Voltage divider. 12 v , to 5v and 3.3v

Hi Guys,
I made a circuit where i need 12v , 5v (for atmega328p and some sensors), and 3.3v(for esp8266). But it doesnt work well, first the 7085 gets too hot despite i have low load. Second after a while when i have some load in 3.3v and 5v the output voltage gets low and lower droping to near 0-2v.
I think its a problem with the capacitors, but i need help i’m chemical engineer, not electric/electronic =(.

I attach photo (edited).
Thank you

The diagram is utterly wrong. You seem to have a common +12V connection and no common ground.

The regulators are positive voltage regulators not negitave voltage regulators.

Sorry i confused colours. where says - is gnd.

2 layer board. red top and blue bottom.

Re attached photo. Now?

Now?

Does not show which way round the regulators are connected. It could be right but from your description of you problems you might have done it wrong.

Also you are still missing a positive connection between the capacitor and the input of your LM7805 regulator.

There is nothing to say if the view is from the top or bottom nor which way round the components are. Try posting a photograph of your wiring so we can see the components and their orientation.

Finally edit the title of your first post to what I have put. This circuit is not a voltage divider but a voltage regulator.

Now i have corrected the diagram. Sorry about that.

Attached with regulators orientation.

Yes that looks fine now.

I think its a problem with the capacitors,

Well the data sheet for the AMS1117 says:-

The addition of 22μF solid tantalum on the output will ensure stability for all operating conditions.

So I would up that value.

The only other thing I would add is a 0.1uF ceramic capacitor across each of your 10uF capacitors.

Yes it will get hot because you are burning off the excess current as heat.

There is still the problem of have you built it right. Given that we have had three goes at getting the circuit diagram correct it is worth posting a picture.

fd0d7fc10526cec3d7f28acd5d9c93a04bac95e3.png

Image fix

Grumpy_Mike: Yes that looks fine now.Well the data sheet for the AMS1117 says:-So I would up that value.

The only other thing I would add is a 0.1uF ceramic capacitor across each of your 10uF capacitors.

Yes it will get hot because you are burning off the excess current as heat.

There is still the problem of have you built it right. Given that we have had three goes at getting the circuit diagram correct it is worth posting a picture.

Ill try that. Thank you

Linear regulators, like the ones you are using, turn the voltage difference (times current) into heat. If you draw 100mA from the 5volt pin, then (12-5)*0.1A= 0.7watt is dissipated in the LM7805, making it ~40C hotter. Whatever else you draw from the 3.3volt pin also has to be supplied by the 7805. Linear regulation is less problematic for the 3.3volt regulator, because it only has to drop 1.7volt.

If you don't want the heat, then replace that 7805 for a switching (buck) regulator. The small (0.5-1A) boards are about the same size/shape as an LM7805. Leo..

Wawa: Linear regulators, like the ones you are using, turn the voltage difference (times current) into heat. If you draw 100mA from the 5volt pin, then (12-5)*0.1A= 0.7watt is dissipated in the LM7805, making it ~40C hotter. Whatever else you draw from the 3.3volt pin also has to be supplied by the 7805. Linear regulation is less problematic for the 3.3volt regulator, because it only has to drop 1.7volt.

If you don't want the heat, then replace that 7805 for a switching (buck) regulator. The small (0.5-1A) boards are about the same size/shape as an LM7805. Leo..

Or just throw a big ol' heat sink on it, cause winter is coming, and who doesn't like a toasty workshop.