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Author Topic: A quick question about diodes and resonances  (Read 3543 times)

Offline d3x0r

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A quick question about diodes and resonances
« on: February 04, 2013, 09:41:55 PM »

@verpies, itsu, et al. (or anyone else)

It seems to me, that if a cap and full wave bridge rectifier is connected to a coil, that the coil cannot experience any sort of resonance.  What is picked up by induction on the coil becomes purely trapped in the capacitor?   


One side goes high, that side, above the capacitor's high will draw from the capacitor; if it's a low side, then it will draw on the capacitor's low side.  If it is not higher than the capacitor positive or lower than the capacitor negative... I guess that part ends up reflecting; and a small capacitor on that side can tune the frequency resonance?










Offline pauldude000

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Re: A quick question about diodes and resonances
« Reply #1 on: February 05, 2013, 04:44:04 PM »
@verpies, itsu, et al. (or anyone else)

It seems to me, that if a cap and full wave bridge rectifier is connected to a coil, that the coil cannot experience any sort of resonance.  What is picked up by induction on the coil becomes purely trapped in the capacitor?   


One side goes high, that side, above the capacitor's high will draw from the capacitor; if it's a low side, then it will draw on the capacitor's low side.  If it is not higher than the capacitor positive or lower than the capacitor negative... I guess that part ends up reflecting; and a small capacitor on that side can tune the frequency resonance?


Resonance as you are inquiring about is an induced vibration which is electrical in origin. (There are other forms.) Every resonance tank (in your case coil and capacitor circuit) has a natural frequency it will resonate at, dependent upon the coil inductance and the capacitors capacitance.


If you feed a coil/capacitor tank with pure unchanging DC, it will not resonate. The capacitor will charge up, but cannot discharge until the voltage applied to it drops. If the voltage change is too gradual, then the capacitor will discharge slowly and the tank will not resonate.


A capacitor itself works by imbalance of charge. An excess of charge on one set of plates, when the applied voltage drops, will follow the circuit to the other set of plates in order to remove the imbalance of charges, thus achieving a neutral balance of charge. Improperly placed diodes can prevent resonation by allowing current flow in only one direction. This could allow a capacitor to charge causing a desired imbalance, but prevent it from discharging and neutralizing the imbalance.


Either AC or changing voltage DC (pulsed) can resonate a tank circuit. The resonation itself is a discharge of electricity which "bounces" back and forth from one set of plates in the capacitor to the other until the energy is either mechanically prevented, or is used up in the creation of various fields and heat. With each "bounce", the stored energy in the capacitor is less than the previous, as with each "bounce" energy is  expended.


Once resonance is possible, then you can talk deliberate energy placement at a specific point in a circuit due to design wavelength, and where specific points of the wave will be at any given time.


I hope this helps.