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Old 14th Nov 2017, 1:25 pm   #1
astral highway
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Wink Transcendental magic: Schade's peak voltage equation and my discovery!!

I have discovered (for myself, by thought-experiment!) a magical property hidden in an equation, which may appeal to those interested in numbers -- just as much as in the maths describing the behaviour of resonant circuits.

I'm pleased to share it, as well as my conclusions.

This is a thematic development from a previous thread where Argus25 (Hugo, thank you again) discussed the equation derived by Otto Schade. It describes the conditions when for a parallel resonant circuit with L and C, with some lumped R, current builds with time to some peak value; Ipk.

Then, when the switch opens, the peak value of voltage across the resonant circuit, given by Schade's 1947 equation is:


Vpk=Ipk * Sqrt (L/C) *e^-Pi/4Q


This is of interest to me in modelling stress conditions in a current project.

For those not familiar, e is the base of natural logarithms, after Napier. It is transcendent but can approximate to 2.71828...

I started by asking: what's the difference between the above circuit, with the exponential multiplicand, and the 'stripped-down' version, which stops with the product of peak current and the square root of the ratio of inductance to capacitance?

I became fascinated by the expression e^-Pi/4Q. I plugged in various numbers for Q, (using the approximation for Pi of 3.1415 and e as above) and was instantly struck by how these converged on values close to unity.

So, what if the whole expression was a remarkably intuitive way of expressing a coefficient?, I asked.

Since, when:

Q=1,000, the expression resolves to 9.9921 e-1, or, as a coefficient, 0.9921

When Q=100, the expression resolves to 9.9218 e-1, or as a coefficient,0.99218...

A factor of ten smaller, and it's not making a huge difference! And

when Q=10, the expression resolves to 9.2447 e-1, or as coefficient, 0.92447

All of which so far shows that Shade's equation has a very subtle way of showing the impact of Q under these conditions.

But here's the exciting thing.

As we go up the powers of 10, the coefficient gets closer and closer to unity. We might think it would never reach unity, given the wonders of the two transcendent numbers, pi and e.

But...

When Q=1 e4, the coefficient becomes 0.99999

And when Q=1 e5, voila! It reaches exactly 1.

And not only that, for all powers of ten above the fifth power, it remains at unity. That is the maximum value of the coefficient.

Why 10,000 should be the theoretical limit of Q is an interesting question, way beyond my knowledge, experience, or powers of conjecture. But according to Schade's equation, this appears to be true.

That's my insight, and I feel quite excited to share it!

And the take away, I suppose, is that in the instance described in the opening paragraph, Q is far less significant determinant than the relationship between inductance and capacitance. Simply, the smaller the capacitance for a given inductance, the greater by far the peak voltage.
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Last edited by astral highway; 14th Nov 2017 at 1:34 pm.
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Old 14th Nov 2017, 1:36 pm   #2
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Default Re: Transcendental magic: Schade's peak voltage equation and my discovery!!

Interesting result, but I think you're probably just reaching the limits of precision of whatever calculating device you're using to get the result. As Q gets larger, so Pi/4Q gets smaller. e to the power of -(small number) is going to be very close to 1, and there will come a point where your calculator doesn't have enough digits to get a sensible result. Indeed, I've just tried the calculation on an ordinary pocket calculator (Casio fx-100D) and it flags an error when I try to do the final e to the x step.

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Old 14th Nov 2017, 2:03 pm   #3
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Default Re: Transcendental magic: Schade's peak voltage equation and my discovery!!

Mathematically you will never get e^-(pi/4Q) to be unity. e^0 is unity (as indeed any positive real number raised to the power 0 is unity), but that is the limit as Q->infinity. You will get very close for large values of Q, close enough for practical purposes I think. What you are seeing, though, is a limit of the accuracy of your calculator.

At least one of my calculators has a function to calculate e^x-1 to avoid this problem. It would give a very small number here, which if you added 1 to it would remain 1 (due ot the limited accuracy of the machine). But most calculators don't have that function and e^(small number) is 1 to the machine accuracy, although not mathematically.
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Old 14th Nov 2017, 2:16 pm   #4
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Default Re: Transcendental magic: Schade's peak voltage equation and my discovery!!

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Interesting result, but I think you're probably just reaching the limits of precision of whatever calculating device you're using to get the result.
Ahah, yes, fair point.

But it doesn't detract from the conclusion that a theoretical Q of 10,000 converges on a coefficient of unity in this expression.

10,000 is, of course, an unattainably high Q in practice!

Also remarkable (to me, anyway) was how such orders of magnitude of Q apart should make so little difference to the resolution of the expression.
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Old 14th Nov 2017, 2:20 pm   #5
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Default Re: Transcendental magic: Schade's peak voltage equation and my discovery!!

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You will get very close for large values of Q, close enough for practical purposes I think. What you are seeing, though, is a limit of the accuracy of your calculator.

Ok so it's lim as Q approaches infinity...

