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| learn:courses:real-analog-chapter-4:start [2016/12/27 23:39] – [4.6: Maximum Power Transfer] Martha | learn:courses:real-analog-chapter-4:start [2022/10/04 11:38] (current) – external edit 127.0.0.1 | ||
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| ====== Real Analog: Chapter 4 ====== | ====== Real Analog: Chapter 4 ====== | ||
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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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| - 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. | - 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. | - 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. | + | - 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. |
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| 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: | 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: | ||
| - | $$v_1 = R_{TH} \cdot i | + | $$v_1 = R_{TH} \cdot i |
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| Superimposing the two voltages above results in: | Superimposing the two voltages above results in: | ||
| - | $$v=v_1 + v_{OC} | + | $$v=v_1 + v_{OC} |
| Or | Or | ||
| - | $$v= R_{TH} \cdot i + v_{OC} | + | $$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// | 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// | ||
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| - | 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. | + | 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. |
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| 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: | 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: | ||
| - | $$i_1 = \frac{v}{R_{TH}} | + | $$i_1 = \frac{v}{R_{TH}} |
| Notice that equation (4.6) can be obtained by rearranging equation (4.3). | Notice that equation (4.6) can be obtained by rearranging equation (4.3). | ||
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| Employing superposition, | Employing superposition, | ||
| - | $$i = i_1 - i_{SC} | + | $$i = i_1 - i_{SC} |
| So | So | ||
| - | $$i = \frac{v}{R_{TH}} - i_{SC} | + | $$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. | 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. | ||
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| === Procedure for Determining Norton Equivalent Circuit: === | === Procedure for Determining Norton Equivalent Circuit: === | ||
| - | - Ordered List ItemIdentify | + | - Identify |
| - Determine the equivalent resistance $R_{TH}$ of the circuit. The approach for determining $R_{TH}$ is the same for Norton circuits as Thévenin circuits. | - 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. | - 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. | ||
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| Equating equations (4.8) and (4.9) leads to the conclusion that: | Equating equations (4.8) and (4.9) leads to the conclusion that: | ||
| - | $$i_{SC} = \frac{v_{OC}{R_{TH}} | + | $$i_{SC} = \frac{v_{OC}}{R_{TH}} |
| Likewise, rearranging equation (4.8) to obtain an expression for v gives: | Likewise, rearranging equation (4.8) to obtain an expression for v gives: | ||
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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: | 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: | ||
| - | $$\frac{\partial P_L}{\partial R_L} = V^2_{OC} \left[ \frac{\left( R_L + R_{TH} \right) ^2 - 2R_L \left( R_{TH} + R_L \right) }{\left( R_L + R_{TH} \right) ^4} \left] = 0$$ | + | $$\frac{\partial P_L}{\partial R_L} = V^2_{OC} \left[ \frac{\left( R_L + R_{TH} \right) ^2 - 2R_L \left( R_{TH} + R_L \right) }{\left( R_L + R_{TH} \right) ^4} \right] = 0 (Eq. 4.15)$$ |
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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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