# Voltage rating vs. power rating of a resistor

My question may sound very basic, but I am very confused by the difference between the voltage and power ratings of a resistor.

Vishay's document says:

Rated Power

The maximum amount of power that can be continuously loaded to a resistor at a rated ambient temperature. Network and array products have both rated power per package as well as per element.

Rated Voltage

The maximum value of DC voltage or AC voltage (rms) capable of being applied continuously to resistors at the rated ambient temperature.

I read this datasheet for a 27Ω, 0.2W resistor. Page 3 of the datasheet shows this formula:

$RCWV = \sqrt{P\times R}$

Where RCWV = Rated DC or RMS AC continous working voltage at commercial-line frequency and waveform(volt)

P = power rating (watt)

R = nominal resistance (ohm)

The above 27Ω resistor on the link has 50V voltage rating and 0.2W power rating, then I place the values into the provided formula

$RCWV = \sqrt{0.2\text{W} \times 27\Omega}=2.32V$

Could anyone explain me why the voltage rating is 50V, not 2.32V?

When I want to calculate the maximum current that the resistor can stand using the power rating of the resistor (0.2W):

$P=I^2 \times R$

$I = \sqrt{\dfrac{P}{R}} = \sqrt{\dfrac{0.2\text{W}}{27\Omega}} = 86\text{mA}$

If I use the voltage rating:

$I = \dfrac{V}{R} = \dfrac{50\text{V}}{27\Omega} = 1.85\text{A}$

By looking at these results, I should use the power rating, right?

The voltage rating is for the resistor series typically and specifies the maximum peak voltage you can apply without danger of damaging the resistor due to corona, breakdown, arcing, etc.

The power rating is completely independent of the voltage rating. It specifies the maximum steady state power the package is able to dissipate under given conditions.

You have to conform to both specs. If placing the maximum voltage across the resistor results in more power than the spec allows you have to reduce the voltage until you meet the spec. Likewise you can't increase the voltage above the rating just because you're not hitting the maximum power limit.

• Nice answer. In practice this means that low value resistors are limited by their power rating (you can't reach their voltage rating) while high value resistors are limited by their voltage rating. Jul 9 '14 at 19:17
• @BrianDrummond: Many resistors will have, in addition to their continuous-power rating, specifications which will allow higher amounts of power to be applied for short periods of time. Holding more than 100V across a 10K 1W 500V resistor would eventually cause it to overheat, but such a resistor could probably survive 125V for a second, or 250V for a quarter-second, or 500V for 1/16 second, if allowed to cool down before and after. Doubling the voltage would cut the time by a factor of four, up to the breakdown voltage, but failure could be immediate above that. Jul 9 '14 at 22:15

The 50V specification is the maximum value allowed to apply on resistor leads. (due to isolation, ...). The datasheet being a general document, knows nothing about your application. So, next I try to show a situation ready to violate both max voltage and max dissipated power (the later related to RMS values). When a pulsed voltage waveform is applied - see the figure below, the corresponding RMS voltage is: $$V_{RMS}=V_{pk}\sqrt{\frac{T_H}{T_H+T_L}}$$

For example, when $$\V_{pk}=50 V\$$, the period $$\T=T_H+T_L=1 ms\$$ and pulse length $$\T_H=2.16\mu s\$$, the $$\V_{RMS}\approx2.323 V\$$ and dissipated power $$\\approx0.2 W\$$. • $$0.2W$$ comes from $$\frac{2.323^2}{27}$$ (in case of someone wonder it) Nov 8 '18 at 6:48

200mW tells you that the resistor cannot disspiate in excess of this value (200mJ/s) continuously or it will overheat and damage itself.

The expression RCWV = sqrt(PxR) gives you insight into the maximum voltage allowed at the max power disspation point. Recall that:

Power = I*V

P*R = IV*R

P*R = V^2

V = sqrt(P*R)

Where we can see that at max power we could have a voltage of 2.32V @ 86mA. However, we could also have 50V @ 200mW/50V = 4uA, or 1.5V @ 133.3mA - the solution set is infinite.

I am not exactly sure the point of the expression above, but just note that the total energy disspiated as heat is taken from:

P = I*V

And simply cannot exceed the power rating of the device.

V=I*R => I = V/R = 50/27 = 1.85A
P = 1.85 * 50 = 92.5W