0

Let's say you have a power supply that at 110 V draws 1.3 A max in its steady state, but has an inrush of up to 50 A. If you want to limit that inrush current to 5 A, you could use a 22 ohm resistor. Easy enough.

What I'm unclear on is if you limit this inrush current to 5 A, will you prolong the length of the inrush? 50 A to 5 A is a factor of 10. Would inrush last 10 times as long? Or is it always a few milliseconds whether current limited or not?

I'm looking to use a timed relay switch to bypass the limiting resistor, but I need to be sure of how much power will be dissipated in the resistor and how long I need to limit current.

I'm thinking a 50 W resistor with pulse tolerance for 5 seconds should be totally fine, but want to be sure. Primary concern is resistor wattage sizing.

Derek Hershman
  • 112
  • 7
  • Probably not exactly 5 times, but yes, it will last longer, because the inrush is caused by charging capacitors up, and ends when they are charged. It's not constant-current, either - that's just an approximation. – user253751 Jul 15 '22 at 17:24

3 Answers3

1

What I'm unclear on is if you limit this inrush current to 5a, will you prolong the length of the inrush? 50a to 5a is a factor of 10. Would inrush last 10 times as long? Or is it always a few milliseconds whether current limited or not?

For a regular power supply, the inrush current is due to charging internal capacitors.

And, no matter what you do, the inrush-limiting resistor will "burn off" the same amount of energy.

Whether it does this in 100 ms or ten seconds, you will lose that energy as heat AND, it will be the same numerical value of energy. The only flexibility you have is to increase the resistance to a higher value and, burn-off energy at a lower power level but, for a longer time.

But, too long might cause your power supply some problems and it might not start or even worse, it might fail.

Andy aka
  • 456,226
  • 28
  • 367
  • 807
  • Yes, I understand why the inrush occurs, and I have no qualms about burning off that energy. My chief concern is sizing a resistor that will a) limit inrush to 5 A or less and b) not fry from overload. – Derek Hershman Jul 15 '22 at 17:55
  • @DerekHershman your title posed a question and, the embodied text in my answer was also a question taken from your original text. They were the only two questions that I could see hence, I tried to answer them. If in fact you have a different question (rather than a concern), you should edit your question and clearly state that you are adding an edited section in case what you write makes any or all of the given answers appear misguided. This is a question and answer site and you need to be clear. I was also in the middle of answering your earlier question when you deleted it. – Andy aka Jul 15 '22 at 17:59
  • The body of the question has contained my proposal of a 22ohm resistor with a 50w power rating limiting the current for 5 seconds this whole time, and no one has addressed those specifics. Sorry for deleting the earlier question, it had been sitting there with few views and no responses so I decided to rephrase. – Derek Hershman Jul 15 '22 at 18:04
  • There is nothing to address. If you'd have linked a data sheet that contains a surge graph or formula and, the DS clearly showed it wouldn't hack the peak power for the 10 seconds you originally stated, then I might have commented. Personally speaking, a 50 watt resistor will probably be OK. – Andy aka Jul 15 '22 at 18:23
  • @DerekHershman if we are done here you should accept an answer. – Andy aka Jul 17 '22 at 09:07
0

If you limit the current from a constant-voltage source into a capacitor with a resistor, the capacitor will charge up exponentially.

https://en.wikipedia.org/wiki/RC_circuit

To reach a fully "topped-off" max voltage will take longer than 5 x (in your example), but the voltage increase per second (or millisecond) toward the end will be much lower than at the start.

To get there faster while maintaining some specified peak current, a constant-current circuit will hold the inrush current to a constant value despite the voltage across the capacitor increasing. More complex than a single resistor, but better performance.

Back to your situation - Note that the 50 A inrush current is for only for the first few power line cycles, and maybe for only the first cycle or even the first half-cycle. It is so short that a 15 A or 20 A magnetic circuit breaker does not trip. Given that, why do you want to limit the inrush current?

AnalogKid
  • 19,998
  • 1
  • 13
  • 32
  • I want to turn the supply on and off using a wifi relay. Not super often, but just often enough that I’m concerned about damaging the relay contacts. – Derek Hershman Jul 15 '22 at 17:43
  • Derek, I= V/ ESR of Caps can be tolerated by derating contact current rating or some other soft start method. It will degrade life, but it depends on number of cycles and overcurrent factor. – Tony Stewart EE75 Jul 15 '22 at 17:54
0

For an RC circuit if we neglect the ESR of the capacitor and diodes, RC=T so by increasing R you increase T [seconds]. But you also reduce the I^2R=P power loss so as to reduce temperature rise, but still need to waste the same amount of energy in heat just spread out over a long time and lower.

Arrhenius Law is used to rate MTBF of anything this way for hours of lifetime vs temperature rise from chemical laws of degradation with temperature rise.

Although there is some variation in chemistry, usually we say 50% loss in life for every 10'C rise or 25% of life expectancy for a 20 'C rise, and so forth. So e-caps may have a rating of only 1000 hours at 80'C which isn't very long then work backward at actual temperature to see expected MTBF.

Although near the short-term failure the Arrhenius rate may double.

For relay contacts, the arc is a visible white light that is hard to measure the temperature of, but let's say 6000'K. But then it's only for milliseconds. So Japanese suppliers of Relays like OMRON would publish (then later delete to promote their electronic switches) the number of life cycles for various loads on current ratings based on a 1 million cycle mechanical life. It does drop many orders of magnitude with either inductive turn-off arcs or capacitive charge arcs. So it is something to pay attention to when choosing any switch; mechanical or electronic, the latter, which is preferred in most cases.

Tony Stewart EE75
  • 1
  • 3
  • 54
  • 185