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Old 14th Nov 2017, 11:33 pm   #16
Argus25
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Join Date: Oct 2016
Location: Maroochydore, Queensland, Australia.
Posts: 2,679
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 11:44 pm.
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