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I have a L78S05V voltage regulator (5V output). The datasheet's maximum ratings suggest the maximum input voltage (35V). I would like to know, is there a minimum input voltage?

I have a circuit that is powered from a 5V power supply. For "forward compatibility" (i.e. if my custom power supply broke) I want to add 5V voltage regulators on the input stage, so I can change the power supply with a 12V one and still have everything powered at 5V. I can add some jumpers to route the 5V input or 12V input (via voltage regulators), but if the 5V voltage regulator works well with a 5V input I can get rid of the jumpers and work with any input voltage between 5V ad 35V without worrying.

Thanks!

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  • \$\begingroup\$ 8 V. It requires 3V more than the output. \$\endgroup\$ – winny Jul 22 '16 at 7:02
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On a linear regulator you can not work with the same voltage at input and output. Basically inside the component there are transitors that are not perfect transistors, because they have a resistance when the current flow through them, and this resistance leads to a loss of voltage in the component. It is called the drop out voltage.

The maximum output voltage you can have with a given component and a given input voltage is : Vout = Vin - Vdropout.

In your datasheet it is wierdly written since Vin is refered at the drop out voltage. The datasheet say the min Vi is 8V, meaning the drop out voltage is actually 3V. Meaning if you put 5V at the input you won't have more than 2V at the output. Then you have to use your jumpers.

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  • \$\begingroup\$ Did some research on the dropout voltage (I mistranslated it... my bad). As you said, it is stated to be "the smallest possible difference between the input voltage and output voltage" (Wikipedia). So I was quite confused when reading 8V (5V+8V=13V !!!). Glad to hear that it means it is actually 3V. And to go practical, I need to use the jumpers, got it. Thanks! \$\endgroup\$ – il_mix Jul 22 '16 at 7:30
  • \$\begingroup\$ @il_mix Another thing to keep in mind is that voltage regulators can get pretty hot. If you want to step down from 12V to 5V you're going to have power dissipation of 7V x Current. If your output is going to be more than 80mA then you might need to look at getting a heatsink \$\endgroup\$ – Doodle Jul 22 '16 at 7:31
  • \$\begingroup\$ Thanks for the hint. Given what you said, I think I'll need a heatsink \$\endgroup\$ – il_mix Jul 22 '16 at 8:08
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You're so close ... with datasheet open ... look for minimum input voltage. Or dropout voltage. It's 8V

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