Circuits I: RLC Circuit Response

hi I'm dr. Kyle Montgomery and in this video we're going to be looking at RLC circuits and a little bit different from some my other videos here I want to just kind of go through sort of the general process by which we are able to determine what in general terms what the responses are for given voltage or current in a variety of different RLC circuits okay so I'd encourage you to maybe pause the video copy down what I have here so we're going to look at two specific cases if we have a parallel combination of resistor inductor and capacitor and a series combination of the inductor resistor and capacitor and both of these you can see that I have no source so what we're talking about here initially is known as natural response okay where natural response is basically indicating that the circuit might initially have some let's say charge some voltage built up on my capacitor or might have some current built up in my inductor on either one of these two cases and but as you can see I've disconnected wherever whatever my source was maybe I had a switch to some other current or voltage source over here and therefore the currents and voltages are all going to eventually just decay down to zero once it reaches the steady state condition and so we want to do is basically evaluate you know how do we go through this process of determining what the final solutions for let's say voltage or a current might be in either one of these circuits based on the various elements that we have there okay so we'll just take a quick pause and allow you to do that and then we'll get right into it okay so what we're going to do here is basically look at each one of these two circuit cases the parallel case in this series case and I kind of want to go through them in tandem sometimes we separate these and maybe only just look at the parallel and then go back and look at the series case here I'm going to kind of try to do them together so we sort of see that there's a lot of commonalities between the approach that we do for both the parallel in the series case and it ultimately gets us to seeing how the solution is very similar to each of these there's only just some relatively minor different okay so if we look at the parallel circuit case here and we say initially just want to write we need some relationship some equation that's going to describe the voltages and the currents happening in this parallel combination of the capacitor inductor and resistor so if we think about just writing a KCl equation maybe from the top node and indicate each current here going down through each branch I see il and I are okay so we know that all those would have to sum to zero by Kirchhoff's current law so instead of let's just start here by saying what we know IC plus il plus IR all have to equal to zero okay so now then how can I describe these currents in terms of some other known quantities that I have so if I say in total let me indicate just the voltage in general terms by lowercase V across each of these different branches so I know that the current I see I could use by just saying C DV DT so we know that that disc C times the derivative of the voltage describes the current through a capacitor so that's what I see is for il here this would involve the integration or one over the inductance times the integration from zero to T times the voltage times D tau that's the general term for any time T and then because we might have some initial current here we're going to add in plus I naught so that's just my initial current condition and again this these two terms are just in general for any inductor telling us what the current through that inductor should be as a function of time and then finally for the current through my resistor I are well here I could just use Ohm's law to say that this is V over R and so again all those terms should still sum to zero okay so here we have still a little bit of a tricky equation that we're having to work with so let's do a couple of things to get it to a form that is a little bit more easy to approach so the first thing I'll do is take the derivative of this equation with respect to time which I can do that on both sides and not to change anything about the problem here so we take the derivative of my first term here then I'll turn this into the second derivative so our d-v on top over DT squared here and for my next term so this will by taking the derivative of this integral basically just leaves me with V over L I naught is a constant if we're taking the derivative of a constant that just goes to zero there and then for my final term which is V over R taking the derivative with respect to time we would have 1 over R times DV DT equal to 0 there ok so we see now already that we can tell that we have a second-order differential equation because I have a second secondary differential here first-order here and then just our primary term in the middle but we'll rearrange this let's divide by C so we can get the second derivative by itself so then I'm left with d squared V over DT squared I've divided by the capacitance C so I have 1 over or let me bring one over RC times the first derivative the voltage and then 1 over LC times V equals zero okay so here then is our final second-order differential equation that we have that's describing in this case we're looking at the voltage as a function of time in this parallel RLC circuit specific specifically of the case of natural response where we have no other sources that are tied in to this circuit so that's why everything here is just equal to zero and it gets us in a nice form that we can look at and then kind of try to evaluate well what should the solution to this second order differential equation look like and then what are some characteristics of it from there so taking note of this I'm going to go and erase this and we're going to go through the same process now for the series circuit again I just kind of want a little more clearly illustrate how there's a lot of parallels between what's going on in the parallel case and the series case it's still going to get us to a second-order differential as you might expect so we'll go ahead and do that now and then go on to the next steps of how to actually find a solution to these equations okay okay so let now let's goes through the same process for writing some relationship around this series circuit that I have here so here instead of applying Kirchhoff's current law make more sense to apply Kirchhoff's voltage law