Help with the 74ls93

necessarynoize said:
i have 2 74ls93...i want make 1 to be divisible by 8 and the other1 by 5..but i dnt know which 1 to connect..
hi,

THe 74LS193 is binary counter so you would use the Qc output, for a count of 8, of the first LS193 to drive the second LS193.

For a count of 5 you would need to detect when the second LS193 has counted to 6 say, thats Qb and Qc outputs, when this count of 6 is detected it drives the clear pin on the second LS193.

An 'AND' gate would be used to detect Qb and Qc.
If the the count rate is low, then a couple of signal diodes and a resistor could make the 'AND' gate.

As the the LS193 is a synchronous counter there are a number of ways to detect and restart for a count of 5.

Do you follow this OK.?
 
There is a way to count by 5 with the LS193 without using an external gate. You load in a count of 5 (Inputs ABCD=1010, A is the LSB) at the count of zero and operate in the count down mode.

Tie the UP and DOWN inputs to ground. Tie inputs A and C to +5V. Tie inputs B and D to ground. Tie the BORROW out to the LOAD input.

When the count reaches zero the BORROW output goes low which then loads the counter with binary 5 and restarts the sequence.
 
yes i follow but...

i really follow but iam trying to do a fruit machine project..i got a 555 timer feeding in to 3 74ls93. 1 is divisible by 9 so i managed to get that to work which then feeds to the 74ls47 then to the seven segment display..but am stuck on how to make the remaining 2 1 to be divisible by 8 and the other1 by 5....help plz..




 

hi,
Do you need to drive 7 segment display drivers 7447 with the output of these dividers.?

If Yes, then the down count suggested by 'crutschow' will give a count of 5 down to 0.

Can you post a block diagram of what you are trying to do with the dividers.?
 
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