Electronics > Beginners
Understanding circuit for inrush current limiting
T3sl4co1l:
What kind of transformer do you have?
Toroidial: high inrush current, low leakage. A double-whammy.
"Shell" (conventional, windings in layers on top of each other on an E core usually): modest inrush, modest leakage (still fairly low). Very common. No one really cared, back in the day; why do you?*
Bank wound (bobbin is divided, windings are side by side): Low inrush, high leakage. Very easy to start even a high capacitance load... but, the voltage droop under load is higher (poorer regulation), and then you need more headroom for a given output voltage at max current.
*Which is a good question, why? If just for curiosity in avoiding that turn-on "clunk" or dimming of the lights... ? (I mean, don't forget that doing it as an exercise, so you know how to do it when you finally need to, has value!) Is it to meet a standard? Is it even more practical than that, e.g. you've had inrush knocking out your household appliances or generator or something, and want to make something that'll play nicely in that limited environment?
Anyway, regarding startup resistors: the power rating needs to be adequate to handle the starting energy. That's contradictory (power is not energy), so bear with me. Generally, more powerful resistors are also bigger resistors. For example, a 100W resistor is usually good for 10-100J (5ms pulse). The dissipated energy is more-or-less equal to the stored energy in the capacitors.
So, 30mF at 30V is 13.5J, so you should expect to dissipate about as much in the resistor during startup.
Probably a 10 to 50W resistor would be adequate, depending. (I'd say, check the datasheet, but often, they don't say -- and then you need to read a lot of datasheets to get a feel for what resistors of a given style/type may be capable of. Yeah, it's tedious; well, in short, that's one of the things I get paid for...)
Now, how to control the resistor? Preferably, it should be in circuit for:
1. Only a brief moment, during startup;
2. Until the output is fully charged (say to 90% of nominal); and
3. Removed from circuit if condition 2 is not met in the expected time frame.
Why 3? Consider what happens if 2 never occurs, or the bypass relay never closes. That poor resistor is left supplying the entire circuit in operation. It quickly burns out (hopefully without making a smoking or sparking mess -- vitreous/enameled resistors are a good choice here), leaving you with a broken supply. Or, it burns out somehow, and the relay (which would be timed in this case) closes anyway, dumping inrush current to begin with, but also slowly damaging its contacts due to the surge current.
This is easily arranged with a few more transistors, resistors and capacitors.
You don't have to go the extra mile in this case -- I wouldn't be afraid to use the first attached circuit, with suitable changes of course (the resistors as all have noted). That implements #1 and #2. This final #3 step is only what you want for industrial reliability. :)
Tim
AngraMelo:
Hey Tim,
Great considerations! I was thinking about using a 5W because Im using a 1.2sec delay to bypass the resistor and in that time the resistor does not show any signs of getting even remotely warm. But given your message Ill stay on the safe side and use a 10W.
Now, I dont think I have enough skills to make a circuit to save the PS in case the relay goes to shit.
Would you point me in the right direction so I could start researching that?
thank you very much
soldar:
--- Quote from: AngraMelo on December 19, 2018, 05:43:30 pm --- On the Texas Instruments website they have the attached circuit but I do not understand it.
--- End quote ---
It is quite simple. Initially R103 limits the inrush current. When the PSU is outputting 12V then relay K100 is activated and shorts the resistor.
D102 and D103 are there to absorb the flyback current when the 12V is switched off. Not entirely necessary IMHO.
IMHO that circuit is good but overengineered and can be simplified quite a bit. In fact, most computer PSUs just use a NTC to limit inrush and, depending on your needs, that may be enough for you too.
AngraMelo:
--- Quote from: soldar on December 21, 2018, 07:19:24 pm ---
--- Quote from: AngraMelo on December 19, 2018, 05:43:30 pm --- On the Texas Instruments website they have the attached circuit but I do not understand it.
--- End quote ---
It is quite simple. Initially R103 limits the inrush current. When the PSU is outputting 12V then relay K100 is activated and shorts the resistor.
D102 and D103 are there to absorb the flyback current when the 12V is switched off. Not entirely necessary IMHO.
IMHO that circuit is good but overengineered and can be simplified quite a bit. In fact, most computer PSUs just use a NTC to limit inrush and, depending on your needs, that may be enough for you too.
--- End quote ---
Soldar,
what is that zener doing there? if you hold both ends of the coil at 12V potential how can there be current flow?
Also, 2 75k resistors in parallel gives the coil a current of 0.3mA, you would need 1.2kV to drive that relay!
I still have no idea of what is going on.
soldar:
Yeah, I overlooked the two 75K resistors which make no sense and should be shorted. :-//
The diode is there to dissipate the energy stored in the relay coil when it is turned off. It is a very common and standard protection. The Zenner serves the same purpose and can be omitted and the circuit would still work correctly.
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