ntroduction to Circuit Analysis Getting Started We analyze circuits for several reasons Understand how they work Learn how to design from other people s work Debug our own designs Troubleshoot circuit or system that may have failed Observe these are same reasons we analyze a system in any field Chemistry Mechanical engineering Civil engineering Computing science Physics Any other field of science We learning to solve problems Concentrating on electronics for the moment Circuit analysis so far has identified circuit comprise Number of branches Number of loops Have commented that the loops not all independent Same holds true for branches Let s try to develop a formalism Establishing a Basis Consider the following generalized simple circuit Note...this does not have to be an electronic circuit Can build identical model in Fluids Mechanical engineering Thermodynamic systems n following table we show the equivalents for electrical and mechanical systems
Electrical idt C d i C dt di i L i dt d L i Mechanical fdt M d f M dt df f D K dt f f K D dt mass damper spring With a simple change of symbol and independent and dependent variables All of the of the tools we learn for electrical circuits Apply to all the major engineering disciplines Nodal Equations Let s analyze from point of view of Kirchoff s Current Law - a KCL analysis a 4 c 4 3 5 3 5 d For this circuit We have 5 branches 4 nodes
We can write 4 nodal equations relating them a. -i - i 0 b. -i - i 3 - i 4 0 c. i 4 - i 5 0 d. i i 3 i 5 0 We observe these are not independent Any one equation can be found by combining the other 3 a. -i - i 0 b. -i - i 3 - i 4 0 c. i 4 - i 5 0 d. i i 3 i 5 0 i i 0 n general The number of independent KCL node equations is given as N -, where N is the number of nodes in the system As noted we can eliminate any of the equations Typically we do not write an equation for the reference node Ground Loop Equations Let s analyze from point of view of Kirchoff s oltage Law - a KL analysis For this circuit We have 3 loops We can write 3 loop equations relating them Loop: - - 3 0 Loop: 3-4 - 5 0 Loop3: - - 4-5 0
We observe these are not independent Any one equation can be found by combining the other Loop3: - - 4-5 0 Loop: - 3 4 5 0 - - 3 0 ntuitively - following reasoning from branch analysis We would think we would have N- independent loop equations Not quite as simple Consider Circuit with B branches Has B unknowns B currents through the branches B voltages across those branches From linear algebra we need B independent equations Where do such equations come from The answer is Ohm s law With B branches we have B equations of following form Kirchoff s node equations give N - As we ve seen Kirchoff s voltage equations must supply remainder B - B - (N-) - #KL equations 0 #KL equations B - N- n general The number of independent KL loop equations is given as B - N -, where N is the number of nodes in the system
For the given circuit B 5 N 4 Therefore the number of independent KL equations 5-4 As we determined earlier n general finding set of independent KL equations equires techniques from linear algebra Fortunately for most simple circuits Task is clear by inspection When dealing with more complex networks Other tools are available Mesh Equations Let s look at what we have been calling loop analysis in more detail Planar Networks Let N be the set of all networks Define subset P of N as Set of all networks that can be drawn in two dimensions without lines crossing Such a network is called planar Most circuits we will deal with care planar For a planar network We can define a special type of loop We call it a mesh A mesh is a loop that does not enclose any other loops The following network contains 3 loops
a 4 c 3 5 d Loop:,, and 3 Loop: 3, 4, and 5 Loop3:,, 4, and 5 Observe L3 encloses L and L L does not enclose any loops L does not enclose any loops Thus The network contains meshes Equal number of independent KL equations This is not a co-incidence From this observation f we confine the KL equations in planar networks to meshes We will always get a complete set of independent loop (mesh) equations We can restate our earlier formula as a special case The number of independent KL loop equations equals the number of meshes in the system n general A network can be completely solved For all currents and voltages using. Ohm s Law and
