Resistor surge rating

Question

I have a circuit with a 220uF capacitor and a resistor is used to limit the inrush current

I stupidly fitted the wrong power rating resistor and (no surprise it failed) its taken me a while to realise my mistake. I believe the correct resistor is fine (it ran for a significant amount of time and is double the power) but I am a little bewildered as to how I calculate the rating and prove it theoretically

The resistors (I used two in parallel) fitted were 3R3 1.5W 2512 package

TE Connectivity CRGS2512J3R3

There is even a graph in the datasheet telling me the surge rating isnt adequate and I am interested in how they calculate these curves so I can apply the calcs to other resistors where they arent so helpful to provide a graph.

Here is the plot:

I measured the inrush with a scope (100VDC supply) and its a little over 40A, the theoretical max is over 60A but there is a reverse polarity protection diode and fuse and PCB trace and cap ESR reducing this.

That's the total inrush which passes through two resistors in parallel so approx 20A each resistor

As can be seen the surge has decreased to 50% of the peak after approximately 0.5ms so I believe I can treat this as a square wave with a width of 0.5ms as a decent approximation (like the EMC standards advise for TVS diodes etc)

There are app-notes online like

Vishays Pulse Load on SMD Resistors: At the Limit (again not enough repuation to add link)

I understand that periodic pulses require more derating than a single pulse (thats logical) at what point a single pulse becomes periodic is another topic as probably all electronics needs to be turned off at some point!

Using the calculation in vishays datasheet for periodic pulses with a 1 second period and using my inrush measurement data

P = (V^2/R)*ti/tp

V =100, R=3.3, ti = 0.0005, tp = 1

Gives me a value of 1.515W (absolute max) and I can see if applied surges more frequently then the power increases (which is what happened when the resistor failed)

Looking at the graph from the datasheet (isnt easy on the eye) but with 20.1A through the 3R3 resistor then thats a peak power of 1333W

The graph from the datasheet doesnt seem to agree for example derating for a 0.001 pulse time (for a convenient point to read off the values) the graph says approx 1kW max peak pulse power where as the calculation says that the average would be approx 3W which is double the rating of the resistor

I think I have spent too much time looking at this and simply need to go to bed and wake up fresh, I am confused if I am doing this right or do I just not have a clue!

I can calculate the energy into the capacitor but not sure what I would do with that, is the method right? is there a better way? Is this the way to do it when the manufacturer dont specify pulse/surge ratings?

Any advice is much appreciated

Answer 1

Heat flow takes time. In cases where nearly all of the energy is devoted to raising the temperature and where little useful portion of the heat is allowed to have time needed to significantly flow into its surroundings, then you can use the "action integral of the pulse" to estimate failures. If you can find a specification in "Joules per Ohm" or "\$I^2\cdot s\$ for the resistor, then you could apply it. If not, you'll have to use those curves to make estimates.

The above kinds of specifications are more commonly found for fuses, because that's the job they do and are therefore specified to do. Resistors, on the other hand, are actually designed to dissipate. So this adds another factor to consider.

Instead, let's look at your 2512 curve. It's flat until about \$t=100\:\mu\textrm{s}\$. At the corner, I'm guessing it can handle a pulse of about \$4000\:\textrm{W}\cdot 100\:\mu\textrm{s}=400\:\textrm{mJ}\$. This increases linearly (on a log scale) to about \$18\:\textrm{W}\cdot 1\:\textrm{s}=18\:\textrm{J}\$ for a pulse of \$1\:\textrm{s}\$. Given the log scales here, I get the following equation for the resistor's ability to absorb one pulse of energy over time:

$$\begin{split}E_{limit}&=4000\:\textrm{W}\cdot t\\E_{limit}&=1.91089572\:\textrm{J}\cdot \ln \left(t\right)+18\:\textrm{J}\end{split}\quad\begin{split}&\textrm{ where}\quad t \le 100\:\mu\textrm{s}\\&\textrm{ where}\quad 100\:\mu\textrm{s}\le t \le 10\:\textrm{s}\end{split}$$

This is a hot-spot calculation and it's probably only good to a few times the chart duration, where other factors allow the dissipation to stabilize at the rated power. They only show the curve going out to a second. But the above equation might work for a bit past the end of that curve. Regardless, it gives you an idea.

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If I did the integral right, the energy delivered into your R, by your RC circuit, is the following function of time:

$$E_{decay}=\frac{V_0^2\cdot C}{2}\cdot \left(1-e^{-\cfrac{2\cdot t}{R\cdot C}}\right)$$

If this value exceeds \$E_{limit}\$ at any time, you might have a problem. Given that you are measuring up to \$40\:\textrm{A}\$, I'm going to say that your \$V_0=132\:\textrm{V}\$ for the above purposes. So if you look at the case for \$t=100\:\mu\textrm{s}\$, you get \$\approx 462\:\textrm{mJ}\$ which exceeds the rating curve you have. To be safe, you'd probably want to be substantially under it, I think. Not over.

