Human Bistable

redart · Apr 17, 2006

Hi,
I've built this circuit (design courtesy of **broken link removed**) and am having trouble with stability. Sometimes it starts ok and the left hand LED flashes twice for each flash of the right hand LED, as per design. Sometimes it doesn't start, or the LEDs will flash out of sequence. Once it has started it will work fine for maybe 30 seconds, then the LEDs will again go out of sequence. However, if I touch the supply line with a finger, it will start working perfectly and will continue to work as intended until I remove my finger. Can anyone offer an explanation of what is going on, and suggest how I can modify the circuit to replicate whatever it is I'm doing by touching it ?. Thanks.

Dr.EM · Apr 17, 2006

I'm guessing then the driver transistor is crowbarring the power supply when it switches. Your finger is behaving as a capacitor, which alleviates the situation. Place 0.1uF ceramic caps across the chips power supplies. Failing that, try a 2N7000 Mosfet driver instead of the 547.

Nigel Goodwin · Apr 17, 2006

Dr.EM said:
Failing that, try a 2N7000 Mosfet driver instead of the 547.

Fine on the 0.1uF, but I don't see the reasonign behind changing the 547 for an FET? - what possible difference can it make?.

Styx · Apr 17, 2006

try resetting the Dtype on powerup (ie using one of yr remaining Schmitt IC's to setup a pulse to reset the D-type )

Dr.EM · Apr 17, 2006

Hmm, I was assuming the crowbarring effect was related to the semiconductor technology, BJTs doing it and FETs not? I was assuming that because the BJT 555 does crowbar, but the CMOS one does not. Mabye its just to do with the topology used in those particular chips then, not the semiconductor type?

redart · Apr 17, 2006

Thanks for the suggestions. I've added 0.1uF caps between pin 14 on each chip and ground, but no difference. Wouldn't the 47uF already be serving to decouple the power supply?.
As for the reset, this wouldn't solve the problem of it falling into an unstable mode after normal operation, so my finger still seems to be the only solution.

audioguru · Apr 17, 2006

Try using a single input on each 4093 Schmitt trigger NAND gate so it operates as a true Schmitt trigger inverter, by connecting the unused input of each gate to the positive supply.

Are the unused inputs of the 4093 and 4013 properly terminated?

docel · Apr 17, 2006

human

1. The 4093 is overloaded by the LED. Not a good idea to put a 'blinking' led as load on a schmitt, hysteresis etc.,. Also try the tremperature of the IC.

2. You may be lucky by introducing a small value 'R' from +9V line to pin 14 of the 4093, with a .1 to .22uf cap to gnd.

3. If using a 9V battery, change it.

Any one of these may(!) help you.

If it worked at doctronics, it should work anywhere...

Nigel Goodwin · Apr 17, 2006

Dr.EM said:
Hmm, I was assuming the crowbarring effect was related to the semiconductor technology, BJTs doing it and FETs not? I was assuming that because the BJT 555 does crowbar, but the CMOS one does not. Mabye its just to do with the topology used in those particular chips then, not the semiconductor type?

A 'crowbar' is a device that shorts directly across the supply rails, generally in order to blow the fuse - as there's no device across the supply it can't possibly occur.

Presumably a 555, as it can both sink or source current, may have a short moment when both are turned ON, as in a badly designed H-bridge. This isn't a failing of BJT, an FET H-bridge has exactly the same probelm.

Roff · Apr 17, 2006

Here is a compilation of suggestions already posted, with one of my own added (paralleling the Schmitts for the LED).
If this doesn't work, your battery is probably too weak. Try a new one, or 6 AAAs.

Dr.EM · Apr 17, 2006

Ah, thanks for the explaination Nigel, that makes sense. Presumably when they released the ICM7555 they redisgned that dodgy output stage. The NE555 definately does it though, it's virtually impossible to run 2 at different frequencies off the same power supply. An "LED mod" is detailed somewhere which is supposed to overcome it, but when I was using them I just decided the 7555 was easier.

Nigel Goodwin · Apr 17, 2006

Dr.EM said:
Ah, thanks for the explaination Nigel, that makes sense. Presumably when they released the ICM7555 they redisgned that dodgy output stage. The NE555 definately does it though, it's virtually impossible to run 2 at different frequencies off the same power supply.

Perhaps a little misleading?, it's common to run multiple 555's off the same supply, but usually they are running at quite different frequencies.

redart · Apr 18, 2006

Thanks for all the suggestions. Circuit is now working. Solution was as per Ron and Audioguru's suggestion of tying the Schmitt trigger input pins 1 and 5 to the supply rail, and grounding the unused inputs of the 4013.
This last point raises a question:- according to the data sheet, the two flip-flops on the 4013 are, quote, 'independant', so why is it important to terminate the unused inputs of the unused flip-flop ?.

docel · Apr 18, 2006

edited by docel

Styx · Apr 18, 2006

Re: Half-bake

docel said:
....and why did'nt you do it, in the first place??
Dontronics is explicit about this!

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.)

Click to expand...

You have wasted the precious time of several brains, not to mention their valuable time by being half-baked.
i'm sorry to be acidic, as a newbie, but this is preposterous!

now was that really needed?

sure this might of been solved by reading the datasheet, but the reading didn't take place.
What has occured is this particular problem as well as the solution (as well as the un-called for insult !!!!) is not documented and can be searched via GOOGLE if someone had a similar problem

no harm has been done, someone has learn a valuable lession abt electronics (and why you should always show all GATES on the schemetic). What wasn't needed was the attack

docel · Apr 18, 2006

You're right Styx, and i've blanked the post. Thanks for the hint.

Roff · Apr 18, 2006

redart said:
Thanks for all the suggestions. Circuit is now working. Solution was as per Ron and Audioguru's suggestion of tying the Schmitt trigger input pins 1 and 5 to the supply rail, and grounding the unused inputs of the 4013.
This last point raises a question:- according to the data sheet, the two flip-flops on the 4013 are, quote, 'independant', so why is it important to terminate the unused inputs of the unused flip-flop ?.

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.

redart · Apr 18, 2006

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.

docel · Apr 18, 2006

unused gates

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!

redart · Apr 18, 2006

Docel,
Check who you are quoting. It was Ron H who wrote "I am also curious about this.....etc".
Also, you seem to be missing my point. The 4013 contains 2 supposedly independant flip-flops. I can understand stray signals affecting unterminated inputs on the flip-flop being used, but if the 2 flip-flops are truly "independant", then signals picked up in one should not affect the other.

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