On page 35 of The art of Electronics Third Edition, when describing the disadvantages of reducing ripple by choosing large capacitors, the author writes:

The very short interval of current flow during each cycle(only very near the top of the sinusoidal waveform) produces more I²R heating.

However, isn't Q = I²Rt? So a shorter interval should result in less heat - why is more heat produced?

  • \$\begingroup\$ Think constant ripple current from a converter at some particular load which the capacitor a filtering. Time becomes irrelevant. \$\endgroup\$ – winny Jul 8 '16 at 6:00
  • 1
    \$\begingroup\$ Because I is squared and t isn't, and their product I·t (or ∫Idt if you prefer) is constant for the same average current. \$\endgroup\$ – Dmitry Grigoryev Jul 8 '16 at 9:05
  • \$\begingroup\$ Or, see it like this, high peak current in small bursts versus steady average current, same average value but much higher RMS current for the pulsed one. Irms^2*R losses. \$\endgroup\$ – winny Jul 8 '16 at 9:31

For a given average current, drawing it in short bursts creates more heating in a resistor than a smooth flow. The reason for this is that current is averaged over time but power is equal to current squared. For example 1A through 1Ω for 1 second = 1 Watt, but 2A for 0.5s + 0A for 0.5s is 4W*0.5s + 0W*0.5s = 2 Watts.

This is why to get the heating effect of AC current through a resistor you must take an rms (root mean square) measurement, not just the average that a meter would normally show.

Using a larger filter capacitor causes current to be drawn in shorter bursts because the voltage doesn't sag as much between cycles, so the rectifier conducts on a smaller peak part of the AC waveform. If the load draws a fixed current then the average current is the same, but the peaks must be higher. Assuming resistances in the power supply circuit don't change, the power loss will be higher due to the higher rms current.

  • \$\begingroup\$ Err... shouldn't that be joules rather than watts? \$\endgroup\$ – Andrew Morton Jul 8 '16 at 9:47
  • 2
    \$\begingroup\$ @Andrew: You're right that W.s = J, but he's really saying $$ P_{AVG} = \frac {4W \times 0.5s + 0 W \times 0.5s}{1s} = 2~W $$. All the seconds cancel out and you're left with watts. Bruce just left out the divisor. \$\endgroup\$ – Transistor Jul 8 '16 at 10:25
  • \$\begingroup\$ What's the definition of average current please? I suppose a) it can't be the average current over a whole cycle, because ∆Q = 0, and thus I = ∆Q/∆t = 0. b) it can't be the average current during the charging period, because at that time, current flows through the capacitor, rather than the load. c) it can't be the average current during the discharging period, because it is the charging current that has "a shorter interval", and this presumably does not have much impact on discharging current. \$\endgroup\$ – nalzok Jul 8 '16 at 17:06
  • \$\begingroup\$ Hmm…maybe it can be b) the average current during the charging period? \$\endgroup\$ – nalzok Jul 8 '16 at 17:09
  • 1
    \$\begingroup\$ Yes. It's actually the current flowing through the source resistance that we are interested in. This is close to the charging current in a large capacitor, but not quite equal because some current flows directly from source to load (how much depends on the capacitor's ESR, which I am assuming is negligible). The capacitor discharges into the load, whose power loss we are not interested in. \$\endgroup\$ – Bruce Abbott Jul 8 '16 at 17:59

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.