CMOS input protection

On a sunny day (Thu, 10 Jan 2013 10:14:17 +0530) it happened "Pimpom" wrote in :

Use opto couplers.

Sounds good. You could even use 100K, as long as the schottkies aren't too leaky.

Old A-series CD4000 chips could latch up, and short the VCC rail to ground, if you forward-biased the ESD diodes a bit. Nasty stuff. With B series, and 100K resistors, just the ESD diodes would be enough; a cap to ground, lowpass filtering, and a Schmitt gate would be prudent. You don't really need a terminator.

If the grounds aren't the same voltage on both ends, you should do some sort of differential receiver. Opamps would work at these speeds.

Ideally: Use open-drain for the sender with a pull-up resistor on the receiver end.

OK: Diode, cathode toward sender, anode toward receiver, pull-up resistor at receiver. 10K should be adequate to limit back-current if receiver powers-up first. ...Jim Thompson

My knee-jerk suggestion is opto-couplers. Even if you have to get phototransistors and LEDs and roll your own.

Anything that you do involving a direct connection is going to open you up to untoward events happening when your grounds don't match or one or the other of your circuits is on when the other is off -- I'd make sure that you check every possible combination of on and off, as well as considering what happens if your grounds don't agree.

Isolators of some sort are much easier.

Almost ok but not quite. Add a few hundred ohms from your external diode and the IC pin. This makes sure that the lion's share of unwanted current flows through the external diode and not the substrate paths. Also, then you can use regular diodes such as the BAV99. Or a BAV199 if you need low leakage.

What can happen is that the signals coming in from the powered unit try to power up the other unit. The VCC there can then slowly creep up. Make sure this situation is handled properly and nothing can go kaboom because it resulted in undefined logic states. It can affect other units in return because the "half-powered" crcuit can send out undefined voltages on its outputs or possibly oscillate somewhere.

Remember the transmit. Maybe a bit naive, but I used to use 1k on the outputs and 10k on the input. Seemed pretty robust. Never did production in those days though.

This kind of vague thinking is the root of much design evil. You know what you mean, but we don't. The audio guy and a guy who builds picosecond laser drivers will have very different perceptions of "critical transition times". If the time is not critical, make it infinite and you don't need the cable. Numbers were invented for a reason. Use 'em.

Must be the phase of the moon. Today, there seem to be a lot of very vague questions with fromthehip answers. It's like a blindfolded quick-draw contest with live ammo.

You've been given some good advice, but there are a lot of issues to consider.

I take a lot of flak here for insisting that people define exactly what they're trying to accomplish.

A typical scenario involves tunnel vision. Some very smart engineer comes up with a solution that has one little problem. She asks about that problem and gets a solution to THAT problem, not realizing that it created two additional problems. It's often worth a few minutes to rethink the system.

The design goals for intended operation are no more important than the specs for what happens when some idiot gets their hands on it and does unspeakable things. Unfortunately, people tend to skip both.

You don't say what your "product" is, but there's a wide range of system requirements depending on the target demographic. A simple catch-all starting point relevant to this discussion is electrostatic discharge.

What ESD protection do you require and how are you gonna verify that you meet it? A 10K resistor sounds like a lot of protection until you start poking a 15KV ESD generator around. The devil is in the details. And the current doesn't always go through the resistor like you'd expect if you looked at the schematic. If you don't believe it, watch a few lightning strikes up close.

A properly deployed opto-isolator can cure a lot of common-mode system ills, but it's not a panacea. It might protect your CMOS input, but now, you have to protect the opto-isolator from ESD...unless you use fiber for the whole run. And that will address a number of other issues like EMC.

Diodes are another potential gotcha. You're gonna need a minimum of two resistors. One to protect the input from the peak overvoltage allowed by the diodes and strays. And another resistor to protect the diodes.

For 5V hobby circuits, I've taken to using 5.1V zener diodes as the protection element...and here's why.

