Am I hot or cold

[1969mako]

Quick question concerning Biasing. I have a Blackface Vibrochamp and I am running the 6V6 at 28 miliamps. I replaced the 470 ohm resistor with a 1k due to voltage these days are high on the plug.

My question, is 28 mili amps to cold, to hot or just right for the 6v6 in this amp

[sluckey]

Not enough info. What's on the plate?
 

[Geezer]

Not enough info. What's on the plate?
 

IOW, what is the plate voltage on the 6V6? (pin 3) 

[1969mako]

Sorry, 380 volts

[sluckey]

Your porridge is a little cool.

[1969mako]

LOL... Cute....

You have me intrigued, how did you figure that out???
I am always trying to learn. Do you have a formula?
What should I shoot for for a reading? In the thirties?
So I am assuming I should drop the resistor from 1k to something lower?
Any info is very much appreciated…

[sluckey]

You have me intrigued, how did you figure that out???

IAW the data sheets, 6V6 PA_max is 14 watts. Many people like to run single ended amps at 90% PA_max (12.6W for your 6V6). Yours is running about 9.9 watts (70%) according to your numbers. Maybe a little cool. But if you like how it sounds, let it be.

I assumed your current figure was cathode current? How did you determine you have 28ma? Based on my assumption and your numbers, 28ma thru a 1K resistor will drop 28v. That's your cathode voltage. Your plate is at 380v but we're really interested in plate to cathode voltage, 380-28=352v. 352v x 28ma = 9.9 watts.

I replaced the 470 ohm resistor with a 1k due to voltage these days are high on the plug.

How high? What are your numbers with the original 470Ω? I would not change the cathode resistor just because wall voltage is high.

[PRR]

Were you burning-up a 6V6 a week? If not, put it back to 470.

[1969mako]

Thanks for the explanation and the time you have taken.

I determined current using a Bias Probe. This is a new amp to me and I have no idea what it ran at with the 470 resistor in there because when I bought the amp, I found the 25 uf cap was bad and when I measured the 470 ohm, it read 760 ohms. I went to a local amp guy for parts and he handed me a 25uf 50 volt cap and a  1k resistor and told me to run that because of outlet voltage being higher today than in the 60’s. Not knowing enough about it yet, it made sense to me.

I am thinking at this point, I should buy a few resistors, a 470 ohm and maybe a 650 ohm or something like that…

Thoughts???

[sluckey]

I am thinking at this point, I should buy a few resistors, a 470 ohm and maybe a 650 ohm or something like that…

Sounds like a good experiment. Measure voltages and do the calculations for each value resistor. And use your ears too. Some like it "hot". Some like it "cold". Most cannot agree on what's "just right". I'm thinking you'll like the 470Ω.

[1969mako]

So I put everything back together with the 1k in there until I buy a few more resistors. I wish the amp had a little more clean headroom. I am assuming that if I put the 470 ohm in there, the tubes will obviously run hotter. I am also assuming that it will stay cleaner longer, am I right on my thinking?
Thanks….

[plexi50]

Use the 470 ohm. It will be fine as it is. These amps have a nice breakup. If im thinking what your thinking then no your not going to get a ton of headroom from this amp.  But plenty enough.

[1969mako]

Perect, thanks for everyting.....

[catnine]

