outputprotection
I know it's important to protect analog and digital inputs. My question is should you also protect outputs, both analog and digital?
For the inputs, right now I'm using clamping diodes, an MOV, a series resistor, and a series ferrite bead for high frequency noise rejection.
If outputs should be protected, how should this protection differ from inputs?
The typical way to protect ADC inputs (and many other high impedance inputs) is to have a current limiting resistor and clamping diodes at the pin. Many chips already have clamping diodes inside the chip, but it is often useful to have some outside as well (especially when the integrated clamping diodes do not have proper documentation on their specs, which is most of the time).
Your resistor-divider is already going to function as a current limiting resistor. As a general rule, you want your resistor values to be as high as is practical-- with something in the neighborhood of 10K to 500K being reasonable.
As the resistors go higher in value you start to get weird effects from the parasitic capacitance (lowers your frequency response) and the leakage current of the ADC and diodes (messes up your accuracy). If you are unsure about these things, then pick something on the low side of 10K to 500K. If you have already figured out the math or are willing to just try it and see if it works then go with something higher.
There would be one diode between GND and the input. And another diode between the input and VCC. If you have a signal that goes + and -, then connect one diode to the V+ rail and the other to the V- rail. Orient the diodes in such a way that it works (turn the diodes around and nothing will work). Put the diodes between your ADC and the resistor divider.
It works like this... If you get a voltage spike coming in, the diodes will redirect it to a power rail and not directly into the chip. The current limiting resistor will, duh, limit the current to something reasonable. You want to current to be limited for several reasons: 1. You don't want so much current that the diodes blow up. 2. You don't want so much current going into the power rails that it effects the voltage regulation of that rail.
With this setup, the maximum voltage on the input pin will be VCC + the forward voltage drop of the diode. And the minimum voltage on the input will be GND - the forward voltage drop.
There are hundreds or thousands of appropriate diodes for this. The BAV99 is just one of many, but is a good one to start looking at.
Better to use silicon diodes for D1 and D2 rather than schottkies, on leakage grounds. I know their Vf is larger, on paper probably exceeding the maximum mux input voltage, but their leakage will be orders of magnitude lower. Some silicon diodes are advertised as low leakage. However, there's little point in striving for a leakage lower than your mux will provide. Note that both diode and mux leakage currents tend to increase exponentially with temperature, sometimes a horrible looking data sheet figure for worst case at temperature will be OK if you will only be using your system at ambient.
Use a resistor between the diode clamp and the mux input to limit the current into the muxes protection diodes, once D1 or D2 are clamping.
Not all multiplexers are equal, some have robust input protection as they're designed for this type of use. Some specify their input protection didoes can tolerate a high current. Do a wide search and read the data sheets carefully.
Do not assume leakage currents will cancel. Leakage is an uncontrolled parameter.
Don't forget to use an adequate voltage rating for R3, your common or garden resistor is usually only good for 200v. Use several in series, or one rated for a much higher voltage, they're not too expensive. Spikes of 1500v are common on mains.
Something like PUSB2X4Y has pulse specifications in the amps. If it can take a microseconds pulse of 4.5A at typically 3.8v across it, then you can assume it will take 10s of mA all day, every day, without embarrassment.
In case you haven't spotted Matthew's comment below, here is what I think he was suggesting, in the left-hand diagram, back to back diodes across a buffer op-amp.
simulate this circuit – Schematic created using CircuitLab
Although the op-amp output could be thought of as more current-capable than the inputs, it still has substrate diodes and a maximum current specification, so is also in need of protection itself. Some amplifiers even have inputs specified to well outside the rail for input protection, but only 0.3v overvoltage and weedy current spec on the output pin.
Going on from that basic idea, my take on the principle is shown on the right. The diode string D3-6 provides voltage clamping to 'a little outside' the rails, R2 protects the diodes, R4 protects the amplifier input, R3 protects the amplifier output and bootstraps the voltage across diodes D4 and D5 so that their leakage to the input is minimal. With such bootstrapping, the diodes could be almost anything, even big rufty-tufty bomb-proof rectifiers.
