help calculating bias resistor

[iknowjohnny]

I have a marshall style build that i recenty changed to cathode bias. It was cathode biased when i first built it and i used 270 ohm resistors with no issues. But now that i went back to CB for some reason the tubes are slightly red plating even tho i'm using the same 270 ohm resistors. How can i determine how much more resistance i need to stop the red plating? I have 2 EL34's each with it's own 270 ohm resistor and the voltages are as follows...

plates:423v
across cathode resistors:23v

Why is this happening and what should i change?

[VacuumVoodoo]

plates:423v
across cathode resistors:23v

Why is this happening and what should i change?

Same tubes or another set since first time?

23V and 270R means you have 85mA through the tube (plate current + screen grid current)
Assume 5mA screen grid current, leaves 80mA for plate.
Then you have 423-23=400V between plate and cathode,
your plate dissipation is 400Vx80mA=32W

EL34 absolute max plate dissipation is 25W at idle.....it's ok to run close to 25W in cathode bias as the proper name is "Auto bias" or "Self bias" and it has an inherent stabilizing feedback mechanism in it.

...[leaving out some maths and ogling plate characteristic curves....]

something in the 360R to 470R should set it right.

Or, if possible, lower B+ to some 375V

[iknowjohnny]

Thanks. I figured maybe a couple 330's would be enough since it wasn't red plating badly with 270's The plate was just starting to glow a bit. But i'll grab a few from 330 to 400 and see what works.

[pila]

If you don't really need that much power, tube life will be longer with the dissipation that Vacuum Voodoo mentioned. Just under the red plating current would still be a bit high.

[iknowjohnny]

i think 330 would put me there, (well below RP'ing) but i intend to try close to 400 too and see if there is any tone loss.

[iknowjohnny]

Looks like VacuumVoodoo was right. I have so far tried up to 390R and according to weber's calculator it's still dissipating high at 31 watts. So i suppose it will end up being closer to 470. Luckily it hasn't lost any tone from what i can tell.

[iknowjohnny]

a 460 got it right to 25 watts with 61 ma plate current. So i guess i'll leave it there unless anyone can tell me why not. VV said you can run at the max in cathode bias, no?

[rhinson]

keep in mind in cathode biasing you can run tubes hotter than you think because all the tube "sees" is the plate to cathode difference. so for example if you have 400v on the plates and a 30v cathode voltage the actual plate voltage you use for calculation is 370v. so keep this tidbit in mind when cathode biasing an amp. rh

[Andy Le Blanc]

same amp,....measured plate volts to ground and then to cathode

for two tubes: 434V with a .086A measured across a 1 ohm resistor works out to around 19W static dissapation

the same amp but with plate volts measured to the cathode

for two tubes: 402V with a .086A measured across a 1 ohm resistor works out to around 17W static dissapation

the bias voltage measured at the cathode at around 31V

it not a bad way ensure a couple watts of safety when biasing
you can still add the dissapation of the screen grid too....

[modern]

i would argue that doing calculations with improper measurements is in fact a bad way to ensure a couple of watts safety.

it is better to understand where the power is being dissipated.

[iknowjohnny]

You know more than I, no doubt. But on the other hand i trust weber's calculator and it shows i'm dissipating 25 watts. I'm using 451 ohm cathode resistors and when it was red plating i was using 270's. So i think i'm probably safe in assuming 1)-i'm safe, 2)-i have 50 watt output. But if i'm wrong, tell me how to determine it and i will do so.



[img304]http://i24.photobucket.com/albums/c39/dazco/bias.jpg[/img]

[Jana]

plate dissipation and power output are not the same thing.

You are biasing this amp class A. This is what we are talking about, where you are biasing it. You do not have a 50 watt amp. From past experiences building amps, my guess would be that this amp puts out about 10 to 15 watts. No way is it more than 20.

[iknowjohnny]

Ok, i didn't realize that. Now what should i do to get the full 50 or whatever it is capable of in cathode bias?

EDIT: by the way, my PA is very close to the matchless chieftain, or WAS when i was using 270 ohm reistors. But mine red plated slightly then. The chieftain has 350-350 at the OT and the, tho a tube rectifier. So i think the voltages should be similar since mine is 300-300. And they both have a 32/32uf can and the same choke. In fact i pretty much copied that at first before i went to fixed bias. And of course now i'm back to cathode biased. I'm sure the chieftain must be more than 320 watts, but i want to know why so i can see how it sounds louder and cleaner. but from the chieftain i can't see what would make for the diff.

[skyboltone]

Please note, this explanation is for education only. Do not attempt this at home. You could be easily killed


Ok. Well, first get an 8 ohm 50watt wirewound resistor. Hook it up where the speaker goes. (If that's your output impedance.)

Hook up the dual trace scope across the plates of the output tubes and set the time base to about .5 millisecond. Set the amplitute voltage as high as it goes for now. One probe goes to one ouput plate, its ground to ground, the other probe goes to the other plate.

Put a signal generator set for anywhere between 400 and 1000 hz and about 500MV out and run it into your guitar input.

Set the tone controls to mid, presence off.

Master volume all the way up, input gain all the way down.

Bring up the input gain until you begin to see two sine waves, 180 degrees apart on the scope. Re-set the amplitude control on the scope so that the waves are a nice size to view. Then turn up the amp gain and turn down the scope, turn up the amp gain turn down the scope. Follow?

You'll notice the characteristic shape of the sinewaves at the lower gain settings. As you turn up the amp gain the shape will begin to subtley change and the wave will begin to flatten at the sides (up and down slope). The is called the "crossover" point. As you raise the gain more, the wave form will alter dramatically, indicating clipping.

When you've set the gain so that the sine waves are just before the crossover stage, disconnect on of the probes altogether and put the other probe across the 8 ohm dummy load. You will need to adjust the amplitude control to get a trace that just about fills the screen. Now, look at the amplitude and probe multiplication factors and count the divisions on your screen and find out how much voltage is across the load.

(for this step I use a VTVM which is WAY easier to measure voltage at higer frequency. Your VOM will not indicate an accurate voltage but may be close enough for the girls we go with. )

Now, apply ohms law to your findings.
[IMG:486:480]http://i37.tinypic.com/x5cn5h.jpg[/img]

So P equals E squared divided by R.

That's the way you do it, money for nothin' and the chicks for free.

[iknowjohnny]

I don't have a scope or a 50 watt wire wound.