Well, you didn't say which particular trace that each board experienced the burn out with, but let's say for the sake of simplicity, you're referring to the main power leg, either coming into, or out of the relay.
There's always one major dominating factor, that everyone forgets with these things. It's that these relays and the circuit boards are pushed to their limits when connected to a heater element. That thing draws more current than any other component in the spa.
Consider the wire size that YOU would use to connect a contactor to a heater element. Then compare that to the size of the switching contacts of that small relay. Next, compare that overall wire size you would use on a contactor, and the amount of copper used on a circuit board trace.
Then you look at those tiny switching contacts on the zettler relay... of course they 'look' burned. It's not really a phenomenon, it's plain physics. It's going to happen... there's a lot of current getting switched through that tiny thing, and it can handle it, which is why it still works. Most of the 'burnt' that you see, is a byproduct of arcing when making or breaking the circuit. The arcing is burning 'stuff' that's in the air around it. Do that say, 10 times a day to keep the spa up to temperature, multiply it times 365, and you bet, you'll accumulate a lot of carbon around the relay contacts. If the contacts were truly burned, they would cease to be able to handle 18-23 amps and end up totally fried. THEN, you lose functionality altogether.
Ok, all that considered, no matter how many times I've iterated it over the last 8 years, the most important thing to do when installing a new circuit board, a new high current relay such as a contactor, or a high current gfci, or any wire connections that are required to 'push the envelope' of their capability, it is physical connection strength and bonding.
IF, your connections are tight, dry, and corrosion free, you should never see this problem in the field. The fact that you did, AFTER the boards were replaced, points to one significant error.... the connections were never made tight enough to the board to begin with.
A dry fire will not cause this problem, neither will a defective thermostat or a high limit switch. The ONLY thing that will cause burned out relays, burned wires, fried circuit board legs on the high current side, is a
direct short in the heater element circuit.
But even that is extremely unlikely. The heater element wire inside the sheath
should melt down too fast to do significant damage to the relay or the circuit. Since you're a new technician, I emplore you to remember this because you will see and experience this kind of situation time and time again in the field.
I'll provide you with an example in the picture below. This system is a classic Balboa Instruments™ design and from the factory, it should rarely fail. But there's one significant difference in this base design in that it was modified to provide a switched high current/low current leg for the heater. As in, in one situation, the heater receives 230 volts, and in another, it only receives 115. Good idea, but the implementation of the design either failed because these modified connections began to deteriorate over time, and the screws came loose, or the nuts attaching the heater element to the copper terminal strip (on the left) worked their way loose.
It only requires
ONE of these failures (bad connection) to start the entire assembly down this road to full burn out. This assembly is about 6-7 years old. Not bad for longevity, but it still shouldn't have happened.
Note: The following scenario is using some very basic electricity calculations, and while it's not exactly the entire story, I've kept the calculations simple to keep this post short! (And this answer is not short!).
Where do I think that this deterioration/burnout started? Most likely with the screw that attaches the heater element leg on the left to the copper strip. That copper strip is VERY worn out and must be replaced (as well as the heater element and high current wiring). It happens like this:
1. Screw attached to copper strip carrying (we'll use 20 amps for simplicity) heater current comes loose.
2. That connection/bonding point starts to heat up because it no longer has the full connection that it used to.
3. Heat causes the copper strip to begin an oxidization process, (burn), and it builds up a small amount of resistance. Let's say that from the factory, this strip would measure 0 ohms.
Initially, let's throw in a figure of .05 ohms. Doesn't sound like much does it? Well, under low current circumstances it isn't significant. But take .05 ohms X 20 amps of heater current. That gives a 1 volt drop across the strip. Take the 1 volt drop X 20 amps now, and instead of a rock solid copper strip delivering perfect current to the heater element, it is now dissapating 20 watts of energy.
Still doesn't sound like much, but if that strip is disappating 20 watts, then it's actually HEATING to the tune of a 20 watt heater element. (Touch a 25 watt bulb sometimes and you'll understand what I'm talking about).
5. Now this copper strip isn't designed to handle 20 watts like a heater element. It's BURNING initially, at 20 watts.
6. A day or two goes by, because the strip is heating/burning, less of the pure copper is available to deliver the 20 amps to the heater element. So, let's just say that on day three, the resistance of the copper strip has increased from .05 to .08 ohms. Still not much, it looks very insignificant on an ohm meter, and this resistance is undetectable with an analog meter. .08 X 20 amps gives you 1.6 volts. 1.6 volts X 20 amps is 32 watts. Now we're up to the energy/heating equivalent of a 32 watt light bulb.
7. Day 4, 5, 6... etc. Say that a week goes by and the resistance of this copper strip (all because of a loose connection) has increased to a measley .1 ohm. This is STILL virtually undetectable... BUT, take the .1 ohm X 20 amps. Now you're up to a 2 volt drop across this strip. (Still sounds like nothing right?). 2 Volts X 20 Amps, and now you've got the equivalent energy of a 40 watt bulb BURNING.
You must remember that a heater element, like a light bulb is designed to handle heat. A copper strip can't! All it can do is keep burning - it's molecular structure continues to break down. (Think about how designing a simple light bulb filament was difficult... to keep it from burning out. Do that with a thread of copper wire and poof! Just a quick melt down).
8. We go on for another week or two, and that same copper strip is really getting fatigued, surface begins to show serious oxidation (as in the picture above). Let's suppose that this copper strip is showing a reading connection to connection of .4 ohms. Just barely detectable with an analog ohm meter, and still insignificant with a low current situation. But we're running 20 amps through it. .4 ohms X 20 amps = a whopping 8 volts! Heh heh, not much when you've got 230 to work with right? uhhh... hmm, 8 volts X 20 amps... YIKES! This seemingly insignificant resistance in that copper strip is now burning with all the ferocity of a
160 watt light bulb!
By this time, this thing is so hot that everything around it is frying. Take a look at the big red wire in the picture above that connects to the relay on the left. Note it's proximity to the copper strip. The heat from the strip is heating up the copper wire also... and now it is beginning to experience some oxidation from the outside heat, and where is that heat also going? The most critical part of that wire, is where it terminates at the relay's quick disconnect. This is the weakest part of the circuit. The yellow quick disconnect (common on this relay) below it has been replaced.
Now how am I just so sure, that external heat from the copper strip caused all of this mess? It's simple. Take a look at the lower two wires on the relay. The small white and black wires.
Those things only carry all of a few milliamps of current, because their sole function is to energize the relay coil... and that's it. They didn't burn or overheat because the coil shorted out, or because of some strange voltage spike, they're looking bad because of the heat from the copper strip below.
Alright enough for the lesson on high current connections, just remember this basic scenario and apply the lessons first and foremost everytime you come across something like this. In this industry, there are many people looking for phantom external reasons for circuit failure, like thermostats and such. The truth is, if the circuit is wired properly, and correct bonding and securing methods are used, then even if a defective thermostat and shorted high limit switch kept the heater running for a week, the electrical connections will survive. But the spa plumbing will melt down, the heater will dry fire, and then poof, no more current is being delivered to the heater element because it finally burned out and left those relays with an open circuit!
I know this is long and probably not what you expected for a reply, but this should not only answer your question, but the questions of many others as well.
