How accurately do you need to sense current? I am concerned about the ability of your MOSFET switch to stop the short-circuit current from a large battery bank. Remember, during the time that the FET is switching off, there is a huge power dissipation because the voltage is rising at the same time that the current is falling. I don't know what kind of battery bank you are talking about, but let's say it's a 24V 50 Amp-Hour lead-acid battery. The short circuit current could be 5000 Amps. At the worst point, there will be about half voltage (12V) across the FET and about half current (2500 Amps) through it. That's 30kW. Let's also say that the FET will switch off in 500 nanoseconds and that the overall average power dissipation is half of the worst case point. So, 15kW for 500 nanoseconds - that's 7.5 milliJoules. Might not seem like much but given that the actual die is quite tiny, and the bond wires are even smaller, it might well be enough to destroy them. This is ignoring the additional heat which will come from just I^2 R heating during the time when the current is ramping up before your system trips and decides to turn the FET off. It is also ignoring the inductive energy in the leads which will show up as a huge voltage pulse across the FET, increasing the power dissipation and possibly exceeding Vds max. Finally, it is also ignoring the difficulty in keeping a FET turned on fully during the very high current conduction - the voltage drop across the FET (drain-source) will likely exceed Vgs max which will prevent you from applying enough gate drive to keep the FET in the ohmic region. What I am saying is that a FET swtich that is designed to catch and stop the short circuit current from a large battery is NOT a small or inexpensive thing :) Sean On Sat, Dec 17, 2011 at 6:40 AM, Forrest Christian wro= te: > On 12/17/2011 3:09 AM, Mike Harrison wrote: >> Schottkys get very leaky at higher temperatures, so probably not a >> good choice. Is the current really so high that you can't use a more >> chunky sensing resistor? If you really want to bypass the shunt, a >> MOSFET is probably a better bet - arranged such that it turns on >> quickly when the measured current exceeds a threshold, witha >> monostable to hold it on for a while, limiting the current duty cycle >> through the resistor to the monostable time / the turn-on time. > Just to be clear what we're typically talking about here... > > The normal load is around 250mA. =A0Inrush currents can be extremely high= , > in the range of amps, but not close to high enough to exceed the > resistor rating. > > The issue comes where our customer is using a large battery bank with > virtually unlimited current available, and decides to dead short the > output. =A0I think we've actually got the firmware in the PIC doing the > overcurrent protection tweaked so failures just aren't likely - It's a > fine line between being slow enough that the inrush doesn't trip the > protection, and a short or long-term overcurrent will. =A0 With some > creative integration (in the mathematical sense), things seem to be > working well - but I still hate not protecting something I know is close > to the edge. > > At this point, I think I might look at using a mosfet on the input of > the whole device to limit current to something reasonable - dozens of > amps for instance - probably a more reasonable solution. > > And yes, we're already using about as chunky of a sense resistor as is > practicable. > > -forrest > -- > http://www.piclist.com PIC/SX FAQ & list archive > View/change your membership options at > http://mailman.mit.edu/mailman/listinfo/piclist --=20 http://www.piclist.com PIC/SX FAQ & list archive View/change your membership options at http://mailman.mit.edu/mailman/listinfo/piclist .