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The cap needs to be larger than that
For 500ma for 1 second, q = 0.5 x 1 = 0.5 coulombs For a 3v capacitor discharge drop during that 1 second, the required capacitor size is 0.5/3 = 0.166f, which is a very large capacitor You don't need a resistor between the cap and module That will just reduce the available energy from.
I was pressed to find a solution for a test automation need