Princeton Reverb Bias Issues

Okay - I'm familiar with adjusting cathode bias designs such as a 5E3 but I'm not sure how to tackle a fixed bias system such as the Princeton Reverb. I'm referring to vintage models, not the reissue.

I know the 22K resistor on the bias board is the "adjustment" although I'm not really sure what I'm adjusting by altering that. I also know the cathode (pin 8) is tied directly to ground so there is nothing to measure at that point.

Some articles have instructed to lift the cathode ground and insert a 1W, 1 ohm resistor to use to measure. (http://www.diyguitarist.com/GuitarAmps/PT-Biasing.htm)

I'd also be interested in adding an adjustment a la the instructions here - http://www.el34world.com/charts/Biascircuits.htm

Let me have your thoughts!

Basically, lowering the value of that resistor will bias the tues hotter, and raising it will do the opposite. I added the adjustable bias control you linked to and it works like a charm. Highly recommended.

To measure the mA draw, set your meter, clip your black lead to the resistor leg where the red OT wire connects and (one hand in your back pocket) touch your other lead to the pins where your brown and blue OT wires lead. Should be in the 21-25 mA ballpark.

Let's do this with pictures - EVERYTHING is better with pictures!

Blown up area

http://hattmonkey.com/pr-tdpri1.gif

Test Locations

http://hattmonkey.com/pr-tdpri2.gif

That's what you meant, right?

If so, I'm not getting anywhere near that. I pretty much get no reading with the meter set to mA.

I should probably state - the amp is working fine/normally/as expected. I just want to know how to check/test/change bias to mess with different tubes.

Yep, those are the right test points, but I'm not sure why you're not getting a reading unless you have a blown fuse on your meter or multiple mA settings or something like that.

Not sure...I get 0.03 mA - numbers are fine on a known-value circuit.

NOte: If you are using the transformer shunt method of measuring current draw, then I would sugget doing things in this manner. WIth the meter set on amperes, the black(negative) probe is the probe that should contact with the plate..pin 3 on those 6V6's. The red(postive) probe should contact the center tap of the the OT, which indeed is that red wire contact circled above.

Wow, I just re-read my post above and realized I reversed red/black. Sorry for the misinformation!

And what you need to do is measure the voltage drop across each side of the OT primary and divide that voltage drop by the known (measured) DC resistance between the OT CT and each end (plate pin) of the OT primary. (You won't get much of a voltage drop across the OT, maybe one or two volts, so it is better with your meter on the tens of volts setting. But be careful.)

okay - here's what I got. We're going to call V1 the first 6V6 and V2 the second.

OT V1 half - 143 ohms - 3.06 VDC across the OT at that point. Gets me about 23.36 mA.
OT V2 half - 152 ohms - 3.01 VDC - 25.24 mA.

Those numbers seem a bit high, don't they? I've got 396V at the plate on both tubes.

OT V1 half - 143 ohms - 3.06 VDC across the OT at that point. Gets me about 23.36 mA.

396V x .02336A = 9.3W = 77% for a 6V6GT, or 66% for a 6V6GTA or JJ6V6S

OT V2 half - 152 ohms - 3.01 VDC - 25.24 mA

396 x .02524 = 10W = 83% for a 6V6GT, or 71% for a 6V6GTA or JJ6V6s

I don't think that is excessively high for a 6V6GT (some people run theirs at 85%), and obviously not a problem for a GTA or the JJ6V6S

So my math was right then, tubeswell?

(143ohm/2) / 3.06 = 23.36 - that's how you get that right?

If so, why exactly do you need to divide the OT resistance by two? I just don't understand where that part comes from.

That is a lot of work just to arrive at a current draw figure, ime. IF you use the transformer shunt method I outlined above, you will read the current draw directly with the multimeter set to amperes. YOu are working with a live circuit in either case, so the danger is not avoided by making those extra measurements, right?
And.....if you are comfortable running that much current draw and plate dissipation in a fixed bias circuit, then please yourself. YMMV, but the only tube mentioned above that I would run that 'hot' would be the JJ because it is considered to be something like; a 16-18 watt max plate dissiaption tube...compared to the 12 or 14 watts for the other two tubes mentioned.

And....with the transofrmer shunt method, you do not have to divide if you are looking at a two power tube circuit. IF you have a 4 tube circuit, then you would divide the current reading by 2. With 6 power tubes, you would divide by 3....and so on. Why??? Because the reading would be the current draw for that side of the OT primary and not for each tube.

So my math was right then, tubeswell?

(143ohm/2) / 3.06 = 23.36 - that's how you get that right?

If so, why exactly do you need to divide the OT resistance by two? I just don't understand where that part comes from.


Except that I didn't check your math when I calculated the dissipation!

