Current measurement protection - Page 1
Forum section: electronics / projects

author: Twistx77
author_rank: Regular Contributor
timestamp: July 15, 2014, 09:46:56 am
content: Hi,I wan't to make a device to measure power and I have already a concept design working in a breadboard but I now I want to protect it from overvoltage, overcurrent, and reverse polarity. Right now is base in aMAX4372.My main problem right know is the reverse polarity since the MAX372 can only withstand -0.3V in any pin. I thought of using a shottky with a resistor to protect each input but doing that I can go under -0.3V and damage the device. I thought also of using a mosfet with a comparator to cut the input path so the circuit is not closed in case of reverse polarity but in simulation doesn't seem to work either. I've thought of things like a mosfet bridge but I want it to be the cheapest and less intrusive solution so having just a diode in series won't do it since it will add too much error to the voltage measurement.I plan to measure form 0 to 30V from 10's mA to 5A. I've attached a simple schematic. If anyone wants the ltspice file I can attach it in another post.Does anyone know how I could do this?

author: mij59
author_rank: Frequent Contributor
timestamp: July 15, 2014, 02:31:01 pm
content: Hi,If you reference the current shunt to ground its easier to make a protection, you can use a low offset voltage opamp.

author: Twistx77
author_rank: Regular Contributor
timestamp: July 15, 2014, 03:38:17 pm
content: Yes but I want it to be high side no low side.Thanks!

author: Twistx77
author_rank: Regular Contributor
timestamp: July 17, 2014, 06:25:00 am
content: Thanks! That might do it!! Also I've thought of adding a comparator and a mosfet to just cut the path of the ground to my circuit when voltages are negative.

author: Twistx77
author_rank: Regular Contributor
timestamp: July 17, 2014, 10:56:57 am
content: The problem I had was that the diode had a too high drop out, more than 0.3V for high currents. So that solution was discarded.

author: zapta
author_rank: Super Contributor
timestamp: July 17, 2014, 10:53:36 pm
content: You can have two shunts in serial. Have the diodes in parallel to both and measure voltage only from one. Also, you can add a fuse in series such that the diode will not need to sustain the high current for long.You are fight, high side is more useful in general though more challenging.Edit: if you plan to A/D the output voltage you can use an IC that also contains the A/D such as LTC2943.

author: Twistx77
author_rank: Regular Contributor
timestamp: July 18, 2014, 08:20:25 am
content: Quote from: zapta on July 17, 2014, 10:53:36 pmSorry I don't get how that will help me.Thank you for the recomendation I've already look at some others like that because that one can only withstand up to 24V and I want it to go up to 30 - 35V.

author: nctnico
author_rank: Super Contributor
timestamp: July 18, 2014, 09:55:26 am
content: I'd use a series resistor depending on the leakage current into the current sense chip and the maximum current into the ESD diodes. That way the ESD diodes inside the chip take care of keep the chip inside it's specified operating range. There usually is a maximum current which can flow through the ESD diodes without damaging them.Overvoltage in general can be handled using a zener diode. But you have to take into account that a zener diode can have a large leakage current (up to 1mA!).

author: Twistx77
author_rank: Regular Contributor
timestamp: July 18, 2014, 11:34:13 am
content: Yes, I thought of using a series resistor to use the internal ESD diode but I wrote to the TI's forum for another amplifier and they reply with different solutions but that one beacuse they said it couldn't pass -0.3V in each input.Now with the diode shared by gazelle I might just use the external one.

author: zapta
author_rank: Super Contributor
timestamp: July 19, 2014, 09:37:56 pm
content: Quote from: Twistx77 on July 18, 2014, 08:20:25 amHere is an example with arbitrary numbers. Rs is whatever shunt resistor you want to use. Your sense circuit can sustain up to 300mv. The protection diode D can result in up to 700mv. You select resistor Rx such that 700mv on the diode will result with 300mv on Rs. This gives you more choice in selecting the protection diode (e.g. one that does not leak in the 0-300mv range of Rs) at the cost of increased burden voltage.

author: nctnico
author_rank: Super Contributor
timestamp: July 20, 2014, 10:00:12 am
content: Quote from: Twistx77 on July 18, 2014, 11:34:13 amThat is not a problem. They make the wrong assumption that the current at -0.3v is not limited. The resistor will limit the current so staying within the 0.3V limit is guaranteed (with a proper resistor). An external diode can have a large forward and reverse leak which could impair the accuracy of your measurement.