STEVE:
For power to be reflected, doesn't the source have to be a transmission
line? A series C can reduce the power delivered to a load just by adding
reactance and increasing the overall impedance thus reducing the current.
BILLY:
But, the series C will be dissipating the unwanted power--not reflecting it.
If you want proof of that use a low voltage cap and feel for the heat.
>From: "Steve Thompson" <g8gsq@qsl.net>
>To: <amps@contesting.com>
>Subject: Re: [AMPS] Conjugate Matching and Efficiency
To: <amps@contesting.com>
>Date: Fri, 25 May 2001 14:47:30 +0100
>
>
>snip
> >I fail to understand your point. What do you mean by the 3 dB down
>point?
>If you mean the 3 dB points in a filter (which is essentially what a
>matching network is), that is something completely different. I assume
>that's what you mean so I'll address that.
> >
> >The 3 dB down point in a filter is typically what is used to specify
>bandwidth. That 3 dB down level is basically the level where the magnitude
>of S21 is 3 dB below that of its maximum value (minimum loss). At the 3 dB
>points, half the power that is transmitted at that frequency by the source
>or whatever, gets passed through to the load or whatever is on the other
>side of the filter.
> >
> >So where does the other 3 dB go? Is it dissipated? Hardly. To
>understand
>where it goes, one must look at the magnitude of S11 and S22 of the filter.
>At the 3 dB point, the values of S11 and S22 are getting large and all of
>that power is being REFLECTED back towards the source. In fact for the
>most
>part at the 3 dB point, S11 will typically be such that half the power at
>that frequency won't even get beyond the input of the filter.
>
>
>For power to be reflected, doesn't the source have to be a transmission
>line? A series C can reduce the power delivered to a load just by adding
>reactance and increasing the overall impedance thus reducing the current.
>
>Steve
>
>
>
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