When I got the 73 SF Champ I had and after I removed the Torres scream machine mod I checked the 6V6 bias. The plate was 374 VDC and 22.59 VDC across a measured 505 ohm cathode resister that ran the 6V6 at 44mA which was 16.71 watts . I was told at the time it was way to hot and after a new cap can and eletrolytics I installed a 1K ohm cathode resister 29.0 mA 410 VDC plate and 29 VDC across measured 106K ohm =11.82 watts . It sounded fine yet I ended up with a 680 ohm 5 watt which brought it close to 14 watts . The amp still had what looked like the original tubes and certainly had the original cap can and resisters ands all other eletrolytic caps , can't say how may years it was run at 16.71 watts. I have written down my walll voltage and it was at this time 115 VAC, this was 2005. I have read and don't recall where that the 470 ohm was proper for a tweed champ because of the lower powered PT but was never changed when they built the BF and SF champs. As I said my wall voltage was 115 VAC and never goes above 119.5 VAC.   The fender schem does not state the actual AC line voltage . Who knows what is was at any given time this is assumed. The fender schem offeres up 14.2 watts if you go by 19.0VDC across 470 ohm and a 350 Plate . I personally would run the 6V6 at not much more than 14 watts. Above that will lower the plate voltage but it makes the amp to grainy sounding but if that's what you are after and tubes are not an issue than go for it. On another note , no one ever suggests over biasing any push/pull 6V6 amp . If you look at the voltages on the Bronco Ab764 the 6V6 is running close to 17 watts and the Vibro Champ near 16 watts , these are based of the voltages listed on the fender schematics. Same cathode reister value and both have 342 VDC plate only diff is the cathode voltage , 24 VDC across a 470 ohm for the Bronco and 21 VDC across a 470 ohm cathode resister and this is when the line was said to be 115 VAC .

[1969mako]

Thanks for the info…
I think part of my problem also is I am running an electro-voice speaker that looks heavily doped (sounds kinda muddy) while I wait for Ted Weber to re-cone the original Oxford speaker.
I think once I make the resistor change and get the Oxford back in there, I will be happier.

I also think I am use to my 72 Deluxe Reverb. Man, does that amp sound sweet….

[HotBluePlates]

... while I wait for Ted Weber to re-cone the original Oxford speaker.

Well, WeberVST, anyway.

If you wait for Ted to recone it, you'll wait a mighty long time. He passed away not too terribly long ago.

[1969mako]

Not him personally but his company. I know he passed, bought many speakers from them

[phsyconoodler]

One mistake lots of people make is measuring the plate voltage to ground.That's wrong.Measure from the cathode(pin 8) to the plate(pin 3).
  So Catnine,if you measured it that wy then you are right,but if not,subtract 22.59v from the plate voltage and then it's right.

[catnine]

One mistake lots of people make is measuring the plate voltage to ground.That's wrong.Measure from the cathode(pin 8) to the plate(pin 3).
  So Catnine,if you measured it that wy then you are right,but if not,subtract 22.59v from the plate voltage and then it's right.

 what i do is divide the cathode votage by the cathode resister value to  get the mA's then multipy the plate measured to ground by the mA to get the watts. So 22.50 VDC cathode to ground divided by 505 ohm  cathode resister = 44.6 mA x 374 plate = 16.67 watts or IPD. This is the way  was taught long ago . If i'm wrong then please let me know. If i use your calculations and then mine 28vdc divided by 1k = 28 mA x 380 plate measured to ground 28 mA x 380 vdc plate = 10.64 watt.

[sluckey]

Yes you're wrong. Read back thru this thread to see the correct way.

[catnine]

Yes you're wrong. Read back thru this thread to see the correct way.

 here is an old post and I spoke to this fellow Vintage Jon/ Jon Bessent . This is just the one post then I will include the link where you phsyconoodler
were also on it at the time.
 

VintageJon03-14-2006, 07:24 PM
Here tis again.

Remove the 12AX7,(s), to remove any plate current being pulled.
(Thus only the screen current will be indicated for the 6V6...)

Measure the actual value of the 1K resistor coming from pin 8 on the rectifier. Write it down as Rs.

Measure the 6V6 cathode resistor, write it down as Rk.

Measure the voltage across the 1K coming from the rectifier pin 8.
Divide this Voltage by the resistors actual Resistance. Write it down as this is the 6V6's Screen current. Let's call it Is.

Turn amp on and let it warm up. Measure voltage across Rk. Write down as Vrk.

Measure plate voltage and write it down as Vp.

Get yer calculator and figure these:

Vrk / Rk = Ik or cathode current.

