I can create a +5.5V witha LDO but i'm not how to go about creating a -5.5V. Can you point me to some resources to do this?

 

At 2 A output current, your 5.5 V LDO will dissipate 80 W. Is there a reason you don't want to use a switching buck regulator?

For the - output, consider an inverting buck-boost regulator (another type of switcher).Beyond that, your only practical option is to purchase a 48 V / 5 V DC/DC converter, one with an adjustable output.

ak

 

As AnalogKid says, a switching regulator is the way to go with this. Its the only real way to get a negative supply in this case (unless you have a VERY lossy virtual ground by using a potential divider.) You can get some specific switching dual supply regulators, some +, some - voltage regulators also or if you check the data sheet they often show how to get a negative supply, for example:


(This is just one I use a lot myself, and much less power than you need!) Also note that in such a case the negative rail does not actually have any regulation in it in this case.

 

much less power than you need!) Also note that in such a case the negative rail does not actually have any regulation in it in this case.

Those are the two reasons I did not mention a charge pump as a "practical" solution.

ak