Ron H said:
I am also curious about this. I understand that floating CMOS inputs can cause excessive power dissipation, but I am surprised that the unconnected FF and gates caused the circuit to operate erratically.
CMOS inputs are very high impedance (~10^12 ohms). These inputs act as antennas for stray signals and shift back and forth between 0 and 1. This is evident, sometimes, when the finger is brought close to the input pin, without touching it. The output will switch erratically, as if a pulsed signal is is fed to it. This is due to our body acting as a "signal source". We do pick up and have induced currents from various EM radiations like radio transmitters, AC mains transitions etc.,
Simple grounding of the pins will eliminate the stray coupling and ensure the remaining signal pins are clean and respont as designed.
redart said:
Well I thought it was a reasonable question and Ron H. seems to agree. I also did read the data sheet and it says nothing about needing to terminate the inputs of the second, unused flip-flop. The Doctronics website only mentions the need to terminate the active flip-flop inputs, and clearly shows all other inputs (of both the 4093 and 4013) as being left floating.
The document is clear in its statement as can be seen:
from "http://www.doctronics.co.uk/4013.htm"
The inputs of the D-type must be connected, either to LOW or to HIGH, and must not be left open circuit. This includes the SET and RESET inputs which are connected to 0 V. To avoid loading the output of the D-type, a transistor switch indicator circuit is used. It is good practice with CMOS circuits to insert a decoupling capacitor, 47 µF or 100 µF, across the power supply. (This helps to prevent the transfer of spikes along the power supply rails.)
No-where does it say 'active flip-flop'. The 'must be connected' implies 'unused' pins of the D type and in italics too, to stress the point!