Inductor testing frequency - Page 1
Forum section: electronics / beginners

author: pete_dl
author_rank: Newbie
timestamp: August 19, 2018, 09:00:50 pm
content: Hi,I want to measure the precise (within 1%) inductance of an inductor sold as 5H. I am looking at hand held LCR meters and the lowest frequency they commonly test at is 100hz. I will be using the inductor in a LC circuit with resonance at 20hz. I have read that inductance does change with frequency, but I would like some advice whether a 5H inductor measured at 100hz will indicate a much different figure at 20hz ?Many thanks,Peter.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: August 19, 2018, 10:09:37 pm
content: Yes. Typically, the equivalent circuit of an inductor is:( DCR + Lnom ) || Lcore || Zwinding( '||' means "in parallel with", and '+' means "in series with".)Lnom is an ideal inductor, the nominal value. Lcore is a lossy inductor corresponding to eddy current losses in the core. Zwinding is an RC element (or higher order networks) representing the capacitance and loss of the winding.DCR + Lnom gives the LF cutoff, where it goes from resistive to inductive. Lcore and Rwinding give the peak impedance (parallel resonance at some midband frequency). Cwinding dominates above the peak, then at high frequencies, Zwinding dominates (if it's just RC, then just Rwinding; more likely it's some complex peaks and dips, which we try to not care about).The equivalent inductance of that network, will vary with frequency. At ~DC, the inductance is large and complex (which is to say, DCR dominates). In mid band, inductance is usually falling gently (due to core loss) -- how much depends on how good the inductor is. Above the impedance peak, inductance is negative (i.e., it's capacitive instead).Assuming you're talking about an iron cored inductor, I would be shocked if you can even find one that is that precise, and I strongly suspect L will depend on frequency enough that doing it at 20Hz will give a different result.You may well get lucky in that the DCR and core loss effects cancel out, and you get the right inductance by coincidence. But that means your measurement is part inductance, part DCR and part core loss, and all those sources of error are in your measurement.What do you need an inductor of that value and tolerance for, anyway?Tim

author: pete_dl
author_rank: Newbie
timestamp: August 19, 2018, 10:50:51 pm
content: Thanks for the reply tim.I am trying to design a very energy efficient circuit to drive a piezoelectric actuator with a 20hz sine wave. I have managed to drive the actuator with a DAC and opamp, but i want to get more energy efficient.I read papers describing energy recovery schemes that dumps the stored charge from the piezoelectric actuator into an inductor in order to create the other half of the sine wave. basically a LC resonator with the actuator being the capacitor.What I was hoping was I could buy an inductor of roughly 5H, measure it accurately using a meter, then tweak the capacitance of the circuit with extra capacitors in order to achieve the 20hz resonance. I calculated that I need to measure the capacitance and inductance to within 2.5% so as to get the resonance within 1hz of my target.If I cant rely on an inductance meter, then should i tweak the capacitance to maximise how long 1 full charge will ring for in the LC circuit with a scope ?Pete

author: T3sl4co1l
author_rank: Super Contributor
timestamp: August 20, 2018, 12:27:02 am
content: Just use a class D audio amp.Tim

author: pete_dl
author_rank: Newbie
timestamp: August 20, 2018, 07:56:18 am
content: A class D amp would be better than my opamp. However, energy will be lost in the low pass filter and also the energy stored in the piezoelectric actuator will also be lost each cycle. The LC resonator seems to be ideal (with my very limited knowledge of electronics) but I have not calculated how much will be lost in the resistance of the components.Pete

author: T3sl4co1l
author_rank: Super Contributor
timestamp: August 20, 2018, 06:27:34 pm
content: Quote from: pete_dl on August 20, 2018, 07:56:18 amLost where? Do you have a proof of that?Yeah, I suspect a, say, 95% efficient (or better) class D amp is not only orders of magnitude smaller, but more efficient overall, than using an inductor. It's a good solution.Tim

author: pete_dl
author_rank: Newbie
timestamp: August 25, 2018, 02:37:58 pm
content: I think I have to swallow my pride and admit defeat with the idea of using an LC resonator to drive the piezoelectric actuator.The 20hz frequency dictates huge inductors and capacitors. Their resistance will mean the circuit is grossly inefficient.So I will build a class D amp. Any neat way to recover the energy stored in the piezoelectric actuator ? Otherwise the stored charge on the actuator will get shorted to 0V when the PWM switches low.I imagine this does happen when a class D amp drives a loud speaker, but that inefficiency is ignored ? The momentum of the speaker is going to create back emf (not sure that is the right term), that will get dissipated as heat in the amplifier ?Pete.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: August 25, 2018, 02:43:01 pm
content: Quote from: pete_dl on August 25, 2018, 02:37:58 pmSpeakers have DCR ~= motional impedance. Poor example. But, for what reactive power remains, yes, it is recycled.Why not do some simulations, or good old pad-and-pen analysis, and figure this out yourself?Only you can say how efficient the electromechanical coupling is, but I would at least guess it's going to be better than for a speaker.Tim

