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Custom 25 A mains inrush current limiter design review

Electrical Engineering Asked on October 29, 2021

I’ve designed an inrush current limiting circuit and would welcome a review from experts with more experience with this sort of thing. The load will be around 50 x 19 VDC @ 3.5 A linear power supplies connected in parallel with steady state load up to around 20 A @ 230 VAC.

There are a few specific requirements that motivated my own design rather than an off-the-shelf version:

  • I need 25 A @ 230 VAC, and most commercial varieties seem to be 16 A (too low) or 30 A (too high).
  • A 25 A inrush current limiter I did find and test didn’t actually limit to 25 Apk, but more (beyond 35 A where my fuse trips).
  • The place where this will go has considerable mechanical constraints; most commercial packages just don’t fit.

Here is my schematic:
Draft inrush current limiter design by Sean Leavey

As is typical with inrush current limiting circuits, to avoid the power loss in the NTCs during steady state (which at 25 A is significant – around 200 W) I short out the NTCs with a relay after a short delay (~0.6 seconds, depending on tolerance).

There is also a thermal trip switch so that if the NTCs get too hot the current will trip and stay off until the power is cycled (so that the circuit does not yo-yo between on and off as the NTCs get hot and cold). This is my own invention (as far as I know) but I have not tested it before. I did simulate it in SPICE and it seems to work.

The intended sequence when power is first supplied to the circuit is as follows:

  1. With no power, the relays are initially open.
  2. When mains power is initially provided, the transformer drops it down to some manageable level (20 VACpp). The capacitors and LM317 regulator convert this AC into 12 VDC.
  3. Initially the non-inverting inputs to N2A, N2B and N2C are 0 V due to capacitors C8 and C9 (current via R7 then R9/R11 has not yet had time to charge C8 and C9). This means these comparators short their output to ground (they are open collector output type), so the gate-source voltage at T1 and T2 is 0 V and the relays remain off.
  4. The inverting input to N2D is initially around 2.4 VDC. The non-inverting input is 4.2 VDC. The LM339 is therefore initially in high-Z output mode so the pull-up via R7/R8 sets the output to 4.2 VDC. After 0.1 seconds N2B’s non-inverting input will exceed the 3 VDC inverting input and go into high-Z output mode, allowing 12 V to develop at T1’s gate-source and switching on the left relay. This allows current to flow to the load via NTCs R14 and R15.
  5. After 0.6 seconds, N2A and N2C also go into high-Z mode, switching on the right relay and shorting the NTCs.
  6. The circuit operates in steady state until such time as the temperature sensor measures above about 150°C. At this point, the inverting input exceeds the non-inverting input and so N2D’s output gets shorted to ground, which in turn switches off the left relay after around 0.1 seconds, removing the load. Because the shorted output is also connected to the non-inverting input, the inverting input’s voltage always exceeds the non-inverting one and so the circuit remains off until the power is cycled.

Some notes:

  • I went for a linear regulated supply over switch mode for simplicity and maximum longevity.
  • The transient voltage suppressor diodes in parallel to the relays (in addition to the normal diodes to snub the induced back emf) are there to shorten the time the relay contacts arc as they close and open, to prolong the contacts’ lives. This idea I took from the new Art of Electronics x-chapters book. This is a trade-off between time-spent-arcing and maximum induced back emf.
  • There are lots of timings to pay attention to here: how quickly the 12 VDC rail comes up in comparison to the comparator inputs, etc. I only need three of these circuits for my application and I can make adjustments to capacitor values etc. to try to get it to work reliably.

I can’t share the PSU datasheet since it’s confidential but it contains the following information:

  • Max input current @ 230 VAC: 0.65 A
  • Worst case inrush current: ≤ 0.25 A²s ( ∫ i² dt ) / ≤ 12 A

I can open up a PSU to look at the input stage but I suspect it’ll be a fuse, toroidal transformer then smoothing capacitors and regulation (somewhat similar to what I have in my design). I expect the transformer is what causes the large inrush current on switch-on, since at that point it has no magnetic field and thus initially acts like a low value resistor. Unfortunately the datasheet doesn’t state the input capacitance/inductance directly, but perhaps this can be worked out from the values above?

Does anyone spot any issues? Do people think my thermal latch, timings, etc. will work?

2 Answers

20V TVS in relays coil flyback makes no sense, Are 21A 200V mosfets in(relatively) bulky packages to drive them really necessary? In space critical application I would rather expect bc817.

LM317 - why not LM7812? same result with much less components.

Answered by zajc3w on October 29, 2021

I've designed an inrush current limiting circuit and would welcome a review from experts with more experience with this sort of thing.

Depending on how the UPS is designed, it may not "play-ball" with the series resistors because they may cause the UPS to try and take a really excessive starting current (which it can't take due to the resistors). The upshot of all of this is that the UPS never really kicks into action until the relay closes (shorting the current limiting resistors) and then, you have the same inrush problem just delayed in time.

So, to design this we really need to know what the front-end circuitry in the UPS is like.

Regarding the contact closure that short the resistors, I'd be much more inclined to activate that contact when the AC supply output has risen to the point when the UPS (if it plays ball) is e.g. 75% of the input voltage. A fixed time delay produced by R11 and C9 is too "open-loop" to be effective.

You also need an input fuse on transformer L1 because most magnetic components like this are not rated to be connected directly across a very resilient mains AC supply. Fuse F1 on the output of L1 won't cut-the-mustard in this respect. Ditto the input varistor U1.

Why are your flyback components two series diodes in series with a zener. I can understand one diode and a zener but two diodes and a zener seems like you may be misunderstanding something.

Does C1 really need to be 1000 uF (1 mF)?

Answered by Andy aka on October 29, 2021

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