Power XFMR/Output XFMR relationship

[Dealbreaker]

Hello folks, I posted this in the "Introduce Yourself" area, but its actually a technical question.  It seems so important and fundamental to tube amp architecture, and yet I cannot find an answer.  Hopefully someone out there can spin this in such a way my brain can capture it. 

Been lurking for some time, and I just want to send my thanks and gratitude to this site for freely passing out knowledge and real world experience to folks like me who have a passion for the tube sound, but not much knowledge of what's "under the hood."
I have a smattering of electronic know-how, I would say above average in home AC wiring, decent solder skills, can navigate a schematic (at least know what's what, and can usually trace a signal path), but there are a couple things I cannot get my head around.  And maybe this is a separate thread in a more focused section?  But here it is:  Why does every tube amp send the hottest portion of B+ voltage to the output transformer?  I realize that this voltage goes to the output tubes, but surely they could be powered with a different discrete path?  What is so vital about connecting the two transformers?  I ask this knowing it has to be so fundamental to folks with actual design knowledge that I'll probably be asked to come back with a harder question next time...But yeah, this keeps me up at night. Why can't the high voltage/low current audio signal leaving the output tubes just go straight to the output transformer?  Why do we need the hot leg of DC going there too?
Thanks in advance.  I'll hang up and listen for your reply.
Thanks!

[PRR]

The "output" tubes do not MAKE power. They modulate the DC power from the power supply.

[dwinstonwood]

One thing that might be counter-intuitive about vacuum tube circuits is that the DC B+ is wired to the tubes' plates (usually through plate load resistors), but the AC (guitar) signal "travels out of the plates," through a DC-blocking coupling cap, and on to the next tube's grid, or in the case of power tubes to the output transformer's primary leads. The DC and AC sort of ignore each other, in a sense; they can both travel along the same piece of wire while doing their own thing.

[PRR]

> "travels out of the plates," through a DC-blocking coupling cap,

But he is asking about a transformer coupled circuit. Introducing the capacitor coupled form too soon may cloud the issue.

[dwinstonwood]

Right, sorry! Thanks PRR. 

[Dealbreaker]

Thanks guys for the replies.  I did struggle with the conundrum of DC and audio signal sharing the same path early on...the same is true in transistor world, so understanding the purpose of the blocking caps was a key moment in my education.  Its the lack of formal schooling that I am trying to make up for!  Thankfully, the internet is an incredible source of knowledge...unfortunately, I am learning things out of order, so topics like the transformer question have stumped me.
PRR, your response has me thinking...please help steer me...So B+ voltage feeding the output tubes is used as the "fuel tank" and is drawn out as the amplitude of the preamp stages increases. If that is correct, then I understand the principle. The same would hold true with the preamp tubes as well, right?  So why do we need to feed the DC leg from the PT to an input leg of the OT?  The preamp B+ voltage comes from discrete wires post filtering.  Unless there is some benefit to sending that "bumpy" DC to the OT? 
I guess I'm still not getting it.  I'll pull up a random schematic and highlight the path that I am talking about.  Every conventional tube amp I have seen uses this architecture, so it must be as fundamental as adding water to cement... I'm just missing it somehow...
Thanks everyone for your patience.  Hopefully, I am not the only one to struggle with this.  Standby for a pic.

[dwinstonwood]

So why do we need to feed the DC leg from the PT to an input leg of the OT?  The preamp B+ voltage comes from discrete wires post filtering.  Unless there is some benefit to sending that "bumpy" DC to the OT? 

The OPT center tap can come from a more filtered node. This is a Vox AC15 I built. The OPT CT connects to "A" in my schematic. But, you're right, in most guitar amps the OPT CT is fed before the choke at the first filter cap node. The way I understand it, in push-pull amps the DC ripple entering the power tube plates gets cancelled out, so more filtering isn't needed there. And chokes that only see the screen current can be much smaller and less expensive. The choke in my schematic is as large as some 15W OPT's, and more expensive that the typical little 50mA chokes used in lots of Fenders. So, cost drives a lot of design decisions.

[PRR]

The power stage does not need real "clean" power. It needs big power. Minimal filtering of the power flow path is most efficient.

However you will struggle to find a PT directly to an OT. There is a rectifier in the way and a capacitor off to the side. So you may be a victim of over-simplification.

