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Posted: September 07, 2015 02:26 am
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Member No.: 2,202
Joined: March 11, 2005
Lets say we have a simple high pass filter, a capacitor followed by a resistor to ground. We run an AC sweep signal into it.
At some point, the amplitude will start to roll off at some frequency. The phase will shift as well. Current flowing through the capacitor will lag in time to the voltage across it by some factor.
Since the current does not reach the resistor at the same time as the voltage, the voltage across the resistor drops.
Hence, the phase shift does not allow all the power transferred from the source to the resistor.
Therefore, at the end of the day, energy loss is not due to heat (resistivity) but due to phase shift (the current and voltage not reaching the load at the same time).
Remember to take your records off the turntable when you are done listening to them, otherwise they will warp.
Posted: September 07, 2015 04:03 am
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Member No.: 73
Joined: July 24, 2002
If the resistor is the load (filters are only meaningful for matched constant resistance source and load), then this is true.
It's kind of dubious to say the phase shift is the reason, but because phase shift and impedance are locked together, this is more-or-less fine:
(There should be a whole article talking about this matter; it's one of those wave-particle duality things, at its root. Suffice it to say: any time magnitude is changing, phase is necessarily changing at least a certain amount as well. Nonminimum phase networks can have more phase shift, but not less, since less would violate causality. Which puts the limit on how "fast" a filter of some design can be, as must be the case.)
If you imagine a source not as a voltage, but as a power source with some constant output and some reflected amount so that the total comes out right, then when the reactance is small, the source and load are connected together and no power is reflected. When the reactance is large, more power is reflected and less transmitted.
And yes, if you keep your mind active in terms of impedance matching, then you will realize: it's not that a battery sitting on the table is "open circuit", it's that it's constantly transmitting a large amount of power, and all of it is being reflected back from the lack of matching!
Whether this is a useful model of the world depends on what you're doing; normally it's not a concern, but it becomes so much harder to work without it, at RF, or when dealing with any kind of passive circuit (where impedances and ratios, transmission and reflection are the important figures).
Answering questions is a tricky subject to practice. Not due to the difficulty of formulating or locating answers, but due to the human inability of asking the right questions; a skill that, were one to possess, would put them in the "answering" category.
Posted: September 14, 2015 03:21 am
Member No.: 9,581
Joined: May 25, 2007
At some point, the voltage will increase across the resistor as the frequency rises. You specified a high pass filter, remember?
At high frequency, the phase shift is negligible because the reactance is almost zero.
Current does not flow. Current exists or is present. Charge can flow, however. "Current flow" is technical slang that is not correctly descriptive.
Charge does not flow through a capacitor. The dielectric is an insulator that prevents that from happening. Charge accumulates on one side of the dielectric and depletes on the opposite side.
No, the phase has nothing do to with the voltage across the resistor. The voltage across the resistance is proportional to the resistance and the current.
The energy loss only occurs in the resistor due to heat dissipation. The capacitor stores its energy and gives it back to the circuit so no energy is lost.
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