For me, it was fascinating to resolve this for myself.
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Old 14th Nov 2017, 3:00 pm   #6
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Default Re: Transcendental magic: Schade's peak voltage equation and my discovery!!

For large values of Q, the amplitude of oscillation in an LC circuit hardly decays between one cycle and the next.

The ratio of successive peaks is actually e^(-pi/Q) which is 97% for Q=100; 99.4% for Q=500; etc.

But, we're not looking at the relationship between one cycle and the next; we're looking at the relationship between maximum current and maximum voltage. And these occur, 1/4 cycle apart. So during this time the exponential decay has only kicked in for 1/4 of the time, and this explains why the exponential term in your equation is e^(-pi/4Q).

With so little decay, the error in neglecting the exponential term is, for most purposes, peanuts! So, the Q=100 case gives you, as you say, 0.9921 as a multiplier. So if you neglect it, you get an error of 0.78% in peak voltage, not very much! In fact, the % error is approximately 25 x pi/Q, or just 78/Q percent!
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Old 14th Nov 2017, 3:07 pm   #7
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Default Re: Transcendental magic: Schade's peak voltage equation and my discovery!!

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But, we're not looking at the relationship between one cycle and the next; we're looking at the relationship between maximum current and maximum voltage. And these occur, 1/4 cycle apart. So during this time the exponential decay has only kicked in for 1/4 of the time, and this explains why the exponential term in your equation is e^(-pi/4Q)....

With so little decay, the error in neglecting the exponential term is, for most purposes, peanuts!

Neat insight, thank you. I wonder under what circumstances it was important to Schade to derive the peak figure, then? It was post-War...
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Old 14th Nov 2017, 3:09 pm   #8
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Default Re: Transcendental magic: Schade's peak voltage equation and my discovery!!

For small values of x, e^(-x) is 1-x. So for high values of Q you are simply calculating 1- (pi/4Q) to whatever accuracy your calculating device can manage. You can easily check this by comparing the two calculations: exp(-x) vs. 1-x. The difference is that you can by hand do 1-x however small x is - you are limited only by the size of your piece of paper; the calculator runs out of significant figures before you do. Of course, exp (0) = 1.

If you really want to get excited by maths, consider that exp (j pi) + 1 = 0.
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Old 14th Nov 2017, 3:26 pm   #9
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Default Re: Transcendental magic: Schade's peak voltage equation and my discovery!!

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For small values of x, e^(-x) is 1-x. So for high values of Q you are simply calculating 1- (pi/4Q) to whatever accuracy your calculating device can manage.

Neat analysis, thank you Dave.

Quote:
Originally Posted by G8HQP Dave View Post

If you really want to get excited by maths, consider that exp (j pi) + 1 = 0.
I do recognise that one, after Euler? But I have nowhere near enough maths to derive it or explain it!
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Old 14th Nov 2017, 3:47 pm   #10
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Default Re: Transcendental magic: Schade's peak voltage equation and my discovery!!

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I wonder under what circumstances it was important to Schade to derive the peak figure, then? It was post-War...
Probably, calculating peak voltage to be sure that insulation, voltage rating of components, conductor spacings, was sufficient to prevent failure or flashover! Schade being nothing if not thorough, worked out the exact formula.

Quote:
Originally Posted by G8HQP Dave View Post
For small values of x, e^(-x) is 1-x. So for high values of Q you are simply calculating 1- (pi/4Q) to whatever accuracy your calculating device can manage. You can easily check this by comparing the two calculations: exp(-x) vs. 1-x. The difference is that you can by hand do 1-x however small x is - you are limited only by the size of your piece of paper.
Good approximation from Dave, which gets more and more accurate the smaller the value of x. And much easier to calculate! In fact it's the approximation behind my % error being 78/Q... as pi/4Q x 100% simplifies to just this!

So, whereas Al's calculator gives up and displays 1 for Q>10,000 we can now do better if we want... so for a Q of 100,000,000 the exponential term would be 0.99999999215.
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Old 14th Nov 2017, 4:43 pm   #11
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Default Re: Transcendental magic: Schade's peak voltage equation and my discovery!!

Oh wie schade
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Old 14th Nov 2017, 4:47 pm   #12
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Default Re: Transcendental magic: Schade's peak voltage equation and my discovery!!

Hey Trevor, I don't have any German... does that mean something in awe?
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Old 14th Nov 2017, 8:15 pm   #13
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Default Re: Transcendental magic: Schade's peak voltage equation and my discovery!!

Oh what a pity (shame)

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Old 14th Nov 2017, 8:36 pm   #14
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Default Re: Transcendental magic: Schade's peak voltage equation and my discovery!!

Ah, still mystified, Ed?! Why a shame?
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Old 14th Nov 2017, 8:46 pm   #15
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Default Re: Transcendental magic: Schade's peak voltage equation and my discovery!!

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Originally Posted by kalee20 View Post
And much easier to calculate! In fact it's the approximation behind my % error being 78/Q... as pi/4Q x 100% simplifies to just this!

