t L : U c ( t ) = U S t a r t ⋅ e − t τ {\displaystyle {t}_{L}\mathrm {\colon } {U}_{c}\left(t\right)={U}_{\mathit {Start}}\cdot {e}^{\frac {-t}{\tau }}} === 1 2 ⋅ U B = 3 2 ⋅ U B ⋅ e − t L R ⋅ c {\displaystyle {\frac {1}{2}}\cdot {U}_{B}={\frac {3}{2}}\cdot {U}_{B}\cdot {e}^{\frac {-{t}_{L}}{R\cdot c}}}
R ⋅ C ⋅ ln ( 3 ) = t {\displaystyle R\cdot C\cdot \ln \left(3\right)=t}
ln 1 3 = − t L τ {\displaystyle \ln {\frac {1}{3}}={\frac {-{t}_{L}}{\tau }}}
ln 3 − 1 = − ln 3 = − t L τ {\displaystyle \ln {3}^{-1}=-\ln 3=-{\frac {{t}_{L}}{\tau }}}
t L = ln 3 ⋅ R ⋅ C {\displaystyle {t}_{L}=\ln 3\cdot R\cdot C}
weil das Laden und Entladen (bzw. umladen) von C symmetrisch erfolgt.
T = t L − t H = 2 ⋅ t L {\displaystyle T={t}_{L}-{t}_{H}=2\cdot {t}_{L}}
f = 1 T = 1 2 ⋅ ln ( 3 ) ⋅ R ⋅ C {\displaystyle f={\frac {1}{T}}={\frac {1}{2\cdot \ln \left(3\right)\cdot R\cdot C}}}
meist nur näherungsweise erfüllt.
Aufgabe Prototypenbau
Schaltung
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