bias: cathode current vs plate current

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Gizzer
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Re: bias: cathode current vs plate current

Post by Gizzer »

It works for me to see it,
but after saving it (png pix), we get this black spreadsheet when you open it.
Roe
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Re: bias: cathode current vs plate current

Post by Roe »

-It is easy to measure screen current if you have screen resistors. a 7v drop over a 1k resistor, for instance, corresponds to 7mA
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martin manning
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Re: bias: cathode current vs plate current

Post by martin manning »

Ok, I’ll see if I can fix it. The point is the correction I proposed above works fine, and you can get that info from the tube data sheets. Beam tubes like 6L6 and 6V6 will have smaller screen current, about 5% of plate current, and therefore a smaller correction factor.
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pompeiisneaks
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Re: bias: cathode current vs plate current

Post by pompeiisneaks »

odd when I was reading it last night at home, I could see it just fine, today from work I can't.

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martin manning
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Re: bias: cathode current vs plate current

Post by martin manning »

Fixed, I think. See original post.
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pompeiisneaks
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Re: bias: cathode current vs plate current

Post by pompeiisneaks »

Looks fine to me now.

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tubeswell
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Re: bias: cathode current vs plate current

Post by tubeswell »

martin manning wrote: Thu Sep 20, 2018 1:18 am That's rubbish. Here's some measured data backing up the correction I proposed above:
Do you know how many tubes were sampled and which brand? I've got about 2-dozen Soviet EL34G in my stash and the plate, screen and bias voltages and currents are all over the place :lol:
He who dies with the most tubes... wins
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martin manning
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Re: bias: cathode current vs plate current

Post by martin manning »

Exactly one. The point I’m making is that the screen current as a fraction of cathode current remains relatively constant. Even if it is off-spec, it isn’t going to make much difference if you just assume a nominal value. And that value is not the max screen dissipation. As mentioned above, if you really want to know the actual plate dissipation, measure the screen current and subtract it out.
Gizzer
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Re: bias: cathode current vs plate current

Post by Gizzer »

Following Martin's 15% screen current, (and for a maximum of 70% of power dissipation)
425V and 25W+3.75 = 28,75
.7 X 28.75W= 20.1 Watts
now for the bias: 20.1/425= 47.3mA

power wattage would be: .0473 X 425 = 20.1 Watts

If you now take the Bias King maximum formula (63.6% maximum), you get a close similar value in Watts
.636 X 33W= 20.98 Watts
bias at 63.6% (Bias King maximum formula)
we got for the bias: 20.98/425= 49.3mA

power wattage would be: .0493 X 425 = 20.95 Watts

Martin, thanks for all this research
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martin manning
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Re: bias: cathode current vs plate current

Post by martin manning »

Almost... in both cases the actual Pa is about 15% less.
Case 1: 20.1/1.15 = 17.5W (70% Pa max)
Case 2: 29.95/1.15 = 18.2W (73% Pa max)

Close, but the Bias meter method is wonky, and doesn’t let you specify plate dissipation directly.
brewdude
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Re: bias: cathode current vs plate current

Post by brewdude »

I’ve never understood why amps aren’t built with 1R resistors at the plates to measure plate dissipation directly.
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M Fowler
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Re: bias: cathode current vs plate current

Post by M Fowler »

Thanks Martin works fine for me now.

Mark
Gizzer
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Re: bias: cathode current vs plate current

Post by Gizzer »

Thank you so very much Martin Manning

So if I go back to the first post and redo the "right" formula, it goes like this :

Exemple:
With a plate voltage of 400Volts (411Volts unloaded)

- hot biasing with plate current : (25W/400) X .7 (70%) = 43.7mA ==> maximum bias (70%) (reading on a plate current bias probe)

- hot biasing with cathode current :25W X 1.15= 28.75W : (28.75W/400) X .7 (70%) = 50.31mV into a 1ohm resistor = 50.31mA (V=RI) ( on a cathode current probe)

now we remove the screen dissipation (15%) : 50.31/1.15= 43.75mV through 1 ohm resistor = 43.75mA => current to calculate the power ( P= V X I )
400 V X 0.04375 A = 17.5 Watts

Voilà ! :D
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martin manning
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Re: bias: cathode current vs plate current

Post by martin manning »

Yes, and of course the expected dissipation is just the desired percentage of Pa max times rated Pa max.

Here are data sheet values for the EL34 and a few more tube types. EL84 is a true pentode like EL34; the others are beam types.

Setting % Pa max using cathode current (Class AB1) including screen current correction:
Ik = (%Pa max desired x Pa max x Ig2 Adj)/Va
Type, Pa max (W), Ia0 (mA), Ig20 (mA), Ig2 Adj = (Ia+Ig2)/Ia
EL84 12, 18, 2, 1.11
6V6GT 12, 35, 2, 1.06
EL34 25, 30, 4, 1.13
6L6GC 30, 58, 2.8, 1.05
6550 42, 50, 2.5, 1.05
MPManning 2018.09.23
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pompeiisneaks
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Re: bias: cathode current vs plate current

Post by pompeiisneaks »

brewdude wrote: Sat Sep 22, 2018 4:05 am I’ve never understood why amps aren’t built with 1R resistors at the plates to measure plate dissipation directly.
Due to the high voltage, it becomes expensive, you need a high quality 1% resistor. Depending on the tube, let's say it's a 6L6GC at 350VDC. That means at 70% diss you need 60mA so 60mA times 350VDC = 21 Watts. It's best to go up, so 25 watt 1% resistors are EXPENSIVE!

https://www.mouser.com/ProductDetail/AR ... 2buw%3d%3d

this ones 3 bucks, most resistors are like .20 or so.

~Phil
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