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Reduce a 9V to 5V

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Nick'

New Member
I have a 9V battery and would like to reduce it to 5V. How can i accomplish this? If i were to know its current(A), i could use the ohm's formula ( V = Ir) to calculate. But now, i don't know. Please help. Is there any other way? Because i would like to convert my 9V battery to 5V to be used with microcontrollers.
 

kchriste

New Member
Forum Supporter
Use a LDO regulator like the LM1117T, LM2931, or equiv. The cheaper 7805 or 78L05 will work, but you won't get the full capacity out of the battery. If you plan on doing much experimenting it would be better to get a power supply, even a wall wart, instead of the typically weak 9V battery.
 
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AllVol

New Member
Can i use a resistor for this? If yes, how many ohm's should i buy?


You could use two resistors in a voltage divider circuit, although it is a pretty dicey way of doing it. Why don't you go to the Play-hookey website for some solid electronic background?
 

Noggin

Member
Can i use a resistor for this? If yes, how many ohm's should i buy?

That is often a very shoddy way of doing it. If you ever increase the load (draw more current) then whatever resistor you chose will have to be changed. I noticed that you have a thread in the programming section so I'll assume you want to drive a microcontroller. The MCU will NOT like having a resistor there as it's current is always changing.
 

Nick'

New Member
That is often a very shoddy way of doing it. If you ever increase the load (draw more current) then whatever resistor you chose will have to be changed. I noticed that you have a thread in the programming section so I'll assume you want to drive a microcontroller. The MCU will NOT like having a resistor there as it's current is always changing.

Yes you are right, So, what is the best way? Buy those LM things?
 

electronspeed

New Member
If u need 2 convert 9V to 5V best way is to use a voltage regulator typically 7805.Voltage divider can also be used but whenever extra circuit will be added loading effect will work & voltage from voltage divider will vary,secondly power dissipation will also be more.
 

Nick'

New Member
Ok. I got it. But what if I use a different input other than 9V. maybe 6V or 12V. Can i still use the same LM7805?
 

hdc090360

Member
Any sort of linear regulator is going to waste about 44% of the power from the fully charged battery in the regulator itself. A switching regulator like the MC34063A will be about 85% efficient so your battery will last a lot longer. See if you can get hold of an old car charger for a mobile phone that has an output of about 5V. You should be able to change one resistor to get exactly 5V. This saves having to source all the chokes etc.

Duncan
 

Nick'

New Member
Talking about power wastage. I would like to know if a device is wasting or not. I have this two situations.
1. A 2V device running on a battery
2. A 4V device running on a battery

Both are running on a 6V battery. My question is will both of them using the power of 6V? Or they will be using 2V and 4V respectively? If it is the first case, both devices will die off at the same time, but for the second case, the second will die off twice as fast.

what does power wastage depend on? Current or voltage or power?
Please explain, i am new, thank you.
 

Noggin

Member
You can apply ohm's law to it if it is a linear power regulator.

Source is 12v, regulator drops it down to 5v, output is 1 amp. That is 1 Amp at 12v (12 watts) input to the regulator and 1 Amp at 5v (5 watts) output from the regulator. That means that 7 watts is wasted as heat inside the regulator.

In your scenario, assuming that the 2v circuit draws as much current at the 4v circuit, the 2v circuit would live longer but only because the device can continue operating at a lower voltage than the 4v circuit can operate. It won't last much longer though because a battery's voltage drop is not typically linear, especially near the end of its life.
 
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