What's this surge protection component?

I read in sci.electronics.design that DaveC wrote (in ) about 'OK, so how do I choose a replacement thermistor?', on Tue, 18 Nov 2003:

You do the calculations this way. You have +/-65 V rails, and you probably lose 3 V at each end in saturation voltages, source resistors and strays. So you have 124 V peak-to peak signal available, which is

43.8 V r.m.s. If the load is 8 ohms, you get 43.8^2/8 = 240 W. This is the short-term average power, where 'short' isn't so short with your BIG caps in place. We can't say what the continuous average power is, because we don't know how much the 65 V rails drop on sustained load.

Why? These things don't operate like diodes, as far as I know. I didn't see a Vf rating (which may or may not mean something) but their time constant is relatively long, from

30 to 130 seconds. They start out as very high resistance so there would have to be current sharing at the beginning. The limiting R would be lower than that of a single thermistor, which would obviously reduce the desired effect - inrush limiting. Even if not perfectly equal, at the beginning, with two in parallel, both would heat and each would lower its internal resistance, but neither would reduce so quickly as to hog all the current, leaving the other in a cold resistance state. If one was faster than the other, conceivably there could be a point in the ~100 second time constant when one hogs the current to the extent that the other no longer decreases resistance - but for inrush, we are interested in the beginning part.

So, if the above paragraph is correct NTC's in parallel would work, but not as effectively as a single NTC. The hogging, if it does occur, would happen after inrush.

All of the above is said to show my thinking on the subject - please correct me if I am wrong. I claim no expertise on this subject - just what I've learned reading about it. You may have extensive experience on this and the ability to correct any misconceptions I may have about it.

I've got a 20 ohm, 30 watt wirewound power resistor. How do I calculate how this would effect inrush current? If it's now 48 A, can we presume that the turn-on load is about 3.5 ohms? Inserting the 20 ohm resistor would limit it to 8.5 A (using 170 V peak in calculations)?

Could I then use some passive components in-line with the relay's coil to delay the closing of the (N.O.) contacts which would short circuit the resistor?

Am I on the right track, here?

I read in sci.electronics.design that snipped-for-privacy@bellatlantic.net wrote (in ) about 'OK, so how do I choose a replacement thermistor?', on Wed, 19 Nov 2003:

You need to re-think what happens at switch on. Suppose one thermistor is 100 ohms and the other 110 ohms, and they both vary with temperature in the same way (which is not normal: the 'law' varies from sample to sample). The 100 ohm one will take more current, get hotter, lower resistance, take more current, get hotter...

The same thing happens if they have equal cold resistances but the resistance of one falls more steeply with temperature.

It isn't quite *inevitable* that hogging will occur. There is a limited range of initial conditions under which the initial drop in resistance is sufficient to reduce the power dissipated in that thermistor to less than that in the other one. But this condition is almost impossible to design for.

I read in sci.electronics.design that DaveC wrote (in ) about 'OK, so how do I choose a replacement thermistor?', on Tue, 18 Nov 2003:

Not quite. You would have 23.5 ohms in the circuit, so the current would be 7.23 A peak. It doesn't need to be as low as that, but it probably won't do any harm.

You need to put a resistor in series with the coil and a large capacitor in parallel with the coil. So that you can use an electrolytic capacitor, you put a diode in series with the resistor. This diode needs to be adequately rated, and a 1N4007 is a good choice. The R and C values depend on the relay characteristics, and you will have to experiment, with just the diode resistor, relay and capacitor, not the whole amplifier. Start with a resistor of about twice the resistance of the coil. You need to get the relay to close firmly. If it doesn't, reduce the resistor value.

Without more data that you give, it's not possible to say for sure, but generally, yes.

He can use the lamp as the resistor in the resistor-and-cutout relay solution. The lamp can be an ordinary 100 W bulb. There is a power amplifier made by Crest Audio that uses a thermistor-and-cutout relay. When the termistor goes bad, i've used a 100W bulb in its place, and it works OK.