meaning I hadn't need to add up all the voltage drops around this loop here again we have no other sources so I know that the sum of these voltages just must be equal to zero if I indicate in general just some current lowercase I as such then initially if I start off with the voltage drop across my resistor R here I know the vet just be I times R by Ohm's law the voltage drop across my inductor would be l di/dt again that's just standard form for what we know about inductors then the voltage drop across my capacitor would be plus here 1 over C so we have the another integral that we're looking at here from zero to T integration of the current the integration term d-town plus the voltage V naught here ok and maybe just to be a little bit more accurate let me indicate this voltage as being plus 2 minus here so that way we're in the right terms here is what we have and then so that's the three voltage drops I have around this series loop all those should equal to 0 by Kirchhoff's voltage law so now then just basically applying kind of the same process we did we were looking at the parallel circuit case the first thing I'm going to do is take the derivative of the current with respect to time that'll allow us to kind of eliminate having an integral mixed in with what we have right here ok so if I have this then I need our di DT okay taking the derivative this ir term plus L take me to the second derivative of the current times my inductance L okay and then here I would just be left with I over C by taking the derivative of this integral and of course again V naught here is a constant so taking the derivative of that just gives me zero so that doesn't factor into the equation any longer here equals zero so now again we already see we have our second order differential equation I'm going to do a little bit of cleanup again I want to kind of isolate the second derivative term so in this case going to divide by the duct it's L and then just kind of rearrange the terms that we have here okay that's my second derivative term first derivative term here so now I have R over L times di DT and plus 1 over LC times the current I equal to 0 here ok so again now what we have here is the second order differential equation as you can see a very similar form to what we saw in the parallel circuit case now one of course big difference is that in this case we're working with the current specifically whereas with the parallel circuit case we're working with the voltages but in any of these cases we will actually end up seeing that the response of the voltage and the current in any given RLC circuit is basically going to follow the same characteristic more or less so it doesn't matter whether we're evaluating things in terms of the voltage or in terms of the current the response the solution that we're going to get to the form of it will take the same the only differences we'll be looking at initial conditions and final conditions and how those obviously would be different for current verses a voltage but here again we just have a very similar form to the parallel circuit case as you can compare in your notes there and so now we can look at how do we use this second-order differential equation to actually find how do we find a solution to the second-order differential equation okay so the basic process that we're going to follow is to basically start with some assumption about what the form of the solution should look like and in this case we're going to make the assumption that the form of the solution is going to have some exponential term just with some coefficient so here we're talking about the voltage as a function of time is going to be just some coefficient a times exponential - a term s times T which will kind of go into finding what this term s is so this would be for the parallel circuit case because it's a voltage specifically but I could say the same thing about the form of the solution for the series case where here we're looking at the currents or I could say that the current is going to take some form of coefficient with some exponential as given here okay okay so now the basic step here is if we were to then take this form of the solution and plug it back into the dip to the differential equation which again that is what we would normally do if we if we're assuming that this is the solution what that means this solution should in fact complete this equation such that it should actually equal out to 0 so if we were to do this plug this back into the second-order differential equation do a little bit of reduction in terms I'm going to kind of skip over some of those steps but we'll see then how we get to what's called the characteristic equation which is going to be an important term that tells us something about this term s here so let me go and clean up my board here and then we'll get into looking at that specifically ok so again if we took that the form of the solution that we had previously plug it into that second-order differential equation we can factor out the exponential from that and what we're left with is basically what's called the characteristic equation which takes the form as follows it looks like s squared plus s over R times C plus 1 over L times C equals 0 now this is specific for the parallel case ok because again this is just taking into account that those we came up with two separate differentially second-order differential equations one for the parallel one for the series and that they're more or less at the same form but there were some different terms that we saw in there so again this would be specific for the parallel case and then what we'd have for the series case very similar but specifically still s squared plus R over L times s plus 1 over LC equals 0 ok so these are what are called known as the characteristic equations for each of the parallel in the series RLC circuit case and then in order to be able to find basically again here s is the variable that we would need to sort of solve for 2 for this equation to be valid and so if we look at what those would look like for both the series and parallel case right just basically applying the quadratic form here so we should have two roots of each of these equations again they're going to be slightly different for this parallel and series case but still very similar for the parallel case it'll come out to be minus 1 I mean sorry minus R over 2l and then plus or minus the root of R over 2 l squared minus 1 over L times C and then for the actually sorry I think I