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Components and Circuits For discussions about component types, alternatives and availability, circuit configurations and modifications etc. Discussions here should be of a general nature and not about specific sets. |
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31st Oct 2017, 2:21 pm | #21 |
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Re: Maths query: What is 'T' in this expression, please?
With things like pi (vs. tau) I think temporal humility is in order; we should not assume that we are smarter than the ancient people who chose 2pi as the ratio between radius and circumference.
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31st Oct 2017, 2:52 pm | #22 |
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Re: Maths query: What is 'T' in this expression, please?
Argus25: ah, that would be great if you’d kindly dig out the relevant equation when you get back next week. Thank you!
David (RadioWrangler) interesting point on Q, you raise there. Can you expand with one example please, to see this in action?
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31st Oct 2017, 10:06 pm | #23 |
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Re: Maths query: What is 'T' in this expression, please?
I've often seen τ (tau) as a time-constant - often CR or L/R - but never as an abbreviation for 2π. It would not seem necessary to do this, as 2π is fairly succinct.
In this case, as others have said, it has to have the dimensions of time (assuming T in the equation is a time of some sort), otherwise the exponential couldn't be done. Looking forward to the answer, as this equation isn't familiar! |
7th Nov 2017, 5:40 am | #24 | |
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Re: Maths query: What is 'T' in this expression, please?
Quote:
This voltage step initiates oscillations and the current peaks a quarter of a cycle into the oscillations at pi/2 radians as one would expect and would peak negative again at 3pi/2 radians later. The exponential part shows how the oscillation decays with time, with a damping coefficient of R/2L. Interestingly, for the peak current calculation, the time this occurs at pi/2w can be resolved to T/4 where T is the period, or 1/f. |
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7th Nov 2017, 8:33 pm | #25 |
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Re: Maths query: What is 'T' in this expression, please?
Argus25,
Thank you for posting this. It is quite a complicated expression and I'll need some time to study it, model it and think about it. But I get the 'take away' that peak current is quarter of a cycle in. Now I can see why transient suppression diodes, etc need to be so fast at RF. I'm sure others will have insightful things to say about it. Has anyone else used it any capacity?
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9th Nov 2017, 12:03 am | #26 | |
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Re: Maths query: What is 'T' in this expression, please?
Quote:
Also as Alpha or R/2L is small as R is very low, the expression in brackets W(squared) + Aplha(squared) divided by w, simply reduces to w. Also the exponential decay can be ignored and the value of that equal to 1, so it simply ends up as: Peak current = wVC when the proportions of R are small. |
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11th Nov 2017, 5:44 am | #27 |
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Re: Maths query: What is 'T' in this expression, please?
Al,
Coincidentally, I just found the equation you were questioning at the beginning of the thread. It was actually published in an article by Otto Schade (being discussed on another thread for helping to calculate rms currents) in Television, volume V, RCA labs 1947-48 pg 107, in an article by Otto on magnetic deflection circuits. For a Parallel resonant circuit with L and C (where there is some R in series with the L) if a voltage is applied (switched onto the circuit) the 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 Otto's equation is: Vpk = Ipk times root(L/C) times e to the power of - pi/(4Q). Where Q is the circuit's Q. So it looks like the "T" in your original equation was in fact supposed to be pi, someone must have slipped up writing it, making it look like a T. And perhaps tau was supposed to be Q but it might resolve to the same thing. Hugo. Last edited by Argus25; 11th Nov 2017 at 5:52 am. |
11th Nov 2017, 1:24 pm | #28 | ||
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Re: Maths query: What is 'T' in this expression, please?
Quote:
Quote:
So...Vpk=Ipk*sqrt (L/C) *e^pi/4Q? I will have a deeper look into the correct expression that you've published here, it's just what I need. Thanks again, Cheers,
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Al Last edited by Al (astral highway); 11th Nov 2017 at 1:32 pm. |
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11th Nov 2017, 10:10 pm | #29 |
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Re: Maths query: What is 'T' in this expression, please?
Al, yes, but it just needs the minus sign
Vpk = Ipk*sqrt (L/C) * e^-pi/4Q That is the peak voltage across the resonant circuit after the switch opens, but the for voltage across the actual switch contacts, the power supply voltage say E, has to be added to Vpk because the power supply that energized the circuit in the first place is in series with the switch and the parallel resonant circuit. |
12th Nov 2017, 12:15 am | #30 |
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Re: Maths query: What is 'T' in this expression, please?
Al,
Also I found out more about Dr. Otto Schade, apart from his genius level analysis on magnetic deflection systems for TV's and the tables he created to help calculate currents in power circuits, he is also the designer of RCA's 6AS7-G double triode valve. This valve acts like a controlled efficiency or controlled damper diode in TV horizontal deflection circuits produced by RCA. It is a super low impedance twin power triode and it appears these too have attracted the attention of the audiophile industry. I don't have a TV set that uses one though. So Otto was very good at electronics engineering, maths, physical analysis, optics and thermionics/vacuum tube design. He received an honorary doctorate and he was responsible for RCA's Nuvistor. here is some other info on him: https://en.wikipedia.org/wiki/Otto_H._Schade So I think you can trust his equation ! |
12th Nov 2017, 7:50 am | #31 |
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Re: Maths query: What is 'T' in this expression, please?
Al,
Now I have figured out the circuit you were applying the equation to, its really pretty straightforward. If you look at Otto's equation, the exponential part of it just allows for the energy losses. But if you calculate it, the e to the minus pi/4Q, for any reasonable value of Q is likely over 0.9 anyway. Since you are dealing with a high Q circuit for your Tesla coil system, if you want to calculate peak voltage or current, the easy trick is to just make the energy stored in the inductor (which has some peak current) equal to that which is transferred to the capacitor. This is a simple technique for problem solving. For example, if you throw a ball into the air vertically and it leaves your hand with some velocity V and you want to know how high the ball goes, you can simply make the kinetic energy of the ball equal to the gravitational potential energy. So just make 1/2 mV(squared) = to mgh to solve for the height of the the ball and ignore any small losses. The same applies here. So if there is some peak current in the coil a time after the switch or switching device has conducted (if you wait long enough its E/R for any inductive circuit) then you just assume there are no energy losses (don't worry about the exponential decay) and just make the inductor's energy equal to that which is transferred to the capacitor later after the switch has opened. See attachment. Then as you see, it is Otto's equation, sans the exponential decay corresponding to the losses. |
13th Nov 2017, 2:57 pm | #32 |
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Re: Maths query: What is 'T' in this expression, please?
Hi Hugo,
Thanks for pointing out the sign of the exponent, I had copied the expression down incorrectly at post no 29 and found my calculations wildly implausible as a result, obviously!! Dr Otto Schade provides very interesting terrain indeed, including this expression. It takes some insight and clarity of thinking to produce such a result. I will take a deeper look at it and apply it around my circuit. It's not just in the obvious places (the power resonator) that I'm interested. I am looking at every single node in the circuit, either where there is an intentional inductor or stopper choke that I have designed, or, criticallywhere there is stray inductance with lead lengths in a fast dI/dT position. I was also fascinated by the very pleasingly linear behaviour of a RF choke that I put in the power supply to the big MOSFET, whose voltage waveform I monitored as the MOSFET hard switched on and off. Calculations time! Thanks again Hugo, Cheers
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