author: pete_dl
author_rank: Newbie
timestamp: August 25, 2018, 04:03:57 pm
content: >Why not do some simulations, or good old pad-and-pen analysis, and figure this out yourself?Well the answer is in the title of the forumI am a beginner in electronics and a push in the right direction from experts will be invaluable.So the potential energy in a speaker coil/cone momentum is recycled by the class D amp ? The amp puts energy into the speaker to accelerate it forward (for example). My understanding in speakers is that the direction of travel is not reversed by a physical spring, otherwise the speaker will have a very undesirable strong resonance. Without some sort of spring, the amp will have to expend energy to decelerate the coil/cone and reverse the direction of movement.At the moment i cant visualize how the class D amp is recycling the kinetic energy. Isnt it continuously spending energy to accelerate and decelerate the coil/cone ?Thanks,Pete.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: August 26, 2018, 02:42:01 am
content: Quote from: pete_dl on August 25, 2018, 04:03:57 pmWell... what would you need to know, to figure it out?The definition of power:P = V*IThe instantaneous product of voltage and current. Power varies over time. For example, a sine wave V = Vpk sin(ωt), and similarly for I, has P varying as a product of sines, which is also a sine at twice the frequency, due to a trig identity.So, you need algebra and either trig or geometry (for sine waves) or calculus (for general waves) to work it through.We don't normally measure or work with instantaneous power, directly; normally, average power is what we mean. That is, the average level of P.Then you'll need the definition of the resistance, inductance and capacitance.We could use the most general definition (instantaneous values), but this will be much easier for the sinusoidal case. That is, at a given frequency F of a sine wave, the properties are:Resistor: V = I RInductor: X = 2 π F LCapacitor: X = -1 / (2 π F C)X is pure reactance, and does not dissipate power. R is pure resistance, and dissipates only power.Reactance works the same as resistance, but rotated 90 degrees. The angle only matters when you put X and R together (i.e., a lossy inductor or capacitor). In that case, you construct a right triangle, with side lengths R and X, and the length of the hypotenuse gives the impedance Z. You can measure this by applying a current and measuring the voltage (V = I Z).Doing this by geometry would get rather complicated to handle, and so far, you've learned nothing about how power works with an impedance.By convention, we write down Z as its two parts: Z = R + i X. Think of 'i' as a label denoting reactance, and then the unlabeled part is resistance. Like you might label points on graph paper with x and y coordinates, saying P = 10x + 20y, or dropping one label because it's redundant and labeling just the other, P = 10 + 20i.i also has the property that i * i = -1. So, it participates in algebra, and anywhere you have products, it can be simplified. Let's do a multiplication, then a division.You wouldn't normally multiply impedances, willy nilly. For context, let's say we're solving for two impedances in parallel. The same parallel formula applies as always:Ztot = Z1 * Z2 / (Z1 + Z2)This has a product in the numerator, and is divided by the sum.Start by labeling the components...Z1 = a + biZ2 = c + diZtot = (some resistance part) + i (some reactance part)We always write the result in this separated form, because it's easier to work with.First, the denominator. This is easy, nothing happens, it's just all written together. We still group the terms, just by convention.Zdenom = (a + bi) + (c + di)Zdenom = (a + c) + i (b + d)Now the product. The numerator is:Znum = Z1 * Z2Znum = (a + bi) * (c + di)Znum = ac + bdi^2 + bci + adiWe're just distributing terms. Simplify i^2 = -1 and group,Znum = (ac - bd) + (bc + ad)iAnd that's multiplication!To divide, we use a trick. Aside: consider,(x + yi) * (x - yi)what does this simplify to?Use the above result, substituting a = x, b = y, c = x and d = -y.(x + yi) * (x - yi) = (xx - yy) + (yx - xy)i = x^2 + y^2the i part cancels out!Why is this important? It means we don't need to come up with some weird definition to divide by this composite number -- we can reduce the problem to multiplication by a well-spotted term, then division by pure, ordinary variables.So we can finally write the quotient,Zquot = Z1 / Z2again we use the same Z1 and Z2 for show. We must later substitute Znum and Zdenom for a, b, c and d, to get Ztot. It'll be easier doing it later, than dragging all those terms along right now!Zquot = (a + bi) / (c + di)Note the denominator is c + di, so multiply top and bottom by (c - di) and simplify.Zquot = (a + bi) (c - di) / [ (c + di) (c - di) ]Zquot = [ (ac + bd) + (bc - ad)i ] / (c^2 + d^2)Zquot = (ac + bd) / (c^2 + d^2) + i(bc - ad) / (c^2 + d^2)Okay, that last step might be less simplified, but the point is to show it's always separable into two terms, a resistance part (no i) and a reactance part (only one i).Incidentally, c^2 + d^2 = hypotenuse^2 is the Pythagorean theorem. So what's actually being done is dividing by the length of the hypotenuse, of that triangle we constructed -- the divisor's triangle. Well, dividing by it twice. (Units must always be balanced -- we're multiplying on top, so we're dividing on bottom. This is a good way to spot algebra mistakes!)So, division is just like multiplication, except we flip a sign on one reactance term (bc - ad instead of bc + ad), and divide by the squared length of the divisor.And hey, congrats, you just derived a definition, for a