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*To*: tesla-at-pupman-dot-com*Subject*: Re: [Fwd: Spark gap not firing]*From*: "Tesla list" <tesla-at-pupman-dot-com>*Date*: Sat, 10 Feb 2001 15:32:21 -0700*Resent-Date*: Sat, 10 Feb 2001 15:42:11 -0700*Resent-From*: tesla-at-pupman-dot-com*Resent-Message-ID*: <Z9dP8D.A.GID.4Och6-at-poodle>*Resent-Sender*: tesla-request-at-pupman-dot-com

Original poster: "Ed Phillips by way of Terry Fritz <twftesla-at-uswest-dot-net>" <evp-at-pacbell-dot-net> Tesla list wrote: > > Original poster: "Luc by way of Terry Fritz <twftesla-at-uswest-dot-net>" <ludev-at-videotron.ca> > > Hi Terry, list, > > I'm in the process of designing a power supply and I try to find a formula to > create a RC low pass filter (Terry NST filter protection ) I look at your > paper, > on the web, I remember I read about it in the past but were ? I know if you > change the value of R and C you change at what frequency the filter start > to cut > and how fast the db cut rise with the frequency. I could apply your value but I > like to understand what I do . Can you help me to find the formula or if my > memory is good the graph to calculated it ? > > Luc Benard The expression for the half-power cutoff frequency of a low pass RC filter is simply Fc = 1/(2*pi*R*C) where Fc is in Hz at which the output voltage/input voltage ratio is 0.707 [1/sqrt(2)] , R is in ohms, and C is in farads. This applies to the simplified case of a filter between a zero-impedance source and an infinite impedance load. In the case of such a filter inserted between a TC primary circuit and a transformer, things may not be quite so simple. Ed

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