Finally finished this board for my 12v led strip light power backup. It works like this: When 12v main power is on, it will charge a 3.7v lithium battery using the TP4056 ic, and when power is out, a relay will turn on to power a dc-dc step up circuit to power the 12 led strip. Everything works fine, except that when testing the power input with 18V, the AMS1117 is overheating and hot to touch (and yes, I use my finger as a testing instrument.). TP4056 can output 1A-130mA, and I am setting it to 580ma using a 2k resistor. according the datasheet of AMS1117, it can output 800ma, so here are my questions regarding this circuit.
1, why it's over-heating? is it because of higher voltage drop?
2, what should be the optimal resistor value (or current) for the tp4056 so that the AMS1117 won't over-heat when powered with 7-15v? (I think max input for AMS1117 is 19v, but I can't find it anywhere in the datasheet I got. It only states input as 15v.)
3, How many watts will be wasted in this AMS1117 if my input is 12v and I am charging a 4000mah 3.7v battery?
Best Answer
The SOT223 package might rise up in temperature somewhere in the region of what is quoted in the data sheet below: -
This may be lower if the copper on the PCB is bigger i.e. it can take heat away but, looking at your PCB this probably isn't the case.
So for every watts dissipated by your linear regulator it will raise the internal junction temperature by 90°C above local ambient. Given that the data sheet states: -
Then you aren't really going to be able to dissipate much power before the device turns off. If you are dropping 18V to 5V at (say) 100 mA (determined by load) then the power dissipated is 13V x 0.1A = 1.3 watts.
So, ask yourself how much current does the linear regulator need to source and what the maximum input-output voltage is and do your own sums. Hint: at 580mA, the device is going to shut down.