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9V to 5V

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Matt D.

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I am new at all this electrical stuff. I can build circuts without resistors or capacitators because I don't quite understand the relationship between curent(I),Volts(V),and resistance(R) in the equation V=IxR. I just have one question: I am building a curcuit with an inpu of 9V. I need it to have an output of 5V. How would I do that?
 
I am new at all this electrical stuff. I can build circuts without resistors or capacitators because I don't quite understand the relationship between curent(I),Volts(V),and resistance(R) in the equation V=IxR. I just have one question: I am building a curcuit with an inpu of 9V. I need it to have an output of 5V. How would I do that?

Without using resistors or caps, put the 9v battery in series with a 4v battery, + to +, or in series with a 4v Zener diode.
 
So you mean that if you connect two batteries + to + you just subtract the voltage. Thats a good thing to know. Thank you very much.

P.S.-What does a regulator look like. I serched it on google but all I could find was regulator test kits.
 
If you're using a 9V battery an LM7805 won't last very long because it will stop regulating properly when the battery voltage drops to 7V, for best results use a low dropout regulator.
 
simple way

Hi Matt D,

Why not simply use 2 resistors in a potential divider netowort?

An 800R (at +9v end) and a 1K (at 0v end) in series with a center tap in the middle will give you 5V in the middle. Current loss wont be much differnt to a linear reg. The voltage out will drop though as the battery drains. Or use a 5.1v (5v1) zener? even better.

oliver
 
i think the zener would be the best and the most efficient since resistors dissipate power and zener dissipates less
 
i think the zener would be the best and the most efficient since resistors dissipate power and zener dissipates less

No, a zener dissipates exactly the same, but gives you a degree of regulation, depending how regulated you require it though, you may be able to lower the wasted current somewhat.

But really it all depends on exactly what you're doing and why.
 
And with an analog circuit you may not need a regulator at all. Opamps take a pretty wide power supply voltage range.
 
Hi Matt D,

Why not simply use 2 resistors in a potential divider netowort?

An 800R (at +9v end) and a 1K (at 0v end) in series with a center tap in the middle will give you 5V in the middle.
800R isn't a standard value, use 1k6 and 2k.

This is not a very good idea anyway as any load in parallel with the 1k will reduce the voltage.
 
If you have the money and need a good relgulated 5VDC voltage, look at the Datel 7805SR-C or other DC-DC converter that don't typically need external parts. (You don't say how much current you need.)
 
Thank you all for the help. I just switched it to AC so the voltage is constant. So I guess my problem is solved.
 
For a 9V battery, it may be a lot wiser to use a switched-mode DC-DC converter, dropping 4V is a lot of wasted energy, especially on low capacity batteries (which is a problem with 9V batteries). Of course this is just for future reference.
 
Switched to AC? I assume you mean a plug in the wall adapter rather than battery. :eek:

On the side of Switching convertors DC-DC, I have personally used the LM2574 with output ranges from 3.3 to 15v from inputs 4.75-40v. There simple to wire-up, quiet and very effiecient compared to linear regs. As with all switching devices, there is some ripple on the output lines, so this all depends on what you need voltages for.

Hero999: Correct, 800R isn't standard but was just for demo purpose as the ratio can be scaled. :D
 
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