Couple Questions about this Resistor Datasheet

[Abdullah Almosalami]

TL;DR Summary

Overload voltage rating
Sqrt in equation

So I have these 10k 5% resistors from Stackpole Electronics and just because I never actually took the time to look at a resistor datasheet beyond nominal resistance, tolerance, and power rating specs, I decided to look them up. Came across a couple things I wanted to ask about:

1) Maximum overload voltage rating and single pulse power plot
Here's a picture of the basic resistor specs (highlighted in orange) (ignore my notes). As you can see, max overload voltage is 700V.

Here is the plot on single pulse power: (my resistor type is highlighted in yellow)

So my first question: It seems if the pulse is short enough (say somewhere between 1us to 10us for mine), my 1/2W 10k resistor can take 430W (holy smokes). At that power, the applied voltage would be around 2kV, which is a lot higher than the max overload voltage of 700V. I'm just wondering if the test to determine the overload voltage was done at a longer pulse duration? And so indeed my resistor can take 2kV pulses of 1us length? I 'lined' out in orange what power level would correspond to an applied voltage of 700V on the graph. Also, is "overload voltage" a single pulse from 0V -> rated overload or more like operating in continuous rated voltage and then pulse overvoltage? Maybe this is a manufacturer-specific kind of question?

2) Repetitive Pulse Data Equations
Elsewhere in the datasheet, we have equations for determining rated voltage, current, and wattage for repetitive pulses:

So I wrote out the calculation for Vp for my resistor and I got 11.3kV, which is definitely wrong. What's up?

Edit: Actually, now I'm thinking the sqrt on the P is actually meant to include the R and T/t... So my calculation should be $$V_p = K\sqrt {P * R * \frac T t } \text{ instead of } V_p = K * \sqrt {P} * R * \frac T t$$That would give 80V, which makes a lot more sense... Hehe

 

[DaveE]

The data sheet is written to apply to all resistor values. When you do calculations using your specific value (10K) you may find that one performance limit occurs well before the others. Suppose your resistor was 1 ohm, then that pulse voltage would be within the limits.

I don't think you can interpret the data sheet as representing experimental results. They are simply communicating how to use the product reliably as clearly as they can for engineers that may have diverse applications. For example, the pulse power limits were probably the result of thermal modeling that didn't really take into account how the heat was introduced.

I would assume that the pulse limits, and the continuous limits, must be derated if there is also continuous power being applied along with a pulse. You will need to do your own thermal modeling to solve the combination as well as maybe getting additional guidance from the application engineers. For example, the maximum hot spot temperature is a key thermal parameter.

You are correct about the square root. That comes from P=V²/R and the RMS calculation for pulses.

 

[alan123hk]

It seems if the pulse is short enough (say somewhere between 1us to 10us for mine), my 1/2W 10k resistor can take 430W (holy smokes). At that power, the applied voltage would be around 2kV, which is a lot higher than the max overload voltage of 700V.

It is only said that the maximum 1us single pulse power of the CF12 type resistor is 430W, but it has never been said that the voltage value can be calculated based on the 430W and applied to the resistor.

To ensure safety, I believe that even if the pulse duration is 1us or less, we cannot apply a voltage higher than the overload voltage or the maximum overload voltage to the resistor.

Sometimes, the manufacturer’s data sheet may not cover all possible situations, in which case, I think we should put the greatest safety first.

In addition, according to the link https://www.koaglobal.com/product/library/resistor/term?sc_lang=en,
the overload voltage is the allowable voltage which is applied in 5 seconds under short time overload test. Overload voltage shall be 2.5 times of rated voltage or max. overload voltage, whichever is lower.

Therefore, please note that we must also not apply a voltage greater than 2.5 times the rated voltage.

 

[Tom.G]

From a practical standpoint, the maximum voltage is the lesser of:

• The flashover voltage.
• The highest voltage that can be applied and remain under the maximum power.

I wouldn't be at all surprised if that's what the manufacturer quotes.

 

[yungman]

For single pulse, it's more like an AND equation, (Vp < 700V) AND (max pulse power), not just look at the max pulse power and think it will take 2KV.

They are two completely different thing, voltage max is is governed by the lead spacing( surface creepage), the material makeup of the resistor etc. You don't need power to get to HV breakdown. Power is heat. You need to obey the HV limit AND the power limit.

I use a lot of Stackpole resistors in building my power amps, BECAUSE it's CHEAP! I sure won't push the limit. I don't trust them at all. I use them at like 1/2 the spec. The resistors are getting smaller and smaller for the power rating. You ever try pumping 3/4W through a 2W rated resistor? You don't touch it! I use Stackpole and another Chinese brand, I use 3W resistor for under 1W to be safe. You look at the size of the 3W now a days, they are same or smaller size as an old style 1W resistor! I am not a physicist, but isn't power dissipation related to the surface area in air?