converting single supply to dual supply

[mamech]

hello

I made a dual power supply by making voltage divider to create a virtual ground, and then I made the a opamp for buffering to prevent this virtual ground to have loading effect.


everything worked great. the only problem is that i have now , is that I am restricted by the maximum current that can be drawn from op amp. for example, i tried to make position controller for a DC motor (similar to this circuit https://www.bing.com/images/search?...7E12BFB923A397D2B164CE6B90103&selectedIndex=0), and the motor hardly starts rotation!

I liked very much the idea of splitting supply using divider and op amp. is there any modification that I can do so the circuit can produce more supply?


Actually from my search , I found some suggestions :

1- to use power amplifier :
i have no experience about power amplifiers, can I replace my lm324 opamp (that I used to isolate the virtual ground) and use TBA820 instead? Can TBA820 to be used as a in normal buffer configuration like normal opamps ??
datasheet : https://www.ram-e-shop.com/ds/ic/TBA820.pdf

2- to use voltage regulator (like 7805 for ex)
The main problem that I noticed in this method, is that if I use it with 9volts dc supply, I think it will split it to +4volts and -5volts, which is not something good to have this difference. is there any solution that makes them equally divided without affecting the maximum current that can be drawn?

thanks for help

 

[dick_freebird]

So long as you have adequate headroom (supply vs signal)
and a decent CMRR I don't think you care a lot about fine
symmetry.

If you used a pair of adjustable linear regulators like
(say) LM317 and LM337, and let them fight it out over
the center rail, you would probably be able to set them
to (say) 4V each on the output at very low currents.
They have an unfortunately high headroom requirement
however. Modern LDOs might be better if you can find
a good negative one.

A solid slab o' ground beats a virtual one in almost all
cases, except when creating that virtual ground is part
of circuit trickery (like feedback summing). Where ground
is supposed to be ground, make it real. Even where your
DC readings are not bothering you, the transient
behavior is liable to meddle when you least want it to.

Now the qualities of your "9V DC adaptor" want some
thought. Maybe if you cracked the case you would
find a handy, perhaps even unused, center tap to
the transformer and could do better if you put the
regulators closer-in to the source bridge rect & filter.
And of course if you happened to have two of them,
making a center tapped stack is easy and you'd have
plenty of headroom for a LM317 (or 7805) on the plus
and a LM337 (or 7905) on the bottom. I have a pair
of wall warts arranged exactly so, with those kinds of
regulators on a "dongle" board, for use at off-site test
facilities where luggage weight is more of a consideration
than ultimate supply quality & accuracy (which tend to
put you over checked-bag weight).

 

[crutschow]

Here's an approach using two low-dropout voltage regulators. It works because the regulators can either source or sink current to maintain zero relative voltage at the virtual ground (for example when the positive load current is higher than the negative current). Most regulators can't sink current.

 

[Chaitanya Varma D]

Here I can suggest you an IC which can convert positive supply to negative supply. It is LTC660. Data sheet @ http://cds.linear.com/docs/en/datasheet/660fa.pdf. The IC can source a max of 100mA, but in you case since its a motor, I shall cascade them so that I can get enough current.

 

[kam1787]

Using an op-amp **broken link removed**
going all discrete **broken link removed**

 

[BradtheRad]

This half-bridge charges two stacked capacitors from a single supply, creating a dual supply. In fact it is a voltage doubler.

I made the loads different ohm values, to show that voltage remains the same on each supply.

The clock at left can be an oscillator made from your present op amp. It needs to sink current to drive the PNP. It needs to source current to drive the NPN.