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HACHEUR
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<Uh> = 1/t * integrale(ut) | |
<Uh> = Vs(2£-1) | |
(a utilisé pour savoir quand | |
tension positive) | |
Uh = Ldi/dt + Ua | |
<Uh> = Ua | |
Tension moyenne induc = 0 | |
Ldi/dt = Vs - Ua | |
(equa diff) | |
Devient : (intégration) | |
Ia(t) = (Vs-Uh)/L * t + im | |
Ia(£t) = IM = (Vs-Uh)/L * £t + im | |
Delta (Ia) = IM - Im = (reponse au dessus) | |
Imoy = centrer | |
Delta Ia = autour de Imoy | |
Imoy = 4A | |
Delta I = 1.6 | |
Imin = 4A - 1.6/2 | |
Imax = 4A + 1.6/2 | |
Freinage | |
Ua > 0 | |
Ia < 0 | |
T2 T4 a contrôlé | |
Uh ou Ua > 0, £ < 0.5 |
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uk = 0 quand fermé | |
Uk = V quand ouvert | |
i = cst dans moteur | |
pour avoir I = maille | |
(valeur VL null sur moyenne) | |
fem flux constant, proportionnel vitesse angulaire | |
E = k* omega | |
E = (loi maille) |
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