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6.01: Introduction to EECS I Lecture 9 April 5, 2011 6.01: Introduction to EECS I Midterm Examination #2 Circuit Abstractions Time: Tuesday, April 12, 7:30 pm to 9:30 pm Location: Walker Memorial (if last name starts with A-M) 10-250 (if last name starts with N-Z) Coverage: Everything up to and including Design Lab 9. You may refer to any printed materials that you bring to exam. You may use a calculator. You may not use a computer, phone, or music player. No software lab in week 10. Review sessions during oﬃce hours. April 5, 2011 Last Time: Interaction of Circuit Elements Last Time: Buﬀering with Op-Amps Circuit design is complicated by interactions among the elements. Adding an element changes voltages & currents throughout circuit. Buﬀers can be used to simplify modular design. 1Ω Example: closing a switch is equivalent to adding a new element. 12V Io 2Ω 8V 8V 1Ω 2Ω 12V Vo This op-amp circuit produces an output voltage equal to its input voltage (8V) while having no eﬀect on the left part of the circuit. There are also other useful ways to deal with element interactions. Today: abstractions to characterize circuit interactions Closing the switch decreases Vo and increases Io . Systematic Changes Check Yourself Altering an element changes voltages and currents systematically. 3Ω Io Example: consider changes in Vo and Io when Ro is changed. 6Ω 90V 3Ω 90V 6Ω Vo Ro Io Vo How many of the blue numbers are wrong? Ro Ro [Ω] 0 2 3 8 ∞ 1 Vo [V] 0 30 36 48 60 Io [A] 30 15 12 6 0 6.01: Introduction to EECS I Lecture 9 April 5, 2011 Current-Voltage Relations One-Ports The current-voltage relation summarizes all possible behaviors of circuit – regardless of what the circuit is connected to. A “one-port” is a circuit that is regarded as a single, generalized component. 3Ω I Io Io + V Vo 6Ω 90V − Vo As with other components, a one-port has two terminals: • current I enters “+” terminal and exits “−” terminal and • produces voltage V across the terminals. It allows us to think about circuit as a single element: a one-port. I = −Io + V = Vo − Current-Voltage Relations Current-Voltage Relations Under what conditions is the current-voltage relation a straight line? Current-voltage relations for resistors and sources are straight lines. 3Ω I Io Io I V R0 Vo 6Ω 90V I V V0 V I0 Vo I I V I V V = R0 I V = V0 V I = −I0 What happens when you combine diﬀerent types of components? Check Yourself Parallel One-Ports If the i-v curves for two one-ports are both straight lines, then the i-v curve for the parallel combination is a straight line. I Graphical “proof”: 2Ω + 5V − I B V I C V V1 Ip I1 + V1 − What is the corresponding current-voltage relation? A I1 V I D V I + I2 + V2 − I2 Vp − V2 Ip = I1 + I2 V p = V1 = V2 V The “sum” of two straight lines is a straight line. 2 6.01: Introduction to EECS I Lecture 9 April 5, 2011 Series One-Ports Current-Voltage Relations If the i-v curves for two one-ports are both straight lines, then the i-v curve for the series combination is a straight line. More generally, any combination of one-ports with straight i-v curves will produce a one-port with a straight i-v curve. Graphical “proof”: Is Example: if a one-port contains only resistors and sources, then the current-voltage relation will be a straight line. + I1 + V1 − I2 + V2 − I1 Vs 3Ω − Vo 6Ω 90V I2 Io Io Vo Is = I1 = I2 V1 V2 Vs = V1 + V2 The “horizontal sum” of two straight lines is a straight line. Linear Equations Thevenin Equivalents If the current-voltage relation is a straight line, then the relation between element current and element voltage is a linear equation. If the current-voltage relation is linear, the current-voltage relation is the same as that for a voltage source in series with a resistor. 3Ω I Vo 6Ω 90V Io Io I Vo V0 V R V V0 1 R0 Linear equations have the form V − V0 . R If I = 0, then V = V0 (the x-intercept of the plot). ai Vi + bi Ii + ci = 0. From the circuit, I = √ (i.e., there are no Vi2 terms, no Ii terms, etc.) V V0 − with V is the slope 1/R. R R If all of the component equations are linear, then the i-v curve for the one-port will also be linear (since the solution to a system of linear components plus associated KVL and KCL equations is linear). The rate of growth of I = Norton Equivalents Open-Circuit Voltage and Short-Circuit Current If the current-voltage relation is linear, the current-voltage relation is the same as that for a current source in parallel with a resistor. If a one-port contains just resistors and current and voltage sources, then its i-v relation is determined by two points. I First point: open-circuit voltage I V 1 R0 −I0 I V R I I0 1 R0 −I0 V0 V V From