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5V out of 9V battery

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ansplus

New Member
I have a small project requires a 5V supply, I was reading about the voltage regulators, the LM7805 and the 7805SR-C, which one is better? I have the LM7805C in the lab, how can I calculate R1 and R2 in the following circuit to get a 5V from a 9V battery??
9068-r8cgn8.jpg
 

Mike_2545

Super Moderator
If you use a 'low drop out' regulator you will get more out of the 9v battery, they will work at lower input voltages than the good ole stand by 78XX.
 

Speakerguy

Active Member
If you are using a 7805, you can eliminate the resistors and just tie pin 2 to GND. The output will be at 5V.

There is very likely no practical difference between the LM7805 and the 7805SR-C. Probably just a different manufacturer. It's a generic part, lots of companies make equivalents.

The 7805 will give you 5V out until the battery drops to about 7V. So there is still quite a bit of life left in the battery when the regulator will drop out of regulation. The circuit will work for a longer period of time if you use an LDO regulator. Try an LM1086 instead. Those are somewhat widely available as they are an older part. If you are in the US or other Western country there are many more and even better LDO regulators out there.
 
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Bilthon

New Member
Well.. I've got a similar issue, but in my case I need 3,15 volts and I have a battery providing me 3,7. I also have access to some LM78XX, but I'm not sure how can I lower the voltage. Should I change R1 or R2?
 

ansplus

New Member
Thanks guys.
I just noticed the difference between the LM7805 and the 7805SR-C, the 7805SR-C has lower quiescent current (3mA vs. 5 mA), higher input voltage (40V vs. 32V), and better output accuracy (±1.5% vs. ±5%), and it has built-in filter capacitors so that no need for external components. but its much more expensive than the LM7805, doesn't worth the cost difference.
To change the output voltage I guess we should use those two equations
Io=Vxx/R1+IQ and then Vout=Vxx(1+R2/R1)+IQ*R2
am not sure how to calculate IQ!
 

mvs sarma

Well-Known Member
If you use a 'low drop out' regulator you will get more out of the 9v battery, they will work at lower input voltages than the good ole stand by 78XX.
but Mike, with all said, a weak battery is a weak battery and as it looses power, though voltage might indicate 7V or 6.5 V , when it is not capable of delivering current , what can an LDO do?
instead any modern converter with better efficiency appears better. as the power during the working life is optimally used as against linear regulation

perhaps 5V LDO is good for a 6V SLA battery etc
 

audioguru

Well-Known Member
Most Helpful Member
I use a low dropout 5V regulator for most of my 9V battery powered circuits. The battery cannot supply much current so the circuits do not use much current.
The battery lasts a long time until its voltage drops to 5.5V.

A new 9V battery voltage drops fairly quickly to 7.2V where an ordinary 7805 regulator is ready to quit working.
 
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