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By 'simple AC to DC converter' I mean the familiar circuit shown in the figure below, where the capacitor C reduces the ripple.

Simple AC to DC converter

A sufficiently large capacitor can be used in order to reduce the ripple voltage to any desired level. But in The Art of Electronics the authors state that this approach has some disadvantages, one of which is

The very short interval of current flow during each cycle (only very near the top of the sinusoidal waveform) produces more \$I^2 R\$ heating.

I don't understand what is meant by this extract. Does \$R\$ here denote the resistance of the load? What is meant by 'the very short interval of current flow during each cycle'? Does a bigger capacitor cause more wasted energy?

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    \$\begingroup\$ Is this listed as a disadvantage of a small or a big capacitor value? \$\endgroup\$ – Eugene Sh. Sep 13 '18 at 17:43
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    \$\begingroup\$ R denotes the load. Yes, a little bit in the capacitor. By increasing the capacitance the duty cycles becomes smaller and smaller while peak current becomes higher and higher, both of which are bad. On top of that your power factor is decreasing. \$\endgroup\$ – winny Sep 13 '18 at 17:44
  • \$\begingroup\$ @EugeneSh., all that is said: 'By choosing capacitors that are sufficiently large, you can reduce the ripple voltage to any desired level. This brute-force approach has three disadvantages.' \$\endgroup\$ – apadana Sep 13 '18 at 17:46
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    \$\begingroup\$ I would say that the "R" refers to the resistance of the wiring between the AC source and the load, and the effective resistance of the diodes in the bridge - all these items will have to carry a large current to charge the capacitor in the short time that the bridge output is sufficient to charge the capacitor. \$\endgroup\$ – Peter Bennett Sep 13 '18 at 18:34
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    \$\begingroup\$ You could use a simulator of your choice and see how exactly the current through the capacitor will look like for various values of C. \$\endgroup\$ – a concerned citizen Sep 13 '18 at 20:36
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I think there is an unspoken assumption that as the capacitor value increases the load voltage and current will also increase, which would be true if the load is truly resistive. In that case, the ac input waveform can only supply energy to the capacitor when the rectified ac waveform has a higher voltage that the capacitor voltage. As the capacitor voltage increases there is a smaller and smaller part of the ac cycle that has enough voltage to forward bias the bridge rectifiers and add charge to the capacitor. Therefore, the average current during these short times goes up, and the \$I^2R\$ heating during those short times goes up.

However, it's not clear to me that the average power due to \$I^2R\$ loss will go up significantly. I know that many people worship the TAoE but I have found it hard to read and confusing in places.

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  • \$\begingroup\$ I don't know about how significant losses would be, but for the same energy delivery per second, decreasing duty cycle must increase instantaneous current, and wasted power will be proportional to the square of that current. If instantaneous current doubles and energy delivery stays the same, losses must quadruple, so the effect could be quite severe especially if you get down to 10% or lower duty cycle... 100x losses aren't good. \$\endgroup\$ – K H Sep 13 '18 at 21:16
  • \$\begingroup\$ @KH Yes, instantaneous power loss is proportional to instantaneous current squared, but the duty cycle is going down at the same time. Furthermore, this is only an issue if the resistance in the rectifier/capacitor wiring is significant compared to the load resistance. If the \$I^2R\$ loss at 20% duty cycle is negligible, who cares if it doubles or quadruples at 10% duty cycle? \$\endgroup\$ – Elliot Alderson Sep 13 '18 at 21:21
  • \$\begingroup\$ You make a good point and come to think of it, losses would only be 10x at 10%, not 100x. I guess it will really depend on switching speed. \$\endgroup\$ – K H Sep 13 '18 at 21:31
  • \$\begingroup\$ Since a larger capacitor has lower reactance in a given frequency, more current goes through it. Can this be what the authors are essentially saying? \$\endgroup\$ – apadana Sep 14 '18 at 1:41
  • \$\begingroup\$ By using Fourier analysis, the output of the rectifier can be written as the sum of a constant voltage and sine waves of different frequencies and amplitudes. \$\endgroup\$ – apadana Sep 14 '18 at 2:07

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