Transformer hooked up wrong

Say you have a CT transformer running into a a bridge then the bridge goes into two series capacitors with the CT into the middle. This is the "standard" dual polarity setup I think. Now, if you end up swapping the CT of the transformer for one of its other leads then, atleast from testing, one gets double th voltage.

something like

Transformer

A--

B--

C--

Bridge

V+out --

ACin1 --

ACin2 --

V-out --

Caps

CL --|| --- CG --- ||--- CH

So Normally A goes to ACin1 and C goes to ACin2 and V+out goes into CH and V-out goes into CL with CG attached to B.

Now if you instead attached B to V+out, which should effectively cut the voltage across CL to CH in half, and A to VG you get a voltage doubler like effect(across CL to CH).

But I can't understand why this seems to work without any ripple(atleast the simulation I did didn't have any ripple(atleast not large ripple).

It seems that connecting A to CG raises the floor by whatever the voltage across the transformer. Is this effectively a voltage doubler? (sorta looks like it but I it seems strange)

Thanks, Jon

Swapping A and B is not the same thing as connecting B to V+out because A was connected to ACin1 before.

When you swap A and B, you get twice as much output voltage, but the circuit is operating as a half-wave rectifier, and the two diodes connected to B (ACin1) never conduct. You might as well leave the CT open-circuit. It's not a voltage doubler in the usual sense.

If you connected B to V+out whilst C was connected to ACin2 in a real circuit, you would get a hot transformer and/or a hot/blown diode, because it would short half the winding.

If you want to see ripple, put load resistors across the capacitors. You'll get ripple at the power line frequency for half-wave, or twice line frequency for full-wave rectification.