For the Center-Tapped Secondary Full Wave Rectifier, the PIV is double the Vout. OK, but I can't see why? Not much luck on Google. Lots of info on what full-wave rectifiers are but not why PIV is double. On Stack Exchange, I found opposing answers that basically said it only needs to be double when there is a capacitor on the output. I can understand that but not when there is simply a resistive load? The negative peaks are 180° apart so why would they be additive? Or... Are the peaks additive due to the secondary ground tap?

 

Not much luck on Google.

Google worked for me....https://www.elprocus.com/full-wave-bridge-rectifier-versus-center-tapped-full-wave-rectifier/

 

D1 conducts and offers almost 0 resistance And what happened to it 1MΩ+ internal resistance?

Vmax is developed across the load. OK

Now the V across non-conducting D2 is the sum of V across 2nd half of secondary plus V across RL This is where they lose me...

Is it due to the inverting of the second sine wave? The Vout isn't doubled but the PIV is? Or is Vout also doubled? I need to look at the V. I think my confusion may be coming from all the V conversions from rms to AVG to Peak etc.

 

When the voltage at the anode of D1 is at the positive peak value of the voltage across the upper half of the winding with respect to the center tap, the anode of D2 is at the negative peak of the lower winding. Its cathode voltage is the same voltage as the cathode of D1 so the voltage across D2 is the peak to peak voltage across both windings.
Of course, the actual voltages will have to take into account the forward voltage drop across the diodes but I hope this helps to clear up your confusion. You have obviously been over-thinking it and confused yourself.
Regards,
Keith

 

I do tend to over-think things... K it's the VPP, that makes sense. Got it, thx.