I’m looking to drop a 9 volt battery to 3.5 volts around 1 amp load momentarily with a resistor. Does anyone know how to do this math ? Or a 7805 regulator but I don’t know if it can go down to 3.5v.

Google ohm's law. Put the resistors in series with the load....also calculate watrage V x I.
That regulator can't go that low. You'd need a LM317 or LM337 for that.

1) How accurate does the 3.5 volts need to be?
2) An ordinary alkaline 9V battery cannot supply 1 Ampere without dropping voltage drastically and rather unpredictably. It might even be below 3.5 volts at such a high current. You need to find a heftier source.

https://www.youtube.com/watch?v=rG2ozSOvxBY&t=3s

1) How accurate does the 3.5 volts need to be?
2) An ordinary alkaline 9V battery cannot supply 1 Ampere without dropping voltage drastically and rather unpredictably. It might even be below 3.5 volts at such a high current. You need to find a heftier source.

https://www.youtube.com/watch?v=rG2ozSOvxBY&t=3s

Forget the 1 amp it’s probably 500ma and the voltage around 3-3.5 volts

You can see from the video that 500 mA is still going to be too much for reliable operation.

A LM7805 with 2 to 3 silicon rectifiers(most of these are a 0.5V drop) in series on the output-should get you in the ball park of the 3.5 VDC. As for the current, how long do you need the 0.5 amp for. A capacitor discharge circuit might get you the surge current you require.

Hi Timmy,

Here is what you need:

https://www.ne555.it/category/schemi/alimentatori/

I think just this one will fit:

https://www.ne555.it/alimentatore-switching-3-3v/#more-3804


Best regards,
TV-collector :stupid:

Next question: does it need to be filtered and how well?