Protecting CMOS Devices?

[Paul Marossy]

With the recent experience with my ART X-15 MIDI Foot Controller having some bad 74HC244 chips in it (CMOS), I was wondering about something:

What can be done to prevent those sorts of chips from becoming damaged from static electricity? In the case of the X-15, the PCB is completely isolated from the metal casing it is in. Would connecting the PCB ground to the case minimize that problem or would it cause more problems?

I ask because I am 90% sure that static electricity is what killed those IC chips and I want to prevent it from happening again.  :x

[wui223]

i think there is a kind of anti-static sticker to stick on the IC. but i dunno where to get it? perhaps some IC in a computer may have

[Peter Snowberg]

How many 244s went out and under what circumstances? What are they connected to in the circuit? ....and finally, is there a schematic on the net?

The traditional protection for a digital CMOS input is a reversed zener from the input node to ground followed by a current limiting resistor between the input node and the chip input. Since CMOS input drive current is so minuscule, you can easily have giant input resistors there to protect the inputs without affecting the operation.

If the circuit is powered by +5V you might be able to do something like drop an LS244 into the same position. That would make it just about bullet proof.

[R.G.]

I highly recommend "The CMOS Cookbook" by Don Lancaster. Unlike some other cookbooks, this one is available.

If the 244's were used as input chips, there may be something you can do short of redesigning the whole thing. CMOS input protection on the chip is designed to protect the chip in handling, not after manufacture. There are exceptions, but the 74C, HC, HCT, etc. are not among them.

The in-chip protection diodes and resistors are fairly fragile. As Lancaster puts it, you have to protect the protection. At every place a CMOS input comes from off the board, add series resistance in the 10K to 100K range and diode clamps to ground and +V. This will insure that only transients within the power supply range plus a diode drop ever gets to the chip. It also protects the chip when the power supply is off from low-impedance transients.

[Paul Marossy]

How many 244s went out and under what circumstances? What are they connected to in the circuit? ....and finally, is there a schematic on the net?

Check out this thread I started a few days ago: http://www.diystompboxes.com/sboxforum/viewtopic.php?t=36393 The schematics are linked there.

I'm not sure if both 244s went bad, but I replaced both of them just to be sure (I don't have a logic probe and wouldn't know how to use one if I did). As far as how the crapped out, it was just in normal use. It was also a gradual thing - it worked, but not quite right and then not all. The 244 I suspected is labeled "U6", and controls what the footswitches on the MIDI footcontroller do. Herein lies the problem, IMO: the PCB is inside this metal case, but is completely isolated from it. So, if your were using this thing on some new carpet on a cold winter day, POW! Can you say static discharge?! Maybe the designers at ART figured that protection from that kind of thing would be provided from the ground on the 9VAC wall wart? Or maybe no thought was given to this potential problem? I know of one other individual that had to do the exact same thing to fix theirs (replacing the 244s).

Anyhow, I am hoping there is something simple to I can do to prevent this from happening again. I have sockets for those two ICs in there now, but I would rather not have to go thru this ever again. It could be a different chip next time...  :?

[R.G.]

That's the chip I'd have suspected. It has inputs that go directly to pushbuttons, but there are *NO* protective components on them. Zap the pushbutton, and some of the zap will get to the CMOS input line. You'd think those guys would know better.

OK, connecting the metal shell to circuit ground might or might not help. It might help by providing a larger hunk of metal to zap on, but there's no real line ground anywhere in here. I don't suppose it will hurt, either.

I personally would protect those inputs that go off to the pushbuttons (pins , 4, 15, and 17). I'd put on each one a series 10K resistor and a diode from ground to pin and pin to +5. Arrange the diodes so they're reverse biased normally. That way, a transient riding in on the KSCAN line will move the outside end of the 10K to above/below the power supply, a diode will conduct, and the excess energy is eaten by the 10K and the power supply, which is where the energy is dumped.

I started to type that you'd think the designers would have done a better job, but I see that they used a zener/transistor power regulator instead of a three terminal. The designers didn't get to do this one. The MBAs did. They were too cheap to put in four more resistors and eight more diodes.

And they got away with it.

[Paul Marossy]

That's the chip I'd have suspected. It has inputs that go directly to pushbuttons, but there are *NO* protective components on them. Zap the pushbutton, and some of the zap will get to the CMOS input line. You'd think those guys would know better.

Exactly my point! I suppose that ART felt the rubber pads over the footswitches would act as a insulator? Or more likely, they probably wanted to save 50 cents on each unit.  :roll:

OK, connecting the metal shell to circuit ground might or might not help. It might help by providing a larger hunk of metal to zap on, but there's no real line ground anywhere in here. I don't suppose it will hurt, either.

I was thinking this would probably be a 50/50 chance that it would work.

