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I want to replace an ICL (inrush current limiting) NTC thermistor in a vintage computer PSU (for Amiga 1200). This question is similar to 'Selecting an NTC for inrush current limiting', except that question was related to a triac control circuit and so the answer was specific to that case.

Damaged Amiga 1200 PSU ICL NTC
Figure 1: Damaged Amiga 1200 PSU ICL NTC

Taken from a working PSU, the ICL NTC marking reads "KCC 30 OHM", which I understand means it should measure 30 Ω at 25 °C.

Intact Amiga 1200 PSU NTC
Figure 2: Intact Amiga 1200 PSU ICL NTC

What I noticed while trying to find a replacement is that there are different current ratings for the same resistance. For example, using Farnell, when I filter the THC ICLs on 30 Ω, I get two in stock; 1.5 A (10 mm; 20% larger than my part) and 5 A (22 mm; about 2x the diameter of mine). The fuse is 1 A fast blow (F1AH250V), so should the THT ICL current rating be less than or more than 1 A? Does the replacement have to measure 30 Ω or can a different resistance be used to limit inrush current?

I noticed the question, 'Thermistor replacement', but that seems more relevant to finding a rare resistance rating rather than deciding on the current rating.

In case you're wondering: the ICL NTC was damaged due to using a battleshort instead of a fuse (yes, bad idea, I know), and the fault in the circuit was 3 of 4 shorted diodes in the full bridge rectifier.

Edit: The measurements for the part are...

  • Diameter: 0.31 in / 8 mm
  • Thickness: 0.11 in / 2.8 mm

One of the 30 Ω replacement parts is 1.5 A which measures 10 mm diameter, so perhaps I can estimate that the original part is rated around 1.2 A? (i.e. 80% of the 10 mm part). Does it work that way? i.e. Can the diameter of an ICL NTC indicate it's current rating? I noticed that 8 mm parts are rated at 10 Ω and 2 to 4 A.

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    \$\begingroup\$ Measure the diameter and thickness of it and match to other offerings if you can’t find a datasheet to match the current rating. \$\endgroup\$
    – winny
    Commented Jan 31, 2023 at 13:03
  • \$\begingroup\$ Aha, good idea. 0.31 in / 8 mm, thickness = 0.11 in / 2.8 mm \$\endgroup\$ Commented Jan 31, 2023 at 13:05
  • \$\begingroup\$ Good. What’s the closest Vishay part you can find? \$\endgroup\$
    – winny
    Commented Jan 31, 2023 at 13:12
  • \$\begingroup\$ Hmm, I don't see any Vishay ICL NTCs on Farnell or Mouser. \$\endgroup\$ Commented Jan 31, 2023 at 16:14
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    \$\begingroup\$ Sorry, I mixed up Vishay and TDK. \$\endgroup\$
    – winny
    Commented Feb 1, 2023 at 19:25

1 Answer 1

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The current rating isn't actually the most important rating for this application. As long as the NTC is rated for the PSU's current draw (1 A), it's fine. What's more important here is the thermal (heat) capacity of the NTC.

During turn-on, the bulk capacitor behind the NTC gets charged, which means that a certain amount of energy is dumped into the NTC. When charging a capacitor from a DC power source through a resistor, only half of the energy consumed is actually put into the capacitor, while the other half is dissipated in the resistor. The consequence of that is that, in the worst case, the NTC dissipates exactly as much energy as is stored in the bulk cap at maximum input voltage.

So, in order to not break the NTC, its thermal capacity has to be high enough to absorb the turn-on energy without overheating.

Let's run the numbers for an example NTC: The B57153S0330M051 by TDK/Epcos, which is rated for 1.3 A.

If I'm not mistaken, the Amiga 1200 PSU's primary-side capacitor is 47 µF 385 V. Under normal operation, it's never going to see anything more than 350 V, so let's go with that voltage.

The energy contained in this capacitor at 350 V is about 2.9 joules (that's quite a lot). The NTC must be able to survive getting this much energy dumped into it all at once during turn-on.

The NTC's thermal capacity is specified as 240 mJ/K. Dividing our energy by its thermal capacity, we get a worst-case temperature rise of 2.9 J / (0.24J/K) = 12.1 K.

That's barely anything. If the NTC starts out at 25°C, it'll be at 37°C after turn-on, which is still practically cold. You could choose an even smaller NTC as long as it can sustain 1 A continuously. Of course, if you rapidly plug and unplug the PSU, the NTC will get a bit hotter with every cycle, so a bit of safety margin is always good. In this case, though, the safety margin is more than 10x (the NTC can operate at up to 170°C), which should be more than enough.

In short: Don't be afraid to choose an NTC that "looks small". Just run the numbers and you'll know for sure whether a particular NTC will work or not.

Here's an interesting fact: Since NTCs limit their own power dissipation, their impact on efficiency is tiny, and you can easily calculate the maximum possible voltage drop across an NTC. The B57153S0330M051 consumes at most 1.4 W during steady-state operation. With its 33 Ω of resistance, it'll drop at most U = sqrt(P * R) = 6.8 V in the worst case (when the downstream circuitry is just barely not pulling enough current to fully heat up the NTC). As the current draw goes up, the voltage dropped across the NTC goes down even further. Even if you chose an NTC that's way too big for your application, the voltage drop wouldn't be that much higher - for example, if you chose an NTC that needs 5 W during steady-state operation instead, it'd drop a maximum of 13 V, which still isn't much compared to the line voltage.

This means that not only is the power dissipation in an NTC limited, but there's also a hard limit on the voltage dropped across it.

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  • \$\begingroup\$ Brilliant answer, thanks. \$\endgroup\$ Commented Jan 31, 2023 at 16:13
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    \$\begingroup\$ +1. You basically covered it, I would just like to add that going too high in current rating would mean that the NTC never gets hot enough to "trip over" and you'll end up with 30 ohm in series forever. For high load, this may cause under voltage condition. \$\endgroup\$
    – winny
    Commented Jan 31, 2023 at 16:55
  • \$\begingroup\$ @winny The fun thing about NTCs is that they limit their own power dissipation and voltage drop - in case of the B57153S0330M051, it will never dissipate more than 1.4W, no matter the downstream power consumption. As a result, the power wasted in (and therefore voltage dropped across) the NTC will be negligible at any power level, even if you chose an NTC that's a little too big. I'll edit the answer to go into more detail on that. \$\endgroup\$ Commented Jan 31, 2023 at 21:01
  • \$\begingroup\$ You have a pretty big margin of error allowed in most power supplies, but if the UVLO triggers faster than the NTC heats up, you will have problems. You want to "trip over" the NTC early. Ideally, bypass with a relay once started to save losses and avoid such dynamic situations afterwards if you go from light load to high load quickly. \$\endgroup\$
    – winny
    Commented Jan 31, 2023 at 21:52
  • \$\begingroup\$ Fair enough. Large, sudden changes in current will be a problem if the NTC is physically too large. \$\endgroup\$ Commented Feb 1, 2023 at 16:29

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