Kapanadze Cousin - DALLY FREE ENERGY

[T-1000]

When i add one more diode (cathode at point A, anode at point B), the signal at my scope is presented rectified as expected. Why is that? I mean shouldn't my scope show a rectified signal even with one diode in the circuit? 

The diode internal capacitance is making AC signal to scope without load. If you attach bleeding resistor for load it will work as expected.

[Jeg]

The diode internal capacitance is making AC signal to scope without load. If you attach bleeding resistor for load it will work as expected.

Shouldn't my probe already have a Mohm resistance internally? Shouldn't this resistance act like the bleeding resistance that you describe?

If i connect my output to a coil instead of a resistance. What will happen? Will one diode do the rectification as someone would expect?

And at the end, why there is the need of a bleeding resistor from the begin? I mean how it is interacted with the coil system and my probe? (yes i know...back to school!)

[T-1000]

Shouldn't my probe already have a Mohm resistance internally? Shouldn't this resistance act like the bleeding resistance that you describe?

The MOhm resistance of the probe is not enough to shunt picofarad capacitance. Also if you attach coil after diode it will be not enough to magnetize coil from the opposite current due limiting current over picofarad capacitance of the diode. Unless you will go up to few MHz range where it will start making influence.

[Jeg]

Tnks T1000
Last question on this.
Why when we connect a second diode as i described above we have a correct dc waveform even without bleeding resistor? It is like the second diode determines the reference ground as this can be "seen" by the oscilloscope. What is your opinion on this?

[verpies]

Tnks T1000
Last question on this.
Why when we connect a second diode as i described above we have a correct dc waveform even without bleeding resistor?

Because it works as a load.  Measuring across a load gives you a current reading (not a voltage reading).
You can accomplish a very similar rectification with a single diode connected across points A and B in your diagram, while the other diode is deleted and replaced by a piece of wire.

Your confusion arises because you forgot that real diodes do not rectify voltage - they rectify current.
Once you set up your jig to measure the current flowing through the diode, everything will clear up.