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Is this the correct way to solve the resistor for the ATMEGA? The ATMEGA can only handle 6.6mA 4.98v @ 16Mhz but my transformer outputs up to 2Amps of current.

This is my first time that I designed a power supply with a fuse on it, do you guys know how to choose the correct fuse for the power supply in Primary and Secondary winding? I am really confused about the ratings of fuses.

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4 Answers 4

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Welcome! First off, I'm a little unsure if what you mean by the "resistor for the ATMEGA." You shouldn't put a resistor between the power supply and your chip since that will cause a voltage drop which will end up meaning your chip won't get the voltage you want it to get. It's ok to connect it directly to the regulator output.

Regarding fuses: think if them as emergency brakes in your circuit. If current exceeds some level, fuses will blow up and stop current from flowing. They don't restrict current flow, they are just protection. And just because current can flow doesn't mean it will. The MCU will take as much current as it needs to run, and is usually only a few milliamps. Even if your supply could provide 2 A, the MCU will still only draw a few milliamps.

You actually don't see fuses very often used to protect ICs because of how little current they draw and how cheap MCUs are (fuses might cost more than the chip!) Putting a fuse on the AC side like your have is a good idea. You want to protect from a short that could blow your house's circuit breaker or start a fire. But you don't have to protect both sides of the transformer. Just protecting the AC side is fine and safe.

Edit: regarding fuse size, calculate what the max current you ever expect to draw, add a little bit to that, and place a fuse in series with the hot input from your power outlet. This way if it blows, your neutral and ground are still connected to what your want them to be for safety. The very first thing the hot of your plug sees should be the fuse, and everything must connect and go through that fuse. Right now your relay is before the fuse, it should go after it (and the fuse must be rated high enough to supply enough current for whatever the relay does).

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Don't use an output resistor - the ATMEGA requires a regulated 5 volt output: -

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Regards the secondary fuse: -

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For the primary fuse, this will be based on what the transformer manufacturer recommends so, obtain the data sheet for the transformer and see what it says.

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  • \$\begingroup\$ The fuse is to protect the wire. If too much current can flow through the wire, it might overheat and cause a fire. So you usually size the fuse to the wire gauge. Depending where you are in the world, it may be a legal requirement. \$\endgroup\$
    – Kartman
    Mar 24, 2021 at 10:07
  • \$\begingroup\$ @Kartman which fuse do you refer to? \$\endgroup\$
    – Andy aka
    Mar 24, 2021 at 10:09
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A fuse has 0ohm of resistance.

On the schematic, it simply looks like a resistor that might be there to avoid inrush current but doesn't really make any sense, this resistor is probably not required at all.

The ATMega consumes 6.6mA, the transformer will output whatever current is consumed but not more, you don't need to limit the current with a resistor.

What is important is that the voltage is not higher than what the ATMega can handle.

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The correct way is to not compute anything and remove the resistor. Regulator supply output should be directly connected to MCU supply input.

However if the transformer output is 25VAC under rated load, it means that with little or no load the AC output is higher. The voltage at regulator input is quite close to the nominal maximum operating voltage already, and depending on which exact regulator manufacturer and model you use it can even exceed the absolute maximum ratings and the regulator can go up in a puff of smoke.

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  • \$\begingroup\$ Do you know any 5 volts regulator that has a higher input voltage? I only have LM7805 which has a max input voltage of 35 volts. \$\endgroup\$ Mar 24, 2021 at 11:08
  • \$\begingroup\$ The real solution would be to use lower voltage transformer. You don't say how much load there is consuming the 5V, but with about 30V volt voltage difference any 5V regulator will heat up just as much. At least you could bring down the voltage in other ways, but for that you need to approximate min and max current consumption of 5V. \$\endgroup\$
    – Justme
    Mar 24, 2021 at 11:24
  • \$\begingroup\$ I have 4 more 5 volts regulator could this help in lowering the voltage of the transformer? I chose 25 volts because I am using a total of 5 5volts regulators 1 for each MCU, LCD, load sensor, and other components. \$\endgroup\$ Mar 24, 2021 at 12:10
  • \$\begingroup\$ Is there a reason why you use separate regulator for each component? \$\endgroup\$
    – Justme
    Mar 24, 2021 at 12:20
  • \$\begingroup\$ each component requires 5 volts, would it be a good idea if only 1 5volts regulator for the 4 components that require 5 volts? \$\endgroup\$ Mar 24, 2021 at 12:26

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