Ik - Is = Ip or plate current.

Ip X Vp = IPD or Idle Plate Dissipation in Watts. This should not exceed 14W.
 
 Now the only difference in my calculations and Jons is I did not remove the pre amp tubes and he measured the screen current which at idle is not worth dealing with . Read the entire thing . I am not debating you here at all this is just where I got my info from . The resister on pin 8 of the rectifier Jons talks about is the 1k first dropping resister to calculate screen current . I just want to get this correct and now I am getting something different.

This is the link to the entire thread .
 
http://www.thegearpage.net/board/archive/index.php/t-138465.html

[catnine]

 Then there is this from webers site .

  Calculate Plate Dissipation Based On Plate Voltage And Cathode Current Readings.
In these calculations, 5% of the cathode current is assumed to be screen current. This calculator is for a single tube. 

Plate voltage
DC voltage measured between plate and cathode. (Actual Plate Voltage)
Example:
Octal - pins 3 and 8
9 pin - pins 7 and 3
Cathode Current (milliamps)
Displays Approximate Plate Current After Calculation
Plate Dissipation (Watts)   
Note: For calculating plate dissipation in cathode biased circuits, subtract the measured cathode voltage from the measured plate voltage and use that as the plate voltage in the calculator above. 


 link here http://www.webervst.com/tubes1/calcbias.htm

[sluckey]

Then there is this from webers site .

  Calculate Plate Dissipation Based On Plate Voltage And Cathode Current Readings.
In these calculations, 5% of the cathode current is assumed to be screen current. This calculator is for a single tube. 

Plate voltage
DC voltage measured between plate and cathode. (Actual Plate Voltage)
Example:
Octal - pins 3 and 8
9 pin - pins 7 and 3
Cathode Current (milliamps)
Displays Approximate Plate Current After Calculation
Plate Dissipation (Watts)   
Note: For calculating plate dissipation in cathode biased circuits, subtract the measured cathode voltage from the measured plate voltage and use that as the plate voltage in the calculator above. 


 link here http://www.webervst.com/tubes1/calcbias.htm

Weber is right. Phsyco is right. You and Jon are wrong. I used red text to indicate the part that you are missing.

[catnine]

Then there is this from webers site .

  Calculate Plate Dissipation Based On Plate Voltage And Cathode Current Readings.
In these calculations, 5% of the cathode current is assumed to be screen current. This calculator is for a single tube.  

Plate voltage
DC voltage measured between plate and cathode. (Actual Plate Voltage)
Example:
Octal - pins 3 and 8
9 pin - pins 7 and 3
Cathode Current (milliamps)
Displays Approximate Plate Current After Calculation
Plate Dissipation (Watts)  
Note: For calculating plate dissipation in cathode biased circuits, subtract the measured cathode voltage from the measured plate voltage and use that as the plate voltage in the calculator above.  


 link here http://www.webervst.com/tubes1/calcbias.htm

Weber is right. Phsyco is right. You and Jon are wrong. I used red text to indicate the part that you are missing.

 That's why I included the weber site info because it conflicts what Jon said and what he told me to do , yes the areas you put in red are why I posted webers link. Here is the thing , Jon worked on hundreds of amp for many years and I am not saying this to say anyone here is wrong or has bad info , it's just at this point I do personally find it difficult to believe Jon was wrong. No matter how you cut it you can still get the mA by dividing the cathode voltage by the cathode resister value is this not correct ? And to subtract the cathode voltage from the plate to ground reading or measure across pin 3 and 8 of the 6V6 will still render the same "actual plate voltage" so in the end I am running about 1 watt less at least on my champ .

 I am not missing anything just getting two types on info and now wonder where the info you offer or weber offers really came from or was arrived at. That's my question. not to add debate it's just you do have to take the cathode voltage reading from somewhere in order to calculate the rest. This is all to say I cannot take the 680 ohm cathode reister out on my champ and install a 500 ohm in it's place and know what the actual will be until I do it and take the readings yet I do know more or less that the end result will be more than 14 watts.

