Bipolar power from unipolar walwart

[nordine]

hi people

i'm trying to get some synth fun (Ken Stone circuits and others), so i need bipolar power.

searching for ways to get it, i came across this:

http://www.generalguitargadgets.com/richardo/distortion/
("bipolar power supply" image, somewhere in the middle)

now, my question:

-does this trick works only for batteries?

i have a 18V DC 1A unipolar walwart... did the little addon, and got (read with multimeter)   +9.2V, 0, and  -9.2V... i thought "great!".. buy..

then i supplied my headphone amp (386 based) (and a CMOS oscillator attached to it) with these +9.2 volts.... however it didn't worked properly... the volume got reduced drastically, and i get this "splurt" (like an abrupt gate, the typical 'misbias' sound) everytime i touch the input capacitor of the head amp

what would it be? ( does a lm386 need a lot amperage?)

[nordine]

anyone?

[Peter Snowberg]

When you have a wall wart with just two wires and you need bipolar power, very often the ground connection is just to provide a reference voltage at 1/2 of the supply voltage.

You can use an op-amp, but TI makes a chip called a rail splitter that should work perfectly for you. Search here for some posts about it.
Also try searching for "artificial ground". That may bring of the same threads.

If you provide links to the circuits you want to power, the people here can probably state the modifications required.

[nordine]

thanks a lot for answering, peter

the circuit can be found here: http://www.cgs.synth.net/modules/cgs52_folder.html ( also, i'd really like to try the Craig Anderton Compressor)

i didn't understand too well what to do with that artificial ground schematic... it outputs 1/2 VCC, but what has to do this with negative power?  i'm very confused....

i tried the way i said first (voltage divider), then i tried  a 555 timer based circuit (for getting negative power), which seemed to work as well ( multimeter measures gave me +10v, 0, and -10v).. but.... still can't power up anything...

by the way, i'm testing it powering up (w/ positive side and ground) said CMOS oscillator... would this be the problem?... do bipolar power work only on bipolar feeded circuits? 

[Peter Snowberg]

Bipolar power can be a little confusing at first. Don't worry about it. 

When you have uni-polar power, we call the two lines power and ground.

When you have bi-polar power, we call the lines positive, ground, and negative.

To adapt uni-polar to bi-polar, the unipolar power wire connects to the bi-polar positive connection, and the positive connection to the artificial ground.

The uni-polar ground wire connects to the bi-polar negative connection, and the negative connection to the artificial ground.

The 1/2V output from the artificial ground then connects to the bi-polar ground connections of your circuit.

The names change, but the job is the same.

You really need an opamp based circuit to make the artificial ground. A resistor divider will not allow you to use very much current.

[nordine]

heheh yeah, that way it's soo clear 
..it worked!

to the date is the first method (virtual ground) that gave me good results (amplified my bass thru said headphone amp -LM386 based-)... however, when i play hard, it distorts in a very ugly way... maybe the 386 sucks too much juice?

[Peter Snowberg]

Yes, it could easily be that the 386 uses too much juice for your opamp, especially when used with bass. Try using a NE5532 or NE5534 opamp if you have one for more current drive capability.

Instead of using a splitter of some sort to lower the voltage to a 386 amp, I would use an LM386-4 or a JRC386BD and run it on the full 18V. 

You can also use a three terminal regulator like the 7809, 7812, or LM317 to supply a lower voltage at much higher current.

[nordine]

of course, the use of the 386 circuit was only for testing purposes, the (now) bipolar walwart will only feed effects and not power amp stages

here: http://tangentsoft.net/elec/vgrounds.html if found great info on how to improve the opamp virtual grounder (for those interested)

i apreciate a lot your help, peter, thanks