Alright, good to know. And by spark, you’re referring to a physical spark within the switch when it is thrown, I’m assuming? Also I’ve done fairly long, steep, uphill climbs (over a mile at greater than 15 degree incline), and I’m pretty sure I was pulling more than 20 Amps there for a long time, so the switch must more robust than advertised I guess. As long as it keeps working, I’ll keep using it and then when it does blow out, just replace it with another $5 switch. Much better in my opinion than spending $60 on a switch.
to destroy you will probably need quite a high current - i mean it needs to melt 
- but it will also not be efficient. On that climb you “lost” quite a lot of energy in heat.
Ah that makes sense. Although I’ve touched the switch several times after riding and it wasn’t even warm. Also, now that I think about it, I think my explanation of why the switch can handle 50 amps may be correct. Because after all what causes the switch to melt is heat dissipation, which is determined by the power through a component, and power is v*i, so reducing voltage reduces the power, allowing an increase in current before the limits of the component are exceeded. I’m not trying to start a massive debate/argument, just trying to get to the bottom of this.
The heat dissipation is P=R*I² Only the current is important and the resistance of the switch. The squared current is also why higher currents are so damn annoying
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Ah, okay, I was confusing the voltage drop across the switch with the voltage coming out of the batteries, which is why I used the wrong equation. Thanks for explaining that to me.
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I’m an electrical engineer and came across this post. I thought I would better explain how the current limiter works.
You are all correct with saying the current limiting comes from the resistor and capacitor network. However, it is R1 and C1 doing all the work, R2 is basically irrelevant.
When the switch is in the off position Vgs (gate-source voltage) is being shorted to zero. All the FETs are in cutoff and not conducting (since Vgs is less than the threshold voltage Vtn, Vtn is roughly 5V). Additionally, the voltage across C1 is at your pack voltage. If you don’t understand why, think of the load across controller + - as a very small resistor. C1 is then effectively connected between ground and Pack+ through a 1K resistor.
Now lets throw the switch…
The voltage across a capacitor can not change instantly, so there is still the pack voltage across C1. This effectively keeps Vgs at ground (the FETs are still off). C1 starts to discharge through R1, R2, and the load. Current actually starts flowing backwards through controller as C1 discharges. As C1 continues to discharge, the voltage across the cap decreases, raising Vgs. Once Vgs surpasses the threshold voltage Vtn (~5V) the FET’s start to conduct (they enter the saturation regime) and current flows through the controller in the correct direction and the voltage across the controller lines gradually increase, slowly turning on the controller.
Note, that in the saturation regime the current flowing through the FETs are roughly proportional to Vgs^2, so the current given to the controller ramps up quadratically with Vgs. At some point the voltage across C1 goes to zero and then C1 starts charging in the opposite direction (hence why C1 must be non-polar). As we said the current the FETs pass is related to Vgs, but Vgs is controlled during switching transients by the voltage across C1. Since C1 is charged up very slowly through the giant resistor R1, the FETs turn on very slowly. Finally, once C1 is fully charged, the FETs are fully on and the controller can can use as much current as it likes.
As for R2, pretty sure all it does it prevent a spark across across the switch when turning it off due to C1 discharging. The zener diode just clamps Vgs so it can never rise above 12V, otherwise it would reach your pack voltage and many fets have a max Vgs of ~20V.
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That was your first post? Wow, and welcome to the community. Great to have an expert clear things up 
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Noob question: Without the inrush current limiter, where do you see the spark?
When you connect your battery
That could be using a connector or a switch.