The curve does indicate that, given a little more time, there should be enough time and therefore no remaining problems. But this does seem to suggest a corner case problem when using a single device.

I gather you are using two of them and still having problems. (I'm not sure how all this is mounted and that could also be important.) In any case, if you plug in the \$1.65\:\Omega\$ paired-resistor equivalent, you get \$812\:\textrm{mJ}\$ for both. Which, divided between the two still exceeds the spec (by only a little.)

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Just an added note because I had to make a correction to the first equation above, for \$t\lt 100\:\mu\textrm{s}\$. I had just made it a constant before, but it really is a function of time. Less time? Less delivered energy. The curve's flat line there makes that apparent. I'd just failed to account for it in the equation.

So with the correction, you can more easily see that for an even smaller period of time, say \$t=10\:\mu\textrm{s}\$, that the \$E_{decay}\$ equation supplies (using my \$V_0=132\:\textrm{V}\$ figure based on the \$40\:\textrm{A}\$ you had written) for about \$100\:\textrm{mJ}\$ of energy into \$1.65\:\Omega\$. But \$4000\:\textrm{W}\cdot 10\:\mu\textrm{s}=40\:\textrm{mJ}\$ as a limit by the curve. So the curve is far exceeded when considering shorter times like this. Even using \$V_0=100\:\textrm{V}\$, I get \$60\:\textrm{mJ}\$ of energy in that short time. So, still exceeds the specification.

I can see why you are having troubles.

Answer 2

Lets compute the temperature rise from one pulse, assuming the heat remains totally INSIDE the resistor. If 5 degree Cent, its OK, right? But if 5,000 degree Cent rise, its (already been turned into plasma) not OK, agreed?

We need to know how much heat the resistor can store internally. Here is a useful number: the specific heat of silicon (like the pure stuff, used as silicon wafers) is 1.6 picoJoules per cubic micron per degree Centigrade rise.

I'll let you convert the resistor size into microns, Length, Width, Height, and compute the total volume. Assume the resistor has a clay/ceramic base on which the metal film is deposited. Heat is generated in the film, and quickly flows into the silicon/clay/ceramic base.

What are the time constants? PAY ATTENTION HERE. The time constants for heat flow are NOT LINEAR with size. The time constants change as the Square of the size.

Size of silicon cube Time Constant

1meter cube 11,400 seconds

10cm cube 114 seconds

1cm cube 1.14 seconds

1mm cube 0.014 seconds (14 milliSeconds) which is about the size of SMT resistors

100 micron cube 114 microseconds

10 micron cube 1.14 microseconds

1 micron cube 11.4 nanoseconds

0.1 micron cube 114 picoseconds (approximately the thickness of the conductive layer of FETS

EDIT In my opinion, the thicker the resistive region, the more survivable is the resistor. In thin films, the heat has to flow into the clay/silicon bulk. In a carbon composition resistor, most of the resistor body comprises the resistor; result is the heat is generated throughout the resitive-bulk and well exploits the entire mass as an immediate heat sink, because the heat has no where to go, except out the leads. With that in mind, examine this diagram:

^{simulate this circuit – Schematic created using CircuitLab}

Let's discuss the energy storage of 1mm cube resistor. You have an energy storage challenge. At 1,000 microns per side, that cube has 1 Billion cubic microns of volume. Assuming the entire resistor---- resistive region, outside protective glaze, and any interior hard-ceramic base ---- have 1.6 picoJoules per cubicmicron per degree Cent, your heat capacity is

$$ 1billion cubicmicrons * 1.6 picoJoule/cubicmicron/degree Cent$$

or 1.6 milliJoule/degree Cent energy storage.

Your energy is 20 amps per resistor (2 Rs in parallel, each 3.3) for 0.5milliSec. What is the Joules? P = I^2 &* R = 20 * 20 * 3.3 * 0.0005 sec, or 1320 joules/second * 0.0005 = 0.65 joules.

Now divide 650 milliJoules / 1.6 milliJoule (for a 1milliMeter cube volume) and the rise in temperature is 400 degreeC. Solder melts; aluminum creeps.

Answer 3

I believe you should have used 1.65 instead of 3.3 ohm in the formula you provide, $$P = v^2/R\frac{ti}{tp} = (100x100/1.65) \text x 0.0005/1 = 3.03W$$ As you can see, this shows you need twice the power. Also, the peak power would now be, $$p = I^2R = 40x40x1.65 = 2,640$$ This shows that each resistor needs to be at least 1.5W, but since you are using two 3.3 ohm unmatched resistors, to be safe, you should make each resistor carry the full load. Therefore you need each 3.3 resistor to be able to dissipate 3W.

EDIT: Additional reasons to double the power rating are: 1) each resistor interferes with the heat dissipation capability of the other resistor, 2) each resistor becomes a heater, for the other resistor.