Derive Peak Current in RLC circuit charging a capacitor - Page 1
Forum section: electronics / beginners

author: AQUAMAN
author_rank: Regular Contributor
timestamp: May 15, 2015, 11:52:41 am
content: If I have a DC voltage source charging a capacitor, can anyone guide me how to derive a formula for the peak inrush current into the capacitor?I have made a peak detector circuit that detects the peak current when charging 10nF-1uF capacitors and would like to have some theory to explain changes in the peak that I can see in real lifeLets say there is a step voltage from 0-25V, R=1.25Ohms, L=50nH and C=50nFHave found some interesting slides here:http://www.ece.sunysb.edu/~oe/Leon/ESE271S13/Lecture23.pdfBut really my maths is not that good

author: T3sl4co1l
author_rank: Super Contributor
timestamp: May 15, 2015, 01:17:37 pm
content: As in step by step?A probably lengthy route would be to solve the time-domain differential equation and find the first local maxima, but it would be, in a certain sense, the most provably accurate.Equivalent, and usually a little easier, is dragging it through Laplace, including the initial condition terms (which in this case will be set to zero, since you're asking the transient from zero to a final condition), and ending up in the same place.If you really just want to know the answer, you should end up at either Ipk ~= V / R or V / sqrt(L/C), depending on Q factor (i.e., the ratio between sqrt(L/C) and R).Tim

author: The Electrician
author_rank: Frequent Contributor
timestamp: May 15, 2015, 08:27:03 pm
content: You haven't shown a switch in your circuit. What are you using for making the initial connection? A mechanical switch, a solid state device? If you're using a mechanical switch, there could be some sparking at closure, which will change the peak current.

author: AQUAMAN
author_rank: Regular Contributor
timestamp: May 16, 2015, 06:40:33 pm
content: The switch is a MOSFET, it is insignificant in my opinionQuote from: T3sl4co1l on May 15, 2015, 01:17:37 pmI have been using Ipk ~= V / R so farI know that the inductance and the capacitance is constant (although I dont know their exact values - but they ARE constant), the only things that are changing are the voltage source and the resistance.Basically I know peak current, and I measure the exact voltage on the voltage source.The only thing I don't know is the resistance.Up until now, just using Ipk ~= V / R seems logical and roughly correct, and it probably is. But it would be nice to have a better approximation of it.

author: Zero999
author_rank: Super Contributor
timestamp: May 16, 2015, 08:08:35 pm
content: I wasn't much good at maths either and never properly learned how to do this at the crappy college I attended. However I have a good understanding of how the circuit works.IPEEKwill be near V/R as the circuit is over damped so the effect of the inductance is negligible.Try changing the inductor to 50mH and you'll see it's totally different.

author: IanB
author_rank: Super Contributor
timestamp: May 16, 2015, 08:29:03 pm
content: Edit: Oops, I goofed on the solution. Redacted until I figure it out. The analysis at least, looks like this:

author: The Electrician
author_rank: Frequent Contributor
timestamp: May 16, 2015, 10:11:43 pm
content: Here's a formula for the underdamped case:

author: AQUAMAN
author_rank: Regular Contributor
timestamp: May 17, 2015, 02:44:32 am
content: Quote from: Hero999 on May 16, 2015, 08:08:35 pmYes I think that me and you are in the same boat.I have assumed that Ipeak is almost the same as V/R.I have neglected I and C because I know that they do not change. The only changes are in the V and the R.My inductance is more like 50-100nH!!The peak current occurs around 50-200ns after the voltage is applied

author: IanB
author_rank: Super Contributor
timestamp: May 17, 2015, 04:58:15 am
content: Here is what the current vs time looks like with the values you gave in the first post.Vs = 25 V at t = 0, R = 1.25 ohms, L = 50 nH, C = 50 nFYou can see that the peak current is about 12 A.This compares with V/R = 25/1.25 = 20 A.The difference is quite significant.

author: IanB
author_rank: Super Contributor
timestamp: May 17, 2015, 05:21:04 am
content: You can generate the graph of the response using the following formula:

author: IanB
author_rank: Super Contributor
timestamp: May 17, 2015, 06:07:40 am
content: Quote from: Hero999 on May 16, 2015, 08:08:35 pmIt turns out the damping ratio found from R / [2 sqrt(L/C)] is 1.25 / [2 * sqrt(50/50)] = 0.625Since the damping ratio is less than unity the circuit is actually under damped in this case.

author: AQUAMAN
author_rank: Regular Contributor
timestamp: May 18, 2015, 08:27:19 am
content: Quote from: IanB on May 17, 2015, 05:21:04 amThank you for all your help, I am beginning to understand betterI am comfortable with calculating the expression for I(t) and then differentiating it and solving to = 0 to find the maximum.This book helped me a lot too:http://faculty.weber.edu/snaik/EE2260/alexander_sadiku_fundamentals_of_electric_circuits_4thed.pdf,page 333Really now what I want is how to justify that if I know both the peak current and voltage swing, I can use R=V/I to approximate small changes (mOhm level) in the resistance in the circuit.This is what I was really doing - trying to get an expression for the peak current so that I could justify removing I and C and just using R=V/II now seem to understand that this is more-so the case when the circuit is overdamped. What is the explanation for why when the circuit is overdamped that the inductance has little influence on the peak current so that you can approximate it with V=IR? Obviously it is something to do with the damping factor.

author: AQUAMAN
author_rank: Regular Contributor
timestamp: May 18, 2015, 09:46:36 am
content: Here I have attached simulation of an RLC circuitI have also plotted the real resistance in the circuit, as well as the 'calcresistance' where I have estimated it by dividing the voltage by the peak current (V=IR ohms law).I have also plotted the 'ratio' between theseAs we can see, when the resistance is small (circuit underdamped), my calcresistance using V=IR is several times the actual resistance in the circuit. However as the resistance increases (circuit overdamped) this ratio tends closer and closer to 1Really I need to find a concise explanation for this.I have since found in this thread that using V=IR to approximate small changes in the resistance is only valid for overdamped circuit configurations (ie the inductance must be kept very low).This fact might also explain why I can't explain some of my results - some of my circuits were underdamped and using V=IR isn't that accurate in this case.

author: IanB
author_rank: Super Contributor
timestamp: May 18, 2015, 02:54:15 pm
content: Quote from: AQUAMAN on May 18, 2015, 08:27:19 amAs the inductance approaches zero the system becomes ever closer to a first order RC circuit. In a first order system the peak current occurs at time zero and is equal to V/R.