how to get 5v, from +-9v??

Lorenzo · Posted: Fri Jul 27, 2012 4:18 am Post subject: how to get 5v, from +-9v??

Hi,
the circuit I'm trying to build need +-9v and 5v to work.

For +-9v I will use two 9v batteries, but for 5v... have I to use three 1.5v batteries? or could I use the two 9v batteries someway?
Is it possible?

I mean to use 5 batteries for a little PCB seems to be a waste! Wink

Suggestion?

Thanks!

bod · Joined: Apr 28, 2009 Posts: 133 Location: Glasgow

the easiest and simplest option imo would be to use a 5v voltage reg and a capacitor on the input and output and supply it from the +9v connection.

the draw on the +9v will be greater than the -9v but would be better all round. Very Happy

corex · Joined: Mar 02, 2010 Posts: 100 Location: Las Vegas

Agreed -- take a look at the datasheet for LM7805.

Lorenzo · Posted: Fri Jul 27, 2012 10:04 am Post subject:

richardc64 · Posted: Fri Jul 27, 2012 12:01 pm Post subject:

fonik · Posted: Fri Jul 27, 2012 2:23 pm Post subject:

another option would be to utilze a Zener diode. schem like divider, but:
R1 = 1K to limit the current, R2 = 5.1V Zener...

Lorenzo · Posted: Fri Jul 27, 2012 3:56 pm Post subject:

Thank you all very much! I will try a voltage divider on the breadboard... then, if the circuit works fine, I'll make things properly using 7805.

JovianPyx · Posted: Sat Jul 28, 2012 12:58 pm Post subject:

The best way to get a steady stable 5 volts at varying currents is to use a regulator such as 7805 or 78L05. The datasheet for the 7805 shows a quiescent current curve that goes from about 5 mA to 6 mA. Quiescent current is the current the regulator draws just by being in the circuit.

9 volt batteries are not high current devices. They are composed of 6 small 1.5 volt cells and were designed for transistor radios years ago (often they were then called "transistor radio batteries"). If the current draw on the +5 rail is fairly significant, it will cause the 9v battery powering the 5v regulator to die faster than the other. In fact, most analog application circuits tend to exhibit a behavior of higher current on the + rail than on the - rail. That said, you might be wise to power the +5 regulator from it's own 9 volt battery.

Also - do not be fooled by the use of CMOS and think that it's going to be low power. CMOS circuits are low power only when they are used in a purely digital fashion, that is, where the outputs are always at one rail or the other. Some very popular CMOS circuits used for audio with parts like 4069, 4049, 4007 and others are used in a "linear amplifier" configuration. The device then operates in some ways like a poor opamp This causes the device to draw much more current than when used in a digital manner. They can draw enough current to get physically warm.

Bottom line: know how much current is needed at +5 as well as the current requirements for +9 and -9. Then you can design a proper system that will not fade out and die when you're trying to get something done with it. And remember that a +5 regulator needs a minimum of +7.5 volts at the regulator input to operate statisfactorily. If the 9 volt supply to the regulator sags below 7.5 volts, the system will not be reliable.

Lorenzo · Posted: Sun Jul 29, 2012 1:54 am Post subject: