Device shutdown: harm from charged capacitors?

Question

In my application, I have 220nF capacitors connected to the inputs of an A/D converter (MCP3204). Now I wonder what will happen when the circuit shuts down and the supply voltage drops below the capacitor voltage.

Similarly, I am sending the output of a D/A converter to the input of an LT1013 op amp. On shutdown, the supply voltage of the D/A would probably drop more slowly than that of the op amp, so it is possible that the maximum ratings of the op amp would be briefly exceeded.

My intuition is that both these cases probably won't cause any issues - the A/D converter will discharge the capacitor through the internal protection diodes at least, and the op amp input has a 10K resistor in front of it - but I'm not sure where the limits would be. After all, it does exceed the absolute maximum ratings, so generally it can harm the devices. How do you figure this out with some degree of certainty?

Answer 1

If the power supply can be shorted at any time, or have a heavy load on it (especially more than tens of mA) you should use an external series resistor on the inputs to limit the current.

If you don't want to protect against a screwdriver across the supply, you can assume a maximum dv/dt at the rails (perhaps from the load and bypass caps) and ensure the current is limited by the cap value.

Answer 2

It most likely doesn't matter at all in your case.

Do check max energy stored in L and C when you design DC/DC converter or in fact any power curcuit.

Let's analyze your case in more detail -- what is the worst that could happen?

Supply voltage drops fast (how fast?) and ADin > Vcc.

Typical A/D converter includes over/undervoltage protection. If yours does not, perhaps you may need to include external components regardless of this scenario, to protect your input from being connected to something of different potential, line noise, interference, EMP, etc...

So let's say you some over/undervoltage protection, most like it's a diode in reverse. Thus when Vcc suddenly and abruptly drops to 0, the capacitor will be discharged through that diode.

You can compute max current through the diode taking into account parasitic resistance (RC chain) and parasitic inductance (LC chain).

I don't know the inner details of MCP3204 and data sheet does not specify protection circuitry, but I suspect the energy stored in 220nF cap at max VDD=7V is just way too small to cause any real damage.

To put this into perspective, if you worked with really fragile components, e.g. unprotected FETs, 220nF could easily be significant.

Keep in mind that this chip is features ESD protection on all pins > 4kV according to spec. That's quite a lot, but it does not actually specify max current at 4KV. I hazard a guess it refers to http://en.wikipedia.org/wiki/Human-body_model in which case peak transient current is 2.67A for a very very short time. Your capacitor stores nowhere near this energy.

Answer 3

Build it and see. Unless I'm missing something, it seems like you're concerned about minutiae that would break nearly every circuit in existence if it were a problem!