This circuit works fine when I replace the 220v power supply with a cheap power supply(I assume they are isolated) from aliexpress. However, when I use my own power supply(schematic below), it causes the 0R resistor, fusible resistor, AMS1117 IC and HLW8012 IC to blow out with smoke.
my power supply circuit is also working fine under load when not connected to power monitoring circuit.
I think there is something wrong with my GND but I don't know how to solve it.
Chinese datasheet for HLW8012 can be found here and some english info about it here
Electrical – What could be the issue with energy monitoring IC (HLW8012) and the non-isolated power supply
power supplyswitch-mode-power-supply
Related Solutions
As long as you don't select the 6, 9, or 12V settings on replacement power supply, the magic smoke won't escape. However, 4.5V may not be high enough to power up the device, in which case it either won't turn on or won't function properly, although it shouldn't cause any damage.
The replacement power supply is rated for a lower current rating, meaning that if the device draws more than 2.5A, it will brown out. This again shouldn't cause any damage. Even the cheap power supplies today have some basic short-circuit/over-current protection.
One thing to note: When you're measuring 5.16-5.18V, you're probably doing this with "no load" (i.e. when the power supply isn't connected to the device). Since the power supply is not perfect voltage source but a Thevenin equivalent, the voltage reading will drop when you connect it to the device.
Another thing to note: Most digital devices today that run off 5V feed that rail directly into a voltage regulator which drops it to 3.3V or lower. Those devices need the input voltage to be a little higher than the target voltage, so maybe 3.6-3.9V at a minimum. Bottom line, you should be fine with the 4.5V supply.
The important thing to remember in this design is that "live" (the hot wire) is also ground to the circuit: -
Live directly connects to the shunt resistor and this means that live is also GND (0 volt reference) for the chip.
You can swap incoming live and neutral connections and it will still work but now neutral becomes the 0 volt reference for the chip.
So, providing you design the "power supply" section to respect the reference 0 volts for the circuit you are fine.
If you choose neutral as the reference you can utilize this design to measure power in two (or more) phases providing you connect the current shunts to a star point at neutral. You should also be able to use a common "power supply" section but, because I don't read Chinese, I have some doubts about chips input common-mode range for the shunt differential inputs. I would be cautious here.
Best Answer
Yes you will have a ground issue with this circuit. The problem is that the GND of your circuit is a local ground. That means local to your circuit. If your circuit is powered on then the GND in your circuit will be at roughly half of the mains voltage.
That means that if you connect this circuit to a PC, it's case would become live at half mains voltage. That is extremely dangerous and that's why PC cases are (or should be) grounded via the mains earth connecttion. This means that the PC's mains earth connection will short your circuit's ground to mains earth. Your power supply circuit is not designed to handle that so it blows up.
You really need to study the subject of mains isolation, why it is needed (safety) and how it is done (transformer, opto couplers, wireless).
Then browse the internet for similar (mains power monitoring) projects to learn how it is done. You will find that monitoring mains current directly via a shunt resistor is always a challenge. It is much easier to use a current transformer or a Hall sensor based module as these already provide mains isolation so they're much easier to use.