Mit einem digitalen Ausgang soll eine LED geschaltet werden.
Gegeben ist:
UB:=12V{\displaystyle {U}_{B}\mathrm {\colon } =12V} , UF:=1.8V{\displaystyle {U}_{F}\mathrm {\colon } =1.8V} , UCEsat:=0.2V{\displaystyle {U}_{\mathit {CEsat}}\mathrm {\colon } =0.2V} , IF:=20mA{\displaystyle {I}_{F}\mathrm {\colon } =20{\mathit {mA}}}
Berechnen der Widerstände:
RC:=UB−UF−UCEsatIF=500.0Ω{\displaystyle {R}_{C}\mathrm {\colon } ={\frac {{U}_{B}-{U}_{F}-{U}_{\mathit {CEsat}}}{{I}_{F}}}={\text{500.0}}\Omega }
U1:=4.6V{\displaystyle {U}_{1}\mathrm {\colon } =4.6V} , UBE:=0.7V{\displaystyle {U}_{\mathit {BE}}\mathrm {\colon } =0.7V} , B:=200{\displaystyle B\mathrm {\colon } =200} , IC:=IF{\displaystyle {I}_{C}\mathrm {\colon } ={I}_{F}} , u¨:=3{\displaystyle {\ddot {u}}\mathrm {\colon } =3}
RB:=(U1−UBE)∗Bu¨∗IC=1.300×10⁴Ω{\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 }