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Author Topic: Q Factor, Volts times Q  (Read 3738 times)

Offline angryScientist

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Q Factor, Volts times Q
« on: June 22, 2008, 12:49:13 AM »
The Q factor, also known as,
Quality factor,
Energy factor,
Magnification factor,
Merit factor,
is the impedance of a coil divided by the resistance of the coil.



I understand that in a series resonant circuit the voltage across the individual components at resonant frequency is the voltage input times Q. So if we have a coil of 150 Q in a series resonant circuit and we drive it with 5 Volts then the voltages across either the coil or capacitor is Q * Vin = 150 * 5 = 750 Volts.

My question is; Does the current in a series resonant circuit get multiplied by Q also as it does in a parallel resonant circuit?

Free Energy | searching for free energy and discussing free energy

Q Factor, Volts times Q
« on: June 22, 2008, 12:49:13 AM »

Offline angryScientist

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Re: Q Factor, Volts times Q
« Reply #1 on: July 01, 2008, 08:42:04 AM »
Quote
Generally the frequency of a tuned circuit is assumed to be determined solely by L and C values,

F = 1 / (2pi sqrt LC)

However, series resistance has some effect on resonant frequency, as shown by the formula

F = (sqrt 1 - 1/4(Q sqrd)) / (2pi sqrt LC)

From this formula it can be reasoned that the resonant frequency is directly proportional to circuit Q. As the Q increases the resonant frequency will increase somewhat.

The Q of a resonant circuit used in communications may be about 5 to 15 in radio transmitters, 25 to 200 in RF tuned circuits in receivers or several hundred in specially designed high-Q filter circuits. By using regeneration in vacuum-tube or transistor circuits, it is possible to obtain a resonant circuit with a Q well over 10,000.


pp 119,120

ELECTORNIC COMMUNICATION second edition

Robert L Shrader
Laney College
Oakland, California

Copyright 1959,1967

Offline angryScientist

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Re: Q Factor, Volts times Q
« Reply #2 on: July 01, 2008, 12:14:34 PM »
I was thinking about metal at one point and how electricity flows in a metal. I learned that there are a certain number of electrons that are shared to fill the outer orbitals of the metal. Also, in order to get an electron flow going an amount of energy needs to be added to the turn the shared electrons from simple bound electrons to conducting electrons. The greater the electro-motive force applied the more free electrons you have available.

The energy it takes to transition a bound electron to a conducting electron is a resistance to the electron flow.

To increase the amount of available unbound or conducting electrons increase the total number of electrons. If you increase the number of electrons you increase the static charge of the metal. The more free electrons to start out with, in a conductor, the less effort required to produce a current flow.

Charging the conductor to a highly negative state makes the metal more similar to a plasma. In a plasma free charges are easy to find and transporting current takes almost no effort to achieve. This is what makes a plasma a super conductor.

Finally, reducing the resistance of a tuned circuit increases the Q of the circuit. What that means is, the more perfect the tuned circuit the longer it will store every little bit of tuned energy added to it.

Free Energy | searching for free energy and discussing free energy

Re: Q Factor, Volts times Q
« Reply #2 on: July 01, 2008, 12:14:34 PM »
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Offline Vortex1

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Re: Q Factor, Volts times Q
« Reply #3 on: July 02, 2008, 03:14:28 AM »
Quote
To increase the amount of available unbound or conducting electrons increase the total number of electrons. If you increase the number of electrons you increase the static charge of the metal. The more free electrons to start out with, in a conductor, the less effort required to produce a current flow.

Charging the conductor to a highly negative state makes the metal more similar to a plasma. In a plasma free charges are easy to find and transporting current takes almost no effort to achieve. This is what makes a plasma a super conductor.

I thought you had an interesting idea here so I set out to test it. I set up a resonant circuit isolated from the measurement side by a small CM choke. I applied (slowly increased) a negative charge on the resonant circuit. I was coupling the resonant frequency by way of a small open ended inductor. The small CM choke was biased with a small magnet so that it presented a rather low impedance to the resonant series LC

I saw no change in the quality or amplitude of resonance with 0 to -7 kV imparted to the resonant circuit. With or without out the HV supply, there were some rather interesting effects as frequency was swept around resonance with the small neo magnet on the core of the CM choke.

With the magnet, the resonance would force the core out of saturation, but only briefly, like a narrow pulse instead of a sine wave. This effect would quickly snap to a much lower amplitude normal sine wave if the amplitude of the pulse became too high, then I would have to start lower in frequency and sweep slightly up to find the effect again.......serendipity. I urge others to try this. Slowly removing the magnet would cause the pulses to grow again, then suddenly snap back down to normal operation.

thanks for the thought........V


 

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