# Unimportant Details Dept: Bias and 1-ohm cathode resistors

Discussion in 'Shock Brother's DIY Amps' started by King Fan, Dec 5, 2020.

1. ### King FanPoster ExtraordinaireAd Free Member

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I've raved here about how great it is to fit 1-ohm resistors between cathode and ground to directly measure plate current in fixed-bias amps. Except of course you're actually measuring *cathode* current, which includes both plate and screen current.

Luckily, of course, the difference is small, and as noted the other day in a nearby thread, it's in the *safe* direction -- the actual bias will be a little cooler than you calculate. And we should bias by ear -- so if you end up at 60% when you think you're at 65%, it still *sounds* good and the exact number doesn't matter.

But Rob pointed out there he lists his formulas on his bias calculator, and I'm the first to say I love that page and use it all the time. He also (typical Rob generosity) has a link to his Excel spreadsheet, which (typical Rob attention to detail) must have been a ton of work.

So, just for curiosity, I took the 5.5% (arbitrary/approximate) offset he uses elsewhere for screen current and applied it to the 'plate current' section of the calculator. I simplified mine since he has a ton of power features, like a lookup table for dozens of tube types.

How big is the difference in terms of scale? Here're some made-up numbers in my baby version that includes the offset:

So a 5.5% offset to the current measurement results in a *less than 5%* difference in terms of percent dissipation (though note this is simply a result of comparing proportions; in fact 8.1/7.7 = 1.05). So folks smarter than me wouldn't need to do all this calculating.

In any case, I can live with that -- at least if 5.5% is a reasonable estimate of the offset for screen current.

Back at the ranch, Rob's bias method 'Using Cathode Resistor Voltage Drop' *includes* the 5.5% offset for screen current; a little thought tells us we can use this method with 1-ohm resistors, too, if we scale the units and recall we're now measuring for just one tube. In fact his "if triode" calculation does what I did above and shows the 'uncorrected' bias value...

2. ### LowerleftcoastFriend of Leo's

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When rounding off 7.6545 watts to 7.7 watts and when rounding off 21.2625mA to 21.3mA the end result may be *less than 5%* difference.
When the figures are not rounded off the 5.5% figure holds true.

7.6545w / 12w = 63.7875%
67.5 - 5.5% = 63.7875%

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3. ### King FanPoster ExtraordinaireAd Free Member

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Right: Good old rounding error.

But, LLC, help me out. Is 5.5% some accurate and universal figure? I assumed the actual difference varied from circuit to circuit and tube to tube, and that 5.5% was just a 'useful, typical estimate.'

At this point, I recalled I have an old Eurotubes bias probe that directly measures actual plate current. On my 5G9, the 1R-derived cathode current was 22mA, and measured plate current was 20mA, so at least for this amp, the screen current looks like 9–10% of the total cathode current. (Though, even so, my calculated percent max dissipation only drops from 67% to 61% -- not a huge difference.)

Or am I just adding good old measurement error (and imprecision) to the rounding error problem?

EDIT: I shoulda said I'm grateful to LLC for reminding me in the first place that cathode current is *not* exactly the same as plate current.

4. ### LowerleftcoastFriend of Leo's

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We must conclude that Rob's calculator is used for many different amps and all the tubes on his list and all the tubes such as Russian 13W tubes which are not on the list. To err on the safe side a 5.5% estimate will cover Rob's fanny (most of the time).
As pointed out in the other thread it was around 18% difference and as you noted the 5G9 about 10% difference.
So, no 5.5% is not a universal number but it keeps Rob out of trouble.

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5. ### King FanPoster ExtraordinaireAd Free Member

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Thanks! And good, I was gonna be weirded out if 5.5% was some kinda tube-amp equivalent of Planck's constant...

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6. ### LowerleftcoastFriend of Leo's

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Now I am waiting for Rob to chime in and say there is a 5.5% numeral from the universe.

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7. ### King FanPoster ExtraordinaireAd Free Member

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Yeah, "Rob's constant." Hey, at least it's not 6.02*10^23...

Actually, all my circular math aside, the fun thing I learned here is when I measure current (as mV=mA) across my 1-ohm resistors, I could use Rob's "Cathode Resistor Voltage Drop" and see the both the 'corrected' (estimated plate) and uncorrected results (in the 'if triode' last line).

Notes:
1. I already did the first section, with tube type and plate voltage (in fixed bias, plate V = plate-to-cathode V)
2. Then, '1' tube, current in volts (mV/1000), and '1' ohm for your resistor....

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8. ### Cruisin HomeTele-Meister

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..of course one can also consider the tolerance of the 1 ohm resistor they are using! haha

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9. ### Doctorx33Tele-Afflicted

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Great explanation. I understand all of it except the part after “I've raved here

10. ### King FanPoster ExtraordinaireAd Free Member

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You're right. I've always used 1R 1% ½ W Dales -- 1% is just nice, and the 1/2W makes them big enough to, like, see -- and clip your grabbers on. But lets say you used 5% tolerance resistors and measured 20.9 or 23.1 instead of 22 -- a full ±5% error. The result in % max dissipation would be 63% to 69% instead of 66% -- not enough to likely change your 'safety zone' calculations -- but hey not as precise as it could be.

And of course if you happened to have only 2ohm or 10ohm resistors, you could do the Ohm math -- or just plug into Rob's cathode resistor calculator above.

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