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To continue.
I use a high brightness LED for a pilot light. With a 25,000 ohm series resistor and 9 V DC applied to the series string the LED voltage drop is 1.7 V. This means the forward current is (9-1.7)/25,000 = 0.3 mA.
If the series resistor can tolerate the power dissipation, then with 120 V DC applied the current would be about 5 mA. I don't remember the reverse voltage rating, that doesnt't matter, but suppose actual breakdown is 15 V, and I apply reverse polarity, then LED power dissipation would be about 15*.005 = 75 mW. If this is too much power for the LED then it would fail.
To avoid this problem I could just put a regular diode, like a 1N4148, reverse biased across the LED. Now with a high enough power rated series resistor I could probably apply 360 V peak AC or DC to the series combination. The resistor would need to be at least a 360*0.015 = 5.5 W rating.
Suppose your LED sign to be added has its own built-in current limiting and is expected to be driven by a DC voltage source. If this has no internal means to protect against application of reverse polarity, then application of reverse polarity may cause damage. If a simple series input diode is included, then a reverse polarity up to the reverse breakdown of the series protective diode could be applied without damage. If a bridge rectifier was at the input, then either polarity of DC or an AC input could be applied and the sign would be protected against reverse polarity, and the sign would function.
You need to find out the limitations of the devices, and how they work.
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