indicated that wrong this should be actually the series the roots of the series characteristic equation and then the one for the parallel would give me the roots are going to be equal to minus 1 over 2r c plus or minus square root of 1 over 2r c squared minus 1 over LC ok so again the only difference here that we see between these root the form of the roots for the series versus parallel case is basically a couple of the coefficient terms here and here specifically are a little bit we stove the one over LC which is the same again that kind of makes sense by just looking to what we have from each of the characteristic equations up top okay but again still basically very similar pathway for both the parallel and series solutions okay so now if these are the roots that we found s 1 and s 2 if we think back to the solution that we've assumed ok which again took the form of say voltage as a function of time was equal to a time's the exponential to the s to the power of s times T time ok so here we're saying these roots s 1 and 2 where it's as 1 & 2 simply indicating that I can either add or subtract this quadratic term but either one of those should be able to plug into the form of the solution and give me in fact a valid solution so as we know from what math tells us with differential equations if either one of these roots is a solution then we also know that the sum of each of these forms of the solution would also be basically a third solution so that's the form that we're going to work with to actually come to a more general generic solution for the RLC circuit case so let me kind of erase what I have here go on to looking at what those forms of those solution actually do look like ok so again what I just described in words was to say that because we are able to distinguish two different roots this s 1 and s 2 that tells us that I can have two different forms of the solution but we also know just from what math would tell us that an additional solution would be the sum of those two solutions and so this is the form we're going to work with where we're basically encompassing both s 1 and s 2 and just differentiating that these coefficients a 1 and a 2 are going to be distinct in generally speaking okay but now the actual form of the response then we can actually break down to being three different forms based on the fact that in taking the roots of the characteristic equation these roots s 1 and s 2 themselves might be have a slightly different form themselves so there's going to be three different cases that we're going to examine that could we could possibly encounter so in one case again talking about the roots roots s1 and s2 that could both be real and distinct and this would be what we call the over damped response okay the second option is that both of the roots are actually complex and still distinct or a net well they're distinct but specifically they'll actually be conjugates of one another thinking about two our complex math and this will indicate as being so called under damped response and then the third and final option that we could have is to say that both s1 and s2 are actually the real values but they're also equal okay and this will be indicated as the so-called critically damped response okay so again this is just indicating that because we're dealing with the characteristic equation we're taking the roots of that characteristic equation those roots themselves could be either real or complex and so based on what the conditions are that we'll fall into one of these three categories and tell us about the type of response over damped under damped and critically damped that we might possibly have so now let's get into kind of how do we evaluate if we're looking at any given circuit how will we know which one of these specific damping responses would would be in without having to necessarily always solve for s 1 and s 2 specifically they're actually a little bit of a shortcut we can look at to determine which one of these categories we kind of fall into ok so now for finding out which of the which of those three different types of responses that we have what we can do is compare two frequencies we're going to define Omega naught and alpha so Omega naught is known as the resonant Radian frequency okay and then alpha here is the so called Knepper frequency so each of these two frequencies actually come into play when we're talking about the roots because the roots will basically what we're doing is making some substitution here in the those of character the roots of the characteristic equation so s 1 & 2 we could basically make these substitutions as minus alpha plus or minus the square root of alpha squared minus Omega naught squared so then you can see if you compare getting your notes back to what we found for these roots we're basically just plugging in the values that we had in those coefficients for what we have now is alpha and omega not so specifically in the parallel circuit case that means that alpha is just 1 over 2 times RC and Omega naught is equal to 1 over the root of LC and then for the series case we come up with alpha being slightly different here of R over to L but Omega naught is actually the same quantity here 1 over root L C so this isn't just one minor difference and so what we'll see then is if we compare Omega nought to alpha whether or not one is bigger than the other or whether or not they're equal that's going to tell us which of the three types of damping responses we will actually have and then which form of the solution that we can use to evaluate what we have and again see here that there is still very much a lot of similarities between the parallel circuit case and the series circuit case the only difference that we really see is between the two forms that alpha takes is where you kind of have to draw your distinction as to understand are you looking at a series RLC circuit or are you looking at a parallel RLC circuit and then you need to apply one of these two forms of these frequencies to come up with a value now just one other note about these two frequencies in terms of the it's the units are because these are both angular frequencies they're in terms of radians per second rather than just Hertz or inverse seconds okay so let me erase this then we'll kind of go through just evaluating based on the two these two frequencies how they correlate will tell us the type of damper in response then let's look at the form of the solution that each of these takes alright so to know again which