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Now I'm working on my project which contains Arduino, LCD, electromagnetic lock, 6 channel 5v relay and 500V isolated converter. My project was designed to power by using a power bank. Now I use eloop E14 20000mAh power bank which can provide 5V 1A and 5V 2.1A so, I use 5V 1 A to powering the Arduino via USB and 5V 2.1A power the circuit. In the circuit part firstly the LCD consumes current about 200-300 mA and the voltage must maintain around 4.75 to 5.25 V. Next the electromagnetic lock consumes 12V 200mA so, I increase 5V to 12V by using a 5-12 converter that can withstand around 1-2 A and the 500V converter that uses the input of 12 V then boost to 500V and this converter consumes 8mA. And the last one 6 channel 5 v relay which consumes around 70mA when power 1 channel and I only power 1 channel in a time The question is when I started to supply my circuit the voltage will drop to less than 4.75V so that the LCD will shut down. Can anyone tell me why is this happen?. I have calculated the total current of 5V 2.1A port which is about 508mA ( less than 2.1A). According to my understanding, I reckon that the 5V will start to drop when we supply the circuit that exceeds 2.1A, so why is this happen. Lastly, I power my circuit via USB-micro USB. Any assistance will be appreciated. Thanks

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  • \$\begingroup\$ It sounds like you've answered your own question. You've overloaded the supply and so the voltage is drooping. What is the question you're trying to ask. PS. You must include a complete schematic. \$\endgroup\$ – scorpdaddy May 20 at 2:27
  • \$\begingroup\$ Resistance can (by which I mean, does) also drop voltage... wires, cables, connectors all contribute some. Then remember V = I * R \$\endgroup\$ – vicatcu May 20 at 2:37
  • \$\begingroup\$ Thank you very much for your guidance but I don't understand why you said that I have overloaded the supply( the total current is around 500-600mA but the supply is 5V 2.1A) \$\endgroup\$ – Panupong Hirunrote May 20 at 2:49
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Next the electromagnetic lock consumes 12V 200mA so, I increase 5V to 12V by using a 5-12 converter

Conservation of energy always applies, so when you boost the output voltage the input current must increase. 12V * 200mA = 2.4W. 2.4W / 5V = 480mA. However the booster may only be 80% efficient, so it might need to draw 25% more current to make up for losses in the conversion, ie. a total of 600mA which is 3 times the output current.

and the 500V converter that uses the input of 12 V then boost to 500V and this converter consumes 8mA

So that's another 24mA at the 12V booster input, for an (estimated) total of 624mA. Now add 300mA for the LCD and 70mA for one relay, for a grand total of 994mA. A port rated for 2.1A should handle that if it is properly designed.

But how efficient is the 12V booster really? does your 500V booster actually only draw 8mA under load? What is the real total current draw under all operating conditions? To find out you need to measure the current draw, not just guess it. And can the '2.1A' port really deliver 1A without dropping below 4.75V? Again, you should test it with a known load current. Only then will you know where the problem lies.

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