EECS 6A Designing Information Devices and Systems I Spring 08 Homework 7 This homework is due March, 08, at 3:59. Self-grades are due March 5, 08, at 3:59. Submission Format Your homework submission should consist of one file. hw7.pdf: A single PDF file that contains all of your answers (any handwritten answers should be scanned). Submit the file to the appropriate assignment on Gradescope.. (PRACTICE) Circuit Analysis Solve the circuit given below for currents and voltages. 0Ω A 0Ω 50Ω 3A Method : Nodal Analysis Applying KCL at Node, we get which gives implying Applying KCL at Node, we get V 0 V V = 0 V 0 0 V V = 0 0 V V V 0 = 0 3V V = 0 () V V V 0 3 = 0 EECS 6A, Spring 08, Homework 7, Last Updated: 08-03- 3:30
which gives implying V V V 0 0 50 3 = 0 5V 5V V 300 = 0 5V 7V = 300 () Writing equations and in matrix form, we get [ 3 5 7 ][ V V ] [ ] 0 = 300 Solving the system of equations, we will get V = 0V and V = 50V. Method : Superposition We define i and i 3 as follows: 0Ω A 0Ω 50Ω 3A i i 3 First, consider the effect of only the left A current source on. Using current divider rule, we have i = A and Therefore, and i = 70 A = 0.875A 0 70 i 3 = A 0.875A = 0.5A V a = i 0Ω = 0.875A 0Ω = 8.75V V a = i 3 50Ω = 0.5A 50Ω = 6.5V Second, consider the effect of only the right 3A current source on. Using current divider rule, we have i = 3A i = 50 3A =.875A 50 30 EECS 6A, Spring 08, Homework 7, Last Updated: 08-03- 3:30
and Therefore, and i 3 = 3A.875A =.5A V b = i 0Ω =.875A 0Ω = 8.75V V b = i 3 50Ω =.5A 50Ω = 56.5V Since the circuit is linear (i.e. we have linear elements and sources), we can use the principle of superposition to get V = V a V b and V = V a V b. Therefore, we get V = 8.75V 8.75V and V = 6.5V 56.5V. Finally, V = 0V and V = 50V. This solution agrees with the solution we obtained using nodal analysis.. (PRACTICE) Superposition Solve the circuits shown below using superposition. 3Ω i.5ω 5V A 3Ω.5Ω i = 3 A Consider the circuits obtained by: (a) Turning off the A current source: 3Ω i.5ω 5V 3Ω.5Ω EECS 6A, Spring 08, Homework 7, Last Updated: 08-03- 3:30 3
In the above circuit, no current is going to flow through the middle branch, as it is an open circuit. Thus this is just a 5V voltage source connected to two 3Ω resistors in series so (b) Turning off the 5V voltage source: i = 5 6 A 3Ω i O.5Ω A.5Ω 3Ω i In the above circuit, notice that the 3Ω resistors are in parallel and therefore form a current divider. Since the values of the resistances are equal, the current flowing through them will also be equal, that is i = i. Applying KCL to node O, we get which gives us i i = 0 i = A Now, applying the principle of superposition, we have i = i i = 5 6 A A = 3 A. 3. Fruity Fred Fruity Fred just got back from Berkeley Bowl with a bunch of mangoes, pineapples, and coconuts. He wants to sort his mangoes in order of weight, so he decides to use his knowledge from EE6A to build a scale. He finds two identical bars of material (M and M ) of length L (meters) and cross-sectional area A c (meters ), which are made of a material with resistivity ρ. He knows that the length of these bars decreases by k meters per Newton of force applied, while the cross-sectional area remains constant. He builds his scale as shown below, where the top of the bars are connected with an ideal electrical wire. The left side of the diagram shows the scale at rest (with no object placed on it), and the right side shows it when the applied force is F (Newtons), causing the length to decrease by kf meters. Fred s mangoes are not very heavy, so L kf. F L A M M M M B A B L-kF EECS 6A, Spring 08, Homework 7, Last Updated: 08-03- 3:30 4
