8
\$\begingroup\$

How much pulse power does a standard (i. e. not further specified) resistor withstand?

Consider for example a MOSFET and a 33R resistor that is connected to its gate. If the switching voltage would be 10 V, the resistor sees up to 3 W every time the MOSFET gets switched - but only for a very short time, until the gate (in my case, about 15 nC) is charged.

Do I really need to specify a pulse withstanding resistor? Or is there any way to tell whether a 0402, 0603 or 0805 standard resistor (without further specification) will suffice?

\$\endgroup\$
14
\$\begingroup\$

Any half-decent reputable resistor supplier will have pulse power dissipation limits such as this one from Vishay: -

enter image description here

It tells you how much power can be delivered in a pulse to a resistor. For instance, an 0603 resistor can take pulses of power up to about 20 watts if the duration is only a micro second. For a milli second the power can be no more than 1 watt.

Or is there any way to tell whether a 0402, 0603 or 0805 standard resistor (without further specification) will suffice?

If you want to do the job properly read the data sheets. The data sheet will also tell you how much copper you may need to use around the resistor to achieve this specification so, you can't really guess with any accuracy.

\$\endgroup\$
  • \$\begingroup\$ Isn't there some information missing in the graph? Like duty cycle? \$\endgroup\$ – Curd Aug 11 '17 at 9:40
  • \$\begingroup\$ @Curd here's the data sheet link: google.com/… - I expect that it details what you ask. \$\endgroup\$ – Andy aka Aug 11 '17 at 9:51
3
\$\begingroup\$

While I agree with Andy Aka's answer, if you cannot get a decent data sheet, I'd estimate based on the following:

The enemy of reliability is the heat load on the resistor, (temperature perhaps).

Based on that, you know how much energy is being transferred to the caps via

$$E=\frac{CV^2}{2}$$

Hopefully you can find a mech-eng for back of the envelope air cooling based vs $$I^2R$$ and your energy per pulse.

You could also do IR inspection on a prototype- but if you have access to IR equipment - why not source a resistor with a decent data sheet?

\$\endgroup\$
  • 5
    \$\begingroup\$ My philosophy is that if you can't get a decent data sheet for the resistor of choice then it gets relegated to the resistor you should not choose but +1 for the decent answer. \$\endgroup\$ – Andy aka Aug 10 '17 at 19:50
2
\$\begingroup\$

The resistor needs to withstand the maximum average power dissipated.

That can be a complicated equation, but basically it means the more often you will be switching the MOSFET, either continuously or in bursts, the higher the wattage of resistor you need.

If it's a simple switch to turn on a relay or lamp it won't matter.

If however, you are pulse width modulating some coil current at hundreds of hertz or kilohertz, the charging currents become more sustained and the power dissipated by the resistor matters.

Note I also mentioned sustained bursts. If you periodically fire a few thousand pulses for a significant duration, you need to use those values as your worst case scenario.

\$\endgroup\$
1
\$\begingroup\$

For the pulse, a short pulse, the only place the heat can be stored in within the ceramic core of the resistor. That core needs to remain cooler than? 100 degree Centigrade?

The specific heat of silicon is 1.6 picoJoules/micron^3 * degree Centigrade.

Assume the resistor has volume of 4mm by 5mm by 5mm, or 100 mm cubed. The # cubic microns is 4,000U * 5,000U * 4000U, or 100,000,000,000 cubic microns.

The specific heat of that resistor is 1.6pF * 0.1 TeraMicrons = 0.16 Joules.

How fast can the resistor dump heat?

The thermal timeconstant of silicon (we'll assume the resistor core is clay/silicon) is 9,000 seconds for 1meter cube, 90 seconds for 0.1meter cube, 0.9 seconds for 1cm, and 0.009 seconds for 1mm. Our exit paths are 1/2 (both ends can dump heat to the PCB trace) * 4mm, or 2mm. The Tau for 2mm is 4X that of 1mm, thus 36 milliSeconds.

Can we use these numbers?

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for?Browse other questions tagged or ask your own question.