This part of the log is to calculate theoretical filter dissipation for the measurements that I have done.

I've done measurements with a transformer that had a voltage ratio of roughly 1:30.

Therefore, all voltages measured in the previous log have to be multiplied by 30 in order to relate them to the true powerline frequencies.

However, we don't really need to consider the voltage ratio of the transformer, because we've measured the transfer function of the transformer, and established that it's essentially flat from 0 to 70 KHz. So we can just take the measurements of the transformer, and use the relative attenuation for the powerline voltages.

For instance, the 3rd harmonic was at -25 dB. That is equal to:

where:

• y = amplitude expressed in V
• x = amplitude expressed in dbV (log10)

x = -25 dB, corresponding to 0.003162

The powerline frequency (50 Hz) has an amplitude of 230 VAC (for Europe), so the third harmonic would have a voltage of 230 * 0.003162 = 0.727324 VAC.

This is of course the RMS (Root Mean Squared) voltage, which is determines how much power would be dissipated in a load for a sine frequency.

That's a small amount.

Now, this device is supposed to dissipate (short out and turn into heat) all frequencies above 50 Hz.

There is a bit of a problem with calculating how much power would be dissipated for these frequencies.

Let's assume that the power station generates only a power frequency of 50 Hz, and that all frequencies above that are generated by devices that are connected to the power line.

Let's say that the powerline can generate up to 13 A (which is the normal type of fuse in the UK).

Then the power generated at each socket at the power line frequency is 230 V * 13 A = 3 kW (approximately). We certainly don't want to dissipate that amount with this device.

Dissipating this 3 kW takes a resistive value of W=I^2/R = 3000/(13A^2) = 17.75 ohms.

The problem is about source impedance: (or output impedance)

Each source has an impedance in series, and attaching a load to that source causes the voltage to be divided between the source impedance and the load impedance, using a resistive divider calculation.

The source resistance of the powerline is (should be) really low, otherwise if you attach a few high wattage devices, the powerline voltage will drop (which is called a "weak net"). It turns out that for the UK this is about 0.5 ohms at 50 Hz [1]

But how much source impedance is there at the harmonic frequencies? The generators for these frequencies are inside your devices.

Also, what if the frequencies generated by the net are not 50 Hz, but also these high frequencies are generated with a low output resistance? Unlikely though.

Theory about power line issues:
http://www.marcspages.co.uk/pq/index.htm

And this also seems relevant:

http://www.marcspages.co.uk/pq/3241.htm

I suppose we just went to do some simulations to look at what ifs and filter topologies.

[1] : http://www.acoustica.org.uk/other/mains_Z.html