Hey people.
I am designing a circuit for my home automation system. To give a gist, I have a 12V 3.5A supply. I need to operate multiple loads off it.
One of the Loads is a ESp12E module.

Esp12e module ----- Max current pull = 350mA. Voltage = 3.3V.

For this purpose, I am using a regulator AMS1117-3.3
Following is my circuit diagram:



My Problem is, with this connection, the AMS1117-3.3 regulator is heating up and hence blowing up. I have read the datasheet and it says "Max output current = 1A"

What can be the problem?

 

I have read the datasheet and it says "Max output current = 1A"

But did you read the rest of the data sheet?
Read the part in the data sheet on Thermal Considerations (page 5 of this data sheet) and how to heatsink for the (12V-3.3V) * 350mA ≈ 3W it has to dissipate.
Without proper heatsinking it definitely will overheat.

 

But did you read the rest of the data sheet?
Read the part in the data sheet on Thermal Considerations (page 5 of this data sheet) and how to heatsink for the (12V-3.3V) * 350mA ≈ 3W it has to dissipate.
Without proper heatsinking it definitely will overheat.

The problem is, I have searched my entire city, and AMS1117 reg is available only in SOT223 package.
The other solution I can undertake is, first bring down the voltage to 5V using 7805 and then use the AMS1117 to bring the voltage down to 3.3V. What do you think?

 

The other solution I can undertake is, first bring down the voltage to 5V using 7805 and then use the AMS1117 to bring the voltage down to 3.3V. What do you think?

Why not just use an LM317 to bring the voltage down to 3.3V from 12V.
That device is easier to heatsink.

 

What can be the problem?

calculate the power dissipation on the 1117. look up the thermal resistance and then calculate temperature rise at junction. you will know the answer.

the solution is to use a power pnp/p-ch.

 

Why not just use an LM317 to bring the voltage down to 3.3V from 12V.
That device is easier to heatsink.

You are right. But, 12V to 3.3V at 500mA current pull, it will dissipate 4.4W of power. Will a single heat sink be able to cool down the reg dissipating 4.5W of power?

 

hi,
Have you considered adding a series power resistor on the Input side of the regulator.?
Chose a resistor value that will drop say 6V at full output load, place the 100uF cap at the junction of the resistor and the regulator.
E

 

hi,
Have you considered adding a series power resistor on the Input side of the regulator.?
Chose a resistor value that will drop say 6V at full output load, place the 100uF cap at the junction of the resistor and the regulator.
E

No. I have not tried that. I will try what you have suggested. But won't a power resistor limit my current capability of the regulator?

 

I will try what you have suggested.

Again, do the math: calculate the power dissipation and temperature rise for your attempted solution and you can save yourself some pain, money and time.

 

Say you require an output current of 0.5A at 3.3v.
Allow 1V drop across the AMS1117, the absolute minimum specified.

Your voltage source is 12v, so you need to drop 12v- 4.3v = 7.7v,,, make that 7v.
The wattage within the resistor will be 7v * 0.5a = 3.5Watts, so choose at least a 10Watt rated 14 Ohm resistor for a continuous power usage.

The voltage across the AMS will be 1.7v at 0.5A =0.85 Watts, check this image for the AMS cooling requirement.

 

Will a single heat sink be able to cool down the reg dissipating 4.5W of power?

It will if it's of sufficient size, which depends upon your highest ambient temperature.