This question is about designing a split power supply from a transformer and a rectifier. I'm trying to understand why a full-wave bridge rectifier with a center-tapped transformer is not affected from loading comparing to a not center-tapped one.

In below schematics both configurations are simulated in LTspice. The only difference is the one on the left side uses the transformer as center-tapped and the one on the right uses the transformer the transformer as usual.

X and Y represents the terminals of the split supplies. Both supplies gives the same output at X and Y terminals without any loads. They both output around +8V and -8V when no loads attached. But as seen in the schematics, when I load both supply terminals with two different loads 1k and 100 ohm, the output voltages at the X Y terminals becomes as:

(please left-click to enlarge the schematics)

For the center-tapped configuration the output at X Y terminals:

For the non-center-tapped configuration the output at X Y terminals:

It seems like one should use a center-tapped transformer to create a split power supply otherwise the loading corrupts the output for non-symmetric loads.

My question is: I can verify this by LTspice but I do not understand the reason behind. How can we explain this difference logically?

In a floating supply (without the centre tap), the same current has to flow through both loads. As the loads are different resistances, they will develop different voltages across them.

When you add a centre tap to the transformer, you provide an additional current path, and create two fully independent power supplies, which can support different currents.

Use LTSpice to probe the current in the centre tap.

The transformer without the centre-tap produces a voltage across the output of the diode bridge (positive to negative) which is exactly the same as for when it has a centre tap. This is indisputable.

The problem arises with the non-centre tapped transformer when you arbitrarily connect both positive leg and negative leg to an otherwise floating 0V via two different value resistors.

A current flows through the 1k and 100 ohm resistors but the asymmetry of those values produces an offset on the output. This barely happens with a centre-tap because the centre-tap provides a low impedance source to your 0V reference.

Both supplies gives the same output at X and Y terminals without any loads. They both output around +8V and -8V when no loads attached.

The output of the 'not center-tapped' power supply is 16V. It only looks like +-8V compared to your grounding point because the voltage splits evenly between the two filter capacitors (since each capacitor has the same value, it will charge up to the same voltage because the same charging current flows through both capacitors). Try changing C4 to eg. 2000uF and you will see quite different voltages.

When you apply asymmetric loads the voltage will split according to their characteristics. Since the transformer's secondary winding is isolated you can place the 'ground' anywhere you like, but it doesn't affect how the voltage is distributed to the loads.

Apart from the grounding point, your non center-tapped 'dual output' circuit is the same as this:-

simulate this circuit – Schematic created using CircuitLab

In the centre tapped schematic both windings provide independant full wave positive (D3 D4) and a full wave negative (D1 D2) rectification. So the bridge has to be seen as 2 sets of diodes. The non centertapped schematic the bridge rectifier is needed to create a full wave rectification. However only one voltage between X and Y. If you use a symetrical load then the voltage over each load will be equal. However the common in this case is not a real common like in the centertapped situation