Spec'ing a PT

[joesatch]

100 watt PT needs.  I see these two with vastly different HT secondaries.   I would think rectifying 345v +345V would result in a B+ well over 1000v ? Help me understand this.  The attachment looks more normal to me (a hammond)

Secondary:            
      345V@400mA      Red
      0V            Green/Yellow
      345V            Red

[SILVERGUN]

Read this:
The Valve Wizard

[shooter]

What he said ^^^

still need a couple pieces;
will it feed full wave, 2 diodes? or
full wave bridge, 4 diodes?


what is your "expected" B+ that feeds the PA section?


100W doesn't tell much, I can get that with 5V and lots of current, or I can get it with 500V and less current

[sluckey]

The first PT is meant to be used with a conventional full wave rectifier. Unloaded B+ will be 345 x 1.414 = 487vdc.

The second PT is meant to be used with a full wave bridge rectifier. Unloaded B+ will be 2 x 175 x 1.414 = 495vdc.

Either one will work just fine, but you gotta use the correct rectifier.

[joesatch]

Gotcha thanks. I have only used FWB in anything i've put together. What does the yellow wire do in the hammond?

[acheld]

What does the yellow wire do in the hammond?

It completes the circuit for each of the half wave rectifiers used with it. 

LOL.   Valve Wizard explains this much better than me.   It took me awhile to figure out what the difference between bridge rectifiers and two phase rectifiers . . .

[joesatch]

From what i know when wiring a FWB, there is no center tap (or no CT grounded) on the PT and the FWB is grounded.  The Hammond has a yellow common wire (a CT?). In a FWB i would leave that yellow disconnected?

[SILVERGUN]

From what i know when wiring a FWB, there is no center tap (or no CT grounded) on the PT and the FWB is grounded.  The Hammond has a yellow common wire (a CT?). In a FWB i would leave that yellow disconnected?

In the snip below see how Marshall attached the CT to the middle of 2 series filter caps. YOU CANNOT ATTACH IT DIRECTLY TO GROUND as you would in the "non-bridge" rectifier version.

[SILVERGUN]

Also good:
Full-wave Center-tapped Rectifier Tutorial | Basic Electronics - YouTube

[shooter]

do the reading
the cliff-notes;
if you wire a full wave bridge correctly, then wire the PT CT to ground, you will be re-enacting the scene from the bridge over the river qui(sp?) 

[scstill]

Kwai
great movie with Alec Guinness (I think)

[scstill]

The first PT is meant to be used with a conventional full wave rectifier. Unloaded B+ will be 345 x 1.414 = 487vdc.

The second PT is meant to be used with a full wave bridge rectifier. Unloaded B+ will be 2 x 175 x 1.414 = 495vdc.

Why did you multiply 175 by 2?
Both examples seem to be center tapped.
But even still, according to http://www.valvewizard.co.uk/bridge.html
Whether CT or not CT; Vdc =Vrms*1.4

Are typical PT specs for Vrms or Vpeak?

[sluckey]

Why did you multiply 175 by 2?

Because 175 is only half the secondary voltage. 350V is across the entire secondary. The FWB wants to be connected across the entire secondary.

175-0-175 is the same as 350 with CT.

[shooter]

Are typical PT specs for Vrms or Vpeak?

depends, but pick ONE, don't have a datasheet with mix n match, we in the field get grumpy 


most everything in my world is/was Vrms, your 120vac i believe is rms, like 156vac p-p

[scstill]

175-0-175 is the same as 350 with CT.

So in like manner, the example was 345-0-345 so shouldn't this be 690vct*1.4 = 966vdc?

[sluckey]

175-0-175 is the same as 350 with CT.

So in like manner, the example was 345-0-345 so shouldn't this be 690vct*1.4 = 966vdc?

Exactly. And this is why that PT should ***NOT***be used with a FWB. But, if you use a conventional (2 diode) rectifier with center tap connected to ground, only half the secondary (345v) is used at a time, so the dc output is 345*1.414.

This can be confusing at first but just think about it. It will click eventually. 

Read that valve wizard page a few times will help.

[scstill]

Yes it does make sense
I only use tube rectifiers which if full wave are 2 diodes
Thus my calcs are almost always Vdc = 0.707*Vct - Vdrop

[sluckey]

I only use tube rectifiers which if full wave are 2 diodes
Thus my calcs are almost always Vdc = 0.707*Vct - Vdrop

What is Vct? What is Vdrop?

[scstill]

Vct = Voltage center tap = 2*345 in our example
Vdrop = Voltage drop of the rectifier tube; for example ~25v for a 5u4
I have seen these terms used before somewhere.

[66Strat]

The voltage drop of the rectifier is not a constant. The 5U4GB datasheet indicates a voltage drop of 50 volts under 275 milliamp load. The tube will drop less voltage with lower current demands. Refer to the operation characteristics  (page 4) in the attached datasheet.

[scstill]

Thanks for the tip
I Typically use the datasheet value
Not sure where I got the value of 20v as that is not even close :-)

[HotBluePlates]

Yes it does make sense
I only use tube rectifiers which if full wave are 2 diodes
Thus my calcs are almost always Vdc = 0.707*Vct - Vdrop

Check out NEETS Module 6, pages 3-12 through 3-18.  It includes diagrams that are helpful in understanding why the different rectifier circuits behave differently.

     Ultimately, the Conventional Full-Wave rectifier allows only 1/2 of the PT winding to conduct at any moment.

     The Full-Wave Bridge rectifier allows the entire winding to conduct all the time.  Hence, "double voltage output."