6 01 Spring Semester 2008 Exploration 9 Issued Thursday April 10 1 MASSACHVSETTS INSTITVTE OF TECHNOLOGY Department of Electrical Engineering and Computer Science 6 01 Introduction to EECS I Spring Semester 2008 Exploration 9 Issued Thursday April 10 This exploration is due April 24 or 25 in your design lab Option 1 More abstract circuit specifications This part of the exploration is worth 5 out of 10 points If you plan to do option 2 you should do it instead of option 1 If you do option 1 in isolation first you are likely to find that your solution cannot be extended easily to option 2 Solving a circuit this way is a lot easier than using a pencil and paper but it s still kind of a pain to specify the circuit We find that we often make mistakes with the signs on the KCL constraints The fact is that once we know what the components in the circuit are and how they re connected together the entire circuit is specified In the language we used for last week s lab we could do our job with simpler specifications c Circuit Resistor ra n1 n2 Resistor rb n4 n1 Resistor rd n4 n3 Wire n4 n2 VSrc vc n2 n3 c solve n3 In this specification we still have to name the nodes but not the currents Then for each component we just specify its value and its terminals The order of the terminals still has to agree with the order of the variables in the underlying constraint function Use the underlying constraint solving code from this week s lab to implement a system similar to the one we used last week Your solution should Provide a convenient way to specify circuits just saying what the components are and how they are connected together but not specifying currents Implement resistors voltage sources and op amps Make it relatively easy to add new components with linear constraints and Display solutions nicely Exploration 1 Hand in your code and demonstrate it on the example in figure 3 of software lab 9 Exploration 2 Explain what someone would have to do to add a new type of component 6 01 Spring Semester 2008 Exploration 9 Issued Thursday April 10 2 Option 2 Defining and using new primitives This option is quite difficult It is worth 10 points You can do it as an alternative to option 1 So far in all of our circuit sepcification methods we have been able to define primitives and means of combining them We haven t however been able to build and use abstractions very effectively during circuit specification We would like to be able to define the pattern of a voltage divider or an inverting amplifier and then use it as a component in future circuits we might design So your goal in this exploration is to implement a system that will Provide a convenient way to specify circuits just saying what the components are and how they are connected together Implement resistors voltage sources and op amps Allow you to take circuits patterns that you have already specified and use them as components in future circuits and Display solutions nicely Note that there is an important difference between the generic idea of an inverting amplifier or a buffered voltage divider and any particular instance of one A particular instance has actual voltages and currents and you can t put two copies of the same instance in a circuit This is a very hard problem that can be approached in several ways Below we show you some structures that we built using our solution to this problem We provide it here to show you the basic idea of what we want and to give you some idea of one way to go about it But it is not crucial to approach it this way for example you might find it easier or more beautiful to make the program more functional and less object oriented than we have here A voltage divider with resistance r1 on top and r2 on the bottom Needs names of top middle and bottom nodes class Divider Circuit def init self r1 r2 nHi nOut nLo Just make a circuit with two resistors connected in the right way Circuit init self Resistor r1 nHi nOut Resistor r2 nOut nLo Two dividers connected together in our familiar configuration that doesn t divide by 4 Uses all the same resistance for simplicity class DoubleDivider Circuit def init self r vPlus vOut vMinus Each new instance of this circuit will have a different internal node the output of the first divider So we have to generate a new name for that node each time through See code for gensym for details middleNode gensym divider Now make a circuit with two dividers connected up appropriately Circuit init self Divider r r vPlus middleNode vMinus Divider r r middleNode vOut vMinus A common wiring pattern for an op amp is a simple follower 6 01 Spring Semester 2008 Exploration 9 Issued Thursday April 10 class Follower Circuit def init self vIn vOut Circuit init self OpAmp vIn vOut vOut A voltage divider with resistance r1 on top and r2 on the bottom Needs names of top middle and bottom nodes Output of the divider is run through a follower This is how we made virtual ground class BufferedDivider Circuit def init self r1 r2 nHi nOut nLo Have to generate a name for the node that is the output of the divider and the input to the follower middleNode gensym divider Circuit init self Resistor r1 nHi middleNode Resistor r2 middleNode nLo Follower middleNode nOut Connecting two buffered dividers does actually divide by 4 class D o u b leBuf fere dDivi der Circuit def init self r vPlus vOut vMinus Have to generate a name for the node that is the output of the first divider and the top of the second divider middleNode gensym doubleDivider Circuit init self BufferedDivider r r vPlus middleNode vMinus BufferedDivider r r middleNode vOut vMinus Inverting amplifiers are a handy pattern We specify input and output nodes and the multiplier we desire so that vOut multiplier vIn class In vertingAmplifier Circuit def init self multiplier vIn vOut Just pick a value for the input resistor r 100 0 Need a name for the node that is the minus input to the op amp vMinus gensym vMinusAmp Put a resistor on the input and another resistor r multiplier on the feedback path of an op amp components Resistor r vIn vMinus Resistor r multiplier vOut vMinus OpAmp gnd vMinus vOut Circuit init self components Testing the inverting amplifier c Circuit VSrc 10 vIn gnd InvertingAmplifier 2 vIn vOut c solve 1 vIn 1 gnd 10 None 1 vIn 1 vMinusAmp2 100 0 i3 0 None 1 vOut 1 vMinusAmp2 200 0 i4 0 None 1 vOut 1 gnd 10000 gnd 10000 vMinusAmp2 0 None 1 gnd 0 None 1 i3 1 i4 0 None 1 i1 1 i3 0 None 1 i4 1 iOpAmp5 0 None gnd 1 46975764892 e 15 i1 0
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