The schematic can be confusing a bit as I can see your questions. There is not 2 different 5V bus. V5-In for ULN drivers, the power coming from the power supply directly. VCC and VDD is regulated 5V which trimmed from 12V by AMS1117 5V fixed, 12V coming from power supply directly again. This regulated 5V is feeding the MCU, power supply is already regulating but it is something guarantee since the mcu is very sensitive. Maybe I will use another suitable reg or remove it totally.

AM jumpers are for measuring the currents only.

And you are right partial schematic might not be enough since any wrong connection about mcu peripherals can short the circuit!

Well all my system is working good if I connect lower power such 6-7 V instead of 12. As we disgussed above before, this AMS ICs are really not suitable for high voltage as I got experince, I dont even tell about high current!

Thanks

 

It is not that the AMX IC is not suitable for the 12V in, but that you did not give it any heat sinking. It would not matter much what linear regulator you used there, the problem would be the same. With linear regulators, you must supply heat sinking when you have a more than a couple of volts differential, and significant current flow. They all would have to dissipate the same amount of heat. On most of my boards I use switching regulators now, or have a pre regulator switcher drop the volts so there are only 2 to 4 Volts across the linear reg. But all my surface mount regulators are soldered down to as large an area of copper, usually both sides connected with multiple feed throughs, to give it some chance of getting rid of the heat produced.

Here are a couple of examples of what I'm talking about from some of my boards.

 

The schematic can be confusing a bit as I can see your questions. There is not 2 different 5V bus. V5-In for ULN drivers, the power coming from the power supply directly. VCC and VDD is regulated 5V which trimmed from 12V by AMS1117 5V fixed, 12V coming from power supply directly again. This regulated 5V is feeding the MCU, power supply is already regulating but it is something guarantee since the mcu is very sensitive. Maybe I will use another suitable reg or remove it totally.

AM jumpers are for measuring the currents only.

And you are right partial schematic might not be enough since any wrong connection about mcu peripherals can short the circuit!

Well all my system is working good if I connect lower power such 6-7 V instead of 12. As we disgussed above before, this AMS ICs are really not suitable for high voltage as I got experince, I dont even tell about high current!

Thanks

Thanks for the clarification. I'm inclined to agree with the others that simply trying a different regulator won't get you anywhere. The issue isn't simply voltage or current capabilities of a regulator - it's a combination of them. Specifically how much voltage you're dropping across the regulator (12 - 5 = 7V drop) times the current running through the regulator (0.05 - 0.08A from what you've said.) So, 7*0.08=0.56W of heat dissipation. That regulator might be capable of 1A, but that would be with a good heat sink and a lower voltage drop - if you were feeding it 6.5V input, it would only have to drop 1.5V, for a total heat dissipation of 0.12W.

What are the actual voltage requirements of your mcu? If it could run happily at 4 or 4.5V, you might be able to run a suitable low dropout regulator fed from the 5V leg of your triple supply, in which case there would be very little heat to dissipate.

 

Try removing the existing regulator and adding a 7805 like this...

A couple of 10uF tant caps added at the regulator for stability. Note the polarity!
Orange = 12V in, Black - common and Red = +5V out.

Here you can see the insulation mounting kit ready for it all to be screwed to a heat sink or the metal case. Check that there is no short from the tab to the heat sink.


And solder the wires to the ends of the caps on your board.

This is worth a try.

I had some more detailed pics of the steps to make it all, but it turns out the SD card had the write protect switch on so the pictures did not save and I missed the warning message

 

What are the actual voltage requirements of your mcu? If it could run happily at 4 or 4.5V, you might be able to run a suitable low dropout regulator fed from the 5V leg of your triple supply, in which case there would be very little heat to dissipate.

That is exactly what I m doing right now: I m using 5V output of triple power supply and the regulator giving 4.65V as output, it is enough for the mcu. Thank you

 

Try removing the existing regulator and adding a 7805 like this...
View attachment 131468
A couple of 10uF tant caps added at the regulator for stability. Note the polarity!
Orange = 12V in, Black - common and Red = +5V out.
View attachment 131469
Here you can see the insulation mounting kit ready for it all to be screwed to a heat sink or the metal case. Check that there is no short from the tab to the heat sink.

View attachment 131470
And solder the wires to the ends of the caps on your board.

This is worth a try.

I had some more detailed pics of the steps to make it all, but it turns out the SD card had the write protect switch on so the pictures did not save and I missed the warning message

Thanks a lot, I m unserstanding what you said from the beginning. I was just did not understand why it is so getting hot although it is saying 1A regulator. Now I understood as @ebeowulf17 said it is a combination. And as @AnalogKid said watts is watts.

I will try to mount an aluminium heatsink.

Thanks you all for everything

 

Thanks a lot, I m unserstanding what you said from the beginning. I was just did not understand why it is so getting hot although it is saying 1A regulator. Now I understood as @ebeowulf17 said it is a combination. And as @AnalogKid said watts is watts.

I will try to mount an aluminium heatsink.

Thanks you all for everything

I'm glad the heat dissipation issue is starting to make sense for you. It can be a weird concept at first. I went through some of the same struggles you're going through not that long ago, so I understand.

 

That is exactly what I m doing right now: I m using 5V output of triple power supply and the regulator giving 4.65V as output, it is enough for the mcu. Thank you

A word of caution - the 5V linear regulator is probably not regulating anything at this point - it's only dropping a fixed voltage. If your mcu really does require tighter regulation (less switching ripple, more stable voltage ref for analog operations, whatever,) than what the triple supply provides on its own, then this configuration isn't going to work with your current 5V regulator.

I'll try to explain this correctly, but I'm still a bit new to all this myself, so hopefully analogkid or some other more experienced member can correct any mistakes I make. Here goes:

Every linear regulator has to drop some amount of voltage from input to output. If the input voltage isn't greater than the required output plus the dropout voltage, then the regulator can't control anything. In that case it basically acts like a saturated transistor, passing as much current as it can while trying to reach the target output voltage. The only thing limiting it is the Vce junction drop or Source-Drain RDS-on, depending on architecture, and this limit is what defines the dropout voltage for any particular regulator.

The dropout voltage varies as a function of current, and surprisingly low dropout voltages can be had, depending on your current requirements. This graph shows dropout voltage as a function of current for the Texas Instruments LP3985IM5-4.7:


I chose this model as an example because it has low dropout capability and its output is 4.7V, which should be achievable with 5V input and low current requirements. You can see that at 100mA, which I believe is more than you need, the worst case dropout voltage is 50mV, meaning that the regulator can output a regulated 4.7V with as little as 4.75V input (although personally I would never push my luck that far.) If we assume your triple supply is usually delivering very close to its 5V target, that means the linear regulator is dropping 0.3V. Going back to the earlier 80mA current figure, your heat dissipation will be 0.3V * 0.08A = 0.024W.

Just to be clear, I'm not specifically recommending this particular regulator. I just used it as an example to demonstrate the concept of dropout voltage and how it varies. Hope this helps!

 

I'm glad the heat dissipation issue is starting to make sense for you. It can be a weird concept at first. I went through some of the same struggles you're going through not that long ago, so I understand.

You are right, I forgot it starts to regulate from 6.5V. In this case it just drops some voltages as you said. It seems the circuit is working good without the regulator then