Any interest in seeing some Dumble Tweed Deluxe photos?

[Delmar Evans]

Charlie

Warmed amp up 10 minutes.

At pin 8 of PI is 47.8 VDC
At 4.7 K side from pin 8 is 47.5
After 4.7 K is 45.8

If you need more voltage checks let me know. I have the back off
And on the bench

[ijedouglas]

What are you guys getting on pin 8 of the PI? I have more to add but I am curious about the cathode voltages on pin 8 and the other side of the 4k7 resistor.
CW

Hey Charlie,

I re-measured again as wall voltage is higher in the AM (123.8 )
416.7
365.3
317.6
3.77
266.3
317.5
212.3
292.8
166
292.8
173.1

PIp8 - 46.32
Top of 4k7 - 44.2

[Charlie Wilson]

OK so here is what I think, the measured 21v on pin 7 is voltage traveling across the 1 meg resistor from the junction of the 4k7 and the 1 meg. That voltage is less with the 10 meg because it is a much larger resistor. I am pretty sure the voltages I put on the layout were from the original Tweedle Dee and that measured 21v is pretty strong evidence that the original resistor is 1 meg but I have looked at all of my photos of the original and the photos of my first replica taken with the same camera and there is no way that is a green band. It is blue. The other possibility, given that I had a very limited amount of time to take pictures and measure things, is that I made a mistake when I measured that voltage. I may get a chance to look in another original Tweedle Dee in the future but for now I'm going with the 10 meg. Delmar, I am bit confused by the reference to unbalancing the voltages. It makes me wonder if you are applying the info on balancing the phase invertor of an ODS to the Tweedle Dee and, well, its not the same. Dumble is using the 10k trimmer to balance between the cathode and plate of the phase invertor and it is to balance not unbalance. Attached is a photo I took of the input jacks on the original taken with the same camera and you can see that green looks green.
CW

[Delmar Evans]

Charlie

Thanks for the reply. As far as Balancing or unbalancing goes , everything that I read from five years ago talked about the Tweedledee and how to balance the phase inverter and or unbalanced to phase inverter. The commonality with what everyone said is that right around 6 V differential creates that second harmonic. I was able to do this and achieve a 5.75 V differential when the one meg resistor was in place. I injected various tones into the system for the D , E Flat and the E notes on the B string, and was able to achieve this Differential on the PI tube and on pin 6 of v3 and v4. So are you saying that this is not the correct way to do this for this amp but the purpose is to get them as close to each other as possible? With the 10 meg resistance in place I noticed that on V3 and V4 pin six the total sweep of the potentiometer was only about 1.6 V AC. So there is no way to get 21v with a 10 meg in that location, so it must me a 1 meg ? If you can shed some light on this, please let me know.

[Charlie Wilson]

Delmar, I think, and I could be wrong, but the 6v(DC) thing was for balancing between the two plates of a ODS PI. What I would do is inject your signal into the amp and look at the AC voltages where the two .1 coupling caps feed the power tubes. If you can get those the same then consider the amp close to balanced. The Output transformer is slightly unbalanced and your power tubes may not be exactly matched and that comes after those measured voltages, but it should be close. Forget about the 21v for now. Also, there may be some confusion between AC and DC voltages going on.
CW

[ijedouglas]

Thanks for the update Charlie.

[martin manning]

You can't measure the grid voltage directly due to the meter's input impedance, so measuring at the bottom of the bias resistor is the better way to go. In a simulation I don't see any difference between 1M and 10M for the bias point, and I believe that is accurate. FWIW, at 310V on the supply node I see 258.8 on the plate, 45.93 on the grid, and 48.12 on the cathode. At the bottom of the bias resistor I get 45.93, so -2.19V Vg-k and 466 µA current.

If you want to balance the PI with a scope you can put a 1Ω between the power tube cathode resistor and ground, apply a sine input, and adjust the trimmer to match the positive voltage peaks on the 1Ω. If a balanced output is the goal, I think a 100k plate load will work better than the 110k shown on the schematic (that's with the 56k + 33k on the cathode).

