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hello,
Here is something for your personnal pleasure. If you look at the datasheet, there is something call “Dropout voltage” that egals 2V.
This mean to have 5V at the output you need 7V at the input.There is a 2V dropout.
7V – vDrop = 5V7V – 2V = 5V
If you apply 6V at the input you will not get 5V at the output. You will get 4,68V That is what I just measured with my 7805.
Most of electronic component will not see the difference of 0,32V but this need to be considered in your future projet.
Good luck and thanks!
Jerome
You’re right. It should be more than 7V at the regulator input. I tested it with a 6V PSU which had 7.5 V at its output.
Thanks for pointing out, Jerome
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hello,
Here is something for your personnal pleasure. If you look at the datasheet, there is something call “Dropout voltage” that egals 2V.
This mean to have 5V at the output you need 7V at the input.
There is a 2V dropout.
7V – vDrop = 5V
7V – 2V = 5V
If you apply 6V at the input you will not get 5V at the output. You will get 4,68V That is what I just measured with my 7805.
Most of electronic component will not see the difference of 0,32V but this need to be considered in your future projet.
Good luck and thanks!
Jerome
You’re right. It should be more than 7V at the regulator input. I tested it with a 6V PSU which had 7.5 V at its output.
Thanks for pointing out, Jerome