Yes Ahmad,

There is no need to use a second positive regulator . You can split the output of any  positive voltage regulator on two equal voltages by using two large equal capacitors say 100 microfarads. In order to force an equal voltages connect 10 kilo ohm resistor across each capacitor. Now you have three terminals. The end terminals and the center terminal. Take the center terminal as a reference then you will get a positive voltage at the direct output of the regulator and a negatie output at the other end terminal. This is a common technique.

You can implement it straight forward.

wish you success

It works as long as the input voltage sources are independent (Vs1 and Vs2). I carried out simulations using the same input source and as result I had +5 V and -6.2 V. Later I will try to make a small circuit on the breadboard and I will let you know about the results.

Thank you Prof Abdelhalim for your guidance. but I already tested this technique before and the output voltage is not too efficient when load is applied. The total resistant from load will change the R1&R2 value thus change the value of positive and negative voltage.

Thanks  Diomadson. I will check that too.

Dear  eng. Ahmad,

You are right. But not give up. For every problem, there are many solutions and you you choose the best one for you. Every power supply has a driving emf and an internal source resistance.If the current drawn from the supply is too high. its output voltage will begin to decrease than its driving emfs.This is the case when the load resistance is made comparable to the source resistance. So long as the load resistance RL >> than the source resistance  RS, Then the power supply will act as a constant voltage power supply. This is valid for any power supply including the proposed one. In order to mitigate the problem, one has to increase the values of the capacitors and relatively decrease the values of the parallel resistors. Another factor is to use a nearly balanced positive and negative loads such as feeding operational amplifiers.

Independent source means another winding of the transformer or a separate transformer. So, the cot will be larger.

wish you success.

I agree with Professor Abdelhalim: with a couple of capacitors you can divide the voltage, and displace the ground. It is essential that power consumption for both sources are balanced. Thinl that your capacitors act as a voltage source, so the larger value, the better. I have used this in order to generate IGBT driving signals and it works fine. It is also true that in this case both consumptions (positive and negative) are very well balanced. There are some ICs that make a similar function. for instance MAX232 generates +- voltage, since it adapts TTL levels to RS-232 level. I do not know your power consumption, and it is essential to tell you something. Keep in mind that these circuits (MAX232 and others) do not supply large current values