1MASSACHVSETTS INSTITVTE OF TECHNOLOGYDepartment of Electrical Engineering and Compvter Science6.01 Introdvction to EECS 1Spring Term, 2008Week 7 LectureMarch 18, 2008Introduction to Electric CircuitsPrimitives:•Voltage•Current•Circuit ElementsMeans of Combination:•Circuit Constraints (KCL, KVL), describing wiringMeans of Abstraction:•Thevenin•NortonCommon Patterns:•Series•Parallel•Voltage Divider•Current DividerTwo primitive notions with respect to a circuitelement:•Current is the flow of charge (electrons)through the element. Current is measured inAmperes. We will usually refer tomilliamperes. (mA)•Voltage is the electromotive force pushingthe electrons through the element. Thevoltage appears ‘across’ the element.Voltage is also called ‘potential’ because itrepresents potential energy change of thecharges from one end of the element to theother. The unit of potential is the Volt.2Conservation laws govern current and voltageConservation of current: is also called KCL,Or Kirchhoff’s Current Lawikk!= 0Sum of all currents entering a nodeis zero, meaning whatever currententers a node must also leave thenode. Current does not build up, butmust go in circles.1824-1887KVL is Kirchhoff’s Voltage Law, and it says that voltage really is apotential that is single valued. The sum of voltages around a loopmust add to zero. So if you take one node as ‘datum’ or ‘ground’,every other node has a uniquely defined potential (voltage)vkk!= 0Primitive Circuit ElementsVoltage Source: Fixes its terminalvoltage, independent of current(Sort of like a battery, but more aboutthat later)Curent source: Fixes its terminalcurrent, independent of voltage(sort of like lightning, but youdon’t want to fiddle with that…)Resistance: Has a fixed ratio ofvoltage to current:The unit if resistance is the Ohm,a volt/ampere.V = RI3Here is a simple problem: what current will flow in the resistor?What current will flow in the source?Around the KVL loop:!Vs+ Vr= 0i =VsR=12v1k!= 12mASo current is:One more thing about this:Power isP = V ! I = I2R =V2RFor 12 v and 1k, that is 144 mWBut if R=100 Ohms, it is 1.44 W.How would a 1/4 watt resistor handle this?4Here is a second problem, with two resistors:KVL can be used around two loops:V1= V2= Vs= 12VAnd then KCL yields, for current OUT OF the source:Is= I1+ I2= Vs1R1+1R2!"#$%&This is a parallel resistor combination:And the same combination gives usthe current divider:i1=vR1= iR2R1+ R2v = iR1|| R2=i1R1+1R2= iR1R2R1+ R25Series Combination:Vs= V1+ V2= i R1+ R2( )V2= R2i = VsR2R1+ R2And the associated voltage divider relationship:Rs= R1+ R2A simple problem: what are the currents?As far as voltages are concerned, this is the same circuitv1= VsR1|| R2R1|| R2+ Ra= VsR1R2RaR1+ RaR2+ R1R2i1=v1R1= VsR2RaR1+ RaR2+ R1R2i2=v1R2= VsR1RaR1+ RaR2+ R1R26So what about a more complicated problem? What isthe output voltage of this one?1Rv1! Vs( )+12Rv1+1Rv1! v2( )= 01Rv2! v1( )+12Rv2+1Rv2! vo( )= 01Rvo! v2( )+1Rvo= 0Here is the formulation of the problem in circuit constraints:It is not that bad: look at serial and parallel combinations to get to theSimplest series combination:vo= v2RR + R=12v2v2= v1RR + R=12v1v1= VsRR + R=12Vsv0=12v2=14v1=18VsNow what would it look like if you were to allow it to supplysome current? Consider what might happen with a currentsource connected to the output:7It is, after all, a linear circuit, as werecognized by TheveninSo output voltage is a linearfunction of current1857-1926The behavior of any circuit of this type can be represented by eitherthe Thevinin or Norton equivalents (which are equivalent)It is perhaps remarkable, and can be proven, that any circuitmade of sources and other linear components may berepresented this way. We won’t prove it, but we will use this fact.So how do we use these things? Consider a photo detector:In dark, Rp= 10k! vo=12VsIn light, Rp= 1k! vo=1011VsIf Vs=10v, Dark Vo = 5 v Light Vo = 9.1 vAnd you can detect thisYou will use a photoresistorin the lab: it may not haveexactly these values8A very useful resistive circuit: Wheatstone BridgeV0= VsRbRa+ Rb!RdRc+ Rd"#$%&'How to handle this one? Voltage across RLHere is the way of solving that:V1= VsRbRa+ RbV2= VsRdRc+ RdR1= Ra|| RbR2= Rc|| RcVo= V1! V2( )RLRL+ R1+
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