MASSACHVSETTS INSTITVTE OF TECHNOLOGY Department of Electrical Engineering and Compvter Science 6 01 Introdvction to EECS 1 Spring Term 2008 Week 7 Lecture March 18 2008 Introduction to Electric Circuits Primitives Voltage Current Circuit Elements Means of Combination Circuit Constraints KCL KVL describing wiring Means of Abstraction Thevenin Norton Common Patterns Series Parallel Voltage Divider Current Divider Two primitive notions with respect to a circuit element Current is the flow of charge electrons through the element Current is measured in Amperes We will usually refer to milliamperes mA Voltage is the electromotive force pushing the electrons through the element The voltage appears across the element Voltage is also called potential because it represents potential energy change of the charges from one end of the element to the other The unit of potential is the Volt 1 Conservation laws govern current and voltage Conservation of current is also called KCL Or Kirchhoff s Current Law i k 0 k Sum of all currents entering a node is zero meaning whatever current enters a node must also leave the node Current does not build up but must go in circles 1824 1887 KVL is Kirchhoff s Voltage Law and it says that voltage really is a potential that is single valued The sum of voltages around a loop must add to zero So if you take one node as datum or ground every other node has a uniquely defined potential voltage k vk 0 Primitive Circuit Elements Voltage Source Fixes its terminal voltage independent of current Sort of like a battery but more about that later Curent source Fixes its terminal current independent of voltage sort of like lightning but you don t want to fiddle with that Resistance Has a fixed ratio of voltage to current The unit if resistance is the Ohm a volt ampere V RI 2 Here is a simple problem what current will flow in the resistor What current will flow in the source Vs Vr 0 Around the KVL loop i So current is One more thing about this Power is P V I I 2R Vs 12v 12mA R 1k V2 R For 12 v and 1k that is 144 mW But if R 100 Ohms it is 1 44 W How would a 1 4 watt resistor handle this 3 Here is a second problem with two resistors KVL can be used around two loops V1 V2 Vs 12V And then KCL yields for current OUT OF the source 1 1 I s I1 I 2 Vs R1 R2 This is a parallel resistor combination And the same combination gives us the current divider v iR1 R2 i1 i RR i 1 2 1 1 R1 R2 R1 R2 v R2 i R1 R1 R2 4 Series Combination Vs V1 V2 i R1 R2 Rs R1 R2 And the associated voltage divider relationship V2 R2i Vs R2 R1 R2 A simple problem what are the currents As far as voltages are concerned this is the same circuit R1 R2 R1 R2 Vs R1 R2 Ra Ra R1 Ra R2 R1 R2 v R2 i1 1 Vs R1 Ra R1 Ra R2 R1 R2 v1 Vs i2 v1 R1 Vs R2 Ra R1 Ra R2 R1 R2 5 So what about a more complicated problem What is the output voltage of this one Here is the formulation of the problem in circuit constraints 1 1 1 v1 Vs v1 v1 v2 0 R 2R R 1 1 1 v2 v1 v2 v2 vo 0 R 2R R 1 1 vo v2 vo 0 R R It is not that bad look at serial and parallel combinations to get to the Simplest series combination R 1 v2 R R 2 R 1 v2 v1 v1 R R 2 R 1 v1 Vs Vs R R 2 1 1 1 v0 v2 v1 Vs 2 4 8 vo v2 Now what would it look like if you were to allow it to supply some current Consider what might happen with a current source connected to the output 6 It is after all a linear circuit as we recognized by Thevenin So output voltage is a linear function of current 1857 1926 The behavior of any circuit of this type can be represented by either the Thevinin or Norton equivalents which are equivalent It is perhaps remarkable and can be proven that any circuit made of sources and other linear components may be represented this way We won t prove it but we will use this fact So how do we use these things Consider a photo detector In dark In light 1 Rp 10k vo Vs 2 10 Rp 1k vo Vs 11 You will use a photoresistor in the lab it may not have exactly these values If Vs 10v Dark Vo 5 v Light Vo 9 1 v And you can detect this 7 A very useful resistive circuit Wheatstone Bridge Rb Rd V0 Vs Ra Rb Rc Rd How to handle this one Voltage across RL Here is the way of solving that Rb Ra Rb Rd V2 Vs Rc Rd R1 Ra Rb R2 Rc Rc V1 Vs Vo V1 V2 RL RL R1 R2 8
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