Voltage Influence on Impedance

[SilverFox]

From the post regarding the use of a 4.3 K output transformer:

A pair of 6L6 running around 500V B+ are almost perfect.

If so, in what way does the B+ relate to the impedance of an output circuit? Does impedance change with voltage in an output circuit?

Silverfox.

[Jana]

Load lines--

http://www.valvewizard.co.uk/se.html

http://www.valvewizard.co.uk/pp.html

[pdf64]

Have a play with this http://bmamps.com/Tech_tds.html and it may be apparent that as VB+ is increased, to keep average (plate) dissipation within the limiting value, p-p load impedance must likewise increase.

[EtherealWidow]

As PDF is alluding to, is not that a voltage change influences the impedance, rather, it's that you're using the B+ and the given impedance to determine if the plate dissipation is within the acceptable range. In this case:

W=E² / R

500² / 4300 = 58W

Just about perfect for a pair of 6L6's

[JMFahey]

Tubes are basically current limited, so if you rise voltage you must rise impedance big way to compensate.

4300 ohms plate to plate at above 500V seems way too low.

As said above, you can calculate optimum load impedance, real power available, etc. , by plotting load lines on the tube plate current and voltage curves supplied in the datasheet.

To avoid retyping from 0, here's my more detailed explanation at our friends' TGP Forum:

http://www.thegearpage.net/board/showth ... ?t=1524165

EDIT: to avoid zigzagging between 2 pages, I just cut and pasted here:

1) to begin with, limiting values are exactly that, what will destroy the tube.
Which means reaching one is just bad enough, don't even dream to combine 2 or 3
Or "don't even dream of applying 800V, passing 150mA and surpassing 25W dissipation, all at once"

So:
* above 800V it will probably arc or short inside
* above 25W it will redplate
* above 150mA (continuous) it won't last much.

there's some extra values, characterized by a "sub 0" sign, which they don't explain in that datasheet, such as Ua0 (plate)=2000V and Ug20 (screen) = 800v.
It might mean maximum voltage with absolute zero current passing or internal arcing voltage, between that element and others nearby, even with tube off/cold (filaments off) .

Not sure about which of these but anyway those are not "working" conditions by any means so we won't even approach them.

2) Back to the design.

They publish a graph showing the relationship between 4 parameters:

* Plate current Ia
* Control grid voltage Ug1
* Plate voltage Ua
* Dissipation. Wa

[img:597:562]http://i57.tinypic.com/v40w76.gif[/img]

Read it knowing what each curve means, now things to make sense and fit together.

What can I "read" here?

One example, suppose I have an EL34 plugged in a socket on my workbench, apply, say, 120V to its plate, and -15V to its grid ... how much current will pass through its plate?

Example in violet:
First start from 0 V 0mA end of graph, then go to 120V on the horizontal line (voltage steps shown are 20V each, so we have 0 .. 20 ..40 .. ........ 120V which is the 6th vertical line) , the rise it until we meet the line labelled -15V, then turn left to reach the plate current, the mA scale and read it.
Current steps are 20mA each so since it's the 4th horizontal line it means 80mA .

Same for any plate votage/current/grid voltage combination.

Why a graph and not a formula?

For once, the formula, which exists, is very complex and confusing for most readers, although perfectly fine for circuit simulators .

And the beauty of the graph is that it's "a map" which shows "all at once".

FWIW I'm an old style Engineer which started University in 1969 and way back then, using graphs was the only Poker game in town.

[SilverFox]

I may have understood this before to some degree. If what I'm reading here on the replys is: As B+ goes up you need a different impedance value for the output transformer to keep the load matched; that makes sense. Although I hadn't given it that direct consideration in the past.

The objective is to keep the plate current within the design limits, so as the voltage goes up perhaps use a different output tap or get a different transformer. Importantly you would not simply add a resistor to the output circuit- Well you couldn't since that would create an LR circuit and change the tone to something bad.

What I thought I was reading- By changing the B+ you change the impedance of the circuit.

Rather, if you change the B+ by too much, change the impedance to keep the plate load matched.

Thanks for the info and let me know if I've still got it wrong.

Silverfox.

[pdf64]

Rather, if you change the B+ by too much, change the impedance to keep the plate load matched

Yes, but matched is a funny word, which may be taken as 'the load impedance matching the tube's impedance'.
Try 'reasonable', 'appropriate', 'suitable' etc, rather than matched.
Also it may be helpful to think of it as more of a continuem rather than discrete steps, even though we're a bit limited with implementation by what's available.

[EtherealWidow]

"Matched" to me has more to do with efficiently mating source impedance to load impedance for optimum signal transfer. For example, a cathode follower going to a tone stack, or loading a guitar pickup with a 1M grid leak.

It's not necessarily to keep plate current within design limits, but plate WATTAGE under signal conditions and making sure that you're getting as much power as you can from a given tube(s).

Also, adding a resistor would not only probably create some weird tones with an LR filter, but it would further diminish your overall power. Some of that signal would be dropped across the resistor rather than being transferred to the voice coil, making sound.

[roberto]

Well, I'm late to the party, anyway you already have had all the info.