Top graph: voltage drop across the resistor vs. time

Bottom graph: current through resistor vs. time

Voltage and current on a resistor over time

This is a resistor charging a capacitor.

If the cap is being charged and discharged 2 times a second, what is the power rating I need my resistor to be? Of course, I want it to be as low as possible because of size.

  • \$\begingroup\$ And now a third graph: voltage times current over time. \$\endgroup\$ – Ignacio Vazquez-Abrams Sep 17 '13 at 21:37
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    \$\begingroup\$ This sounds like homework, so partial answer: Multiply the two curves together, point by point. What is the peak power? What is the average power? \$\endgroup\$ – Dave Tweed Sep 17 '13 at 21:38
  • \$\begingroup\$ This is actually not homework, it's a resistor on a board I'm designing. I'm wanting the resistor to be as small as possible. \$\endgroup\$ – brett s Sep 18 '13 at 1:45
  • \$\begingroup\$ If the period is on the order of 2 seconds, do you really need the charging time constant to be on the order of a few ms? If you could make it a few hundred ms, the peak power would be much lower (mW instead of several Watts), and you'd be able to use a very tiny resistor indeed. \$\endgroup\$ – Dave Tweed Sep 18 '13 at 3:44
  • \$\begingroup\$ You probably realize this, but (if you size the resistor power rating this way) make sure that there are no conditions where the capacitor discharging circuit will remain on (while the resistor is powered) or your resistor may burn up. Take special notice that this does not occur during reset conditions. \$\endgroup\$ – Tut Sep 18 '13 at 15:26

All you need to know is the worst case scenario, when the current is at its peak. Use that current and the value of the resistor to calculate the power dissipation using I^2*R

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    \$\begingroup\$ That's certainly a safe answer; but for something cyclic over a short term it may be possible to factor in the duty cycle and find that, at least for a thermal power rating, the practical number is much lower. \$\endgroup\$ – Chris Stratton Sep 17 '13 at 22:02
  • \$\begingroup\$ @ChrisStratton: In some designs it might be acceptable, but in many of the designs I've used always the worst case must be taken into account. Sure, you could calculate the amount of heating in a short period until it reaches safe levels, but when boards are placed in a situation where the environment is already hot, you will take less chances \$\endgroup\$ – Gustavo Litovsky Sep 17 '13 at 22:06
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    \$\begingroup\$ Adjustments for ambient temperature and adjustments for duty cycle are both potentially important, but they are each unique topics - one doesn't explain the other. \$\endgroup\$ – Chris Stratton Sep 17 '13 at 22:08
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    \$\begingroup\$ @ChrisStratton All Gustavo is saying is that in situations with a hotter environment, the tolerance needs to be higher. You have a good point though. In this case, the maximum power looks like it's roughly 2ish watts, yet a much smaller resistor could be used (1/2W) since current drops off very quickly. I still have to say Gustavo's answer is solid though, by using the worst case value. That's a good practice and habit that is necessary in lots of things. \$\endgroup\$ – krb686 Sep 18 '13 at 2:51

The energy dissipated in a resistor charging a capacitor is equal to the stored energy in the capacitor, 1/2 * CV^2. When the cap is discharged, its stored energy is dissipated in the resistor. So the total energy in the resistor for one charge/discharge cycle is CV^2. Multiply this by the frequency and you get the average power in the resistor.

This will give you the lower limit for your power requirement. In practice you will want to go with a higher rating.

  • \$\begingroup\$ Does this apply to polarized caps, which is what I'm using? \$\endgroup\$ – brett s Sep 18 '13 at 20:08
  • \$\begingroup\$ It does apply to polarized caps. \$\endgroup\$ – user28910 Sep 19 '13 at 12:53
  • \$\begingroup\$ Since your time constant (5-6 ms) is short compared to your charge/discharge period (500 ms), you may want to know the average power during charging or discharging interval. This can be approximated by dividing the resistor energy 0.5 * CV^2 by 3 time constants (time constant = R * C). \$\endgroup\$ – user28910 Sep 19 '13 at 13:00
  • \$\begingroup\$ 15 ohm with 330uF cap= 4.95ms time constant.So (0.5*330u*14.75^2) / (3*RC) = 2.4W ?? \$\endgroup\$ – brett s Sep 19 '13 at 19:06
  • \$\begingroup\$ Yes, 2.4 watts average power in the resistor during capacitor charging, if we agree that the charging period ends at 3 time constants, which is at 95% fully charged. \$\endgroup\$ – user28910 Sep 23 '13 at 20:10

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