Protecting low side current sense resistor from shorts to V+ - Page 1
Forum section: electronics / projects

author: mwsoft
author_rank: Contributor
timestamp: May 07, 2015, 10:46:43 am
content: HiOn an automotive module, I have a low side current sensing input. It's just a small (~0,1ohm) resistor between module ground and input pin. Normal operating currents are small, in 1-2A range.I'm looking for ideas, how to protect this input from shorts to 24V, so that resistor doesn't vaporize.The problem is, that short can come from any circuit in the car, including those with e.g. 35A fuse. Short circuit current can easily shoot up to 100-200A for 10-50ms.No reasonably small (0,5-2W) resistor can survive this on it's own (they can handle up to 100-200W peak power in 10-50ms pulse).First that come to mind is PTC in series, but they have typically 40-100A current limit. And resistor can still vaporize, because they aren't especially fast.Larger sense resistor (~1ohm) limits short circuit current to 20-40A that PTC can handle, but in time it requires to trip (~10ms), resistor is still taking 500-1000W peak power.And of course with larger resistor come larger voltage drop and dissipated power.Any suggestions?

author: Psi
author_rank: Super Contributor
timestamp: May 07, 2015, 11:04:55 am
content: what is this module do, other than monitor the current, Is it a switched output like a mosfet?With the current sense resistor in the line to also check the current through the fet?If so, you could swap the mosfet for an automotive low or high side driver IC.You can get them with built in short circuit protection as well as built in current sense output. So you dont need the resister at all

author: erihog
author_rank: Newbie
timestamp: May 07, 2015, 11:27:14 am
content: Can you place a diode that takes the current if the 24V is connected to the ground pin?The diode should be rated to handle the current until the external fuse blows.

author: akis
author_rank: Frequent Contributor
timestamp: May 07, 2015, 12:44:34 pm
content: Quote from: mwsoft on May 07, 2015, 10:46:43 amWhy can't you have a large 0.1R resistor that can take whatever you throw at it? After all it is not magic, for the current to reach the resistor it must first pass through the connecting wires. What kind of wires are you using to connect this device? Engine cranking leads? I doubt it. So use something like this:http://uk.farnell.com/vishay-sfernice/lto100fr1000fte3/resistor-100w-0r1-1/dp/1357106And it is still 0.1R and so you will not waste a lot of power.

author: jeroen79
author_rank: Frequent Contributor
timestamp: May 07, 2015, 01:47:44 pm
content: A .1 ohm resistor will draw 240A from 24V.That equals 5760W.Can't you rely on the existing fuses?The 240A will come from the shorted circuit and that circuit's fuse will blow.Then the sense resistor will no longer be shorted to 24V.You could also give the sense resistor it's own fuse that will blow at the maximum current you will allow through the resistor.But that would disconnect all circuits leading to that resistor.If needed you can use a fuse that will blow faster than the normal fuses.Another option is to compare the voltage over the resistor and when it reaches a limit you use a (latched) relay to bypass the resistor.But that will not stop the original short and even more current will flow.Which brings me back to the second paragraph of this post.And what is wrong with vaporizing the resistor if it is properly shielded? That is what is happening in a fuse.And you would have to replace either afterwards.

author: DanielS
author_rank: Frequent Contributor
timestamp: May 07, 2015, 02:06:57 pm
content: I would simply whack a fuse in there. If blowing the circuit open is not acceptable, then you can put some diodes in parallel with you shunt resistor to shunt the excess current until whatever fuse is upstream blows.

author: Mechatrommer
author_rank: Super Contributor
timestamp: May 07, 2015, 04:03:57 pm
content: intelligent "fuseless" or "no need to change burnt fuse" or "the lazyman" automotive grade system. 10 minutes to draw, probably hours to make it work O0

author: Niklas
author_rank: Frequent Contributor
timestamp: May 07, 2015, 09:14:16 pm
content: I assume that a fuse is not an option due to some demand similar to "all input pins must be able to withstand short to B+ without damage and return to normal operation after the short is removed". Correct?One option would be to add a transistor/low side switch between the input pin and the resistor and to turn it off when the voltage drop over the current shunt is over a certain threshold. The current can be limited for a short period of time and then use SW to take action to limit the amount of heat dissipation. The resistor must still be able to handle a couple of Amps, and don't forget the derating above 70'C, but a transistor in a DPAK can do the heavy lifting.Just to give you an idea of the basic circuitry. Invert the polarity, i.e. use NPN instead of PNP, N-channel instead of P, "+4-6V sisään" is GND, "+ Ledille" is the input. Maybe add filtering between R2 and the bipolar transistor's base and some feedback to the micro about the fault condition so it can turn of the control signal (the switch).http://320volt.com/en/1w-350ma-power-led-surucu-devresi-mosfet/

author: Andreas
author_rank: Super Contributor
timestamp: May 07, 2015, 09:17:40 pm
content: Quote from: Mechatrommer on May 07, 2015, 04:03:57 pmFor shure:Try to get a relay that opens reliably at 120 Amps short cirquit current.With normal relays the contacts get welded. So the resistor will blow before the relay opens.With best regardsAndreas

author: Mechatrommer
author_rank: Super Contributor
timestamp: May 08, 2015, 02:29:24 am
content: maybe add a capacitor big enough to avoid instantaneous weld. give more time for comparator and relay to act. protection diodes are not included in the drawing also just so he knows. the hard part to think is how to deal with diminishing contact performance overtime, ie increasing resistance due to build up deposit due to repeatetive shorts.