What is the Vdc of a center-tapped full-wave rectifier once you consider the potential drops across the diodes?

We have learnt that the Vdc for a half-wave rectifier is o.318(Vm - Vk). Vk being the diode's cutoff voltage. That is understood because for every cycle, input voltage is across one diode.

Also, for a full-wave bridge rectifier, Vdc=0.636(Vm- 2Vk). This is also understood because in every half-cycle, the input voltage is across two diodes.

I was unable to find the Vdc for a center-tapped full-wave rectifier specified clearly anywhere. Based on the previous deductions, my guess is that it will be 0.636(Vm - Vk). Correct me if I am wrong.

 

hi S.
How would this center-tapped full-wave rectifier be connected.?
If the centre tap of the TXR was considered as 0V common, the two outputs of the FWB bridge would be Negative and Positive voltages relative to the CT.
E

 

How would this center-tapped full-wave rectifier be connected.?

The full wave resistor load would be the one. It says Vdc is 0.45*Vac. I thought it is (2/(pi))*Vac. (which would be 0.636*Vac). They haven't mentioned the effect of the potential drop across the diodes has on Vdc.

 

hi,
This is what LTSpice shows, its a 10Vpk out, so divide the Avg and RMS by 10.
E

 

Hi,

The average DC is 2/pi times the peak once the peak is adjusted for the diode drop(s).

In a full wave setup with four diodes and one output, there are two diode drops so the adjusted peak is: Vpk-2*Vd.
In a full wave setup with two diodes and center tap transformer, there is only one drop so the adjusted peak is Vpk-Vd.

It is very rare to see this in real life though, as there are almost always capacitors used for filtering the pulsating DC into a more smooth DC. That changes everything except the peak and even the peak changes a little because of extra diode voltage drops with higher current peaks.

There are also drops caused by line and transformer resistances and inductances which i assume we are ignoring for now.