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I have made a home made power supply using an LS-A7632-PT transformer I yanked from an old electronic. It takes 120v input and steps it down to somewhere around 6v and has two 6v legs. I then have a full bridge rectifier and a 6.3v 2200uF capacitor for leveling out the peaks. This goes to a smaller 354 ceramic capacitor then to the l7805CV 5v voltage regulator out to another 104 ceramic capacitor then to a connection terminal.

I read 4.98v and 4.97v on both of the terminals which is really good. I try to power my raspberry pi up with it, the light comes on as though it is working then it turns off then back on, then off repeatedly. I have read that the PI needs at least 2 amps for certain types of functionality and at a minimum 1 amp to work.

I have been unable to find the amp output of the LS-A7632-PT but I imagine it can probably put out plenty of amps as I measure the gauge of wire from the 6v output it has 22AWG copper from what I can tell.

I believe this is rated for max 7 amps. If this is the case, I would only be limited by the 1N4001 diodes and they are capable of carrying 1 amp from what I can tell. Am I doing something wrong for the needs of the power supply?

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    \$\begingroup\$ The 7805 is rated only 1 to 1.5 A and it needs a beefy heat sink at its maximum current. It probably goes into thermal shutdown which explains the behaviour you see. It also needs a voltage differential of about 2 V, which means an input voltage of at least 7 V. \$\endgroup\$
    – StarCat
    Commented Nov 9, 2023 at 21:04

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The design has many issues.

The combination of that transformer, that capacitor, that linear regulator, and the need to supply a minimum 1A current at 5V to a load is simply impossible.

  1. 6VAC RMS has amplitude of 8.48V. If that is rectified with a full bridge, under no load conditions the capacitor will be charged to 8.48V and under light load conditions the two diode drops may cause the capacitor to peak at 7V or so. Which obviously is too much for a 6.3V capacitor.

  2. The L7805CV is a linear regulator, and it's an old linear regulator and not a low drop out regulator. To be able to work, it needs at least 7V on the input. Which means the 7V or 8.48V is fine under no load or very light load conditions. Under those conditions it could even provide 1A output current, if there is enough voltage on rectifier capacitor. But regulator is not even rated to handle more than 1.5A under any useful conditions.

  3. Assuming you live in a 60 Hz land where mains voltage charges the capacitor to 8.5V every 8.333 milliseconds, there is only 8.333 milliseconds during which the capacitor voltage is allowed to drop down to 7V or the regulator stops working. That's 1.5V of voltage change in a 2200uF capacitor. So if the allowed time is 8.333 milliseconds, you can only draw about 400mA of current or the capacitor voltage droops below 7V and regulator stops working. You need more than double the current you have available, and for that the capacitance must be at least 5600uF, or the transformer output voltage larger to allow more ripple from higher voltage down to 7V.

  4. Even if you did have larger 5600uF capacitor that allowed 1A to be drawn, at 1A the regulator would also be dissipating between 2 and 3.5 watts at 7V to 8.5V Which is a lot. Too much in fact. The IC temperature rise is at least 50 degrees C per watt, so even at 2W dissipation without heat sink, it will rise at least 100 degrees above room temperature. Assuming room temperature is 25, the device has hit the limit for operating temperature range and will go into thermal shutdown. You need a big heat sink, or regulator bolted on to a large metal case, maybe even a fan. It is not intended for such high power dissipation.

So all in all, the components used are not able to provide 5V at 1A current into a load from many points of view.

Which is why people use switch mode regulators such as USB chargers to supply their Raspberry Pis.

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I just looked up L7805CV and they are capable of 1.5 A. So that will be the limit of your power supply. If your Pi really does draw 2 A the result may be what you are observing.

Also--these current ratings hold only if the heatsink is attached to something other than the air around them.

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  • \$\begingroup\$ I couldn't find it, do you have a link to where you found that information and thank you so much! Also, I don't know if I will need 2A, I doubt I will need much of anything because I am only switching two SSR's-25 DA for heating coils. powering a 4 digit 7 segment LED display and then 2 LED's I don't think these will take even 1 amp. \$\endgroup\$
    – Xendel
    Commented Nov 9, 2023 at 21:31
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    \$\begingroup\$ While it is true the 7805 is only capable of 1.5A, it does not matter as it also has a power dissipation limit, and the transformer output voltage and capacitor won't even allow for 0.5A. \$\endgroup\$
    – Justme
    Commented Nov 10, 2023 at 6:19
  • \$\begingroup\$ The datasheet is here: eu.mouser.com/datasheet/2/389/l78-1849632.pdf. It shows 1.5A max, and the thermal transfer numbers which imply it will need a big heatsink voor 1,5A. It also specifies the 2V drop that @Justme mentioned. \$\endgroup\$
    – Paul
    Commented Nov 10, 2023 at 9:14

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