So, whereas Al's calculator gives up and displays 1 for Q>10,000 we can now do better if we want... so for a Q of 100,000,000 the exponential term would be 0.99999999215.
Yes, neat work!

But hey, I was wondering...

There must be a practical limit to Q, taking any definition?
David's (RadioWrangler)'s for example?

'The reciprocal of the fraction of the resonator energy per radian of time at the resonant frequency...' seems particularly appropriate...

Can it really anywhere much higher than, I'm guessing, for an air cored inductor, a very optimistic 1,000, using idealised materials and design?
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Old 14th Nov 2017, 10:33 pm   #16
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Default Re: Transcendental magic: Schade's peak voltage equation and my discovery!!

Al,

The e ^- pi/4Q is just a special case of specifying when the peak gets measured.

In the resonant circuit, after the switch opens it oscillates as you know and the time between current peaks, or voltage peaks is half a cycle or pi radians (2pi radians is a whole cycle). The voltage and current peaks are positive and negative peaks.

Initially after the switch opens the current in the inductor starts to fall towards zero in a sinusoidal fashion with a shape that looks like it had already started on the peak of a sine wave. By the time its hits zero current, the inductor (magnetic field) energy is now zero and the voltage on the capacitor has peaked, because now all the energy is stored as an electric field in the capacitor. This voltage peak is 1/4 of a cycle or pi/2 radians into the oscillation. After that the capacitor returns its energy to the inductor and a 1/4 cycle later there is a negative current peak and the capacitor voltage is again zero.

The general way the exponential decay part of Otto's equation is often specified for resonant circuits is e ^ -tR/2L. (as shown in the previous equation I gave you on the other link for a decaying resonant circuit). This is probably a better way to specify it as you can select and time time t into the oscillation to find the correct amplitude.

For example if you did what Otto did and you want to know the amplitude of the voltage peak at 1/4 of a cycle in (when the capacitor voltage peaks), the angle into the oscillation is wt, where w = 2.pi.f and t is time, so:

you make wt = pi/2, this obviously makes the time t = pi/2w, which is the time from when the switch opens to the voltage peak 1/4 of a cycle in to the oscillations.

So if you substitute this time of pi/2w seconds into the general form of the exponential decay equation where the value is -tR/2L

you get:
- piR/4wL

Which you can see equals Otto's equation where it was -pi/4Q because the R/wL part of it is 1/Q.

So in Otto's form of the equation it is just a specific case of a value 1/4 of a cycle into the oscillation. It is actually often easier to use e^- tR/2L if you select the correct time into the cycle when you would like to know the value.

As you have already figured out and I mentioned on the other thread, with the circuit Q's you are dealing there is very little loss to the first peak and the decay is only very significant over many cycles. So likely the e^-pi/4Q will be over 0.9 for your circuits with any luck.

Hugo

Last edited by Argus25; 14th Nov 2017 at 10:44 pm.
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Old 14th Nov 2017, 11:33 pm   #17
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Default Re: Transcendental magic: Schade's peak voltage equation and my discovery!!

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Ah, still mystified, Ed?! Why a shame?
Just a fun (multilingual) pun?

Made me laugh, anyway.
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Old 14th Nov 2017, 11:54 pm   #18
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Default Re: Transcendental magic: Schade's peak voltage equation and my discovery!!

Al,

I could have mentioned, that the basic function e^x is the only function where it equals the value of its own derivative, so if y = e^x then dy/dx = e^x, so the function value at any point equals the rate of change at that point too. So I think that qualifies it as your "Transcendental magic"

Hugo.
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Old 15th Nov 2017, 12:40 pm   #19
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Default Re: Transcendental magic: Schade's peak voltage equation and my discovery!!

Achievable Q depends on the technology being used. Discrete L and C puts you in the low 100s - the limit is set by resistance, dielectric losses, ferromagnetic losses (if a core is used) and radiation losses. Cavities in the high 100s. Ceramic resonators (as used in some AM sets) maybe in the 1000s? Quartz crystals in the 10,000s. More exotic resonators (rings?) go higher.
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Old 15th Nov 2017, 1:42 pm   #20
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Default Re: Transcendental magic: Schade's peak voltage equation and my discovery!!

Quote:
Originally Posted by Argus25 View Post

For example if you did what Otto did and you want to know the amplitude of the voltage peak at 1/4 of a cycle in (when the capacitor voltage peaks), the angle into the oscillation is wt, where w = 2.pi.f and t is time, so:

you make wt = pi/2, this obviously makes the time t = pi/2w, which is the time from when the switch opens to the voltage peak 1/4 of a cycle in to the oscillations.

So if you substitute this time of pi/2w seconds into the general form of the exponential decay equation where the value is -tR/2L

you get:
- piR/4wL
Hey Hugo,

This is incredibly useful and detailed analysis. Thank you, I will turn my attention to using the relevant equation to find the voltage 90 degrees into a cycle!
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