supply through series resistor problem - Page 1
Forum section: electronics / projects

author: brownt
author_rank: Regular Contributor
timestamp: July 16, 2018, 10:48:55 am
content: How can this be done.I have 5 volts coming out of an enclosure and entering another enclosure through a patch cable. The voltage needs to be current limited so that a user does not inadvertently short the supply as the patch cable and the various devices that can attach to it are designed to be plug and play and so it is quite possible that a user might plug something unexpected in, or do something to short the supply.At times the 5v is simply across a potentiometer, with the wiper feeding back to the other enclosure, and for that application the limiting resistor is not a problem. However, the five volts is sometimes used to power an op-amp and uC circuit (about 40mA total) that doesn't like working through the series resistor.What actually happens is that the output of the op-amp circuit , which is configured with some gain and a 2.5v swing around zero, starts to pulsate. The uC circuit seems to work ok, though I guess it is unstable.What about a charge pump, would that make any difference, or what else can be done.thanks

author: sourcecharge
author_rank: Regular Contributor
timestamp: July 16, 2018, 11:07:52 am
content: I'm not an expert, but what about just a mosfet switch that is regulated by the ucurrent (or something similar) circuit voltage output?The ucurrent circuit is just a great shunt type of amplified current sensor, and it can work on the high side and low side.If the voltage output reaches the equivalent current you are trying to limit, the switch will increase the resistance until it completely turns off which limits the current.

author: brownt
author_rank: Regular Contributor
timestamp: July 16, 2018, 11:18:38 am
content: I am limited in what can be done in the supplying enclosure. Is that where the FET circuit would go?What is that type of circuit called, so I can search for it.

author: sourcecharge
author_rank: Regular Contributor
timestamp: July 16, 2018, 11:34:15 am
content: how much current are you trying to limit it to?I was referring to a linear regulator, and a high side current sensor.Although on second thought, why not just use a fuse?

author: Siwastaja
author_rank: Super Contributor
timestamp: July 16, 2018, 11:53:39 am
content: Off-the-shelf linear 5V regulator IC with current limit and output short protection. Which means any regulator on the market, such as 78M05. If you don't have a higher voltage rail available to allow for the regulator drop-off, but have to use the existing 5V bus, then a high-side switch IC with current limit and output short protection:https://www.digikey.fi/products/en/integrated-circuits-ics/pmic-power-distribution-switches-load-drivers/726. Make sure the output current limit worst case maximum value is less than your 5V regulators worst case minimum current limit + worst case maximum required by other circuitry driven by that regulator, so that the short on the 5V output will never droop the 5V line and reset the microcontroller/whatever.Polyfuse may be appropriate as well.

author: brownt
author_rank: Regular Contributor
timestamp: July 16, 2018, 12:14:25 pm
content: The 5 volts is coming from a AMS 1117, which has short circuit protection. But a short means the supply device turns off, and then the regulator thermally cycles, shutting down at 165 degrees and then restarting. That can't be good. But anyway, the unit switches off when there is a short, so the current must be limited.A polyfuse means the system will still go off.I will look at the FET option, but I don't know much about it. how does the circuit work?Also, I know this might not be the usual way, but what if I sized the limiting resistor such that half the voltage was dropped across it and used a 40mA charge pump in the receiving device. I don't think the current for the uC and opamp circuit varies much. If the uC and opamp circuit is drawing 40mA, then can I assume the input resistance is 125 ohms?So, I size the limiting resistor at 125 ohms/ 250mW and use a 40mA charge pump in the receiving device. Unorthodox, but ....?

author: IDEngineer
author_rank: Super Contributor
timestamp: July 17, 2018, 04:01:47 am
content: Quote from: brownt on July 16, 2018, 10:48:55 amNo surprise there. You've created a supply rail with quite a bit of impedance. That's never a good thing.I haven't looked it up, but are you certain the source 5V will, when shorted, cycle based on die temperature? I would expect it to detect the excessive current and self-limit without waiting for the die to heat up. A temperature-based limit would also slow the turn-on when the short was removed, since it would have to wait for the die to dissipate its heat. Finally, the chip designer would know this would stress the device both structurally and thermally. Far better to rapidly detect the current itself than to wait for a second-order effect like temperature while the current is destroying the device during the heating process.Let's confirm this temperature-based current detection theory before progressing. If it's false, and I hope it is, then your problem just got a lot easier as long as the source enclosure's circuitry can tolerate the shutdown of its 5V rail. (If it can't, then you're going to need to develop an entirely separate power supply that can supply the same voltage as its own input, which will likely be a combo buck-boost SMPS.)

author: Siwastaja
author_rank: Super Contributor
timestamp: July 17, 2018, 09:14:26 am
content: Quote from: brownt on July 16, 2018, 12:14:25 pmYou know, nobody forces you to run your to-be-shorted output and the internal 5V from the same regulator, or from the same polyfuse. That was my point: use a separete protection. Your system will not go off, just the output you are protecting. Just copypasting the existing AMS1117 solution would probably work, although if you don't need that much current, use a smaller reg.No need to design your own using discretes when these integrated solutions exist.

author: Alex Nikitin
author_rank: Super Contributor
timestamp: July 17, 2018, 10:04:23 am
content: Quote from: brownt on July 16, 2018, 10:48:55 amA charge pump won't help. I would rather use a current protection chip, like LTC4361 or (depending on the 5V supply capabilities) a USB over-current protection chip - there are some with adjustable current limits, like TPD3S716, and some with 0.85A fixed limit, like TPD3S014 for example.CheersAlex

author: brownt
author_rank: Regular Contributor
timestamp: July 17, 2018, 11:33:29 am
content: That's an excellent point. As it turns out, the PCB has a print (blank holes) for a regulator that comes after and in series with the 5 volt regulator. If I used a 3.3 volt regulator there, that might be enough for the op/amp uC circuit. Or I could probably cut a track and rewire a 5 volt regulator in parallel.Can I determine from the data sheet for the AMS1117 what type of short circuit protection it has. From my tests when I shorted it, it just seemed to heat up until I couldn't touch it, I assumed it would then move into thermal protection and switch off and then cycle again. Which would not be good. I suppose the practical test is proof enough but as I understand it that type of thermal cycling should not be the definition of short protection.

author: brownt
author_rank: Regular Contributor
timestamp: July 17, 2018, 10:17:53 pm
content: Quote from: IDEngineer on July 17, 2018, 04:01:47 amI can't see anything in the datasheet that explains exactly what 'short circuit protection' is for this device. All it says is "protected against short circuit and thermal overloads. Thermal protection circuitry will shut-down the regulator should the junction temperature exceed 165°C at the sense point.There is a graph attached