4-bit shift register wont work (unknown state)

Question

I must design a 4-bit shift register circuit. I am pretty sure that I need 4 D flip-flops sharing the same clock. right??

So I did that and here is the result:

As you the LED lights are neither completely off (0) nor completely on (1) rather they seem to be in this, uh, indifferent state? Anyway, the point is, it is not working no matter where i put the switch.

If I go to each ff and remove the preset/clear lines the circuit works! but I do not understand why that is an issue? Why doesn't it just work the way I drew it?

Answer 1

In a logic circuit, you typically cannot leave any pins floating (not connected to anything).

If you include preset and clear control lines, they need to be set to a logic level, either high (your logic-level voltage) or low (circuit ground). If they are left floating, then the flip flops are in an undefined state. When you remove the preset and clear lines from your simulation, then the flip-flops will operate normally because it doesn't expect any input from them. D flip-flops can come with or without these additional control lines.

Take a look at the truth table for D flip flop with preset and clear control lines.

The preset and clear lines are considered "active low" which means they perform their functions when pulled low. When they are pulled high, the clock and D control the outputs as you would expect. Notice, if they are both pulled low, the outputs are in an undefined state (represented by 'X').

• Preset - sets the output (Q) high
• Clear - clears the output (sets Q low)

Answer 2

You could get away with leaving pins open in early bipolar IC families. It was always considered bad practice. Things could go wrong, but rarely did. Even so the only excuse for doing so was to save the expense of a resistor. It was another bad practice to tie input pins directly to Vcc. A resistor to Vcc provided protection from transients on the power bus. The maximum voltage rating for Vcc is higher than that for inputs. (The 74LS family was the first one that allowed direct connection for pulling up unused pins, because its inputs were diodes rather than emitters.)

It has never been allowable to leave CMOS inputs open. Unlike bipolar logic, an open input is not seen as a logic one. In some cases it's seen as an intermediate state, P-channel and N-channel MOS transistors are simultaneously ON, and the increased power can damage the IC.

All of that is TLDR, but your simulator has good reason to disallow open input pins. If it allowed specifying the flip-flops by part number, it would probably show the shift register working for 7474, but not for 74HC74.