I personally would protect those inputs that go off to the pushbuttons (pins , 4, 15, and 17). I'd put on each one a series 10K resistor and a diode from ground to pin and pin to +5. Arrange the diodes so they're reverse biased normally. That way, a transient riding in on the KSCAN line will move the outside end of the 10K to above/below the power supply, a diode will conduct, and the excess energy is eaten by the 10K and the power supply, which is where the energy is dumped.

That would actually be fairly easy to accomplish. I think I will do this to make the thing as bulletproof as possible!  

I started to type that you'd think the designers would have done a better job, but I see that they used a zener/transistor power regulator instead of a three terminal. The designers didn't get to do this one. The MBAs did. They were too cheap to put in four more resistors and eight more diodes.

I have a sinking feeling that you've hit the bullseye on that one!  :x

[Paul Marossy]

One last (dumb) question: Wouldn't it be enough to just add the 10K series resistors on those four inputs - I mean wouldn't any energy traveling along those routes be dissipated by the 10K resistor by itself? Or would it not be - because it's that sudden high voltage / low current presented by the static electricity at the pins which can damage a CMOS device when it's in a circuit?

[R.G.]

10k's by themselves? Sure, better than nothing.

Here's the reasoning.

Consider a 10kV spike which jumps into a keyscan line, rips down it for a while and hits the CMOS input. The CMOS input is designed to withstand peaks up to 15kV. Why are we worried?
(1) The 15kV is probably the "human body model" which is a certain number of pF through a given resistance - limited energy in other words.
(2) the input may have already eaten several hundred of these.

When the spike hits the chip, the chip input is pulled t'ords heaven or hell by the spike, as the polarity may be, and is only reined in by the on-chip protection. The on-chip protection is a silicon resistor in series and parasitic diodes from along the body of that resistor to chip ground or +V. The chip "ground" is a metalization layer that's 20ma thick; likewise the +V. From there, it has to get to power supply +V and ground, and it does this through pins and PCB, which are about 2 to many amperes thick.
There's a certain chance that spikes will eat away some of the 20ma metal on the chip with each spike. It's like the t-shirts: "I survived Three Mile Island - I think..."

Putting a 10K resistor outside the chip limits the current to ... um... 10kV and 10K ohms... one amp??? Yep, if the chip does the clamping to +V and ground. It will probably do this fine - for a while.

If you put the clamping diodes outside the package, you now have much beefier parts to handle the transient current and it never eats chip metalization nor wire bonds. You get to pick the level of protection you like. After all, it lasted a while to get to here.

If you are sharp, your next question is "Well, how come I don't just do the easy thing and buy a few extra '244's and put a socket on the board? Wouldn't that be easier?" It sure would. Maybe tape a few extras inside the covers. That IS a valid way to handle it, the price of '244's being what they are.

[Paul Marossy]

If you are sharp, your next question is "Well, how come I don't just do the easy thing and buy a few extra '244's and put a socket on the board? Wouldn't that be easier?" It sure would. Maybe tape a few extras inside the covers. That IS a valid way to handle it, the price of '244's being what they are.

Well, I did put those chips in sockets at the time of replacement.  :wink:  Thanks for the explanation on the mechanics of the IC chip, now I can see why all those things need to be done.  You're right, those are cheap IC chips -  I could go either way on that...

[austin]

This is interesting... most of the push buttons I've seen are isolated (the button surface is not electrically connected to the conductors).  Are these push buttons different?  Can an electrostatic discharge jump from a plastic button surface into the circuit?  Nice job on troubleshooting it.

[davebungo]

I have used these in the past for clamping surges to the rails:
http://www.fairchildsemi.com/pf/BA/BAR43.html

They are SMD packages of 2 diodes in a 3 terminal device.  I haven't actually used them for static protection; more for clamping reflected pulses in digital circuits.  They have quite a low forward voltage drop (in the order of 300mV).  I used them simply because we had some available and didn't have much space on the board.  You still need the dissipating resistor though.

I stumbled on this random article also:
http://www.ce-mag.com/archive/01/Spring/Lee.html

[Paul Marossy]

This is interesting... most of the push buttons I've seen are isolated (the button surface is not electrically connected to the conductors).  Are these push buttons different?  Can an electrostatic discharge jump from a plastic button surface into the circuit?  Nice job on troubleshooting it.

I know, it's kind of a paradox, isn't it? I hear what you are saying and I don't know exactly how it gets damaged from static electricity, but it happens. Including myself, I know of two instances of this happening to an X-15.

[Dirk_Hendrik]

Hold it on one point.

Since when have TTL IC's (74 series) become CMOS (40 series)??

[Paul Marossy]

Hold it on one point.

Since when have TTL IC's (74 series) become CMOS (40 series)??

See the data sheet: http://focus.ti.com/lit/ds/symlink/cd74hc244.pdf

[GFR]

Hold it on one point.

Since when have TTL IC's (74 series) become CMOS (40 series)??

SInce they put "HC" after the "74", like "74HC".