Quick Impedance Question - Page 1
Forum section: electronics / beginners

author: mart1n
author_rank: Contributor
timestamp: February 06, 2017, 10:23:57 am
content: Hi EEVBlog,I've been working on wrapping my noodle around impedance and I've got a question or two...With a simple DC circuit, if resistance goes up, more voltage would be required to achieve the same current flow...Question #1With a AC circuit and a capacitor, a lower frequency will have a higher impedance and a higher frequency a lower impedance (Is that correct?)...can we also increase voltage to overcome the impedance and achieve the same current flow?Question #2When people/books/texts say that a capacitor resists changes in voltage and an inductor resists changes in current...Does that just mean...if there is a drop in voltage that the cap will supply it from it's stored charge? Like when adding a cap in parallel to the VCC/GND pins of an IC?Thanks for taking the time to help.

author: capt bullshot
author_rank: Super Contributor
timestamp: February 06, 2017, 10:31:32 am
content: Quote from: mart1n on February 06, 2017, 10:23:57 amFirst one: yes, second one: yes (assumed you speak of lowered frequency and increased voltage)Quote from: mart1n on February 06, 2017, 10:23:57 amYes

author: Vtile
author_rank: Super Contributor
timestamp: February 06, 2017, 10:57:47 am
content: I'm not correct person to give you the answer of these consept, I'm merely trying to understand them myself, but..First of the ideal inductance and capacitance is not measured in impedance Z, but reactance X (which in some old literature is written in form of reactive resistance if that is more expressive term).The definitions for ideal devices are in steady state sinusoidal models. Where L is Henries and C is FaradsInductance (Inductor) = L*wCapacitance (Capacitor) = 1/(C*w)Where w is angular velocity (2*Pi*frequenzy) in form of radians per second (revolutions per second) where frequenzy is 1/T, where T is time of cycle in seconds.This is the vertical part of the impedance if we stay on geometric algebra (sin, cos and pythagoran theorem).The impedance you were referring to is what you really get with the real devices.It is combination of typical resistance and reactance.https://upload.wikimedia.org/wikipedia/commons/thumb/c/c4/Complex_Impedance.svg/200px-Complex_Impedance.svg.pngThe so called ESR is what creates the horizontal part of impedance.Impedance in circuit calculus is then handled as resistance.So based on these models your assumptions seems to be correct (like answered while I were slowly writing this) atleast for question #1. For Question #2 I only can say that that is also my understanding of the work done. The difference in components can be seen on the Henry and Farad definitions: F=(A/V)*s, H=(V/A)*s. Both components store energy, so they fill the power depletion in given amount of time (as long as there is still energy stored in magnetic field or electric field).Now when we enter the high frequenzy world these models deteriorate (per my current understanding) while we climp up the frequenzy.PS. The charge, is under depate.Edit: Some corrections and correction of inductance, impedance mix.

author: TimFox
author_rank: Super Contributor
timestamp: February 06, 2017, 04:07:25 pm
content: The first thing I was taught in electronics class about reactance (ca. 1968):"You cannot change the voltage across a capacitor instantaneously, and you cannot change the current through an inductor instantaneously."This maxim is a good place to start understanding the function of reactances.

author: Benta
author_rank: Super Contributor
timestamp: February 06, 2017, 04:15:39 pm
content: Quote from: TimFox on February 06, 2017, 04:07:25 pmYep, it can't be expressed much shorter and to the point, and is easy to memorize.

author: orolo
author_rank: Frequent Contributor
timestamp: February 07, 2017, 07:29:51 am
content: Quote from: TimFox on February 06, 2017, 04:07:25 pmBut if I connect an ideal charged capacitor to an ideal grounded switch, it will discharge instantly upon closing the switch.

author: bson
author_rank: Unknown
timestamp: February 07, 2017, 07:38:00 am
content: Quote from: orolo on February 07, 2017, 07:29:51 amOnly because you're dealing with ideal components that can handle infinite currents.

author: capt bullshot
author_rank: Super Contributor
timestamp: February 07, 2017, 07:39:19 am
content: Quote from: orolo on February 07, 2017, 07:29:51 amCongratulations, you've just invented the dirac current pulse generator. I want one, please

author: mikerj
author_rank: Super Contributor
timestamp: February 07, 2017, 08:38:56 am
content: Quote from: orolo on February 07, 2017, 07:29:51 amYou'll also need some ideal wire that has no resistance or inductance to connect the switch and cap together.

author: orolo
author_rank: Frequent Contributor
timestamp: February 07, 2017, 09:33:23 am
content: Quote from: bson on February 07, 2017, 07:38:00 amAh, so the opposition to changes in voltage, which is the ideal definition of capacitance, stems from nonidealities. Therefore, if I put a resistor connected to a nonideal switch, with nonideal wire, and close the switch, the voltage doesn't change at once. From there I infer the resistor is a capacitor.

