74HC163 wiring diagram

Q0 output gives divide by two and Q1 output gives divide by 4,

...but 74HC163 needs that you set its other inputs properly. You have to refer to the timing diagram for this.

If it is not mandatory to use this counter, use other binary counter without presetting facility. If it is educational, then find out, elaborate your doubts and ask.

Gajanan Phadte
 
Q0 output gives divide by two and Q1 output gives divide by 4,

...but 74HC163 needs that you set its other inputs properly. You have to refer to the timing diagram for this.

If it is not mandatory to use this counter, use other binary counter without presetting facility. If it is educational, then find out, elaborate your doubts and ask.

Gajanan Phadte

I will use it to reclock the TDA1541A S1 dac. 11.2896mhz/2=5.6448mhz for the pin 2+4 and 11.2896mhz/4=2.8224mhz for dem clock.
Thanks
Noyan
 

FoMoCo

Member
2012-12-04 10:04 pm
Just read the datasheet:
P0, P1, P2, and P3, are pre-load inputs. You don't need them. Tie them to ground. (They cannot float!)
PE and TE are count enables. Tie them to supply. I recommend using a 1K (or whatever you have handy) resistor in series.
!SPE is the master reset. It resets when low. If you have a reset pulse that goes low briefly, use that. Otherwise, tie it high through a series resistor and use a small capacitor to ground. (R=1K and C=0.01uF or whatever you have on hand that is close.)
Your clock pulse goes in on CP.
Q0 divides by 2
Q1 divides by 4
Q2 divides by 8
Don't forget the obligatory supply bypass of 0.1uF or 0.01uF.
 
Just read the datasheet:
P0, P1, P2, and P3, are pre-load inputs. You don't need them. Tie them to ground. (They cannot float!)
PE and TE are count enables. Tie them to supply. I recommend using a 1K (or whatever you have handy) resistor in series.
!SPE is the master reset. It resets when low. If you have a reset pulse that goes low briefly, use that. Otherwise, tie it high through a series resistor and use a small capacitor to ground. (R=1K and C=0.01uF or whatever you have on hand that is close.)
Your clock pulse goes in on CP.
Q0 divides by 2
Q1 divides by 4
Q2 divides by 8
Don't forget the obligatory supply bypass of 0.1uF or 0.01uF.

I think now i can do it :) Thank you very much for your explanation!
Noyan