(a) Let R AB be the resistance between nodes A and B. Write an expression for R AB as a function of A c, L, ρ, F, and k. The length of each spring as a function of F is L kf. The combination of and has a resistance R AB = = ρ(lkf) A c. (b) Fred s scale design is such that the resistance R AB changes depending on how much weight is placed on it. However, he really wants to measure a voltage rather than a resistance. Design a circuit for Fred that outputs a voltage that is some function of the weight. Your circuit should include R AB, and you may use any number of voltage sources and resistors in your design. Be sure to label where the voltage should be measured in your circuit. Also provide an expression relating the output voltage of your circuit to the force applied on the scale. One possible solution: use a voltage divider. R fixed V in R AB V out V out = V out = R AB R fixed R AB V in ρ(lkf) A c ρ(l kf) R fixed ρ(lkf) V in = R fixed A c ρ(l kf) V in A c 4. Resistive Voltage Regulator In this problem, we will design a circuit that provides an approximately constant voltage divider across a range of loads. We will use a resistive voltage divider circuit. The goal is to design a circuit that, from a source voltage of, would yield an output voltage within 5% of 4V for loads in the range of kω to 00kΩ. (a) First, consider the resistive voltage divider in the following circuit. What resistance R would achieve a voltage V out of 4V? 0kΩ a R V out b EECS 6A, Spring 08, Homework 7, Last Updated: 08-03- 3:30 5
The current in the circuit is I = R0kΩ which means that the output voltage can be calculated as R V out = I R = R0kΩ. By constraining V out = 4V, we get 4V 4V = R R 0kΩ (R 0kΩ) = R ( 4V 0kΩ = 4V ) 4V ( 4V ) 0kΩ = R 4V 4V 0kΩ = R 0.00kΩ = R (b) Now using the same resistor R as calculated in part (a), consider loading the circuit with a resistor of kω as depicted in the following circuit. What is the voltage V out now? 0kΩ a R V out kω b The value from part (a) is R = 0.00kΩ. This means that the effective resistance is (0.00kΩ kω) = 0.00 0.00 = 0.95kΩ. The current is therefore I = 0kΩ0.95kΩ, which yields: 0.95kΩ V out = 0kΩ 0.95kΩ V out = 0.5V (c) Now using the same resistor R as calculated in part (a), consider loading the circuit with a resistor of 00kΩ, instead, as depicted in the following circuit. What is the voltage V out now? 0kΩ a R V out 00kΩ b EECS 6A, Spring 08, Homework 7, Last Updated: 08-03- 3:30 6
The value from part (a) is R = 0.00kΩ. This means that the effective resistance is (0.00kΩ 00kΩ) = 0.00 00 0.0000 = 6.67kΩ. The current is therefore I = 0kΩ6.67kΩ, which yields: 6.67kΩ V out = 0kΩ 6.67kΩ V out = 3.75V (d) Now we would like to design a divider that would keep the voltage V out regulated for loads for a range of loads R l. By that, we would like the voltage to remain within a 5% window of 4V. That is, we would like to design the following circuit such that 3.80V V out 4.0V for a range of loads R l. As a first step, what is the Norton equivalent of the circuit on the left? Write and G e f f in terms of conductance values G = and G =. = G a a = G R e f f = G e f f b b In open circuit, the current through the resistors is I =. This means that the open circuit voltage R is V T h = I =. We now calculate the short circuit current. In short circuit, the current is simply I =. Therefore, the effective resistance is R e f f = V T h From this, we can write = G and G e f f = G G. = = = ( ). (e) For the second step, using the Norton equivalent circuit you found in part (d), what is the range of G e f f that achieves 3.80V V out 4.0V in terms of and G l? a R e f f = G e f f V out R l = G l b From Ohm s law, V out (G e f f G l ) =. This means that V out = G e f f G l. From the bound 3.80V V out 4.0V, we have 3.80V G e f f G l 4.0V. This yields the following two inequalities: 3.80V G e f f G l 4.0V, G e f f G l EECS 6A, Spring 08, Homework 7, Last Updated: 08-03- 3:30 7
which is equivalent to which is equivalent to G e f f G l 3.80V 4.0V G e f f G l, or concisely G e f f 3.80V G l 4.0V G l G e f f, 4.0V G l G e f f 3.80V G l. (f) Translate the range of G e f f in terms of and G l (that you found in part (e)) into a range on G in terms of G and G l. By plugging in the values of G e f f and into 4.0V G l G e f f 3.80V G l, we get or equivalently which is equivalent to G 4.0V G l G G G 3.80V G l, G 4.0V G l G G G 3.80V G l G, ( ) ( ) G 4.0V G l G G 3.80V G l. (g) Say we want to support loads in the range kω R l 00kΩ with approximately constant voltage as described above (that is, 3.80V V out 4.0V). What is the range of G in terms of G now? Translate the range of G in terms of G into a range of in terms of. Note: The unit of conductance is Siemens and is denoted by S. We plug R l = kω (equivalently G l = ms) into the bound found above and get G ( 4.0V ) ms G G ( 3.80V ) ms. Similarly, we plug R l = 00kΩ (equivalently G l = 00mS) into the bound found above and get ( ) G 4.0V 00 ms G G ( ) 3.80V 00 ms. The intersection of the above two sets of inequalities is ( ) G 4.0V ( ) 00 ms G G 3.80V ms EECS 6A, Spring 08, Homework 7, Last Updated: 08-03- 3:30 8