author: mikerj
author_rank: Super Contributor
timestamp: February 07, 2017, 10:16:57 am
content: Quote from: orolo on February 07, 2017, 09:33:23 amIt stems from reality. You should try it

author: orolo
author_rank: Frequent Contributor
timestamp: February 07, 2017, 10:45:00 am
content: Quote from: mikerj on February 07, 2017, 10:16:57 amI'm not convinced. A good ideal definition should suffice. Everyone knows the definition of an ideal op-amp, or an ideal filter, and they are consistent. I'm surprised that such definition of capacitance requires nonideal considerations. Even more, there is a non sequitur: A cap is defined as a component that resists steps in voltage. Then, steps in voltage are defined as "unreal". From there, anything follows.Edit:Okay, seriously. Capacitance has nothing to do with the impossibility of step changes in voltage. Let's formalize this.The right ideal definition of a capacitor is a device governed by the linear equation I = C·dV/dt. If we apply an ideal step pulse, \$ V = V_0\cdot H(t)\$, for H the heaviside function, as a poster above said, \$ I = C\cdot V_0\cdot\delta(x) \$. We get an infinite current pulse with total charge \$ Q = CV_0 \$. This is 100% consistent in the mathematical and logical sense. In the real world, the previous computation can be repeated changing H(x) for an arbitrarily steep voltage transition, and the result will be an arbitrarily short current pulse. The only limitations are given by the quality of the switch, the parasitics in the circuit, and the propagation speed of the charges and signals. All these factors are external to the definition of capacitance, and apply to other components as well.So, to summarize:1. If we are talking about an ideal situation, there is no mathematical nor logical obstacle for a capacitor to experience step voltage changes.2. If we are talking about a real situation, a capacitor can experience arbitrarily fast voltage transitions (steps in the limit), the only limitations being factors external to the definition of capacitance.That's the reason I think the "volage doesn't change instantaneously" definition of capacitance is highly misleading. As far as it's ideal it's wrong, and as far as it's real it's a non-definition. It's akin to defining a massive body in newtonian mechanincs as being impossible to move instantly: ideally you can, in reality you can come as close as your means allow.

author: TimFox
author_rank: Super Contributor
timestamp: February 07, 2017, 02:25:50 pm
content: The circuit including the ideal capacitor and ideal switch will still have inductance because of its finite area. Unless your ideal components are also Euclidean points that coincide. Without resistance, closing the switch will cause a ringing resonant circuit which will radiate energy away from the circuit. The energy stored is always (1/2) C V^2 , which will occur even in the limit as the resistance goes to zero.

author: Vtile
author_rank: Super Contributor
timestamp: February 07, 2017, 03:00:02 pm
content: Quote from: mikerj on February 07, 2017, 08:38:56 amBatteroo+ now with ideal wire.Tbh. I don't follow the ideal joke, did I manage to start it. :/

author: djnz
author_rank: Regular Contributor
timestamp: February 07, 2017, 04:37:03 pm
content: I think impedance becomes a lot more clear when you look at it as both a complex number and the real world behavior that that number captures. Staying away from the math is not a very good idea here. Plus, once you become comfortable with the math, all kinds of cool and practically significant stuff (stabilty analysis etc.) becomes accessible.

author: Benta
author_rank: Super Contributor
timestamp: February 07, 2017, 05:52:00 pm
content: Could you guys stop the pissing contest? This is the "Beginners" forum.We can all agree that real-life capacitors, inductors or resistors are not ideal.I still think that the reply from TimFox in reply #3 is the most helpful for beginner understanding.All the rest is for another discussion.

author: rstofer
author_rank: Super Contributor
timestamp: February 07, 2017, 06:42:29 pm
content: Quote from: djnz on February 07, 2017, 04:37:03 pmThis is 'electronics', you simply can't avoid the math. Engineering IS math. A lot of insight, but mostly math.One of the best around here for doing the math is Orolo - when he isn't messing with people. Note that he did present the definitive equation for capacitance I=C dv/dt. Right off the bat, we need calculus or at least some way to express dv/dt (change in voltage divided by change in time). And we're usually talking about dt (change in time) as approaching 0. If dt approaches 0, dv/dt approaches infinity even for small voltage changes. Thus the current also approaches infinity as dt approaches 0. Switching in zero time takes infinite energy.The thing is, that equation is in the time domain and not entirely useful in the frequency domain where we want impedances in either complex or vector notation. Again, it's the math. Complex numbers (and especially matrices of complex numbers) are a PITA absent a computer solver. Complex numbers are easy to add, more difficult to multiply. Vectors are easy to multiply, more difficult to add. So, we convert back and forth - often!Reactance of a capacitor Xc= 1 / (2*pi*f*C) for frequency (f) in Hertz and capacitance (C) in farads. This is the equation we want, in most cases. But it is NOT the impedance of the capacitor.Note the difference between reactance and impedance in the attached:http://www.allaboutcircuits.com/tools/capacitor-impedance-calculator/QuoteAnd that's how we get to complex numbers. Just throw in a -j and everything gets complex.

author: orolo
author_rank: Frequent Contributor
timestamp: February 07, 2017, 06:55:19 pm
content: My apologies for the trolling. I got carried away.

author: Vtile
author_rank: Super Contributor
timestamp: February 07, 2017, 07:01:51 pm
content: I'll want to add that complex numbers are not complex and there is nothing imaginary on them! They are just a bit different what "old & ancient" mathematicians were used to so they pulled these terms to descripe their annoyance (add some idiom here).

author: rstofer
author_rank: Super Contributor
timestamp: February 07, 2017, 07:56:51 pm
content: Imaginary numbers and, hence, complex numbers, were first described around 70 AD but it wasn't until the late 1700's and early 1800's that they became accepted as anything other than some odd thing that resulted from taking the square root of a negative number.https://en.wikipedia.org/wiki/Imaginary_numberIt took mathematicians 1800 years to understand complex numbers and I was expected to learn them in a small fraction of a semester!In any event, 'complex' doesn't mean 'difficult', it simply means that there is a real component and an imaginary component. And 'imaginary' doesn't mean 'ficticious', it just means a number in a different plane.For the EE, math is everything!