of these three types of responses that we fall into we're just going to take a comparison between those two frequencies that we just defined so for the over damped case this would be if Omega nought squared is less than alpha squared then we would know that we have an over damped response case for the under damped response that would just be the exact opposite Omega naught squared is greater than alpha squared and for the critically damped response this would come into play if these two frequencies are exactly equivalent okay so now if we've defined which type of response that we have we had a general form of the solution but what we actually find is because the roots in each of these cases were a little bit different remember the over damped response we're talking about we having real and distinct roots the under damped we're dealing with complex quantities and critically damped still real but they're equal that kind of comes into play then for the general form of the solution that we have to work with in each case and what it works out to be is as follows so if we're talking about in the over damped case here we're basically left with what we already saw previously where let's say the voltage is a function of time again we could be talking about current and in either case is still just the sum of these two exponential terms with the coefficients a 1 and a 2 okay so that one's basically what we saw before but now for the under damped case we're dealing with complex roots so that changes the form a little bit so here the voltage would take the form I'm going to change the coefficients just indicate these are a little bit different so we'd have B 1 times exponential minus alpha T times cosine of Omega D times T we're going to want to define what this Omega D that's a different frequency that we're going to include now I'll define that in a minute plus second term which is going to include a second coefficient B two times the same exponential minus alpha T times the sine of Omega DT so this Omega D is indicated or noted as the damping frequency and is defined as such Omega D is equal to the square root of alpha squared minus apologizing alpha squared Omega naught squared minus alpha squared okay so basically we would if we have already solved for what Omega naught and alpha are we could use that to solve for Omega D our damping frequency and plug that into the form of the solution that we have in the under damped case and then finally for the critically damped case the form of the solution would be as follows so the voltage is going to be a different coefficient here just D 1 times T with an exponential minus alpha t plus D 2 with the exponential minus alpha T okay so these each of these three solutions are what we should basically end up with based on you know and again remember that Omega not an alpha are going to be based on your elements your RL and C quantities so if I make a change to maybe my resistance that may change the type of damping response that I have and therefore that's going to change which of these solutions that I need to be working with okay so that basically gets us to what we wanted to have in the end but now the only kind of question left is well certainly we know how to find the roots s 1 and s 2 we know how to find alpha we know how to find Omega D but the what we don't quite know or we haven't discussed yet is what these coefficients a 1 and a 2 B 1 B 2 D 1 and D 2 are so what kind of map out how we get to that but the basic idea here is looking at you'll conditions okay so if we think about maybe in the overdamped case if I'm talking about the voltage at time T equals zero well that means I plug in zero to time T here well that means both of these Exponential's go to 1 which means I'm left with the voltage at time T 0 is equal to a 1 plus a 2 so then I would want to look at my circuit okay if I maybe have the parallel RLC circuit as such and maybe the voltage again here indicated as across the capacitor C well that initial voltage V of 0 if I have some initial voltage here that is what I would use to set that equal to a 1 and plus a 2 so that's the general idea the second initial condition that we'll need because we have two unknowns here we need to in two equations is to actually take the derivative of one of these of any one of these response equations after taking the derivative set time T equals zero and again solve for what you have left that will get you a second equation to solve for your second unknown and in that case you'll just need to know what the the change the DV DT or di DT depending on the case is a time T equals zero in your specific circuit okay so I clean this up and again we'll kind of evaluate specifically how to find those coefficients for each of these three different cases ok so again we're talking about now how am I going to go about finding the coefficients a 1 and a 2 or B 1 B 2 or D 1 and D 2 depending on the type of response that I have and again I just said that for doing this basically we're going to take that form the solution that we just had plug in T equals zero we're left in this in the over damped case that the voltage at time T equals zero is simply a 1 plus a 2 so again if we look at our whatever the circuit that we had might be evaluate what that voltage is at time T 0 that we can plug that in here the second one then is to say what is DV the derivative of the voltage at time T equals zero so again taking that same equation taking the derivative plug in T equals zero this ends up in this case just being equal to a 1 times s 1 plus a 2 times s 2 ok now again here for the finding the derivative of the voltage as a function of time in my circuit well here I might want to look at in this case let's say the current through my inductor let's say could tell me something about the derivative of the voltage again just looking at the standard relationships for inductance or if we're doing the opposite if we had to find di DT we might look to the capacitor because we could evaluate what di DT might be in that case okay so now for the under damped case again if we just plug in T equals zero to the solution that we just had up on the board here we would just have this would tell us immediately what our first coefficient b1 was and then taking the derivative of that function and again setting equal to zero would leave me with minus alpha times b1 plus omega D times b2 okay so that would be the second equation that we would need to solve for both of those coefficients and in the case