[Delmar Evans]

Charlie

So what I did was I injected a signal into the amp and then checked AC voltages at the power tubes pin 6. In order to get the tubes balanced , I had to turn the potentiometer completely counterclockwise , Out of travel point, that put one tube at 29.04 and 29.6. There was no discernible difference in tone at this setting. I also rechecked the ohm reading on the potentiometer
The lower leg is 9.4 K and the upper is 1.1 k. So with those readings it doesn’t match what the layout shows at 6.6 7K on the upper and 3.536 on the lower. My thoughts are that with those ohm readings in your drawing in the readings coming off of the one mag resistor, are correct. I think that resistor is a one Megger because it lines up with your potentiometer readings and your voltage readings. But then that’s my opinion I tried two different meters a fluke 88 and Klein

[Charlie Wilson]

Couple of things. The meter used has a 10 meg input impedance. Still wont measure accurately? The 1 ohm resistors will work with am amp sharing a single 250 ohm cathode resistor on the power tubes? In fact, in a cathode biased amp, a 1 ohm to ground seems like a bad idea.? Never mind, I think you are saying disconnect the wire on the cathode pin of the tube socket and put a 1 ohm between it and the wire going to the 250 ohm shared cathode resistor. Got it.
CW

[martin manning]

The meter used has a 10 meg input impedance. Still wont measure accurately?

If you see any significant difference between voltage measured at the grid and the voltage measured at the bottom of the 4k7 bias resistor, the meter loading is a problem. If you have two meters, put one on the cathode and then measure grid voltage with the other and watch the cathode voltage.

The 1 ohm resistors will work with am amp sharing a single 250 ohm cathode resistor on the power tubes? In fact, in a cathode biased amp, a 1 ohm to ground seems like a bad idea?

Yes, and no problem; it's 0.4% of Rk. You can take it out after balancing if you like.

[martin manning]

I think you are saying disconnect the wire on the cathode pin of the tube socket and put a 1 ohm between it and the wire going to the 250 ohm shared cathode resistor. Got it.

You could put the 1Ω between the joined cathodes and the shared cathode resistor, or between the cathode resistor and ground.

[Charlie Wilson]

Charlie

So what I did was I injected a signal into the amp and then checked AC voltages at the power tubes pin 6. In order to get the tubes balanced , I had to turn the potentiometer completely counterclockwise , Out of travel point, that put one tube at 29.04 and 29.6. There was no discernible difference in tone at this setting. I also rechecked the ohm reading on the potentiometer
The lower leg is 9.4 K and the upper is 1.1 k. So with those readings it doesn’t match what the layout shows at 6.6 7K on the upper and 3.536 on the lower. My thoughts are that with those ohm readings in your drawing in the readings coming off of the one mag resistor, are correct. I think that resistor is a one Megger because it lines up with your potentiometer readings and your voltage readings. But then that’s my opinion I tried two different meters a fluke 88 and Klein

Delmar, all I can say is that in looking at all of the photos of the original and my replica taken with the same camera, I see no distortion of the color bands on the other resistors so I am going with the 10 meg. The voltages and other measurements are a bit of a mystery to me but I will say that measuring voltages and resistances in one amp with a different PI tube, power tubes, and output transformer not to mention voltages and expecting the same exact result in another amp by setting them the same is, well, a long shot. Plus not sure how Dumble measured the balance. He might have used a harmonic distortion analyzer which measure from the speaker output of the amp.
CW

[Charlie Wilson]

Martin, just to be clear, a 1 ohm resistor from the cathode pin of the power tube socket to ground in am amp that shares a 250 ohm cathode resistor will become the path of least resistance and bye bye power tubes. This amp does not have a cathode resistor for each tube so the 1 ohm could go from the cathode pin on the power tube socket to the wire that connects to the shared 250 ohm resistor in series. Yes?
CW

[martin manning]

Two options:

[Charlie Wilson]

Yes but option #2 between 250 ohm and ground, won't tell you anything about the balance between the two tubes.
CW