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I've made a quick test where I put 12V from a dc power supply to a 15V regulator - so the "input" was lower than the voltage regulator output.

When I measured it, I had a 5V output, and it was what I'm looking for intially, and the regulator wasn't hot.

So I would like to refer to your expertise and ask you if it is a good way to get 5V without any issue like a current drop for example, which is crucial for the working of my components?

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    \$\begingroup\$ It is likely that the voltage will change with the current - measuring the voltage without a load is not useful. Simply use a 5 V regulator instead, and use a heatsink if your calculation of the case temperature shows it will be too high. \$\endgroup\$ – Andrew Morton May 4 at 16:03
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Absolutely no. Closed loop regulators use an internal voltage reference (e.g. 1.25V) and compare it to a fraction of the output to provide load and input voltage regulation.

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That is not a good way to get 5V.

The behaviour of such a regulator is unspecified when the input is below the rated output voltage.

The regulator cannot operate in those conditions. The output you get is a result of the internal resistances and the state of various transistors.

Since the regulator can't actually regulate its output, you cannot expect it to stay at 5V when you put a load on it.

So, don't do it.

Get a 7805, or better a switching regulator than can reduce the voltage more efficiently.

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  • \$\begingroup\$ Got it !!! thanks a lot \$\endgroup\$ – Mehdi May 4 at 17:01
  • \$\begingroup\$ Vote for answers that helped you. Click "Accept" for the answer you think answers your question best. \$\endgroup\$ – JRE May 4 at 17:02

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