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PTCs come with a maximum voltage rating which, according to the answer for this question is essentially because of instantaneous power across the device.

Usually, a PTC is placed across a low-resistance voltage source. However, in my case, the PTC is in series with a resistor which is much greater than the R1max of the device, and the possible voltage across the circuit is greater than the nominal Vdc of the PTC I would like to use.

Example with significant inline resistance

Whilst the PTC is conducting, the voltage across it does not violate the Vdc, since most of the voltage will be dropped across the resistor. Once the PTC trips, most of the voltage will be across the PTC since it will be in a high-resistance state, which would likely violate the Vdc rating of the PTC.

Is it safe to operate it in this configuration? My theory is that it might possibly be safe, since the power dissipated is limited by the in-line resistance, but I don't know what calculations I might be able to use to show this is/isn't true.

For a concrete example, the PTC is a Bourns MF-NSHT016KX with a Vdc of 30V, an I_trip of 0.8A and an R1_max of 6R, and is in a circuit with a series resistance of 22R, meaning the PTC will trip somewhere about Vin=20V (as desired). I would like to be able to expose the circuit to Vin=40V.

(p.s. choosing a PTC with a higher Vdc rating introduces other compromises that I am not willing to make. Similarly for resistor power ratings).

Damien
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  • Simulate/calculate worst case scenario. – winny Dec 19 '17 at 07:52
  • @winny, do you have an example calculation or a reference to an app note that does? It would make a good answer. – Damien Dec 19 '17 at 08:44
  • Not really. Find out what the lowest PTC resistance could possibly be. Calculate current and voltage stress during inrush. Find out worst case highest PTC resistance when the user is plugging and unplugging the device at just the wrong period to have the PTC hot but the input capacitor bank discharged. Ask me how I know. – winny Dec 19 '17 at 11:56
  • This won't work at all because Ihold is only 160mA. You have a high probability of tripping way below 20V. – user57037 Dec 19 '17 at 19:16
  • A PTC is guaranteed to pass Ihold for a lengthy time period without tripping. A PTC is guaranteed to trip at Itrip within a short time period. There is a large spread between Ihold and Itrip. This is why PTC's are not used for over-voltage protection. – user57037 Dec 19 '17 at 19:22
  • Under normal operations, Vin is no more than 4V. The resistor survives up to the ptc trip over tolerance and temperature. My question is whether the ptc survives beyond Vin=30. – Damien Dec 19 '17 at 21:07
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    That would have been good information to put in the original question. And it is not too late. You can edit it. – user57037 Dec 19 '17 at 22:55
  • If you attempt to make a product out of this, "under normal operations" is of little significance. Worst case scenario is what you design for, especially for a saftey related issue like this. – winny Dec 20 '17 at 07:50

1 Answers1

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Most likely, if you provided a lot more information about what you are trying to do, a better way could be proposed. I really think that using a PTC for over-voltage probably a mistake. But, considering that the normal working voltage is only 4V (based on your comment) and that you are targeting a 20V cut-out, I think this has a chance of working.

One way to look at it is this. As the PTC starts to trip, its resistance will increase in a short time from some low value to some high value. We don't know exactly how fast this process is, but we can be sure of one thing. The maximum power dissipation in the PTC will occur when Rptc = Rload.

Since the load is 22 Ohms, the highest PTC dissipation will occur at that instant when the PTC's resistance is 22 Ohms. At that moment, the voltage across the PTC will be half of Vin. So that is 20V across the PTC in the worst case. That is actually fairly promising, because it is lower than the maximum allowable voltage. Also, the current will be roughly 0.9A at that point. So the instantaneous dissipation will be 18W (V * I). Now, prior to this instant, current may be higher, but the voltage will be lower. And after this instant, the voltage may be higher, but the current will be lower.

I can't guarantee anything, but it seems to me that 0.9A is pretty reasonable, and 20V is below the maximum allowable voltage. So I believe that the PTC will survive this.

I have not tried to validate your basic scheme. I didn't do any calculations with Itrip and Ihold. The cutoff of this circuit will definitely vary from unit-to-unit and also over temperature.

I also want to point out that the normal use of fuses is to make sure over-current events don't overload conductors and start a fire. The PTC you have selected may not provide reliable protection for faults other than over-voltage input. For example, if the output is shorted. In that case, unlike your proposed scenario, the current will not be limited by the load resistor, and the PTC may be permanently damage.

user57037
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