Mit einem digitalen Ausgang soll eine LED geschaltet werden.
Gegeben ist:
U B : = 12 V {\displaystyle {U}_{B}\mathrm {\colon } =12V} , U F : = 1.8 V {\displaystyle {U}_{F}\mathrm {\colon } =1.8V} , U C E s a t : = 0.2 V {\displaystyle {U}_{\mathit {CEsat}}\mathrm {\colon } =0.2V} , I F : = 20 m A {\displaystyle {I}_{F}\mathrm {\colon } =20{\mathit {mA}}}
Berechnen der Widerstände:
R C : = U B − U F − U C E s a t I F = 500.0 Ω {\displaystyle {R}_{C}\mathrm {\colon } ={\frac {{U}_{B}-{U}_{F}-{U}_{\mathit {CEsat}}}{{I}_{F}}}={\text{500.0}}\Omega }
U 1 : = 4.6 V {\displaystyle {U}_{1}\mathrm {\colon } =4.6V} , U B E : = 0.7 V {\displaystyle {U}_{\mathit {BE}}\mathrm {\colon } =0.7V} , B : = 200 {\displaystyle B\mathrm {\colon } =200} , I C : = I F {\displaystyle {I}_{C}\mathrm {\colon } ={I}_{F}} , u ¨ : = 3 {\displaystyle {\ddot {u}}\mathrm {\colon } =3}
R B : = ( U 1 − U B E ) ∗ B u ¨ ∗ I C = 1.300×10⁴ Ω {\displaystyle {R}_{B}\mathrm {\colon } ={\frac {\left({U}_{1}-{U}_{\mathit {BE}}\right)\ast B}{{\ddot {u}}\ast {I}_{C}}}={\text{1.300×10⁴}}\Omega }
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