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Balancing PP output - which?

lgo51

Member
2002-03-10 9:29 am
MileHigh
I'm fooling around with a PP amp, trying to learn as much as i can by tweaking and measuring. I'm almost certain the answer to my question is simple, but i've got head all twisted up thinking about it and thought ya'll would be able to straighten me out.

Here's my question... When inspecting (adjusting) the bias of a cathode-baised PP amp is it better to balance the current flows for the tubes or the cathode voltages?

Here's where i got twisted up. This particular amp has OPTs whose primary windings are about 6% imbalanced (center tap isn't really in the center). This, and tube variances, would account for the imbalance in the idle current i was seeing. So, i thought, no problem, just skew the cathode resistors (amp uses separate resistors for each side) to balance up the idle curents. That worked just fine and their now matched quite nicely. However, since the cathode resistors are now dissimilar, and the currents are equal, the cathode voltages are not aligned. This, would then change the grid-cathode differential. Which, at 0v input is now balanced. But what about with signal? Assuming the driver stage is outputing symetrical out of phase signals, won't the two tubes now be operating with different transfer characteristics and therefore the OPT will 'see' not so nicely matched signals from each tube? And, sure enough, as i approach full power output, there's a nasty cross-over distortion quite visable on the scope.

So, were my good intentions misguided? :confused:

LarryO
 
I'm not completely sure about this, but I believe that the cathode to grid voltages should be equal. If the plate currents were slighly unbalanced, I don't think i would matter since the output after the transformer will have no dc anyway... But I think a few more people should comment on this to make sure of what is better (balanced cathodes/balanced current) Anyway, I hope you find out what you need to know;)
 

lgo51

Member
2002-03-10 9:29 am
MileHigh
Duo

Thanks for your reply. I think you understand my dilema well. You're so right about the OPTs not 'caring' about the DC component, but isnt' it the plate current (controlled by the grid-cathode voltage) that determines the voltage the OPT "sees" based on the Zin of the OPT? If so, i'm right back to the balanced current question. :(
 
Hi there,

I am not 100% sure I recall this correctly, But I seem to recall that there is a reason that the center is askew. It is because the characteristics of half the primary is different from the other half with relation to the distance from the core. That is, half the primary is closer to the core. So the imbalance that has been built into the transformer you speak of compensates for the difference in distance from the core.

I think that what is more important than primary current balance is core saturation balance. In other words, one half of the primary winding, as it is built, will saturate the core more than the other if currents are balanced, as you have proved with your experiment. That is why you have the distortions.

Now, if you left everything balanced voltage wise, then the outer windings, which is what likely has the slightly more current, will cause a balance (a signal balance that is) with no crossover or other distortions, because now there is a symmetrical core magnetization balance, which will be reflected in the secondary of the transformer.

Does this make sense? If not, sorry.

But I think that is right.

I think.


Gabe
 
Seems to be a valid point there. But also, it is the grid to cathode voltage that controlls the anode voltage. The farther negative the grid is in respect to the cathode, the less plate current there will be. That's why varying the cathode resistors causes a change in plate current, not only because the resistors will pass more or less voltage at some current, but the fact is, if you have a higher resistance at the cathode, it becomes more positive than the grid. Thus, the higher your cathode resistor is, the less current flows through the tube between cathode and anode. Also, any current flowing through the anode must also flow through the cathode. So, technically, you'd be better off at making the cathode - base voltage equal for each tube to cause a transformer balance like Gabevee just intelligently noted.
If less current is to flow in the closer coil, then there will be a balance between the coils. In other words, leave the resistors on the cathodes the same and let the current be unbalanced and you should have a good working amplifier...;)


EDIT: I was just reading back to your questions. It is the plate current that determines the voltage the OPT sees depending on the impedance, but that will balance out when the currents are unequal, since one side of the CT'd coil is closer to the core than the other, it'll have a greater effect on the saturation and must be fed less current to keep it in equalibrium with the other, farther, part of the coil after the CT.
 

lgo51

Member
2002-03-10 9:29 am
MileHigh
Duo

Yes, i understand the issues with the OPT much better now. I assume that those are some of the reasons they invented the bifilar winding technique.

But, setting that aside for now, and maybe rewinding a bit. What we've said here is that the CT is intentionally not truely in the center in order to skew the currents through the near and far coils of th OPT just enough to ensure that the fields generated by the same voltage across the primary windings will indeed be balanced and therefore grace the secondary winding with a signal of lower distortion. So, in real life, we want the voltages applied to the OPT primary windings (upper half and lower half) to be the same. This will, because of the CT-skew, result in slightly different current flows from the plates - which come from the cathodes in the first place.

IF, i've got that all correct, then the proper thing to do is monitor the plate to B+ voltages (i.e. that across the OPT primaries) when performing a "balancing act". Now, given matched tubes, that condition would also occur when the grid to cathode voltages are "in balance". But, there are no such matched tubes -- well not for long anyway. So doing grid or cathode balancing is really an engineering folly... right??? Or, am i off in space here?

LarryO
 
Yeah, you're on the right track here. When the voltages on the transformer on either side are the same, the currents are different slightly because of the different winding differences. In other words, the center tap isn't necessarily off by a bit, it's just that it appears off because of the different inductance of the winding on either side of it.

So, yes, do make equal voltages from plate to B+ accross the transformer and everything should be just fine...:)
 

lgo51

Member
2002-03-10 9:29 am
MileHigh
Duo

OK. I equalized the plate voltages at idle (G1=0v) and then rechecked the full-signal output waveform. MUCH better. It still wants to go into cross-over distortion just before clipping. Maybe that's the price one pays for allowing the two tubes to run more independently - as opposed to strapping the cathodes together and using a shared bias resistor.

