Basically I have a 4 Bit Shift Register that I am trying to get to work. From my understanding when I send a pulse of 1-1 to inputs S0 and S1..it is suppose to do a parallel load.

I have ensured that the pulse coming is clocked exactly with the clock input (1hz) connected to the Shift Register. Unfortunately, every time I load 0000 inside, the value somehow erractically changes! It goes from 8-4-12-2 to 0, despite the fact that it was suppose to stay at 0!

I have no idea why this is happening please help!

  • \$\begingroup\$ A schematic showing your register, output and clock would be useful. \$\endgroup\$ – Trygve Laugstøl Oct 26 '12 at 9:43

You neglected to say what part number you are using, which makes other people do unnecessary googling. Is it the 40104/40194?

Anyway, it is not true that you send a pulse to S1 and S0. The values 1 1 should be presented, and take effect on the rising edge of the clock.

If you try to set the inputs to a sequential logic device at exactly the same time as the clock edge, you will get erratic behavior. In general, inputs must already be settled by the time the clock arrives.

For instance, take a look at the timing diagram on page 8 of this datasheet: http://www.datasheetcatalog.org/datasheet/philips/HEF40194BD.pdf

There is a minimum "set up time" and "hold time" before and after the clock edge.

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  • \$\begingroup\$ Sorry I was using a 74LS194.. \$\endgroup\$ – Raaj Oct 26 '12 at 9:04
  • \$\begingroup\$ But that should not be a problem yes? Since I am sending the input signal at the same time as the rising edge..it will still register should it not? \$\endgroup\$ – Raaj Oct 26 '12 at 9:05
  • \$\begingroup\$ @user1436508 - Setup time means just what is says. The input signals to the shift register [S0, S1, Pn (n=0-3), Dx (x=R or L)] must be "setup" and valid _before_ the rising edge of the clock. The data sheet for your particular part will tell you what the minimum required setup time is needed. Some inputs like the mode select lines may require different minimum setup times than the parallel inputs or the left/right serial inputs. Many CMOS type parts that are capable of operation over a range of voltages will specify different minimum setup times for specific voltages of operation. \$\endgroup\$ – Michael Karas Oct 26 '12 at 11:20

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