. Kirchoff s Current Law for branch variables or 3. Kirchoff s oltage Law for loop variables Example 5mA i 0 3 0 40 3 4 i 30 4 Use branch variables:,, 3, 4 Node a: s - 0 Node b: - - 3 0 Node c: - - 3 0 We also know.. 3. 3 4 b b c 4 3 4. s 0 b b c 5. 0 b c c 6. 0 3 c 4 3
Some algebra gives 7. s 0 8. ( ) 0 3 3 b c 9. ( ) 0 4 b 3 4 c Substituting variables and solving b 9 v c 5 v From which we get 3 4 5mA.78 ma. ma. ma Equivalent Systems Circuits Often times in engineering we use tools to simplify our analysis eplace a complex circuit or system Equivalent but simpler Can make such replacement as long as behaviour is same n simple arithmetic We can replace the two integers in the following equation by a single one X 3 5... s equivalent to X 8... We think nothing of such a substitution n electronics we do the same thing Let s look at a couple of simple circuits i i
n the first circuit the two resistors are connected in series These give a total resistance between nodes A and B as f we inject a current into node A We will have a voltage drop from A to B of / ( ) n the second circuit notice we have two resistors in parallel n parallel those two resistors will have a resistance of Combining all three resistors We see that we have a total resistance between nodes C and D of f we inject a current into node C We will have a voltage drop from C to D of / ( ) Observe that the behaviour of the two circuits - from their ports - is identical That is they have an identical - relationship Based upon such a relationship we can say the two circuits are equivalent Let s generalize First consider a one port abstraction of a (electrical) system i N Based upon the contents of the system For each current we inject into the port We measure a voltage across the port
Once again note this same analysis holds as well for Mechanical systems Fluidics Thermal systems We can plot the voltage vs current We call this the - characteristic When any two networks have identical - characteristics, we say they are equivalent For the following two networks i i N N N may contain only a few components N may contain thousands None-the-less if the - characteristics are identical The circuits are considered equivalent Example Let s examine a more complex circuit 3 4
Analyzing 3 and 4 are in series - replace these by C C and C are in parallel - replace these by B B and B are in series - replace these by A A The resulting circuit is equivalent to the original network
Sources We can apply a similar analysis to voltage and current sources oltage Series oltage For the following circuit eq We know from Kirchoff s voltage law that - - 0 Thus we see - Now we have For this circuit we have 0 Thus we see - ( ) For this circuit we have - 0 Thus we see ( - ) For this circuit we have
- 0 Thus we see - Current For the following circuit eq We know from Kirchoff s current law that at Node a - 0 Node b - 0 Therefore and Thus we have violated KCL if oltage Parallel oltage For the following circuit eq
We know from Kirchoff s voltage law that - 0-0 Thus we see and Thus we have violated KL if Current For the following circuit eq We know from Kirchoff s current law that at Current sources in parallel add as voltage sources did in series Source Transformation Consider the following two networks s s N N
Analyzing N N S - - 0 S - / - 0 S - S - For these two networks to be equivalent The following must hold S S That is S S / Thus we see: s s /
Similarly s s Example Let s examine the following circuit K 0 0mA K 5 ma 5 K 75 5 K Example Let s use source transformation to solve the following circuit 5 in 0 0 30 3 4 o 40
Find out. Convert in and into a current source a (in/) 30 3 4 0 0 40 o. and a 0/3 30 3 o a a 4 40 3. Convert a, a to voltage source b (0/3)(/) 0/3 b a 3 4 o 4. Convert a and 3 b a 3 0/3 o b a 4
5. Convert b, b to current source c b / b (0/3)(3/0) / o c b 4 6. Convert b and 4 c b 4 c 440/3 o c c 7. out c c (/) (440/3).9 oltage and Current Dividers oltage Dividers Let s look at the following circuit s Using KL s s From Ohm s law 0
Substituting s s Thus we see The supply voltage divides between the two resistors According to the formulas gives Example n the following circuit 0 o 5 f we want out 5 for in 0 We know O in Select K Ω Current Dividers Let s look at the following circuit o Using KCL s
s S Substituting S S Thus we see The supply current divides between the two resistors According to the formulas gives Example n the following circuit We want to find and such that 4 ma s 0 ma S 0mA 4mA 4( ) 0 4 6 let 0 30