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?

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The SOT223 package might rise up in temperature somewhere in the region of what is quoted in the data sheet below: -

SOT-223 package \$\phi_{JA}\$= 90°C/W

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: -

Operating Junction Temperature Range

  • Control Section -40°C to 125°C
  • Power Transistor -40°C to 125°C

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.

  • \$\begingroup\$ Thank you. I never know about temperature vs power dissipation like that until now. I did some research and tested my board for these several days, and Burned out few IC in the process, and I come to the conclusion that this board can't take more than 12V. I am setting the charger ic to 130ma, and 12V-5V=7V >> 7Vx0.13A + 7v X 0.025A=1.085 Watts. (the second part is for the relay.) so With my board, this IC can safely disparate around 1W and the charging current to 130ma with 12v input? The IC is warm to touch, but not too hot. \$\endgroup\$ – Atmega 328 Nov 18 '15 at 3:24
  • \$\begingroup\$ This is very enlightening. \$\endgroup\$ – Atmega 328 Nov 18 '15 at 3:25
  • \$\begingroup\$ also, do you think I should add a diode for the relay? I think I have burned the TP4056 because of the relay flyback voltage, but I am not sure yet because I think I may accidentally shortened the 12v input to the output, thus burned the TP4056. \$\endgroup\$ – Atmega 328 Nov 18 '15 at 3:28
  • \$\begingroup\$ You should have a flyback diode across the relay coil/regulator output, yes. \$\endgroup\$ – Andy aka Nov 18 '15 at 8:16
  • \$\begingroup\$ Thank you. Two last question. My circuit doesn't have a capacitor on the Vout point which is where the relay is. I did some test. After adding a 1uf cap there, repeatedly switching this circuit on/off doesn't effect the TP4056 at all, but after I removed this cap, the TP4056 is dead. do this mean that this tiny cap can assort the flyback voltage from the relay? Last question is: can I connect + of power-in to + of power out directly? The power out terminal is only being switched on when power-in is out. and It's a boost converter setup with a diode. It's used as a power failure backup. \$\endgroup\$ – Atmega 328 Nov 18 '15 at 15:11

1: Yes, I assume you are using 5V version of 1117, so that brings your power dissipation to (InputVoltage-OutputVoltage)*Current=(12V-5V)*580mA=4W and that's a lot of Heat and you'll need good Heat Sinking. My suggestion: Use DC-DC converter instead of Linear Regulator, use something like this one http://uk.farnell.com/xp-power/tr05s3v3/dc-dc-converter-0-5a-3-3v-sip/dp/2319829 (Note this is a 3.3V version you have to search for 5V). or you can use circuit from Car Mobile charger which mostly has MC64063 based buck Converter.

2: Now you can figure the optimum Current to bring power dissipation down, I don't know what will be safe dissipation in current setup. you may need to somehow add a heat sink.

3: Power dissipation will only depend upon current drawn and voltage drop across 1117, capacity of battery doesn't matter here.

Now if you don't want to go with DC-DC converter since you already made PCB, I suggest that you can divide Power dissipation by putting a high Watt resistor in series with the Input of 1117. Lets see for Eg: Dropout Voltage for 1117 is 1.2 So we need to keep Input Voltage to 1117 atleast 6.2 V for 5V ouput. So if your Total Current draw is let's say 600mA. and you have 12V supply you can drop few watts across series resistor and max we can drop is 5.8V (since we need 6.2 at 1117 Input) and So resistor=5.8V/600mA=9.66Ohm, so we choose 10Ohm and P.D across this resistor will be= 10Ohm * (600mA)^2=3.6 Watts. Hence you'll need atleast 5W resistor. and voltage input to 1117 will be 6V(since we chose 10 ohm resistor and hence Voltage drop across it is 6V) P.D across 1117=6V-5V * 600mA= 600 mW.

Edit1: 6V input to 1117 instead of aforesaid 6.2V is also fine since datasheet says 1.2V drop at 800mA and our current is less.

  • \$\begingroup\$ 1, so what is the power dissipation of this 1117? it's surface mount, so can't add heat sink. \$\endgroup\$ – Atmega 328 Nov 14 '15 at 10:42
  • \$\begingroup\$ 2, Power dissipation=V-drop * current; There is a programmable resistor for the tp4056 to control the charging current. so what should this current be? When looking at the datasheet of the 1117, I can't find the max power dissipation of this device nor a max input voltage. \$\endgroup\$ – Atmega 328 Nov 14 '15 at 10:48

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