biasing without 1Ohm resistor on cathode; Is this correct?

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jezzbo
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biasing without 1Ohm resistor on cathode; Is this correct?

Post by jezzbo »

Hi all,

I'm trying to restore my 1966 Bandmaster (see "Bandmaster blues topic)
So far I managed to make the normal channel to work again.
(tremolo ticking later)

But I was wondering about how hot this amp is biased.
In my search for knowledge I found this explanation on this site:
http://www.aikenamps.com/Biasing.html
Quote: "The plate current can also be measured by first measuring the resistance across each side of the output transformer primary (it will usually be different on each side) with the power off. Make a note of the resistance on each side, and then, with the amplifier on, measure the DC voltage drop across each side of the output transformer. Divide this number by the previously measured resistance, and you end up with the plate current for the tubes on that side. Again, if there is more than one tube on each side, you must divide the total current by the number of tubes. This method is extremely accurate, and much safer than the shunt current measurement method, because a slip of the probe won't short anything out due to the high resistance of the voltage measurement setting on the meter compared to the very low resistance of the current measurement setting. You can also make a safer measurement by clipping the negative side of the voltmeter on ground, and measuring the center-tap voltage of the output transformer and the voltage at the plate of each output tube. Subtract the plate voltage from the center-tap voltage and you have the voltage drop across each side, and can then use this to calculate the current in each tube, again dividing by the number of tubes on each side." End quote.

So please can you tell me if I done right when you read the following:

Powertubes are 6L6GC (JJ)
Uct (Centertap voltage) = 422V (measured to ground)
Ua1 (Plate tube 1) = 421V (measured to ground)
Ua2 (Plate tube 2) = 421V (measured to ground)

Resistance from:
Plate from tube 1 to centretap of OT = 70,8Ohm
Plate from tube 2 to centretap of OT = 79,6Ohm
Voltage drop measured from plate tube 1 to centretap of OT = 1V
Voltage drop measured from plate tube 2 to centretap of OT = 1,23V

So: U/R=I

For tube 1= 1V/70,8Ohm = 0,0141242 (A? = 14,1mA?)
For tube 1= 1,23V/79,6Ohm = 0,0154522 (A? = 15,4mA?)

Disipation = U x I (right?) (edit: I first wrote U x R and that's not right :wink:)
So that would be
for tube 1 = 421V x 0,0141242 = 5,946W(?)
for tube 2 = 421V x 0,0154522 = 6,503W(?)

to what I found in the JJ 6L6GC spec (found here: http://www.jj-electronic.com/index.php? ... &Itemid=11) The max disipation = 6,5W (please confirm)

Thank you for checking my thoughts.
Last edited by jezzbo on Fri Jan 15, 2010 9:51 pm, edited 1 time in total.
breakfastbuddy
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tube dispensation

Post by breakfastbuddy »

thats single tube . it says 17,5w on pp but normal is 30w on a 6l6 gc
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Structo
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Re: biasing without 1Ohm resistor on cathode; Is this correct?

Post by Structo »

Yeah, I believe that is the Transformer Shunt Method.
It can be dangerous because you are directing the full amperage through your meter.
So if you happen to slip off of the pin or joint when you are probing you can get a big arc and damage the amp, your probe and meter or you eyes from flying slag.

Really, the easiest way to do it is the one you kind of dismissed.
That is soldering a precision resistor from the cathode to ground and then measuring the voltage drop across the resistor.

Most use a 1 ohm 1% or even .1% resistor here.

The formula for fixed bias is easy as well.

Power is E * R, or E2/ R or, R * I2

(W * .7) / V = I

Where W is watts, the .7 is 70%, V is volts and I is current.

Here W is multiplied by .7 to give us 70% of the tube's maximum plate dissipation.

This number is divided by the actual B+ voltage (plate voltage) in the amp to determine what idle cathode current to adjust the tubes for in AMPS.

An example:
In an amp with 480V B+ and using EL34 output tubes (25W max plate dis.).

(25W * .7) / 480V = .036A

17.5 / 480 = 36mA

Cathode current at idle should be adjusted for 36 milliamps.

This does not take into account the screen current but it is so small that it is usually ignored and actually provides a bit of cushion for the bias you set the tubes to.
Screen current varies but usually it's only 2-4ma so as you can see if you set the bias at 36ma, in reality it may be 32-34ma.

