How to clamp the short-circuit voltage of a shunt resistor? - Page 1
Forum section: electronics / projects

author: insanoff
author_rank: Contributor
timestamp: November 10, 2022, 09:41:30 pm
content: Hello everyone,I am currently playing with a short circuit protection circuit. I wonder how to clamp the voltage so that it does not exceed the supply voltage of the comparator. I assume that in case of a short circuit, the voltage can be high enough to damage the comparator. Apart from that, the protection may not work properly if the voltage is higher than the supply voltage of the comparator.I would like to know how such problem is solved. If there is a reference design to study, that would be nice.Can the problem be solved by using TVS diodes? Because I (hobbyist) have not yet seen TVS diode in combination with a shunt resistor anywhere. Also, I will use 3.3V for the protection circuit and TVS diodes with such low voltage are not common either.For reference I attach some screenshots and the actual LTspice simulation.Thanks,Adam

author: ajb
author_rank: Super Contributor
timestamp: November 11, 2022, 12:01:01 am
content: A tvs diode (or a pair of steering diodes to the supply rails) between R9 and the conparator's input will work. Make sure the series resistor will limit the current through the diode(s) to an acceptable level.A TVS diode across the shunt *might* work depending on what sort of short circuit currents you're dealing with and how long the short might persist (TVS diodes have very high peak current ratings, but will cook themselves given enough time). Be sure to look at what the clamping voltage of the TVS diode is at your anticipated current, this will be notably higher than its standoff voltage.

author: langwadt
author_rank: Super Contributor
timestamp: November 11, 2022, 12:17:45 am
content: afaict from the datasheet the input already have a 1k series resistor and esd diodes rated for +/-10mA, so that's good for ~ +/-10Vand you have another 10k so more like +/-100V

author: Geoff-AU
author_rank: Frequent Contributor
timestamp: November 11, 2022, 04:12:41 am
content: Another important thing to check is the surge rating on the resistor. Depending on construction type, surges will change the resistance value or cook it completely.

author: Le_Bassiste
author_rank: Frequent Contributor
timestamp: November 11, 2022, 09:26:40 am
content: you have shown one particular overload scenario, but is that really the only one to consider?think about a simple fuse: it has a max holding current rating , an I2t rating, and a max voltage rating. so, before you consider a particular protective method, first define (or, assess) the range of overloads in terms of current, voltage and time (and sometimes number of repetitions) that you want to protect your circuit against. if you haven't done that, any suggestion for a protective method is just a shot in the dark.

author: tszaboo
author_rank: Super Contributor
timestamp: November 11, 2022, 10:14:37 am
content: Two diodes in series parallel with the shunt is a kinda-standard way of protecting shunts. Normlly you have 100mV of so on the shunt, if n overload occurs, then it is clamped to 1.2V. The diodes make a small error, which might be acceptable to your specification. Otherwise, a resistor in series with the sense line, and BAV99 or 199 to ground and power supply.

author: insanoff
author_rank: Contributor
timestamp: November 12, 2022, 09:37:39 am
content: Thank you very much everyone for your answers!My concern is not only the risk of damaging the comparator, but also ensuring that the circuit works as it should and protects the main circuit, which will be an inverter. Yes, the inputs of the comparator are protected against ESD. So the 10k input resistor must limit the current.If the surge current gets too high, the measurement voltage will no longer be in the 3.3V measurement range of the comparator. I guess the comparator can mafunction and will not interrupt the power source.The target response time is less than 1 microsecond, so I assume the paralel TVS diode shouldn't be a problem and it should handle the surge. I wonder, if SMF3.3A would solve the problem here?

author: insanoff
author_rank: Contributor
timestamp: November 12, 2022, 04:02:11 pm
content: By using 2 BAT54S barrier diodes at the input, the maximum voltage will be clamped to 3.6V.Probably I can just use a higher voltage comparator and the circuit will work fine. But for TLV3201 the simulation reveals a problem. Can it be that this will not be the case in the real application?

author: insanoff
author_rank: Contributor
timestamp: November 13, 2022, 04:42:17 pm
content: I think I have solved the problem. The steering diodes and R4 keeps the maximal voltage below 3.6V. The resistors R9 and R7 form voltage divider when U2 Schmitt inverter output is low, thus keeping the comarator input within an acceptable range. When a fault occurs, the output of the U2 Schhmitt inverter goes high and latches the protection. A screenshot is attached.

author: TheMG
author_rank: Frequent Contributor
timestamp: November 13, 2022, 05:49:09 pm
content: What kinds of currents are you measuring? 0.1 ohms is a pretty high value for a current shunt. Ok if you're measuring only a few mA but much too big a resistor if you're measuring an amp or more.Smaller resistance value not only reduces heating and energy loss, but also gives you additional headroom when it comes to overload and short currents, so protecting the comparator might not even be necessary.

author: insanoff
author_rank: Contributor
timestamp: November 14, 2022, 05:56:50 pm
content: Quote from: TheMG on November 13, 2022, 05:49:09 pmYou are right. It certainly makes sense, the resistance is too high. But this is just a proof of concept. In the second prototype I will use a 10 times lower resistor and amplify the shunt voltage. I have a 7W shunt resistor (GMR100HTCFAR100) in stock and it should be plenty for the prototype. The maximum operating current will be 3A and the resistor will dissipate about 1W. At first I just want to test the functionality.

author: fourfathom
author_rank: Super Contributor
timestamp: November 14, 2022, 07:07:49 pm
content: Do pay attention to the maximum current through those over-voltage clamp diodes. If the fault current through the diode to the supply rail exceeds the operating current of the circuit, then VCC will rise. Make sure that VCC will stay in a safe range.Admittedly, the shunt voltage is unlikely to be enough to cause this problem, but in other situations this can cause trouble.

author: insanoff
author_rank: Contributor
timestamp: November 15, 2022, 08:38:17 pm
content: Quote from: fourfathom on November 14, 2022, 07:07:49 pmI am also concerned about the current through the clamping diodes. My approach is to use a TVS diode near the clamping diodes to clamp the power rail. I hope this is sufficient.