In terms of that range is (by taking the reciprocal of all sides and using G = R ) ( 4.0V ) 00kΩ ( 3.80V ) kω (h) Note that conductance is always non-negative. From the bounds on G you found in the previous part, derive a bound on G that ensures that G is always non-negative and non-empty (that is, the whole range of possible G values is non-negative and is not empty). Translate this range into a range of possible values. Hint: In addition to the conductance being non-negative, also make sure that the range for G is nonempty. Since conductance has to be positive, we need the range above to consist of positive values (that is, the lower bound has to be positive and the upper bound is larger than the lower bound) which yields: and which is equivalent to and which is equivalent to and The intersection of both ranges yields 0 G ( 4.0V ) 00 ms ( ) G 4.0V ( ) 00 ms G 3.80V ms, ( ) 00 ms G 4.0V ms ( 00 ms G 3.80V ), 4.0V 0.033mS = 6.5835mS = 4.0V 00 ms G ms 00 ms 3.80V 4.0V 6.5835mS G, G.. which in terms of is equivalent to 0.589kΩ (i) Pick the values of and that achieve 3.80V V out 4.0V for kω R l 00kΩ while minimizing the power consumed by the voltage divider circuit in open circuit (when there is no load attached to the output). What are these values and? How much power is consumed in this case? Calculate and report this power consumption using both the original circuit and the Norton equivalent circuit. Are the power you calculated using the original circuit and the power you calculated using the Norton equivalent circuit equal? EECS 6A, Spring 08, Homework 7, Last Updated: 08-03- 3:30 9
V The power in the original circuit (no load) can be calculated as. In order to minimize the power, we should pick the largest possible. This can be achieved by choosing = 0.589kΩ (the highest possible) and = = 0.35567kΩ (the upper bound). The power in 0.589kΩ( ) 4.0V 00kΩ () this case is P = 0.589kΩ0.35567kΩ = 70.976mW. If we use the Norton equivalent circuit, we have R e f f = 0.064kΩ which yields the power consumption P = INo R e f f = 66.086mW. The powers are not equal since Thévenin and Norton equivalents don t preserve power consumption of the circuit (think about the Thévenin equivalent in open circuit: the power consumption is always 0W). (j) Now using the same values and from the previous part, load the circuit with a load of 5kΩ. How much power is consumed by each of the three resistors,, and R l (use the original circuit to compute the power)? The effective resistance in the output is ( R l ) which is equal to 0.353kΩ. This means that the current from the source is I = ( R l ) =.8789mA. Therefore, the power consumed by is P R = I =.4330mW. The output voltage is V out = I ( R l ) = 4.957V, which means the power consumed by is P R = V out = 49.4953mW and the power consumed by R l is P Rl = V out R l = 0.345mW for a total power of P = 7.735mW. Note: Try doing the whole problem using resistance directly without going through conductances. You will see that the math will not be as pretty" as the one presented here (using conductances). Sometimes using conductance yields easier derivation than the derivation using resistance. 5. Multitouch Resistive Touchscreen In this problem, we will look at a simplified version of the multitouch resistive touchscreen. In particular, rather than measuring the position of two potential touch points in both dimensions (i.e. a pair of coordinates (x, y ) and (x, y ) corresponding to two touch positions), let s think about a version where we are interested in measuring only the vertical position of the two touch points (i.e. y and y ). Therefore, unlike the touchscreens we looked at in class, both of the resistive plates (i.e. both the top and the bottom plate) would have conductive strips placed along their top and bottom edges, as shown below. Top View E 4 Side View (no touch) E E E E H E 3 T H E 4 E 3 W (a) Assuming that both of the plates are made out of a material with ρ = Ωm and that the dimensions of the plates are W = 3cm, H = cm, and T = mm, with no touches at all, what is the resistance EECS 6A, Spring 08, Homework 7, Last Updated: 08-03- 3:30 0
