digital logic circuits negative voltage regulators are 1/2 the currents output

I have noticed at my work that the digital logic circuits negative voltage regulators are 1/2 the currents output of the Positive voltage regulators

My manager said that the negative voltage regulators don't need a lot of current output because it's just a reference to turn on the OP amps and IC chips. For logic circuits it only uses the Positive swing/cycle mostly so you have to have a 1 amp current output on the Positive Voltage Regulator.

So the Negative Voltage Regulator is just used as a negative DC offset as a reference and to energize the Op amps and Logic Gates, Etc.?

Since The Positive Voltage Regulator has to be 1 amp current output it's sinking and sourcing more for all the inputs and outputs of the Logic chips, the Pull up resistors or voltage dividers branches?

Also They put a .01uf and a 1K resistor in parallel with the ground to EARTH, this is a Noise/common mode rejection filter which takes out the Noise on the ground rail.

My Manager said that when you have difference IC chips analog op amps , Digital IC chips going to ground they have different potential differences on the ground and there is noise on the ground rail

Why would there be different potential difference on the ground from using different IC chips? ground is zero volts , so if you ground the IC chip to zero volts , how can there be a potential difference between the IC chips? and why is there noise on the ground rail?

In a perfect world, all the grounds would be at exactly the same potential and these problems would not exist. In real life, everythng has resistance, including ground wiring so when a current passes through it, a voltage is developed across it. In digital circuits there is a threshold difference between logic low and high voltages so they are relatively immune to small differences in their ground rails but in an analog circuit the differences may be amplified and become a problem. If you look at PCB layouts for high current circuits you will usually find the grounds are not connected to the nearest pont but returned to a common 'star' point for this reason.
Consider that even 1 milli-Ohm of resistance drops 1mV when 1A flows through it. 1mV doesn't sound like much until you amplify it a few hundred times!

That covers the DC voltage differences but wiring also has inductance and is prone to magnetic and capacitive coupling to nearby noise sources. Although digital circuits may be clocked relatively slowly, the rising and falling edges of the waveforms are considerably faster and produce spikes of current into and from the ground wiring. The inductance of the wiring prevents these spikes form being 'dumped' to somewhere with no signal so each devices ground pin may have different signal voltages on them. This is why it's good practise to use a 'decoupling' capacitor across the supply and ground pins of ICs, they act like a local reservoir of charge to feed the switching circuits without drawing spikes along the whole length of supply and ground wiring.

Brian.