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I'm working through the exercises of Art of Electronics and exercise 1.21 asks to calculate how big a fuse a circuit (full-wave rectifier using a center-tapped transformer) needs to be. The current is a square wave alternating between 0 and 2A with 4K Hz with a duty cycle of 50%. The exercise makes a remark that the because of a long thermal time constant, the fuse will respond to the value \$I^2\$ averaged over time. That should be 2A then.

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However, if I look at the answers at https://milesdai.github.io/TAoE3Solutions/data/taoe3-solutions.pdf, it states that the answer should be calculated using the RMS function:

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My understanding is that the point of the RMS function is to calculate the area under the wave, and that the standard RMS function applies to the sine waves. Wikipedia seems to confirm this.

Could you help me understand what I'm missing?

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    \$\begingroup\$ RMS applies to any waveform and no, it doesn't calculate the area under the wave at all. \$\endgroup\$
    – Andy aka
    Commented Jan 6, 2023 at 10:14

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The rms current of a square-wave signal is obtained by multiplying the peak value by the square root of the duty ratio \$D\$ as shown in the below picture:

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The average value is the peak multiplied by the duty ratio \$D\$.

The rms value is the one you should use to size the fuse you need with some margin of course. This is all about power dissipation linked with the fuse resistance \$R\$.

I think the term averaged creates the confusion in the text. It does not refer to an average current but more to the final rms value seen by the fuse, in steady-state if you wish, once the converter or the equipment has started up and is stabilized. That would be my interpretation at least.

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  • \$\begingroup\$ Ok, my misunderstanding is in the simplifications of the formulas. Your formula differs from the one in the explanation, but they give the same result. I got lost between \$\sqrt \frac{I^2}{2}\$ and \$\frac{I}{\sqrt{2}}\$. \$\endgroup\$
    – Pieter van Ginkel
    Commented Jan 7, 2023 at 8:06
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    \$\begingroup\$ The given answer is highly confusing as the \$\sqrt{2}\$ is a particular case of the 50% duty ratio for their square signal. Change the duty ratio to let's say 25% and it's wrong. \$\endgroup\$
    – Verbal Kint
    Commented Jan 7, 2023 at 8:30

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