P: ¿Mi trabajo es correcto, encontrar la frecuencia del diente de sierra (voltaje a través del capacitor) ?:
Mi trabajo:
- Vdd (t) y Vss (t) son voltajes de CC y: $$ \ text {V} _ {\ text {ss}} (t) = - \ text {V} _ {\ text {dd}} (t) $$
- Supongamos en t = 0, Vc (0) = 0V y Vout (0) = Vdd (t)
- cargando el condensador: $$ \ text {V} _ \ text {C} (t) = \ text {V} _ {\ text {dd}} (t) \ left (1- \ exp \ left [- \ frac {t} { \ text {C} \ text {R} _1} \ right] \ right) $$
- cuando: $$ \ text {V} _ \ text {C} (t) > \ frac {\ text {R} _3} {\ text {R} _3 + \ text {R} _2} \ cdot \ text {V} _ {\ text {out}} (t) = \ text {V} _- (t) $$ luego Vout (t) = - Vdd (t) = Vss (t)
- descargando el condensador: $$ \ text {V} _ \ text {C} (t) = \ text {V} _ {\ text {C}} (0) + \ left (\ text {V} _ {\ text {dd}} (t) - \ text {V} _ {\ text {C}} (0) \ right) \ left (1- \ exp \ left [- \ frac {t} {\ text {C} \ text {R} _1} \ right] \ right) $$
- cuando: $$ \ text {V} _ \ text {C} (t) < - \ frac {\ text {R} _3} {\ text {R} _3 + \ text {R} _2} \ cdot \ text {V} _ {\ text {out}} (t) = - \ text {V} _ + (t) $$ entonces Vout (t) = Vdd (t)
Entonces, cuando queremos encontrar la frecuencia, podemos decir eso, resolviendo para 't':
- $$ \ text {V} _- (t) = \ text {V} _ {\ text {C}} (0) + \ left (\ text {V} _ {\ text {dd}} ( t) - \ text {V} _ {\ text {C}} (0) \ right) \ left (1- \ exp \ left [- \ frac {t_1} {\ text {C} \ text {R} _1 } \ right] \ right) \ Longleftrightarrow $$ $$ t_1 = \ text {C} \ text {R} _1 \ ln \ left (\ frac {\ text {V} _ {\ text {C}} (0) - \ text {V} _ {\ text { dd}} (t)} {\ text {V} _- (t) - \ text {V} _ {\ text {dd}} (t)} \ right) $$
- $$ - \ text {V} _ + (t) = - \ left (\ text {V} _ {\ text {C}} (0) + \ left (\ text {V} _ {\ text {dd}} (t) - \ text {V} _ {\ text {C}} (0) \ right) \ left (1- \ exp \ left [- \ frac {t_2} {\ text {C} \ texto {R} _1} \ derecha] \ derecha) \ derecha) \ Longleftrightarrow $$ $$ t_2 = \ text {C} \ text {R} _1 \ ln \ left (\ frac {\ text {V} _ {\ text {C}} (0) - \ text {V} _ {\ text { dd}} (t)} {\ text {V} _- (t) - \ text {V} _ {\ text {dd}} (t)} \ right) $$
Entonces, el tiempo de un período viene dado por:
$$ \ text {T} _ {\ left [\ text {s} \ right]} = t_1 + t_2 = 2 \ text {C} \ text {R} _1 \ ln \ left (\ frac {\ text {V} _ {\ text {C}} (0) - \ text {V} _ {\ text {dd}} (t)} {\ text {V} _- (t) - \ text {V} _ {\ text {dd}} (t)} \ right) $$
Entonces, la frecuencia está dada por:
$$ \ text {f} _ {\ left [\ text {H} z \ right]} = \ frac {1} {\ text {T} _ {\ left [\ text {s} \ right] }} = \ frac {1} {2 \ text {C} \ text {R} _1 \ ln \ left (\ frac {\ text {V} _ {\ text {C}} (0) - \ text {V } _ {\ text {dd}} (t)} {\ text {V} _- (t) - \ text {V} _ {\ text {dd}} (t)} \ right)} $$