[sluckey]

The whole point of this calculation exercise is to determine the static or idle power dissipated by the TUBE ANODE. When you measure plate voltage to ground, you end up with voltage across the tube (plate to cathode) PLUS the voltage across the cathode resistor. So, when you calculate power using the plate to ground voltage, you're really calculating the power dissipated by the tube PLUS the cathode resistor. We're only concerned with exceeding the power rating of the tube, NOT the power dissipated by the resistor. And that's why you must subtract the cathode voltage from the plate to ground voltage to get the true plate voltage so you can calculate the true power dissipated by the tube.

And I know that to be absolutely accurate, we must consider the screen current too, but we don't seem to be in disagreement about the screen factor.

[catnine]

The whole point of this calculation exercise is to determine the static or idle power dissipated by the TUBE ANODE. When you measure plate voltage to ground, you end up with voltage across the tube (plate to cathode) PLUS the voltage across the cathode resistor. So, when you calculate power using the plate to ground voltage, you're really calculating the power dissipated by the tube PLUS the cathode resistor. We're only concerned with exceeding the power rating of the tube, NOT the power dissipated by the resistor. And that's why you must subtract the cathode voltage from the plate to ground voltage to get the true plate voltage so you can calculate the true power dissipated by the tube.

And I know that to be absolutely accurate, we must consider the screen current too, but we don't seem to be in disagreement about the screen factor.

 I do understand what you are saying in words but not theory. I know the cathode voltage controls the current to the plate which the plate is the output part of the tube and the control grid and screen , it's all do to the flow and attraction of electrons yet I don't understand how the cathode voltage is an actual voltage that can become a direct connect with the plate to be added to it.

[sluckey]

Do it your way. You wont hurt the amp and since you're running less plate dissipation than you think, your champ 6v6 will last longer.

[catnine]

Do it your way. You wont hurt the amp and since you're running less plate dissipation than you think, your champ 6v6 will last longer.

 My line voltage varies from 115VAC to 119.7 VAC and when I measure my way @ 119.7 my amp is just above 14 watts so with that it is closer to 13 watts down to 12.1 watts which is fine by me. But thanks for your and others time on my diving into this thread and the added info.

[sluckey]

So, when doing it my way, where did the lost 1 watt go? Do the math. You have a 505 ohm cathode resistor. The voltage across it is 22.5V.

Power = Voltage x voltage divided by resistance. (P = E²/R)

P = 22.5 X 22.5 / 505 =  1 watt.

And that 1 watt dissipated by the cathode resistor is NOT part of the power dissipated by the tube. Again, that's why you must subtract the cathode voltage from the plate to ground voltage to get the actual voltage across the tube.

To calculate power dissipated by any circuit element, whether it's a resistor, tube, stove burner, light bulb, etc., you must know two of three things--- voltage dropped across that element, current flowing through that element, or resistance of that element. In the case of tube dissipation, it's usually more convenient to measure the voltage ACROSS the tube and calculate the current THROUGH the tube.

[1969mako]

sluckey,
So if I follow all this logic correct, for my Vibro Champ, I will use the webervst’s “Calculate Plate Dissipation Biased On Plate Voltage And Cathode Current Readings” section on the WEB site AND since the Vibro Champ is Cathode Biased, I will subtract the cathode voltage from the plate (Measured from pin 3 to 8) and then enter that value as the plate and then hit the calculate button to get wattage, correct???? Also, to get the Cathode Current, I can simply use my BIAS Probe, am I correct there as well?

Thanks and sorry for the Newbee questions but I am trying to get it straight in my mind…

[sluckey]

I've never seen the Weber calculator, but the info cat9 posted about it is correct. I wouldn't even bother with a bias probe to get cathode current. Just measure the actual resistance of the cathode resistor and measure the voltage across it. Calculate the current using I = E/R.