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Cálculo de los componentes resistivos
y Circuito de Retardo :
Nuestro PLL básico tendrá una Frecuencia Máxima y una Frecuencia Mínima a la cual va a trabajar .
La frecuencia mínima la tenemos , pero la máxima no . Por ende , ten e mos que :
f mín = 1000 Hz = 1 K Hz.
Con el siguiente dato , podemos calcular las resistencias y capacitores n e cesarios para el integrado 4046 B. Dichos cálculos se detallan a continu a ción :
Primero adoptamos una R 2 entre 10 k ð y 1M ð , para nuestro caso la R 2 será igual a 10 k ð :
f mín = 1_______ = 1 __________ =
R2 .( C1 + 32 pf ) 10 kð .( C1 + 32 pf )
1000 Hz ( C1 + 32 pf ) = 1 ____
10 kð
C1 = 1 ________ - 32 pf
10 kð . 1000 Hz
C1 = 100 nf - 32 pf = 99,968 nf
Una vez obtenido el C 1 , podremos calcular en base a los datos que pose e mos la f máx :
f máx = 1 _________ + f mín
R1 ( C1 + 32 pf )
Adoptamos una R 1 = R 2 = 10 K ð , y calculamos la f máx a la que trabajará nuestro PLL :
f máx = 1 ___________+ 1 KHz
10 Kð ( 100 pf + 32 pf)
f máx = 757,58 KHz + 1 KHz
f máx = 758,58 Khz = 0,758 MHz
Cálculo del Filtro Pasabajos o de Retardo :
El filtro pasabajos es el siguiente =
El filtro Pasabajos debe tener una frecuencia de resonancia que sea p e queña , para que ante las variaciones de la frecuencia del PLL se vuelva a enganchar . Por lo tanto , adoptando la R 3 y el C 2 ( o sea la constante de carga del capacitor : ð ) , obtendremos la frecuencia de resonancia .
Adoptamos un valor de Capacitor igual a 0,01 ð f y una Resistencia igual a 1 M ð , por lo tanto la constante de tiempo es igual a 10 ms :
fo = 1______ = 1____________
2 ð . R . C 2 . 3,14159 . 10 K ð . 10 nf
fo = 15,92 Hz .
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Enviado por: Mario Taguorian Y Otros Idioma: castellanoPaís: España