I read in sci.electronics.design that El Meda wrote (in ) about 'OK, so how do I choose a replacement thermistor?', on Wed, 19 Nov 2003:

Maybe it did in your case, but it won't work in all cases. You can finish up with the lamp half on, the relay not energised yet and the amp trying to work from half voltage or less. Remember that the amplifier's no-signal current is not very large, even for a 240 W unit.

Right. And while the 100 ohm is doing that, the 110 ohm is also taking current, getting hotter, lower resistance, take more current, get hotter, etc. The heating in the higher resistance unit will be less than the heating in the lower resistance unit, but it will occur. Conceivably you can arrive at the point where the first NTC has lowered resistance enough so that the inrush voltage can no longer force enough current through the other NTC to cause it to heat, but that takes appreciable time.

Let me do some numbers. At the start, both thermistors are cold. One is 110 ohms, the other is 100 ohms. Say the load is 20 ohms. Ok, with only the 100 ohm thermistor and the load in series, with a 120 volt source, the initial current will be 1 amp, and the thermistor will drop 100 volts. Add the 110 ohm thermistor. The total resistance of the two thermistors in parallel and the load in series is ~72.38 ohms. At turn on, the initial current will be ~1.67 amps, and the thermistor pair will drop 87.3 volts. The current through the 110 ohm thermistor will be ~.79 amps and the current through the 100 ohm thermistor will be ~.87 amps. Thus both will heat, but there will be more heating in the 100 ohm thermistor. If by "current hogging" you are describing the above scenario, then we are in agreement. The question becomes one of rate of change of resistance over time in a thermistor, but initially, the parallel NTC will reduce the current that the first NTC has to handle, at the expense of more total current (less limiting) to the load. I haven't seen any curves of delta R vs time for these limiters. If you have a url that shows the curves, that would be helpful.

Looks like you're on the right track. Your 8.5 A should be lower, as John said, but you have the concept.

You can connect the relay coil across the power supply through a resistor, as you suggest, but you may have a problem. As the capacitors charge, the supply voltage rises. When it is high enough, the relay operates and its contacts bypass the limiting resistor. But the voltage will continue to rise, possibly to the point where it will force too much current through the relay coil. So you have to be judicious in component selection, such that the relay is protected from too much current, yet gets enough to operate at the proper time.

An alternative is to use the power supply voltage level as a signal, rather than a source - run it in to a comparator, or use a zener & transistor, or whatever, which in turn operates the relay from a fixed supply. Or, you could use a turn on delay timing circuit, independent of the supply you are limiting. After N seconds, it operates the relay without regard to the voltage on the power supply. The point is that there are a number of possibilities to choose from - there are even time delay relays that contain all the necessary circuitry that are made for this purpose.

I read in sci.electronics.design that snipped-for-privacy@bellatlantic.net wrote (in ) about 'OK, so how do I choose a replacement thermistor?', on Thu, 20 Nov 2003:

Ratio of currents = 1.1

You need to look at the next step. The thermistor that WAS 100 ohms gets hotter due to the greater current, so its resistance goes down more. Suppose it is 50 ohms at a certain time, and the other thermistor is 60 ohms. We get 47 ohms total, 2.55 A from the 120 V source and 69 V across the thermistors. This gives 1.38 A through the 50 ohm and 1.15 A through the 60 ohms.

Ratio of currents = 1.2

Eventually, one thermistor gets to carry almost all the current, which of course decreases with time as the inrush falls away, while the other gets cooler. This actually happens experimentally, so there is no point in arguing that it doesn't.

Not with time, primarily, but with body temperature.

Time is not significant if both thermistors have the same thermal mass. Since they have to be nominally identical (otherwise current hogging is instantaneous), they do have that property.

Now I'm getting somewhere! This is the part where I could only speculate. I've never experimented with them in parallel.

Thank You!

You can choose a proper wattage lamp by trial and error (that´s what I did). The Crest amp was a 1200 W unit, with big filter capacitors and with my first choice -a 60 W lamp- it did exactly what you describe. With a 100 W bulb the relay energized correctly.

Plus, every time the amp was turned on, the bulb glowed beautifully inside, not unlike an old tube unit. :')