My thought is to use 7805 and 7905 regulators. Input of 7805 = +12V, input of 7905 = 0V, connect the two regulator grounds together to give a floating 0V output. Would I need a voltage divider to stop that 0V from floating?
Obviously, that 0V cannot be connected to ground as it will be at about +6V WRT the 12V ground.
Would it work and, if not, is there a better way?
What are the current requirements.
Leo..
That wont work.
Get yourself a LM7805 and a LM7810. Connect them both to the 12V and 0v of the supply. Then call the output of the 7805 the GND, the output of the 7810 the 5V (it is 5v higher then the output of the 7805 
) and the 0v of the supply the -5V (it is -5V lower then the output of the 7805). The 0v output of the supply may not be connected to GND! It's all about reference 
I didn't take in consideration the current draw. The current of the -5V and 5V combined may not exceed the 1A (because both output go trough the 7805). You didn't tell us what you want to power so you're on your own.
That will give only two 5V supplies it will not give a -5 0 +5 supply.
You need to generate a virtual ground. For low currents you can use two resistors in seriese but for high currents you need something else, like a dual output switching regulator.
Mike, that will definitely give -5V, GND an +5V. It's all about reference. If you pick the output of the LM7805 as your reference, aka GND, the output of the 7810 is 5V. The 0v line of the supply is -5V with reference to the 7805. You don't have to pick the 0V output of the supply as your reference
Yes sorry it is amazing how much clearer a schematic is than words, I was not picking that up from your words.
Problem is that 78xx regulators can only source, not sink.
Imagine a load on the +5volt rail....
Leo..
septillion:
That wont work.Get yourself a LM7805 and a LM7810. Connect them both to the 12V and 0v of the supply. Then call the output of the 7805 the GND, the output of the 7810 the 5V (it is 5v higher then the output of the 7805
I didn't take in consideration the current draw. The current of the -5V and 5V combined may not exceed the 1A (because both output go trough the 7805). You didn't tell us what you want to power so you're on your own.
Doesn't work I'm afraid - +ve voltage regulators can only source current, so you won't be able to
sink current into the 7805's output without pulling its voltage up.
How much current do you need? There are chips that do this poperly for low current, about the 0V of the
input supply. For higher current you could use two DC-DC converters (with greater efficiency but more
analog noise). At least one of the converters needs to be fully isolated so it can generate -ve voltages w.r.t
0V.
If you just want to power opamps you only need a low current stiff virtual ground which is done with
an opamp.
MarkT:
Doesn't work I'm afraid - +ve voltage regulators can only source current, so you won't be able to
sink current into the 7805's output without pulling its voltage up.
That's why I was thinking of using a -5V regulator, a 7905.
How much current do you need?
Not much. Just enough to run a couple of stereo amplifier chips. I don't need HiFi.
I'll look up their power requirements and get back.
There are chips that do this poperly for low current, about the 0V of the input supply.
What are they called?
If you just want to power opamps you only need a low current stiff virtual ground which is done with
an opamp.
Thanks.
I'll look up their power requirements and get back.
I'm going to be using a PGA4311
Quiescent current is slightly less than 50mA. I presume this rises, depending on the power of the output.
I'm going to use it to power 4 small (3" ?) 8 Ohm speakers, not a rock concert or HiFi, but loud enough to be clearly heard in a small room.
EDIT. Not all the speakers will be active at the same time. I want to fade the sound from one speaker to the next to simulate the source moving.
That chip is not a power amplifier, only needs 50mA max on the analog rails.
If does require that -5V really is -5V since the chip also have digital +5V supply.
Since you mention not needing HiFi you could just use an isolated 5V->5V DC-DC
converter to generate the -5V.
If you want HiFi, then use two 5V->12V DC-DC converters to generate +/12V rails, and
use 7805 and 7905 from those (switch mode supplies are usually too noisy for HiFi, linear
regulators are normally needed).
Having said that more recent higher frequency DC-DC converters may have almost all
of their noise in the ultrasonic range, so might sound OK.
My thought is to use 7805 and 7905 regulators. Input of 7805 = +12V, input of 7905 = 0V, connect the two regulator grounds together to give a floating 0V output. Would I need a voltage divider to stop that 0V from floating?
Obviously, that 0V cannot be connected to ground as it will be at about +6V WRT the 12V ground.
Would it work and, if not, is there a better way?
It sounds like you are asking for a dc to dc converter with +12V input and +/- 5V output.
The simplest way to achieve this is buy two 5V walwarts and connect the positive output of one to the negative lead (GND) of the other. The negative lead (normally the GND) of the ac/dc converter that has it's positive lead connected to the gnd of the other becomes the -5V output.
AC = +5V =>>==============> +5V out
0V =>>--|=============>GND
+5V =>>--|=============>GND
0V =>==============> -5V out
The part in bold is the jumper from the +lead of PS2 to the -lead of PS1 , creating a common ground, and turning the second PS into a negative output (-5V ) PS. Almost ALL AC/DC converters are 2-prong devices, so the output is "floating" which allows you to do the above. I have never seen a 3-prong AC/DC walwart. (not that they don't exist but clearly they are the exception)
A more complicated solution is the following:
Your original idea won't work because you don't have a negative input for the 7905. If you had two 12V ac/dc walwarts you could do the same thing I described above but use the -12V output as the input for the 7905 regulator. The MAX232 is not a bad idea either.
Not much. Just enough to run a couple of stereo amplifier chips
Really ? Like what ? Are you sure you don't mean LM386 chips ?
MarkT:
That chip is not a power amplifier, only needs 50mA max on the analog rails.If does require that -5V really is -5V since the chip also have digital +5V supply.
Since you mention not needing HiFi you could just use an isolated 5V->5V DC-DC
converter to generate the -5V.
Do I connect the +5V output of the converter to ground and the 0V of the converter becomes the -5V?
Henry_Best:
Do I connect the +5V output of the converter to ground and the 0V of the converter becomes the -5V?
Of course, if its an isolated supply.
I've found the LT1173-5 which can convert from a +5V input to a -5V output at 75mA. Will that be OK?
Henry_Best:
Do I connect the +5V output of the converter to ground and the 0V of the converter becomes the -5V?
Yes but it must be an isolated converter.