Zero-Current-Switching Circuits

Question

I'm basically driving an AC fan from 120V AC wall power (use of classic magnetic mechanical relay or solid-state relay is fine, but I'm weary of triacs because of their EMI and their wasted voltage drop) and want to reduce the flyback spark (due to inductance of fan windings) when I turn it off every few minutes.

The current peaks at about 5.0A, and getting a zero-crossing turnoff below 0.5A would be possible with an MCU (60 Hz power gives 120 opportunities for exactly zero current every second, and I would just need millisecond-precision) and solid-state relay, but all this seems excessive.

How else can I automatically open my switch only when the current is below 1A? Is there a standard solution for this?

I allow triacs (anything is ok - this will not become an "XY problem"), but I hope not to see their characteristic 60Hz or 120Hz "triggering current EMI jolts" (maybe some additional circuitry could bypass the triac except for the last few cycles before turnoff?).

Answer 1

This isn't really a well-formed question, but I think answering with the background information will address it well regardless.

For EMI, the rate of change matters. So, switch turn-on and -off are very important events.

Between thyristor and mechanical contact, mechanical is much worse: the edge rate can be under a ns, launching waves into the 100s of MHz range. What's more, during turn-off of inductive loads, the terminal voltage can rise (via load inductance charging wiring capacitance) to breakdown voltage, spark, rise again, etc., producing a machine-gun burst of sparks (EFT, electrical fast transients). Semiconductors don't do this; at worst, a thyristor re-triggers then is on for another half-cycle.

Thyristors typically turn on in the 100s of ns time scale, and turn off in µs. So it's already not too bad, but definitely still in the range of concern.

As for switching, there are two circumstances: a high-impedance element going low, and a low-impedance element going high.

These are independent cases: anything we connect in parallel with the low impedance, will be shunted by that impedance and therefore become irrelevant; and anything we connect in series with the high impedance, will be series by that impedance and therefore become irrelevant.

So the general case is, we must have both series and parallel elements around the switch, to have complete effect, say for filtering.

To deal with both edges, we might do something like this:

^{simulate this circuit – Schematic created using CircuitLab}

(Using a switch to represent the general case, since CircuitLab doesn't have a thyristor symbol.)

However, turn-on is generally less severe (not so much in EMI terms, but thyristors have excellent surge current capacity so in practical terms at least--) and inductors of adequate rating are expensive, so, we often settle for just the R+C snubber.

Note that this reduces the on/off ratio, reducing fault current and increasing leakage current; this may be a concern for various purposes.

The leakage can be avoided by putting the R+C across the load, and a larger capacitor across the mains inlet:

^{simulate this circuit}

(Note that a simulation source is ideal, so the capacitor is meaningless as shown, but real mains has a modest impedance at EMI frequencies: a few 100 µH and 10s Ω, such as modeled by a LISN for example.)

The inductive load also illustrates why we might care less about the turn-on edge: if it's still inductive at short time scales, then the peak turn-on current will be low, at least near the edge. It might still rise to inrush or even fault current levels, but over longer time scales (~ms), where inrush limiters and fuses are applicable.