If you are worried about the mod to the amp to do this, it is easily reversed but many feel it is a good mod to do since it makes setting the bias so easy.
Tom

Don't let that smoke out!
vibratoking
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Re: biasing without 1Ohm resistor on cathode; Is this correct?

Post by vibratoking »

There is nothing magic about the 0.7 multiplier. In my experience, it is a tradoff between tone and tube life. In some amps I bias as high as 80% max dissipation and as low as 55% in others. It is very dependent on topology. For instance how hard will you run it, do you get overdrive from stomp boxes, preamp tubes, or the power amp section. Experiment with different bias points and find out what your ears like for that particular amp.
Last edited by vibratoking on Fri Jan 15, 2010 10:27 pm, edited 2 times in total.
jezzbo
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Location: Lebbeke, Belgium

Re: biasing without 1Ohm resistor on cathode; Is this correct?

Post by jezzbo »

Thanks for your answers.

I don't think this is Shunt Method (I'm not really sure though)
There is no current going through my meter. (as far as I know)
I just measured resistance from centertap to one side of the OT (with the power off) and then the Volts (with power on). I did that for both sides of the OT. Then used these values and Ohm's law to calculate the current.

I don't have 1Ohm resistors and just wanted to see how the amp was biased. (I don't want to dismiss the Cathode/1Ohm methode)
But I don't know if my callculations are right.

btw those 1 Ohm resistors; is 1/2W enough?


thanks guys, great info. 8)
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Phil_S
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Re: biasing without 1Ohm resistor on cathode; Is this correct?

Post by Phil_S »

This is what I like to call the modified shunt method.

Using the real shunt method, you simply put the meter in the circuit in parallel to the OT and read the mA from the display. This is a dangerous procedure, IMO, reserved only for pros who are very clear about what they are doing and who are using a pair of Fluke meters.

What Aiken says, and what I do sometimes, is measure the voltage drop on each side of the OT, and use the Ohms of the winding, apply Ohms Law, and solve for mA. Review: V = I * R. Solve for I = V/R. This is considerably safer because it is not nearly as dangerous to measure volts.

You know or will find R. Pull the tubes and measure the resistance between the plate pin of the tube socket and the center tap. Do both sides. The result is different on each side because the outer winding, while it has the same number of turns, uses more wire to make the turns compared to the inner winding.

Now, you need V. To get V, you don't subtract one voltage from aonther. That's not accurate at the 1V tolerance. Place the black meter probe on the + side of the first filter cap -- this is the place where you connected the CT of the OT. Then place the red probe at the plate pin of the tube sockect (tubes in the amp, fully powered in operating mode). Now you'll get the voltage drop -- first one side, then the other. You'll get a number resolved to 2 decimal places. This is the voltage drop reading you need.

Here, I'll give you the example of the amp I'm just finishing. The OT on V3 is 56.6 Ohms and V4 is 61.8 ohms. On V3, I read a drop of 2.95V. On V4 I read a drop of 3.30V. Math V/R:
V3: 2.95/56.6 = 52.12mA. Plate voltage is 362V. 52.1mA * 362V = 18.9W.
V4: 3.30/61.8 = 53.40A. Plate voltage is 362V. 53.4mA * 362V = 19.3W.
Guess what? I think 18.9W and 19.3W are close enough to confirm I've got a reasonably matched pair and about 20W per tube would be OK for this pair of 6L6. I could go a little hotter...this is a choice to make, but you should get the idea.

For sure, a 1.0V drop isn't what you are working with. If it really is in that range, either your tubes are pooped, the windings on your OT are in the 20-something Ohm range, or you bias voltage is way too negative. I think 20-25 ohms on the OT is unlikely, so look at the other two things.

Go back and try this again. Get accurate voltage drop readings "across the winding". No math here. You need to see it on the meter.

Another note, I can understand you may be reticient or just haven't got to it yet, but put in the precision 1-Ohm or 10-Ohm cathode resistors. They do nothing to the tone of the amp and make it all so much safer and easier. You just have to remember to subtract the screen current. If the amp has 1K Ohm screen stoppers, place the meter probes on each side of the screen stopper and that's screen current in mA. If not, just guess -- it is likely 2 to 4 mA.

Good luck.
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