Electronic – What would a professional use to switch sprinkler solenoids

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.

This isn't a direct answer but is pointing out a alternative.

You don't necessarily need a double ended drive like a H bridge. That would certainly work, but a push/pull driver with a capacitor in series with the relay coil would do it too. Here is a possibility to think about:

In the steady state, the capacitor charges up to whatever the drive level is at the left side of the relay coil. It then provides the opposite polarity for the right end for a while immediately after the left end is switched. 22 µF will charge up to only 1.4V after 300 mA for 100 µs. 22 µF and 16 V can be had from a 1206 ceramic capacitor, like this one from Mouser. The cap can be polarized since the top side will always be at or above the bottom side.

The double emitter follower can be driven directly by a CMOS logic gate. There are some that can handle this voltage. The input to that could be driven by a open collector with pullup. Since the CMOS logic input is high impedance and you don't need really fast switching, the pullup can be quite high. 100 kΩ should be low enough to work well, but high enough that the quiescent current in the low state is small.

Of course you could replace the transistors with a half bridge drive chip that takes logical level input for higher integration, but also likely higher cost.

Added:

You are asking about driving this from a single open drain output. As I said above, I'd use a CMOS gate that can run from 15V. The high impedance of the CMOS gate input allows for a high value pullup resistor, as I mentioned previously. Here is this concept shown explicitly:

Q3 is a switch. When off, R1 pulls the input of IC1A high for one relay state. When Q3 is on, the input to IC1A will be low for the other relay state. Q3 could instead be the output transistor in the driver chip you mentioned. However, it only takes one NPN and one resistor to replace each channel of that chip. The left side of R2 can be directly driven by your microcontroller output. The driver chip could be less board space, but the NPN and resistor will be cheaper. The whole circuit from the micro up to C1 could be replaced by a half bridge driver chip, which again will be more cost but maybe less board space. Everything is a tradeoff.

I also flipped the relay coil and C1. Since these are in series, it doesn't matter to the operation of the circuit. However, it may be convenient to tie one end of all the relay coils to ground. This second circuit allows you to do that.