The two diodes in series suggests that care is needed in layout if protection is to extend to fast pulses. Consider the SOT-23 BAV99, two series diodes in one package, to implement the pair D3,4 and the pair D5,6. They are specified continuous >100mA, typical 10mS pulse 800mA, which for any reasonable R2 sounds adequate. BAT754S is an alternative in schottky. Similar currents, but much lower clamping voltage.
You do not actually need an op-amp per channel if your multiplexer leakage is low enough. The circuit at the bottom shows the single buffer following the mux driving all of the input protection diodes. Note that the multiplexer leakage appears at the amplifier input, whereas using a per-channel buffer eliminates mux leakage.
The 'on' channel is receiving the correct bootstrap voltage. The 'off' channels will probably be getting the wrong voltage, and the 'inner' protection diodes may well conduct. This is not a measurement problem, as the channel we want is correct. It may, or may not, be a problem to what is driving those inputs, to have our nominally high impedance inputs yanked off to a different voltage. If we assume it's a very feeble current source (we are concerned about leakage, so we know it's not a low impedance source) with a large capacitance to ground, it may take a long time after selecting that input before the voltage has returned to its correct value.
Actual leakage measurements for diodes at 15C.
diode -2/-5v leakage slope resistance over +/- 10mV
----- -------------- ------------------------------
1N4148 4nA 30Mohm
BAT42 35nA 1Mohm
BAS116 <10pA (30v) >>20Gohm
The BAS116 conduction continued as 40pA 300mV, 45nA 450mV, 16uA 640mV. The BAS116 typ/max spec at 25C is 3pA/5nA, and 3n/80n at 150C.
That means, at that temperature, and making the assumptions of reverse leakage varying by a factor of 2 up and down, and 3mV voltage follower offset, you could assume the following leakage
diode no bootstrap bootstrapped
----- ------------ ------------
1N4148 6nA 1pA
BAT54 50nA 3nA
BAS116 <10pA <<10pA
I made those measurements with a £8 meter with 10M input resistance and a 200mV range, so 10pA per LSB, not difficult (obviously can't tell the difference between 0 and 10pA!). I suggest you do the same with your chosen diodes and at higher temperatures.
Best Answer
As mentioned in the comments, it will depend on your system and it's requirements. It's up to you to evaluate your system and its intended use and decide whether or not to include the protection. Here are some examples of things to consider:
Arguments for Protection:
User Error
Users can find wonderful and borderline magical ways to plug things in incorrectly, so adding protections to every electrical connection that the users have access to will make your device more robust. This is especially a concern in consumer electronics which have a lot of users that aren't very technologically sophisticated (or who have children).
ESD
As mentioned in the comments, even properly used devices that aren't always grounded can build up a static charge that can damage your device.
What is the output driving?
If you have an analog output driving an inductive load (such as a simple DC motor or solenoid) you have to protect against inductive kick and Back EMF. If your device is subject to these kinds of events during normal operation, protecting your outputs isn't really an option.
Other
There is always the chance that unexpected conditions may damage a device, even if users are doing the correct thing and ESD isn't a concern. For example, consider an audio mixer (which are always grounded). Even if an expert plugs all the cables in the right way, there may be two worn out cables in a bundle of 30 of them that causes two outputs to be shorted together. If they drive at different voltages, one of those two outputs is going to be very unhappy if it's left unprotected.
I'm going to bundle the arguments against additional protections into one group, because they're pretty much the same as any other electronics projects: increasing the number of components increases design and validation time and complexity, all kinds of costs, and board and packaging size.
In the end, you have to balance the level of protection required for your product against the various costs associated with implementing those protections.
It's hard to answer your second question without a specific application in mind. For some low voltage analog and digital outputs, a simple diode placed in series with your output may be sufficient. Other situations may require more complex protections like an RC or diode snubber for inductive loads. It just depends on your application.