the circuit, V = (I + I0 )R. If V = 0, then I = −I0 (the negative of the y-intercept of the plot). Set I = 0 in the circuit. Then V = V0 = 1V. The rate of growth of I = −I0 + V /R with V is the slope 1/R. 3 2Ω 2Ω 1V 6.01: Introduction to EECS I Lecture 9 April 5, 2011 Open-Circuit Voltage and Short-Circuit Current Open-Circuit Voltage and Short-Circuit Current If a one-port contains just resistors and current and voltage sources, then its i-v relation is determined by two points. If a one-port contains just resistors and current and voltage sources, then its i-v relation is determined by two points. Second point: short-circuit current The equivalent resistance is R0 = I 1 R0 I I −I0 V V0 Vo . Io 2Ω 2Ω V 1V 1 R0 1 Set V = 0 (wire = short circuit). Then I = −I0 = − A. 2 R0 = I −I0 V0 V 2Ω 2Ω V Vo 1V = 1 = 2Ω Io 2A Thevenin and Norton Equivalents Thevenin Example Equivalent circuits. Find the Thevenin equivalent of this circuit. I I I − 12 A 1 2Ω V 1V 2 V 10V 1Ω 3Ω + − V 1 Open-circuit voltage (i.e., I = 0) 3 V0 = V = × 10 = 7.5 V 3+1 I 1 2Ω 1V I − 12 A V 1V 2 V 1/2 Thevenin Example Thevenin Example Find the Thevenin equivalent of this circuit. This is the Thevenin equivalent. I 10V + − 1Ω 3Ω I V 10V Open-circuit voltage (i.e., I = 0) 3 V0 = V = × 10 = 7.5 V 3+1 Short-circuit current (i.e., V = 0 → add a wire!) 10 V = 10 A I0 = −I = 1Ω Equivalent resistance V0 7.5 V R0 = = = 0.75Ω I0 10 A + − 1Ω 3Ω I V + − 7.5V 0.75Ω V These circuits are “equivalent” in the sense that their i-v curves are the same. I 7.5 −10 4 V 6.01: Introduction to EECS I Lecture 9 Check Yourself April 5, 2011 Conceptual Value of Equivalent Circuits Equivalent circuits have conceptual value. Determine the equivalent circuits viewed from 3 diﬀerent ports. 10Ω 10Ω 10Ω 10A 5Ω 10Ω 10A 5Ω A 4Ω 10Ω B 10Ω 10A Example: Will closing the switch increase or decrease I? C +20V 5Ω How many of the following are wrong? 1. 2. 3. A 20V 5A 4Ω V0 = I0 = R0 = B 40V 5A 6Ω + − 2Ω 4Ω I 2Ω 2Ω We could just solve two circuits questions – one with switch open and one with switch closed – and compare currents. C 60V 10A 6Ω But this question can be answered without doing any calculations! Conceptual Value of Equivalent Circuits Superposition Replace the parts to the left and right of I by equivalent circuits. If a circuit contains only linear parts (resistors, current and voltage sources), then any voltage (or current) can be computed as the sum of those that result when each source is turned on one-at-a-time. 4Ω +20V + − 2Ω I 4Ω 2Ω I 2Ω R1 V0 R2 I0 Equivalent circuit to right of I depends on state of switch. switch open 4Ω I +10V + − switch closed 4Ω I 2Ω +10V + − 1Ω Closing the switch decreases equivalent resistance to right of I. Therefore, closing the switch increases I. Superposition Superposition If a circuit contains only linear parts (resistors, current and voltage sources), then any voltage (or current) can be computed as the sum of those that result when each source is turned on one-at-a-time. If a circuit contains only linear parts (resistors, current and voltage sources), then any voltage (or current) can be computed as the sum of those that result when each source is turned on one-at-a-time. I V0 R1 I I0 R2 Turning oﬀ I0 (i.e., I0 = 0) is equivalent to an open circuit. I1 V0 Then I1 = R1 R2 R1 V0 R2 I0 Turning oﬀ V0 (i.e., V0 = 0) is equivalent to a short circuit. I2 I0 = 0 V0 = 0 V0 . R1 + R2 Then I2 = − 5 R2 I0 . R1 + R2 R1 R2 I0 6.01: Introduction to EECS I Lecture 9 Superposition April 5, 2011 Check Yourself If a circuit contains only linear parts (resistors, current and voltage sources), then any voltage (or current) can be computed as the sum of those that result when each source is turned on one-at-a-time. Determine V using superposition. I R1 V0 R2 1. 2. 3. 4. 5. Combining: I1 V0 I = I1 + I2 = R1 R2 I2 I0 = 0 1Ω 1V V 1A I0 V0 = 0 R1 R2 I0 1V 2V 0V cannot determine none of the above V0 R2 − I0 . R1 + R2 R1 + R2 Check Yourself Summary Consequences of linearity. Determine I using superposition. If a one-port contains just linear elements (resistors, voltage sources, and current sources) then I 1Ω 1V 1. 2. 3. 4. 5. • the current-voltage relation will be linear, and 1A • it can be represented by a Thevenin or Norton equivalent circuit. Linear one-ports can be characterized by two points on their i-v curve (e.g., open-circuit voltage and short-circuit current). 1A 2A 0A cannot determine none of the above Responses of multiple sources can be superposed to ﬁnd electrical responses of linear circuits. 6 MIT OpenCourseWare http://ocw.mit.edu 6.01SC Introduction to Electrical Engineering and Computer Science Spring 2011 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.