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*To*: tesla-at-pupman-dot-com*Subject*: Re: Mot DC Ps*From*: "Tesla list" <tesla-at-pupman-dot-com>*Date*: Sun, 18 Feb 2001 11:35:15 -0700*Resent-Date*: Sun, 18 Feb 2001 11:41:47 -0700*Resent-From*: tesla-at-pupman-dot-com*Resent-Message-ID*: <KYfRv.A.5tE.ndBk6-at-poodle>*Resent-Sender*: tesla-request-at-pupman-dot-com

Original poster: "by way of Terry Fritz <twftesla-at-uswest-dot-net>" <Mddeming-at-aol-dot-com> In a message dated 2/17/01 11:42:58 PM Eastern Standard Time, tesla-at-pupman-dot-com writes: > > Original poster: "Luc by way of Terry Fritz <twftesla-at-uswest-dot-net>" < > ludev-at-videotron.ca> > > Tx Matt > > You lost me after the first paragraph but R=2*sqrt(L/C) is simple if I > don't have > any real inductance except for parasitic could I use something like 10ˆ-6 H > for > value of L in the equation. > > Luc Benard > > > > > > > > I have a question too when you discharge a cap though a resistance for a > > > certain value of R you have an oscillating circuit (because of parasitic > > > inductance ), if you increase the value of R to a certain point the cap > > > will just discharge whit out oscillation how do you calculate this > > > value. > > > > > > Tx > > > Luc Benard > > > > Hi Luc > > The critical point is at R=2*sqrt(L/C). If R is greater than this > > value, no oscillation takes place. The frequency equation > f=1/((2pi*sqrt(LC) > > is actually a simplifications of: > > F=sqrt(1/LC-(R/2L)^2)/2pi If the frequency F of the oscillations is known for a given value of R, the effective parasitic L can be calculated from the equation F=sqrt(1/LC-(R/2L)^2)/2pi and then the critical value of R found from R=2*sqrt(L/C). In the absence of a way to measure F, yes, you can "guesstimate" L and try the resulting values of R until oscillation stops. Matt D.

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