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I was reading the question here with some interest, because I am in the end stages of construction on a project that involves 16 solid-state relays. I'm using non-zero crossing type, because that's what I happened to find on a schematic by someone who completed a very similar project to mine.

The datasheet for my SSRs mentions that a snubber circuit is recommended, especially when driving inductive loads (which I am, since my loads are AC solenoids). I thought I understood that this is to give somewhere for the energy stored in the inductor somewhere to go if the SSR switches off right as the voltage peaks. When I read about ZC-type SSRs, I thought to myself, "self, that would eliminate the need for a snubber circuit, right?"

I then dug up a datasheet for the ZC version of the SSR I'm using, and I found this:

Particular attention needs to be paid when utilizing SSRs that incorporate zero crossing circuitry. If the phase difference between the voltage and the current at the output pins is large enough, zero crossing type SSRs cannot be used.

As well, the snubber circuit continues to be recommended for the ZC-type SSR.

The phrase "phase difference between the voltage and the current" doesn't make sense to me. What does that mean?

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Mark
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2 Answers2

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For reactive loads (capacitive or inductive) the current sine does not coincide with the voltage sine. For inductive loads the voltage leads the current, for capacitive loads it's the other way around.

phase diagram

That means that the zero-crossing of the voltage for the inductor comes some time before the zero-crossing of the current, and that the current is not zero when the voltage is too.
And that's the reason why zero-crossing switching is not recommended for reactive loads.

image from this excellent site.

stevenvh
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    Many SSRs only turn off when there is an absence of carriers and continues to generate their own carriers as long as they are on. This means that when you attempt to turn them off they will stay on as current is still present to generate carriers. Then you stop attempting as they are no longer zero crossing and then current finally crosses zero making it possible to turn them off. You also need a snubber as high dV/dt can false trigger and dI/dt can damage a snubber. – Kortuk Aug 05 '11 at 18:28
  • So, the voltage at the inductor drops, and the magnetic field collapses, and that causes current to continue to flow, but without creating voltage? How does that work? How can current be >0 when voltage =0? – Mark Aug 05 '11 at 18:28
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An old question, but it's a cool topic for beginners to wrap their minds around, so I'll answer it. To answer the last question first, remember that voltage appears across a load, while current is measured through the load.

It may be easier to visualize the phase lag concept if you think of a capacitor rather than an inductor. You're probably familiar with the fact that when you charge a large capacitor, it looks like a short circuit at first. At the instant of connection, current is flowing through the cap, but no voltage appears across it because, hey, it's a short circuit, right? As the cap charges up, the voltage across it rises and the current through it falls. This is all that's meant when people say that "the current leads the voltage" in a capacitor.

With an inductor, we say the voltage leads the current because at the instant of connection the inductor looks like an open circuit. A perfect inductor connected to a voltage source at time=0 will have the whole supply voltage across it, with no current flowing through it. During the 'charging' process the inductor stores energy in its surrounding magnetic field, which cannot happen instantaneously any more than a capacitor can be charged instantaneously. So the voltage "leads" the current in this case.

What's interesting about an inductor is what happens when the source is disconnected. A capacitor will just sit there at the same voltage, slowly losing its charge over a long period of time if there is no load across it. But with an inductor, the magnetic field collapses as soon as the power supply is removed, and this happens quickly. A recently-disconnected inductor will try to maintain the flow of current through the circuit rather than the voltage across itself.... but wait, there is no circuit anymore, because we just opened it.

A perfect inductor would generate an infinite voltage in an attempt to keep the current flowing. Even an imperfect one can turn a few volts into several hundred for a short period of time after disconnection. This is why a zero-crossing switch is not the same thing as a snubber. The snubber's job is to give the inductor a load it can drive when the source is removed altogether -- usually a capacitive one since you don't want it drawing current the rest of the time. It keeps the voltage from rising to levels that could hose semiconductors, burn relay contacts with arcing, or otherwise cause trouble.

John Miles
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