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I will need to calculate that I am still within the power rating of the resistor (1/4 W?). I mean if I am using 5V x 200mA that is already 1 Watt, right?
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When I start swapping out the existing resistor I can do an actual current draw test and know better!
You are right arithmetic-wise, but wrong application-wise.
5V is the voltage across your resister AND the LED in serial. So 5v at 220 mA is the power of
BOTH resister and LED together.
To get the wattage dissipated by the resister, use the Vdrop across the resister only, that would be the right way of calculating the resister's wattage.
But your problem started sooner than that,
at 5v, you can't get 200mA current with a 220ohm resister.
With 220ohm at 5v, the most current you can get is:
5/220=0.0227
0.0227A is 22.7mA
So even if your LED drops 0 volt,
the max current is 22.7mA. You just can't get 200mA with that setup.
Leaving that behind since we know 220ohm is not right (or 200mA is not right)
Whatever
current you really need (at minimum keep awake), let say it is 20mA for your lights+resister to keep-awake, since you have the
current and the resister value, you can calculate the power with that two known value:
Power = I * V (V is unknown, but we know V=IR, so replace V with IR)
Power = I * I * R
So with
current = 20mA (0.02A) and resister=220ohm
power = 0.02*0.02*220
= 0.0880 watt
A 1/8 watt in decimal is 0.125watt. You have a good margin left over. A 1/4 watt is 0.250 and that is 2.84 times the wattage needed (for 20mA and 220ohm).
As to determining the minimum keep-awake current:Try using a Variable Resister (or trim-pot) if you have one. Use a 500ohm pot or a 1K pot in series with 10ohm-ish. The 10ohm in series is there just to make sure you don't have a dead-shot when you lower the pot too much.
When measuring the required resistance for the resister, just don't forget to measure the total (ie: trim-pot plus the 10ohm) instead of just the trim-pot.
By the way, I am quite sure your "keep-awake" current for the power bank should be below 200mA and probably at or below 100mA. It would be a rather bad design for the power bank to go to sleep when it is still charging a battery at 200mA+. At 100mA, it is early but is arguable. One of my two 18650 charger typical cut out at just below 100mA, another charger is at (around) 50mA.
EDIT: adding this:
By the way, don't waste time increasing/decreasing a bit at a time. Newton's method works here too...
Start with say 400mA, if it stays awake, cut it in half to 200mA, then half to 100, then half to 50, then half to 25... keep halving until you hit when it goes asleep.
If it stays awake at 50 but goes to sleep at 25, now you know the right number is between 25 to 50. Go 1/2 way between 25 and 50 for the next cut: (50+25)/2 = 37.5
Depending on if 37.5 is awake or asleep, choose it as high or low.
Say it is 37.5 is awake and 25 asleep, next 1/2 way is (37.5+25)/2=31.25.
But say at 37.5 it is asleep (but 50 is awake), you know it is between 37.5 and 50, (37.5+50)/2=43.75
Knowing which half to use, you can cut it again. Here, you have in between 37 to 50 is 13 integers. Next cut will be 13/2=7 rounding up, then 4, then 2, then 1 single number (if you can indeed be that precise to a single mA). I would be quite please with +-5mA there (ie: a 10mA range).
Cutting it in half and choose the right half is a much faster way to narrow in on the target.
You don't really need the exact decimal each cut, you could have use 31 or even 30 for 31.25, or 40 for 43.75. Since you are determining which next half to use, you don't need it to an exact 50/50 half.