Anyone here done a load line calculator for output tubes?

[tubeswell]

I'm looking for a quick'n'lazy way to check the load of a SE 6K6GT in centre-bias class A.

(I know what the ideal load resistance should be for the given plate voltage I have in mind - I just have an OT that provides somewhat less that this load resistance and I want to fiddle with the bias to get it working reasonably)

[Firestorm]

Cathode bias or fixed? If you don't drive it to cut-off, it,'s still Class A. If it cuts off with cathode bias, AB1. You can't get to AB2 with cathode bias (grids won't ever go positive wrt the cathode.)

[Andy Le Blanc]

You can print off a couple Ip-Ep sheets, line them up on top off each other to represent push-pull, and use some thumb tacks and red yarn to set up a load line. you use 1/2 to visualize SE.

[martin manning]

For a SE load line, a little algebra will get the following:

Io = 2 * (Pa max/Za)^0.5 - Vo/Za,

where Io is the idle Ia, Pa max is the max dissipation on the load line, Za is the load impedance, and Vo is the idle Va-k.

You'll have to iterate a bit to close on Vk for a given B+. Use B+ for Va-k to start, get an Io from the equation, then use the triode-connected curves to get the Vk for that Io, subtract that Vk from B+ to get better Va-k and repeat once or twice. Get a value for Rk from the final Io (adjusted up for screen current) and Vk.

[Phil_S]

Merlin has posted some "instructions."
http://valvewizard.co.uk/se.html
http://valvewizard.co.uk/pp.html
I do not have the technical background for this and am still trying to wrap my mind around the SE article, which I managed to copy into a Word document and format to print in only 5 pages.

[tubeswell]

For a SE load line, a little algebra will get the following:

Io = 2 * (Pa max/Za)^2 - Vo/Za,

where Io is the idle Ia, Pa max is the max dissipation on the load line, Za is the load impedance, and Vo is the idle Va-k.

You'll have to iterate a bit to close on Vk for a given B+. Use B+ for Va-k to start, get an Io from the equation, then use the triode-connected curves to get the Vk for that Io, subtract that Vk from B+ to get better Va-k and repeat once or twice. Get a value for Rk from the final Io (adjusted up for screen current) and Vk.

TFT Martin.

The parameters measured from the build at the moment are:

B+ = 339
Za = ~5k (SE) OT
VO = 310.5V (Va = 338, VRk = 27.5) (Rk = 995R)
VRg2 = 338V
Pa = 8.5W (6K6GT)

Plugging these into your formula for the idle plate current I get:

2 x (8.5/5000)^2 – 310.5/5000 =

2 x .0017^2 – .0621 =

.00000578 – .0621 = -.06209422

Funny that the Io is a minus value?

By my calcs, the ideal load resistance for centre-biased Class A with this tube under the idle voltages I have got is

Va/(Pa/Va) = 338(8.5/338) = 13,440R.

The 5k OT I have hooked up at present is not ideal (I also have a 15k OT which is probably better for this situation, but I wanted to try nudging the bias cooler so that the tube can be operating in a 'safe area' of operation (albeit not centre-biased Class A) with the existing OT to see how it sounds first. This is in a 6G16 reverb unit circuit)

I could go through the palava of extrapolating average plate characteristics for a 6K6 load line with a Vg2 of 338, but I can't be fagged (The 6K6GT .pdf datasheet I got off Frank's internerd tube data page is password-protected so I can't even make a copy of the average transfer characteristics into a separate document and annotate it etc, and my printer has run out of paper to do this manually, and I have to wait for the next payday to get some more printer supplies - pathetic ain't it? ) and I was hoping someone had done a pentode load line calculator already.

I could do ear tests with different Rk values to see where it sounds good-enough without red-plating the 6K6, but I was wanting to be a bit more scientific today.

[jjman]

Cathode bias or fixed? If you don't drive it to cut-off, it,'s still Class A. If it cuts off with cathode bias, AB1. You can't get to AB2 with cathode bias (grids won't ever go positive wrt the cathode.)

Class AB on a SE output? When a SE output tube reaches cutoff the output (to the speaker) will clip, thereby ending the range at which class is definable. My understanding is that class is definable only while the output (to the speaker) is not clipping.

AB PP doesn't have to clip (to the speaker) during the moment of cutoff (in one tube) since the other tube is still conducting at that moment during signal.

Whether a SE output clips on the top 1st or on the bottom 1st (saturation vs cutoff) is useful to know and I’ve spent time looking at that on my amps. But I don't think "class" is intended to quantify that for SE.

[martin manning]

The parameters measured from the build at the moment are:

B+ = 339
Za = ~5k (SE) OT
Pa = 8.5W (6K6GT)

With those values the formula Io = 2 * (Pa max/Za)^0.5 - Vo/Za gives Io = 0.0147. (sorry, typo above; its square root, not squared)
At 15mA and 340V I read the triode curves (1953 GE data sheet) and find about 45V Vk so new Va-k (or Vo) is 339-45=294 (to use the triode curves I'm assuming that Va and Vg2 are close to the same, and your data shows only 1V difference)

Back to the formula to get Io = 0.0237. At 25mA and 300V the triode curves show Vk = 31V so new Va-k is 339-31=308

Back to the formula to get Io = 0.0209. At 21mA and 308V the triode curves show Vk ~35V so new Va-k is 339-35=304

Back to the formula to get Io = 0.0217. At 22mA and 304V the triode curves show Vk ~34V so new Va-k is 339-34=305... close enough.

Transfer curves show about 3mA Ig2, so Rk = 34/(21.7+3)=1.38k ohms

[tubeswell]

The parameters measured from the build at the moment are:

B+ = 339
Za = ~5k (SE) OT
Pa = 8.5W (6K6GT)

With those values the formula Io = 2 * (Pa max/Za)^0.5 - Vo/Za gives Io = 0.0147. (sorry, typo above; its square root, not squared)
At 15mA and 340V I read the triode curves (1953 GE data sheet) and find about 45V Vk so new Va-k (or Vo) is 339-45=294 (to use the triode curves I'm assuming that Va and Vg2 are close to the same, and your data shows only 1V difference)

Back to the formula to get Io = 0.0237. At 25mA and 300V the triode curves show Vk = 31V so new Va-k is 339-31=308

Back to the formula to get Io = 0.0209. At 21mA and 308V the triode curves show Vk ~35V so new Va-k is 339-35=304

Back to the formula to get Io = 0.0217. At 22mA and 304V the triode curves show Vk ~34V so new Va-k is 339-34=305... close enough.

Transfer curves show about 3mA Ig2, so Rk = 34/(21.7+3)=1.38k ohms

Wow! TFT

I found a 1367R power resistor in my stash and the results are:


With Rk = 1,367R

Va = 337 , VRk = 29.6V

Ig2 = 3.16mA (Vg2 = 338, Rg2 = 474R dropping 1.5V)

Io = .02165A - .00313A = 0.02299A

Vo = 307.4, Pa diss = 7W


A lot better.

[martin manning]

Pretty good agreement! Note my numbers above would put Pa at Io and Vo at 6.6W: max dissipation of 8.5W on the load line would be at about 207V Va-k.