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#1
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Dropping resistor
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.
Last edited by timmy; 04-15-2025 at 03:17 PM. |
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#2
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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.
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Tom C. Zenith: The quality stays in EVEN after the name falls off! What I want. --> http://www.videokarma.org/showpost.p...62&postcount=4 |
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#3
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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 |
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#4
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Quote:
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#5
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You can see from the video that 500 mA is still going to be too much for reliable operation.
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| Audiokarma |
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#6
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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.
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#7
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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-sw...-3v/#more-3804 Best regards, TV-collector
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Scotty, beam me up, there is no more 4/3 Television and AM radio in Germany! |
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#8
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Next question: does it need to be filtered and how well?
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