between terminals E and E (which would be the same as the resistance between terminals E 3 and E 4 )? R = ρ L ( ) H A = R EE = ρ W T ( 0 ) m R EE = Ωm 3 0 0 3 m R EE = 4kΩ (b) Now let s look at what happens when we have two touch points. Let s assume that at wherever height the touch occurs, a perfect contact is made between the top plate and the bottom plate along the entire width of the plates (i.e. you don t have to worry about any lateral resistors), but that otherwise none of the electrical characteristics of the plates change. Defining the bottom of the plate as being y = 0 cm (i.e. a touch at E would be at y = 0cm), let s assume that the two touches happen at y = 3cm and y = 7cm and that your answer to part (a) was 8kΩ (which may or may not be the right answer). Draw a model with 6 resistors that captures the electrical connections between E, E, E 3, and E 4 and calculate their resistances. Note that for clarity, the system has been redrawn below to depict this scenario. Top View E 4 Side View ( touches) E E E H H y T E y E 3 y y E 4 E 3 W = 3cm cm R EE = kω 7cm 3cm = cm R EE =.667kΩ cm 7cm = R EE = 3.334kΩ cm EECS 6A, Spring 08, Homework 7, Last Updated: 08-03- 3:30
E E 4 ma E E 3 (c) Using the same assumptions as part (b), if you drove terminals E 3 and E 4 with a ma current source (as shown below) but left terminals E and E open-circuited, what is the voltage you would measure across E 4 E 3 (i.e. V E4E3 )? E 4 ma V E4E3 E 3 The equivalent resistance between E 4 E 3 is R E4E3 = = R = 6.667kΩ V E4E3 = ma R E4E3 = V E4E3 = 6.667V (d) Now let s try to generalize the situation by assuming that the two touches can happen at any two arbitrary points y and y, but with y defined to always be less than y (i.e. y is always the bottom touch point). Leaving the setup the same as in part (c) except for the arbitrary y and y, by measuring only the voltage between E 4 and E 3, what information can you extract about the two touch positions? Please be sure to provide an equation relating V E4E3 to y and y as a part of your answer, and note that you may want to redraw the model from part (b) to help you with this. EECS 6A, Spring 08, Homework 7, Last Updated: 08-03- 3:30
E E 4 E E 3 For general = y cm 8kΩ = y y cm 8kΩ = cm y cm 8kΩ So R E4E3 = R ( = cm y y y = ( cm y y ) 8kΩ cm y ) 8kΩ cm V E4E3 = ma cm y y cm 8kΩ = cm y y cm This means that by measuring V E4E3, we can only measure the distance between the two touch points (y y ). (e) One of your colleagues claims that by measuring the appropriate voltages, not only can they extract what both y and y are in this system, but they can even do so by formulating a system of three independent voltage equations related to y and y. As we will see later, this will allow us to gain some robustness to noise in the voltage measurements. In order to facilitate this, write equations relating V E4E and V EE3 to y and y. (The third voltage we ll use is V E4E3, which you should have already derived an equation for in the previous part of the problem.) 8V EECS 6A, Spring 08, Homework 7, Last Updated: 08-03- 3:30 3
E E 4 ma ma ma E E 3 V E4E = I = cm y cm 8V V EE3 = I = y cm 8V 6. Homework Process and Study Group Who else did you work with on this homework? List names and student ID s. (In case of homework party, you can also just describe the group.) How did you work on this homework? I worked on this homework with... I first worked by myself for hours, but got stuck on problem 5, so I went to office hours on... Then I went to homework party for a few hours, where I finished the homework. EECS 6A, Spring 08, Homework 7, Last Updated: 08-03- 3:30 4