[Dealbreaker]

Hopefully this describes better what I am talking about.  Please chime in.  I welcome all comments!

[Dealbreaker]

Forgot something...

[Dealbreaker]

However you will struggle to find a PT directly to an OT. There is a rectifier in the way and a capacitor off to the side. So you may be a victim of over-simplification.

Quite right, sir.  The rectified output was what I meant to say is what connects to the OTP.

[tubeswell]

... Why does every tube amp send the hottest portion of B+ voltage to the output transformer?  ...

Its not so much that its 'sending B+ to the OT' Primary - it's sending DC to the output tubes via the 'tube plate end(s)' of the OT primary winding. The tubes do the work of moderating/increasing/decreasing current flow through the OT primary, but the 'original' current is 'sourced' from the B+ via the OT Primary. The tubes are electron 'valves', kinda like tap valves - the valve controls how many electrons 'flow' at any given point in time. This 'alteration in flow' is the signal. The B+ current is supplied to the tubes via the OT primary, because the OT primary forms part of the 'reactive load' for the tubes. This 'reactive load' works by electromagnetic induction in the OT, in which the alteration in current/signal current through the OT primary gets inducted (under signal conditions*) and transformed into a higher current/lower voltage in the OT secondary winding which is connected to the speaker.


* Electromagnetic Induction (which was discovered by Michael Faraday, and put into math by James Clerk Maxwell)  only works when a signal is present in the OT. It relies on 'change in current over time'. When the current is constant (i.e. no signal), induction cannot occur.

[Dealbreaker]

Looking at Tubeswell's response, that all makes sense to me....Except, couldn't the same thing be accomplished powering the tubes along the Blue lines in my drawing?  You hit on all the aspects of output tube/OP XFMR interaction, but my question is why can't the tubes be powered with discrete wires to the same plate pins?  Why does the power need to pass through the OP XFMR?

[sluckey]

Except, couldn't the same thing be accomplished powering the tubes along the Blue lines in my drawing?

No. The tube must have a plate load connected to it. Otherwise the tube may melt down. Your blue line puts a short across the OT.

Why does the power need to pass through the OP XFMR?

The tube plate to cathode acts like a variable resistor. It's resistance changes due to an audio signal on the control grid. When the audio signal swings positive the plate to cathode resistance decreases and allows more current to flow from cathode to plate. When the audio signal swings negative the plate to cathode resistance increases and allows less current to flow. IOW the control grid controls the current through the tube. The source of the plate current is the B+ supply.

You must have a resistance in series with this current flow to prevent the tube from burning up. This resistance (plate load, in this case the OT primary) develops a voltage drop that is proportional to the audio signal at the control grid, but much larger. This changing dc current (definition of AC current) flows through the primary of the OT and is coupled to the secondary. An OT is just a big step down transformer. The big ac signal on the primary is stepped down to a small voltage but high current. The speaker needs a lot of current to operate. This is just how it works.

Now your blue line bypasses the OT primary and puts the B+ directly to the tube plate. Since current no longer flows through the OT primary, no signal will be sent to the speaker. Also the tube may melt down because there is no longer any series resistance to limit current flow.

[dwinstonwood]

sluckey, that's a great explanation. It clears up an oversight/blindspot that has confused me.
I know that transformers operate with, and pass, AC. And the rectified voltage from the reservoir is DC. It just never clicked - because it's too obvious - that the signal on the control grid causes the DC B+ connected to the OPT primaries to behave like AC by swinging it up and down.
I don't know why that was hard to see. Probably, because I thought of DC and AC as two distinct things, when their difference is really just a matter of whether or not they swing positive and negative. Thanks.

[Dealbreaker]

OK, now I am on the page.  Thank you sluckey.  When I think about how the preamp tubes function, this makes sense.  They get their B+ voltage and use that to create a high voltage (and low current) audio signal to feed to the next gain stage.  The mechanics of that operation is more or less the same as you describe for the output tubes.  The Output Tube(s) next gain stage is effectively the Output XFMR, so it stands to reason this is the only way to do both:  Power the tube and create a suitable output condition for the tube to function properly.  I have been struggling with this problem for quite a while, and now it finally makes sense.  Thank you all for the help.  Its a great community here!