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Human Bistable

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hi redart,
It should be true, if they are independant. Unfortunately they both share the same power bus. The switching and other transients (AC buzz, motor etc.,) will be conveyed via the power bus. CMOS devices are very sensitive and hence the problem of erratic behaviour.

This is a very troublesome handicap that troubles a lot of people, believe me1
 
Re: Half-bake

Styx said:
by being half-baked.

now was that really needed?

(as well as the un-called for insult !!!!)

What wasn't needed was the attack
Aw, I think I missed the fireworks.
 
Re: unused gates

docel said:
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!
Well, I know all that. I design DRAMs for a living, for God's sake. I'm just having trouble believing that one CD4013 FF (slow logic) rattling around can disturb its housemate. The typical shoot-through current for CD4000 series outputs on a 9V supply is on the order of 15mA - say 30mA max. Double that for Q and Qnot, and it ain't enough to disturb a power supply with 47uF and 100nF decoupling it, unless the power distribution or battery resistance is atrocious! The chip might get a little warm, but that shouldn't cause a problem. I certainly won't buy an argument that random rail-to-rail switching (at normal transition times) can cause a problem.

Please note that I am not advocating leaving the inputs floating. I'm just baffled as to why it didn't work with them floating.
 
Ron,
You were absolutely right. When I re-did the circuit I tied pins 1 and 5 of the 4093 to the supply rail AND grounded all the unused inputs of the FF before testing it. Only one of these was necessary. The circuit works just fine with pins 8, 9, 10 and 11 of the FF floating. I can poke it and prod it and no mysterious stray signals magically jump from one FF to the other. I should have changed one thing at a time....a lesson this half-baked newbie had to learn the hard way. Apologies to you for the misdirect.
Now to tackle what I intially set out to do - get my audio A-B switch box relays to latch on and off using a momentary push-for-on switch and the 4013 FF. Thanks again to everyone for the help.
 
redart said:
Ron,
You were absolutely right. When I re-did the circuit I tied pins 1 and 5 of the 4093 to the supply rail AND grounded all the unused inputs of the FF before testing it. Only one of these was necessary. The circuit works just fine with pins 8, 9, 10 and 11 of the FF floating. I can poke it and prod it and no mysterious stray signals magically jump from one FF to the other. I should have changed one thing at a time....a lesson this half-baked newbie had to learn the hard way. Apologies to you for the misdirect.
Now to tackle what I intially set out to do - get my audio A-B switch box relays to latch on and off using a momentary push-for-on switch and the 4013 FF. Thanks again to everyone for the help.
Redart, thanks for going to the trouble of testing that. Changing one thing at a time gives you insight into why the circuit is acting strangely, but you needed to tie all the unused pins down anyhow, so it wasn't a rash move on your part.
BTW, when I was researching this, I ran across a Fairchild app note about unused inputs on used CMOS gates.
The accepted wisdom is to tie unused inputs to the positive rail on a NAND gate (or GND on a NOR gate), because it minimizes input capacitance. However, if you tie them to a used input, you will get higher pullup capability from a NAND gate, or higher pulldown from a NOR gate. This little tidbit could come in handy if you are driving a heavy load, like an LED.
 
Ron H said:
BTW, when I was researching this, I ran across a Fairchild app note about unused inputs on used CMOS gates.
The accepted wisdom is to tie unused inputs to the positive rail on a NAND gate (or GND on a NOR gate), because it minimizes input capacitance. However, if you tie them to a used input, you will get higher pullup capability from a NAND gate, or higher pulldown from a NOR gate. This little tidbit could come in handy if you are driving a heavy load, like an LED.
Higher output current worked on original Cmos gates. New (for a long time) B-suffix devices have a buffered outputs to make the current the same for pull-up, pull-down or any number of used inputs.
 
audioguru said:
Ron H said:
BTW, when I was researching this, I ran across a Fairchild app note about unused inputs on used CMOS gates.
The accepted wisdom is to tie unused inputs to the positive rail on a NAND gate (or GND on a NOR gate), because it minimizes input capacitance. However, if you tie them to a used input, you will get higher pullup capability from a NAND gate, or higher pulldown from a NOR gate. This little tidbit could come in handy if you are driving a heavy load, like an LED.
Higher output current worked on original Cmos gates. New (for a long time) B-suffix devices have a buffered outputs to make the current the same for pull-up, pull-down or any number of used inputs.
That must have been an old app note. :cry:
 
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