I once built a GPIB interface from a PIC processor. I unplugged the power to the chip and the GPIB kept right on controlling. I immediately started writing a paper on how I'd written code that tapped into zero-point-energy and ordering parts for my water-powered car. Then, I realized that the GPIB was powering the chip through the forward-biased input protection diodes on the PIC. Not any part of it met spec, but it worked just fine.

Depending on the circuit, a diode clamp to VCC can let an input transient take out every IC on the board. That's probably not the protection you expected. It's easy to assume that you can shunt arbitrary current into VCC, but the math might surprise you.

The zener diodes can protect the circuit from power supply injection.

You've identified power sequencing as a potential problem. Consider when the protection diodes put just enough voltage on VCC to cause a lot of dissipation in some external power switching device and let the smoke out. Then, there's the problem that the robotic welding arm kills two workers if you turn power on in the wrong sequence.

We haven't even started talking about what happens when your system clock happens to be just the right frequency to resonate the cable and makes the FCC very unhappy. Unless your operator has a ham license and 40-meters is the electrical length. ;-)

Write the specs...ALL the specs including the design verification procedures and the customer acceptance test procedures and the third party certification test procedures. Consider customer abuse.

Details matter...that's where the devil lives...right next door to Murphy.

Something like the NC7SZ05 claims high impedance inputs and outputs when powered down. I'm not sure this protects external circuitry if the NC7SZ05 power sequencing is not ground-contact-first and ground-contact-last, which may be an issue you should check for with remotely located subassemblies.

You can't always guarantee that power, signals and their references will be connected in a specific way, unless you design them to do so.

RL

Is there a possibility this cable with generate voltage due to a piezo effect?

There will probably be scenarios where the far end is powered through protection diodes via the signal. Open drain would solve that.

I'd go opto.

...

and that's a good solution, especially if used with a CD4050-style CMOS gate which lacks input clamp-to-VCC diodes, because it prevents your signal current from (for instance) charging a nonrechargeable battery. This presumes, of course, that your two CMOS gizmos on the wire share a common ground which is their negative power supply terminal.

Another item that hasn't been mentioned yet, is a pullup or pulldown resistor on any CMOS input that might become disconnected when you unplug the long cable. You cannot ever let CMOS inputs float.

Yet another concern: if the circuit isn't double-insulated (isolated from ground) on either end, your grounding conductor in the cable could carry large currents in case of a fault. If that's an issue, the optoisolator DOES help quite a bit; otherwise, it's just a power and complexity bunion.

The CD4000 series should not latch up if the currents 'forced' through inpu ts and outputs are less than something on the order of 10ma. That is DC, bu t there are unspecified hazards with dV/dt induced internal currents on tho se terminals too. Since your signaling frequency is so low, a simple RC wil l eliminate any possibility of latch-up due to power supply sequencing or E SD, no BATs necessary. An internal loading of Vcc wil ensure that external powering of the dead circuit does not produce a working voltage for any oth er part. Please view in a fixed-width font such as Courier.

. . . . . VDD . | . .----+ . | | . | | . | |\| . >----[1M]---+---------| >---> . | | |/| . | [100K] | . === | | . 0.0033u | '----+ . | | . --- --- . /// /// . . . .

Those type of devices will work as desired.

The way they work:

Outputs use a P-channel structure that allows the body to float up along with the output node potential, thus it won't support current flow in the Output-to-VDD direction... an "almost perfect" diode.

Inputs don't use a diode to VDD. They instead use an avalanche mode N-channel device (to ground) to "catch" ESD. ...Jim Thompson

Probably, so long as there's no hot plugging of assemblies going on. Although power may be remote, there's always a ground either showing up, or missing, when least expected. Battery-powered devices are the worst in this regard, particularly if a charging feature is attempted. That gives a potential positive rail pre-contact. Ouch.

RL

The OP implied the cable was intact, but power could be cycled. Hot-plugging can be pretty wild, and takes a whole lot of specialized circuitry to make everything clean and happy. ...Jim Thompson