(And you don't need to divide the OT resistance by 2 - not sure where you got that from! - unless you are running 2 output tubes in parallel on each side of a PP amp? For a Princeton reverb (with 2 x 6V6s in PP), you simply measure the DC resistance of each half of the OT primary, and then measure the voltage drop across each half to calculate the current being drawn across each half. Voltage/resistance = current. After that, you multiply the current by the respective plate voltage to get the dissipation)

So:

3.06V/143R = .0214A (21.4mA), .0214A x 396V = 8.5W, 8.5W/12W = 71%

3.01V/152R = .0198A (19.8mA), .0198A x 396V = 7.8W, 7.8W/12W = 65%

(which is even less to worry about.)

tubeswell, could you describe this method one more time...because if you read what you worte and what Corliss has done...and do some figuring by the methods that you show above in post #8, then the figures that have been arrived at by corliss just don't come up correct, ime.
Ex: 143ohms with a 3.06 volt drop..... as you instructed in post #8...
3.06V/143ohms = .02139 amps or 21.4ma. IS that correct?
3.01/152 = 19.8ma, right?

Corliss, nowhere in tubeswell's description does he instruct a division by 2 of the resistance on the primary windings.

Corliss, if you go back and use the transformer shunt method outlined above, you will be reading milliamps directly. Wehn you arrived at a reading of .03, you were using an incorrect method as acknowledge by Jimmy. Unless I have lost track of what tubeswell is suggesting as a method of determining current draw and plate dissipation, I would suggest that corliss take a step back and reevaluate the whole thing. The transformer shunt method is direct and definitive, ime. Imho, corliss has become confused about what is goingon here. Jimmy's incorrect instructions coudl be the source for some of that confusion,. but a correction was quickly made. That correction was ignored, and now there is a totally new andmuch more complex method at work...adn I don't htink that it is being used correctly, either.
IF I am mistaken, feel free to explain in depth a6t what point I have lost track of what is going on here. I am not sayign that he method that tubeswell is suggesting is incorrect, but it is more complex than it need be, imho.

tubeswell, could you describe this method one more time...because if you read what you worte and what Corliss has done...and do some figuring by the methods that you show above in post #8, then the figures that have been arrived at by corliss just don't come up correct, ime.
Ex: 143ohms with a 3.06 volt drop..... as you instructed in post #8...
3.06V/143ohms = .02139 amps or 21.4ma. IS that correct?
3.01/152 = 19.8ma, right?



Yeah see my post just before now

Awesome - that makes sense now.

So - slightly off topic, is this why some people scorn the "YOU MUST HAVE MATCHED TUBES OR ELSE" theory? It seems the parts inside the amp itself (halves of OT) certainly aren't matched. I've got a 6% difference on the OT, so who cares if the tubes are matched within 1%?

tubeswell, your post hit my email just as I hit the 'Post' button on my last post. SEe what I mean about the confusion??? I am going to remember this method that you have outlined here, and I am going to check it against the transformer shunt method to see if the results are the same. I am going to think that Iwill never put this method to use, though...becuase it is toomuch trouble unless someone can convince me that it is more accurate. I am gonig to doubt that...because I have never heard of anyone using this method before.....perhaps because of what are probably unnecessarily complex steps.

Well I certainly don't find it much trouble at all to pull out the output tubes and put the R-meter between pin 3 of each tube and the OT centre tap, and then with the tubes in and the power on, (carefully) measure the voltage drop between the same points. The rest is just math and ohms law. What could be simpler? :-)

(Actually, come to think of it you don't even need to pull the output tubes to measure the DC resistance on the OT primary)

tubeswell, with the transformer shunt method, one does not have to remove and reinstall the tubes. ONE also does not have to measure OT primary resistance and voltage drops and do that figuring. One merely needs to measure the plate voltage of the power tubes, measure the current draw for each side of the OT...and then use those Laws to figure the plate dissipation. The transformer shunt method is simpler even if you don't have to pull and reinsert the tubes in your method because you don't have to measure that voltage drop and do that equation to find the current draw, right? The current draw measurement is direct and real time....and therefore is simpler. I suspect that this is why I have never heard of this convoluted method that you describe. IF it is not more accurate...and I don't see how it can be...then I see no reason to use it. YMMV.....
Be that as it may, it matters not to me what method someone uses; but I did correctly sense a confusion on the part of corliss1 that was impeding his ability to accurately figure the plate dissipation in that amp. And...ultimately, it doesn't matter to me if someone burns their amp down unless I am the one that gave the advice that was either incorrect or confusing. There was one incorrect post---acknowledged by Jimmy....and some confusing instructions...or at least they were confusing to corliss. I suspect that the instructions were confusing becuase there were no graphic equations posted....and as corliss noted, he likes 'pictures'...which are what equations are compared to verbal directions.

Good luck with it, Corliss. IMe, folks are more familiar with figuring and adjusting fixed bias baising than they are cathode-biased adjustments. IT is easy to change that cathode-biasing resistor, but the equations that allow one to understand what is going on are more complex than are the figures for fixed-bias, ime. Tubeswell's method makes fixed-bias computations about as complex as cathode-biased computations, imho. I see no need for it. AGain....YMMV....and have a good one....

I do apologize for the lead mixup in my original post. I was trying to respond quickly on my phone and didn't see that I had lead you in backwards.

Transformer shunt isn't fancy, but it's a good, quick method.