the critically damped response what we'd have is the voltage at time zero T equals zero it's going to tell us d2 and then looking at the derivative of the voltage at time T zero is going to be d 1 minus alpha times d2 okay so again it's still a process of you know you need to know initially what the type of response that you have is that tells you the form of the solution and then these relationships are going to what you use to help you to evaluate what your coefficients are that would then allow you to write the final solution and if you're curious I do have some other videos and look at specific problems with RLC circuits that will maybe tell you a little bit more specifically how we can go about finding these coefficients but this is what we can use in general okay so let me finally wrap up with just one other topic here defining or looking at really briefly the difference in the step response versus the natural response here where step response means what I'm when I'm asking the case of if I were to add in a voltage source let's say in the series circuit or maybe a current source in the parallel circuit well that's going to have an impact then on the forms of the solution but actually we see that it all works out to be quite even wirless so I'll just make a couple of quick comments about that and then we'll be done for this video okay so now just a couple of quick comments about the opposite the alternative case that we could have we're just talking about natural response that's where we're looking at both edges and currents that are going to decay down to zero the opposite case that we could be evaluating is a so-called step response case this would where would be where we're actually charging up a capacitor or an inductor in an RLC configuration again we could have a parallel combination or the series combination here I've just kind of evaluated maybe I have different sources and there's going to be some switch so let's say time T equals zero we're going to close that switch that's going to start to charge up my inductor capacitor with this current source or with this voltage source here in the series circuit case so again we won't go through the whole process that we just did for the natural response case but the idea is really still the exact same in that in thinking about the parallel circuit case if we applied KCl at this top node evaluate all the currents and still be the same the only difference now is I have a current here I'll just indicate this as being the current I let's say and what this would tell me then is if I do the same process and evaluate my KCl equation do the same kind of manipulations taking the derivative dividing by the capacitance see what I end up with is the second order second order differential equation that looks like this we're doing d squared IL DT squared plus 1 over RC times first derivative in here roots we were specifically kind of evaluating in terms of the current il so 1 over RC and then plus I L over L C equals I the current I from the source over LC so you see here that the form of this equation is more or less the same the only difference is that I'm no longer equal to zero on this side because of the fact that I now have a actual current source in the mix and I could do the same thing for my series circuit now here I would be applying more of a KVL approach but kind of coming out to again a very similar form to what we had previously the only difference is going to be that I have to add in this voltage V right here so I got the second derivative term plus R over L DV C if we're evaluating with respect to the voltage on the capacitor plus VC over LC all this equals V over L times C ok so now there's a whole process again we could go through for evaluate evaluating the forms of the solution here it's a little bit more tricky in terms of this differential equation because it's don't longer equal to 0 it's actually equal to a constant value over here so the math involved is a little bit more involved but the main idea if we were to go through that whole process we would actually see that for the step response the form of any of the solutions let's say for the first one for the parallel circuit case here that il of T would be equal to whatever let's say I final was so here we were talking about the current through the inductor here we would evaluate what should be the final value of the current through the inductor and then add on to that the form of the natural response okay and then similarly if I were thinking about the series circuit case so I would say VC of T it's going to be equal to whatever the final was here I'm talking about the voltage across the capacitor what's the final voltage there and add it on to that would be whatever the form of the natural response dictated okay again I'm not going through the full analysis how we get to that but again this is basically saying that you can take everything we just talked about and how you find the natural response and I could do the same thing in these step response cases and said well the only difference I have to do is add on to that the final value of each of these now one other difference will be when you're doing the natural response those initial conditions you know might be a little bit different and also you'll need to think about that when I'm taking like let's say the current at time T equals zero well that's obviously going to be whatever I F happens to be plus then maybe your coefficients that you had over here and then if I was taking the derivative of this di DT evaluating at time T equals zero again in this case taking the derivative here the final value would cancel out so I want to have to worry about that there they just need to pay attention to that again when you're thinking about doing finding those coefficients for these natural response cases okay so again I hope that just gives you a real broad idea of the general process we can follow for any kind of RLC specific to the parallel and series case again if you want to see some examples I have some other videos posted up that work through some actual problems here this is just to give you an idea of how we can approach any any given general problem alright so I hope you enjoyed that and I look forward to seeing you on the next video

Channel: Dr. Montgomery

Views: 203,630

Rating: 4.8758621 out of 5

Keywords: education, university of california davis, engineering, circuits, eng17

Id: 61t8Z6ba1vM

Length: 37min 7sec (2227 seconds)

Published: Thu Jun 04 2015