The resultant difference in cathode resistors is a bit troubling, 545 on one side and 520 on the other. This results in about a 2 mA difference between the upper and lower sides. That'd be about 4%. Reasonable?

Oh, BTW, the OPT's display an average 7% difference in DC ohms between the two sides of CT. Is this typical for what we were discussing about inner vs. outter windings? Too much?

Oops, i may have exceeded the #?/post rule :eek:
 
Unless the tubes are matched pair, the difference in this case may be mismatched tubes. Otherwise it is in the transformer. Since now you see the crossover at clipping, this is kind of acceptable. At normal listening levels it wil be fine. I see the same with mine and they do share one cathode resistor.

As for the measured difference, that is because one half of the primary has less windings than the other, so that there is less wire used. Again, this is to make both halves equal in inductance (the distance from the core). That is what makes the signal at the plates.

I have also seen this with power transformers for the same reason.

Just to reference your mention of bifilar winding... bifilar winding is used to in essence parallel inductances. This helps with high frequency and low amplitude performance.

However, some transformers are wound with both halves at the same level, that is, both halves are wound at the same time so that the inductance imbalance does not exist (extremely difficult and found on expensive transformers). That way you can have your cake and eat it too. But... there are problems with that too. As they all say, this technology is a series of compromises.
 

lgo51

Member
2002-03-10 9:29 am
MileHigh
Oh YES

Bypass caps are 140 uF / 60v tantalum, one per side. Found the tantalums in a surplus store, JAN spec, heavy little dudes. Not experienced enought (yet) to critique the sonic affects -- any thoughts?

I've also installed the "AC Cathode Shunt" [don't know what it's really called]. It goes like this: (2) 260 uF / 50v (spec was 6.3v) Non-Polarized electrolytics (T.I.'s) arranged in parallel and strung cathode to cathode.

I have also seen a pair of 6.2v(?) / 1w zeners wired in series (with their cathodes butted together) strung between the tube cathodes to protect the caps in the event of a internal short in the output tubes. Not sure about this one.

I've also noted that in Allen Wright's amp, the entire cathode bias setup (resistors and bypass caps) is 'floating' off of ground via a 68 ohm resisitor.

Comments?
 
Good! You halready have the bypass caps. And I see what you mean be cathode shunt, that would act similar to bypass caps
but in a more active way, quite a neat thing if I say so myself...
though I've never seen caps from cathode to cathode, I do see what it would do and I think I'll try it on my next amp...

And in the allen wright, that 68 ohm resistor is probably where
you'd check the bias from for both tubes... In my dynaco MKIII
the cathodes are connected right together and coupled to ground with an 11.2 ohm bias resistor and you're to measure 1.56 volts accross it for proper bias, which is about 70 mA per tube which is right for class A with KT88 tubes:) the reason the resistor is so small is because the amp will output 60 watts RMS which is moderately high for a lot of tube amps...
 

lgo51

Member
2002-03-10 9:29 am
MileHigh
Hhmmm...

In the A.W. amp, he uses separate bias resistors and bypass caps for each side. But, instead of running to ground, they all run to a single 68 ohm resistor, and then to ground. Bias resistors are 550 ohm if i remember correctly. Tubes are EL34s.

BTW, any thoughts [sonic experiences] on bias current for EL34s? I've seen as low as 65% Pa(max) and as high as 80% Pa(max), about 50 mA and 65 mA, respectively.
 
bias is dependant generally on the class of operation you want and the types of tubes used... Usually between 50% and 80% is
used for class a/ab amplifiers so you really do have some room to play... In general, you tubes will last longer the colder you bias them, that is, less current means longer life. But as you reduce the bias of a puch pull stage, you get out of class A and into AB and if you drop it low enough, you'll get class B which has moderate crossover distortion. BTW, does your amp have any sort of bias adjustments or is it just set by fixed resistors???
If set by resistors and the amp works fine/sounds good with original values and tubes are matched you'd do good to leave it that way since it was designed to work that way...
 

lgo51

Member
2002-03-10 9:29 am
MileHigh
So, where's the line between A and AB(1,2)??? How can i measure it? Sounds like (ooops, bad pun) comfortably inside "A", but not aggressively is a good target.

My amp is cathode biased. I thought the 'original' setting was a bit cool (about 50 mA, 600 ohms), especially since i've often seen 470 ohms spec'd (albeit by tube books that, ultimately, want to sell you more tubes ;-). Thoughts?
 
Well, the "line" betwee A and AB isn't really a specified amount of current, but is more or less the amount of signal the active devices(tubes) operate for... For example, in Class AB, the devices are biased so that they will work for up to between 51% and 99% of the audio waveform... In class A, any device(s) will work for 100% of the waveform and in class B, the device(s) will work for only 50%of the waveform... So, in class B with two tubes(push pull) one tube conducts for the positive half of the signal while the other conducts for the negative half, but you end up with "crossover" distortion when the signal switches from the high to the low tube and vice versa. In your case, you probably want class AB, where each device will conduct for a portion of the signal positive or negative before turning off, which prevents crossover distortion. So essentially, you should bias so that both tubes have about 50%of Pmax or less... But if you've gone as cold as class B, you'll probably hear distortion at low volumes.
Another factor is grid voltage... If your current is 50ma and your cathode resistor is 600 ohms, then you have a grid voltage of -30 volts. If you fuss with the bias too much, you'll need to redo a lot of the amps original numbers... So, I would really recommend leaving the amp at its 50mA if it sounds fine there. If your amp was designed for class AB and you move it into A, the tubes will go bad quick and you stand the chance of overloading some other components as well.
:cool: