met, the circuit and the load are said to be matched. When the load and the circuit are matched, 50% of the power generated in the circuit can be delivered to the load – under any other circumstances, | ||
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| + | ==== Practical Power Supplies ==== | ||
| + | Practical power supplies were discussed in section 2.4. It was seen that the presence of an internal resistance in a voltage or current source limited the power that could be delivered to a circuit connected to the source. Practical power supplies are a special case of the results presented above; we use them below as examples of the application of the above principles. | ||
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| + | In section 2.4, practical voltage sources were modeled as an ideal voltage source $V_S$ in series with some internal resistance $R_S$, as shown in Figure 4.18. This corresponds exactly to a Thevenin circuit with $V_{OC} = V_S$ and $R_{TH} = R_S$. The practical voltage source provides maximum power to a circuit connected to it when the input resistance of the circuit (the equivalent resistance of the circuit, seen at the terminals to which the power source is connected) is equal to the internal resistance of the voltage source. Under these circumstances, | ||
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| + | $$P = \frac{V^2_S}{4R_S}$$ | ||
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| + | The same amount of power is converted to heat within the power supply; this is the reason many power supplies contain a fan to actively disperse this heat to the atmosphere. If the circuit’s input resistance is not equal to the source resistance, less power is transmitted to the circuit and a correspondingly greater amount is dissipated within the power supply. | ||
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| + | Practical current sources were modeled in section 2.4 as an ideal current source $I_S$ in parallel with some internal resistance $R_S$, as shown in Figure 4.19. This corresponds directly to a Norton equivalent circuit with $I_{SC} = I_S$ and $R_{TH} = R_S$. The current source provides maximum power to a circuit connected to it when the input resistance of the circuit is equal to the internal resistance of the source. A source transformation in conjunction with equation (5) indicates that the power delivered to the circuit by the current source is: | ||
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| + | $$P= \frac{R_SI^2_S}{4}$$ | ||
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| + | Again, a reduced percentage of the power generated by the source will be delivered to the circuit when the circuit and source are not well matched. | ||
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| + | Often, it may not be feasible to match the load with the power supply. For example, when we are testing circuits in our lab assignments we do not generally attempt to maximize the power delivered to the circuit – this is typical when prototype circuits are being tested. One simply recognizes that excessive power is being dissipated within the power supply and that the overall system is not functioning efficiently. If, however, the power supply and associated circuit are being designed as part of an integrated overall system one will generally attempt to match the power supply to the rest of the system. | ||
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| + | One problem which can occur during circuit testing is that extremely poorly matched power supply-load combinations may result in so much power being dissipated within the power supply that insufficient power is available to drive the load. This can result in the load apparently behaving abnormally, __unless__ power delivery effects are considered. | ||
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| + | ==== Section Summary ==== | ||
| + | * The maximum power that a circuit can deliver to a load resistor occurs when the load resistance is equal to the Thévenin equivalent resistance of the circuit. | ||
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| + | ---- | ||
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| + | ==== Exercises ==== | ||
| + | - Determine the resistance R which will absorb the maximum power from the 7V source. | ||
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| + | ---- | ||
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| + | [[{}/ | ||
| + | [[{}/ | ||