How Much DC is OK on a Power Line?

How DC Appears on the Mains
There are any number of different machines that can create a mains supply DC offset. Most will be totally outside your control, many DC "events" will be transient in nature, but one common theme applies - they will all load the mains supply asymmetrically for a period of time that ranges from a couple of cycles to minutes at a time. Figure 1 shows a typical (small) example that you may even have in your house - the transformer (shown within the dotted line) is your toroidal transformer. Many older hairdryers (and some heat guns as well) had a switch for "half power" that simply switched a diode in series with the mains. For a 240W element at 240V, that equates to a resistance of 240 Ohms (example only - actual power will vary widely).

If a diode is switched in series with the heating element, this reduces the voltage and hence the power (actual power will be almost exactly half). However, by half-wave rectifying the mains in this manner, there is an inevitable interaction with the mains impedance.

Figure 1 - Half-Wave Rectified Appliance, Transformer & Mains Wiring
The arrangement shown above assumes that the mains has zero impedance. Actual impedance is shown as R1, which varies from one house to the next. The value of 800 milliohms was chosen because this is what I measured at my workbench. Your mains may be better or worse than this.

After the asymmetrical load has done its job, a simulation shows the positive peaks of the 240V AC waveform reach 338.35V, but the (unloaded) negative peaks reach the proper value of 339.28V. This is a tiny bit less than the theoretical value of 339.41V because of the transformer load resistance and simulator resolution. The difference between the peak voltages is 0.93V, but the mean (average) DC voltage is -275mV. It is the mean value that appears as "DC" on the mains. It can also be measured, but to do so requires that one works on live components. This is not recommended as it is inherently dangerous. However, if you must (and PLEASE take extreme care), you need a 100k resistor and a 10uF non-polarised capacitor, wired in series. Connect this circuit across the mains (power off!), and connect a DC voltmeter across the capacitor. This attenuates the AC enough to prevent the front-end of the meter from being overloaded, and the DC voltage is easy to measure. Expect to see the DC vary around the zero voltage, with a normal variation of +/-25mV or so (typical - residential areas). The alternative method is to measure the DC across the diode/capacitor network in the circuit of Figure 3. Do not connect or disconnect the meter with the circuit live, and use alligator clip leads to make the connections.

With a half wave rectified load, the mean DC level is 275mV as described above - polarity is not important, because either polarity will be as bad as the other. If a transformer has a primary DC resistance of 2 ohms, there will be an effective DC current of 137.5mA in the primary. This is many times the current needed to cause the core to saturate during the negative half cycle of the AC waveform. Remember that with a toroidal core, saturation is a "hard limit". Because there is no air gap (intentional or otherwise), when the saturation limit is reached, inductance falls and current rises rapidly.

Tests were done using a 500VA toroidal transformer with very close to the example values given above. With 240V AC mains, 50Hz, 264mV DC offset created by DC injection (see Figure 6), and at no load, the current was seen to rise from 16mA to 218mA. The test was done at no load because this is the worst possible case. As load increases, the effective primary voltage falls - the voltage dropped across the winding's resistance is "lost" to the transformer. 264mV DC offset causes a current of 132mA DC in the transformer primary. This is probably the maximum offset that you will encounter in real life, although some areas may be worse. I have no data on this.