Maybe it's not that tough. All periodic waveforms are made up of a sum of sine and cosine functions. The voltage wave is just an amplitude V multiplied by a unity wave function. When the V = L(di/dt) is integrated for a sine wave, we integrate

di = V/L cos(wt) dt

so

i(t) = V/wL sin(wt)

the current becomes V/wL (the amplitude of the current) multiplied by another unity wave function. Since all the components of a periodic wave are sine or cosine, all integrate as another sine or -cosine with the w coming out as 1/w and you still get V/wL and a unity wave function. So the reactance is always wL.

For example, a unity square wave is just

square(t) = 4/pi * [sin(wt) + 1/3*sin(3wt) + 1/5*sin(5wt) + ...]

This produces a square wave from -1 to +1, and the voltage formula would be just

v(t) = V * square(t).

Every term of the unity function would integrate to 1/w multiplied by a sine or -cosine and the w would come out and combine with the V/L as V/wL from the distributive property.

Unfortunately, you have a snag - your current cannot go negative because you have a diode, so instead of getting a nice triangle wave for the current (being the integral of a squarewave) the negative side is clipped.
You then have two separate situations with different solutions know as "continuous current" and "discontinuous current".
If your waveform was driven to the positive supply by another MOSFET, instead of allowing the stored energy to do it for you via the diode, the current could go negative (energy put back into the supply from the inductor) and your maths would work in all cases.

Actually, if all periodic waves are explained by a Fourier series, then even a DC-offset wave should have a Fourier series. Perhaps X=wt for any periodic wave.

Don

I think someone mentioned that back in post #73.

 

I already posted the exact series of coils I'm using. Don't be so mean.

Here it is again:

Coilcraft series RFC1010

COLC_S_A0010704786_1-2573496.pdf (mouser.com)

Don

 

Unfortunately, you have a snag - your current cannot go negative because you have a diode, so instead of getting a nice triangle wave for the current (being the integral of a squarewave) the negative side is clipped.
You then have two separate situations with different solutions know as "continuous current" and "discontinuous current".
If your waveform was driven to the positive supply by another MOSFET, instead of allowing the stored energy to do it for you via the diode, the current could go negative (energy put back into the supply from the inductor) and your maths would work in all cases.

But isn't a clipped wave still periodic?

Don

 

But isn't a clipped wave still periodic?

Don

Yes, but it is no longer a linear system, so that the current waveform cannot be derived from the voltage waveform using only the reactance

 

I already posted the exact series of coils I'm using. Don't be so mean.

Here it is again:

Coilcraft series RFC1010

COLC_S_A0010704786_1-2573496.pdf (mouser.com)

Don

If I were you I would rephrase your question with as many details as possible.

Users now have to sift through many posts to try and make sense of the problem which was unclear to begin with.

Good luck...

 

To be honest, I'm deliberately cryptic because this is an invention. I have two patents on it already, but I intend to apply for a third soon and I have to be careful what I say online--not so much because I think anyone will steal it, but because my patent attorney constantly warns me that what I say in an online forum might be construed by the PTO as "publishing" my ideas, which would render them unpatentable.

I'm not a crackpot. My invention has been covered by Popular Science, the New York Times, NPR radio and magazines and radio stations all over the world. I'm in the final stages of trying to reduce cost and boost performance before it goes into production.

Sorry if I sound rude. I didn't mean to.

Don

 

patent attorney constantly warns me that what I say in an online forum might be construed by the PTO as "publishing" my ideas, which would render them unpatentable.

He's absolutely right!
Post your existing patent numbers so we can look them up! (they are already in the public domain, so you won't be giving anything away)

 

OK, but I'd rather the discussion didn't get too side-tracked. I'm going to remain cryptic.

The patents are:

US6559369B1 - Apparatus and method for self-tuning a piano - Google Patents

US9190031B1 - Piano string tuning using inductive current pumps and associated method of use - Google Patents

Don

 

What I want is to select an inductor with enough impedance that I don't need current/voltage control at all. Again, a formula for reactance under a DC square wave would let me easily do this.

Don

At pulse DC, Inductors have impedence, not reactance.

 

Yes, the data sheet gives me the DC resistance and I can find the vector sum with the reactance to get impedance...if only I had a formula for reactance.

Don

 

Yes, the data sheet gives me the DC resistance and I can find the vector sum with the reactance to get impedance...if only I had a formula for reactance.

Per my previous replies, and as to what I was always taught, Reactance only occurs in an inductance/capacitor circuit with AC power only.

 

Can you cite where you got this information? My college physics textbook gives the universal differential equation

v = L(di/dt)

Which should handle any function of v or i.

Don

 

Can you cite where you got this information? My college physics textbook gives the universal differential equation

v = L(di/dt)

Which should handle any function of v or i.

Don

So where is reactance in that equation?

As I understand it, reactance is what you get when you apply the Laplace transform to turn the differential equation into a linear algebraic equation in the frequency domain. Easily done fir a sine wave, not so easy fir a discontinuous waveform.

Bob

 

You have to solve the equation for di (using your function for voltage v(t)) and then integrate it to get the current equation i(t). Reactance is V/I.

Don

 

There should be some background to isolate what an automatic piano tuner does exactly, I am not certain, the sound board has the complete sound.
Filtering for one note for each string sequentially comparing it to a piano model standard is my guess.

The pulse mechanism of a piano is combined with the inherent tone of the whole assembly with all of its selected parts and the room acoustics.
Over time the acceptance of an automatic tuner with the better pianos the value adds to the claim of being new and useful, just as wine ages.

The theoretical question asked originates from an aspect of mechanical design, but it deviates into electronic piano.
The same claim if detected post soundboard in the case of an electric piano a question may arise.
Is the sound the same sound as a mechanical piano for that detection process?

 

The 1-st link at the search link @ #40 hints the formula might be \(X_L=2\pi Lf\sqrt{2}\sum_{n=0}^{\infty}{\frac{{\left({-1}\right)}^n}{1+2n}}\) , but this is for the (symmetrical) ±x volts square wave at relatively low current for the ideal inductor L . . . and this does not apply to instant response - but to a settled DC levels (in this case the 0V DC)

 

That summation converges to pi/4, making the formula

X = (pi^2 * f *L)/√2

Interesting.

Don

 

Interesting.

no - it's a bit optimistic/dumb speculation from my side (interesting would be if it actually applied Xb)
Sorry -- it's apx. about the wave packet propagation - i'm not much at home with

 

Well, it's not offset to pulsed DC, but still interesting.

I'm just going to bite the bullet and order an assortment of those inductors and try them out. Thanks everyone for the help.

Don

 

I'm just going to bite the bullet and order an assortment of those inductors and try them out.
Don

CoilCraft will often send you a few as a trial sample .Free.