Power supply current limiting - Page 1
Forum section: electronics / beginners

author: AxleD
author_rank: Regular Contributor
timestamp: March 28, 2015, 07:50:41 pm
content: Hi,I have put together a simple regulated power supply consisting of a current regulator (one LM317 with pot in series with the output of the 317) and a voltage regulator (another LM317 in the standard configuration). A few silly questions but I unfortunately don't have a power supply (and have never used one) so i need to confirm:1. Is it normal for the output voltage of the current regulator to decline as i dial the current down?2. I presume that as soon as the voltage starts declining I am then effectively in current limiting mode?Thanks

author: T3sl4co1l
author_rank: Super Contributor
timestamp: March 28, 2015, 08:30:53 pm
content: The current limiter should be upstream, not downstream. The one on the load side always takes priority; so you've made, more properly speaking: a voltage limited current source.Tim

author: AxleD
author_rank: Regular Contributor
timestamp: March 28, 2015, 08:51:04 pm
content: Sorry absolute noob here. What does upstream refer too?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: March 28, 2015, 10:11:43 pm
content: In this case, on the rectifier-filter side, not the load side.Tim

author: AxleD
author_rank: Regular Contributor
timestamp: March 29, 2015, 04:25:33 am
content: Hi Tim,Thanks.I have included the setup i have in the attached (straight from the datasheet). I believe the current limiter is on the rectifier side.When dialing down the current, the output voltage of the first 317 falls, causing the output voltage of the second 317 to fall. Not sure if that's how its supposed to work.

author: Zero999
author_rank: Super Contributor
timestamp: March 29, 2015, 11:29:26 am
content: Yes, that's what you should expect to happen.The circuit can only regulate either the current or voltage due to Ohm's law. It can't regulate both current and voltage simultaneously.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: March 29, 2015, 03:57:43 pm
content: Most importantly, it should do it in a relatively sharp manner. If you use a large number of different load resistances, and graph the output (V, I), you should find a square, or something near that. This method won't give a perfectly square corner (where it goes straight from voltage to current limit), but it's pretty good for most purposes.Tim

author: Zero999
author_rank: Super Contributor
timestamp: March 29, 2015, 04:05:02 pm
content: A big disadvantage of this design is the current limiter has a large drop-out voltage: 1.25V, plus voltage drop of the LM317 at full load, then there's the drop-out voltage of the second 1M317.

author: AxleD
author_rank: Regular Contributor
timestamp: March 29, 2015, 07:14:57 pm
content: Thank you all. That's exactly what i was looking for.With that working i am allot closer to finishing off this project. From my limited knowledge, the design does seem inefficient but it will have to do for now.If i can be so bold and ask another related question...I was thinking of adding a current limit led that lights up whenever the supply is in current limit mode.I was thinking of using two sense resistors - 1 after the current limiter 317 and one after the load and comparing the voltage drop across bothThe attachment show's the circuitWill this work? Any ideas on how best this could be achieved?

author: mikerj
author_rank: Super Contributor
timestamp: March 29, 2015, 07:28:19 pm
content: Quote from: AxleD on March 29, 2015, 07:14:57 pmNo it won't work. Remember the current in a series connected circuit is the same through every element. The current coming out of the second LM317 will be equal the the current going into it, plus the quiescent current of the LM317 itself.You could get a crude indication just by looking at the difference between the input and output voltage of the first LM317. Under normal conditions, this voltage drop will just be the normal dropout voltage, but when it's current limiting the output voltage will quickly drop. However, it will be difficult to get anything like an accurate indication because the dropout voltage is not constant, it increases with current, there is a temperature dependance and there will likely be small differences between parts.

author: Zero999
author_rank: Super Contributor
timestamp: March 29, 2015, 07:34:57 pm
content: There's no need for two sense resistors. Looking at the voltage across Iset and turning on the LED when it exceeds 1.2V would work.

author: Rupunzell
author_rank: Frequent Contributor
timestamp: March 29, 2015, 07:43:13 pm
content: Constant current or constant voltage. Most bench power supplies work this way by a switch over between current control or voltage control (regulation).It is possible to make a limited current voltage regulator or a limited voltage current regulator, these are often specialized for a specific application.Suggest have a look at the service manuals for a typical bench supply to see how they work before proceeding. Other examples can be found in data sheets like the uA723 or LM723. These older IC regulators have more user driven configurations, but the user needs to know something about how power supplies in general work.Three terminal regulators appear easy to apply for a given requirement until the requirements exceed their original intent, then it can get messy real fast.Bernice

author: AxleD
author_rank: Regular Contributor
timestamp: March 30, 2015, 05:44:49 am
content: Thanks!I will check the voltage drop across the first LM317 as suggested.Appreciate all the help and the suggestions regarding further reading.Cheers