fundamental arithmetic operator, in a new(?) algebra system!By the way, these impedance numbers, or composite numbers, are actually called "complex numbers", and i is called the "imaginary unit". (A number or term without 'i' is called "real", and a part with, is called "imaginary".) I hate these terms because they make people think there's something intangible or difficult or cryptic about them. They're none of the sort. Well, "complex" just in the observational sense, sure -- you can see it's a lot longer, written out, than a mere "a * b" is! -- but abstracting that away as the multiplication or division that it is, it's no different. (Indeed, this system is more general, because you can take the sqrt of any number, positive or negative, and always get a valid and consistent result. Yes, your calculator lied to you: the sqrt of a negative number is not error; it exists, it's just written with a little i in front!)(Also, to be perfectly clear, "composite numbers" are actually their own thing, in number theory. I was just using the term descriptively above. Yes, math, like any other domain, has its fair share of badly named things...)So, after that introduction -- let's handle power.If we apply some current I, to an impedance Z = a + bi (yes, we're using two i's here, unfortunately -- for this reason, EEs traditionally use 'j' instead of 'i', which I probably should've done in the first place but I'm not going to edit all this now..), we get some voltage V = I Z.Now we can write that out as V = I (a + bi). So V is also a complex number. We might as well call I a complex number, too; we'll just define it as all-real (zero imaginary) for convenience.And then we do P = V*I.For convention, we take the conjugate of I, which I will write I' here. That's the minus-imaginary-part we did before. Well, we're defining I as real, so that's trivial, but keep that in mind when starting with a voltage, and using I = V/Z.P = (I Z) I' = I I' Z = I^2 (a + bi)Well, that was rather boring. It's literally just I^2*R over again. As well it should be!Say we start with real voltage V, and get some current I = V/Z. Then we have,P = V I' = V (V / Z)' = V V' / Z' = V^2 / Z'Now we've done something a little different. Again because V is real, the conjugate doesn't matter (for a real number x, x = x'), but be careful when it isn't. We are, however, dividing by conjugate impedance.So what is the meaning of real and imaginary P?Real P is real power (oddly enough!). Work. Heat.Imaginary P is reactive power: energy stored and recycled, back and forth.We normally write S = P + iQ, where S is called the apparent power (what I've been calling P above), P is the real power (a real number, not complex), and Q is the reactive power (also a real number).Clearly(?), we only get reactive power from reactive impedance, and real power from real impedance.Tying it all together, say we apply 10V AC at 20Hz, to a 1uF capacitance.S = V^2 / Z'Z = R + XZ = 0 + -i / (2 π F C)Z ~= -i7957 ΩS = (10V)^2 / (-i7957 Ω)'S = 100 V^2 / i7957 ΩS = -i 0.01257 VA (note that 1/i = -i by the multiply-both-sides-by-conjugate rule)S == P + iQ tells us Q = -0.01257 VAR (by convention, we label Q in "volt-amps reactive")Reactive power is not dissipated power. In the same way that pulling on a load (in the direction of motion) produces work (W = F.x, units of N.m == J), and pulling on a load (perpendicular to the direction of motion) does not (a centripetal force or torque, units of N.m), so too we have watts (same direction) and volt-amps (mixed or perpendicular).The angle of the triangle constructed from P = S + iQ is called the phase angle, and the cosine of that angle is the power factor (PF). PF = 1 when P is real, PF = 0 when imaginary.On an instantaneous basis, reactive power looks like current going constantly between positive and negative current draw, in quadrature with voltage going positive and negative. So it's always chasing its tail and no real work gets done. A class D amplifier recycles the reactive power into its supply rail, so that current is drawn at first (and the supply voltage drops momentarily), then returned (so it rises momentarily). This only causes losses to the extent that the draw and return have losses (ESR, dielectric loss, switch resistance).The downside is a ~constant adder due to switching losses. That is, it's always dissipating some power, necessary to keep the circuit bouncing back and forth. A very slowly varying output (nearly DC) won't be able to recover more reactive power than is lost due to operation alone, and in that case you might as well stick with a linear amplifier (that can be made to have lower idle/bias current).You can recover some switching loss by simply operating at a lower frequency, in which case you need a proportionally larger inductor, but not as large as operating at 20Hz directly.Tim

author: pete_dl
author_rank: Newbie
timestamp: August 27, 2018, 07:54:37 pm
content: Thank you for your detailed notes.So if my actuation frequency was not so low, the reactive power could be recycled into the class D amp.I need to drive multiple Piezoelectric actuators simultaneously. I am thinking I could recover half the energy by driving them in pairs. While one actuator is at Vmax, the other is at 0V. You PWM between them to get a sinusoidal change in voltage until both actuators are at Vmax/2, then push the pair to the whole voltage swing using the power supply. Could that work ?Pete.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: August 27, 2018, 08:41:20 pm
content: You can connect them in parallel, and bias one differently from the other, sure.That doesn't automatically recover power, power is only about the change.Tim