? on CD4027BE

It is VERY unlikely that esd is your problem. I've built many projects with

You've got something else wrong. Post your schematic and someone here will locate your problem.

Bob

Wow. Never damaged any? Honestly, I suspected I had something else wrong, but ran out of ideas several times. A new chip would work for a very brief time, and then fail as described. I blamed myself for taking the breadboard away from the workstation. Hate to say it, but I hesitate to post the schematic as I am hoping to copyright it. May do so anyway. I just can't figure out any differences between my first prototype and my second one (on a larger breadboard, with more space for added features...) Sorry if I sound paranoid, I'm just staking a lot of hope on this. Thanks for the input though. You have me thinking again...

Dave

Hmmm. Thank you, John. More food for thought.

Think about other failure modes.

Are you exceeding any pin current limit or driving the pin voltage outside of the supply rail band (which can cause the chip to turn into an SCR shunting the supply)?

If in doubt, can you add resistance to each pin, without preventing the circuit from functioning. Think about how high that resistance should be able to be and still allow normal function, based on the current you expect each pin to pass. Once in place, these resistors allow you to find unexpected currents (by the difference in the voltage at the two ends of each resistor, and an application of Ohm's law).

Realized this AM that I most likely do have something wrong with the design of my circuit. But it is probably a variance between what the schematic says I should have and what I actually have hooked up. Thus, posting the schamatic would probably be of limited usefulness. I need to go over the breadboard and make sure that I am following the schematic, which does work, as it did in my first prototype (which I have half-cannibalized by this time. May rebuild it along the way.)

Thanks, all for the feedback, and for giving me something else to think about.

Best,

Dave

"...copyright it." Bwahahahahahahaha! Your description of "operation" shows that you have no clue

...Jim Thompson

put some pull resistors on the in's/outs or use some TVS diodes..

Dave; In your first post, you mentioned +5volts and -5volts. Normally, I would connect +5v to pin 16 (Vdd) and ground (common) to pin 8 (Vss). Is this what you are doing, or do you have -5v connected to the chip?

-Dave Pollum

First rule of design: Circuits are not patentable.

Second rule of design: On CMOS **ALL** unused inputs have to go somewhere, ground or Vcc.

Third rule of design: Has anybody used 4000-series CMOS with split supplies?

And I've never damaged a 4027 either, and have used several thousand of them.

Jim

-- "It is the mark of an educated mind to be able to entertain a thought without accepting it." --Aristotle

I have. I have flown the supply of CD4XXXs up at the plus rail of a

The CD4007 could be used as a MOSFET array to do analog switching.

But the OP is trying to use a CD4027, which is a J-K flip-flop. Yet talks of driving an NPN and a PNP :-(

...Jim Thompson

If you don't know how to do that, just ask for help. ;-)

MINUS five volts???

Copyright is automatic, so you already have copyright on a circuit that is your original work, provided it has the usual Copyright and date line on it when you publish it. If you really mean Patent instead of copyright then you would be wasting your time and money.

Dave.

Change the name of the circuit, change two resistor ratios, change the goddamned DATE and it is no longer the same copyright.

Jim

-- "It is the mark of an educated mind to be able to entertain a thought without accepting it." --Aristotle

Sure, and 0/+12 works just fine along side my op-amps hoisted on a pedestal with a virtual ground at +6V. ;-)

I think I've strung a 4001 between +12/-6 as well, since the op-amps were biased to ground in that particular incarnation of the same circuit.

Tim

Hello Dave,

Thanks for the serious question. :) No, I do not have -5V connected to the IC. I have it connected as you describe. The referrence to -5V was just my description of how it was acting. But I have since figured out what I am doing wrong. Stupid mistake, which I can hardly believe I made, but I did. Not to worry though. All is well again. Just had to change the way I use Q and Not-Q from the flip-flop.

Take it easy...

Probably directly from Q/Q' to the base without as much as a base resistor.

Not a problem if they are connected common collector.

-- Regards,

John Popelish