MY builds voltages ?

[catnine]

Basically if you look at the schematic of a 6G2 fender princeton this is what my build is . The only difference is I did not add the trem circuit and for the reason I wanted to use a 12AY7 instead of a 12AX7 preamp I wired V1 as the preamp only. V2 is just the second gain stage and the split load PI so that part is like a 5E3. I didn't like the gain a 12AX7 provides and since the 6G2 for V1 uses both triodes one for the preamp and one for the second gain and V2 one is the PI and the other triode is the tremlo .

Now I know the fender schematics voltages are some average and I used an Allen Amps PT which is a beefed up version of a 6G2 princeton or SF champ same unit in the 6G2 and SF champ . The Allen Amps PT has 325-0-325 VAC at the HT secondaries @180 mA and the fender PT is 325 -0-325 VAC @70 mA .
Yet my build with a 5Y3 rect puts out about the same and even a bit higher voltages across the entire amp than later fender princetons that use a GZ-34 rect tube.

Instead of the 1K and 10K dropping resisters the 6G2 uses in the filter stages I have what a 5E3 had 5K and 22K yet even with these higher values my plate is 407 VDC B+ is 410 VDC the screens on the JJ 6V6s are 385 VDC and the grid being fixed bias is -35volts so each 6V6 is @ 18.5mA V2 is pin #1 208 VDC pin #3 1.57 VDC pin#6 254 VDC pin#7 21.9 VDC pin #8 56.1 VDC. This is a 12AX7

V1 12AY7 pin#6 151.5 VDC pin#8 2.37 VDC .

The only reason I can see that the voltages are higher than a 6G2 is that the PT has 180Ma secondaries and a stcok 6G2 has 70 mA secondaries and my PT is not being loaded down enough.

[Austin]

If the pin 8 is 56.1 volts then the pin 7 grid should be about 2 volts more negative at 54 volts, not 34.2 volts more negative like 21.9 would imply. If thats the case then the tube is in cut off and would explain the high plate voltage. The rest of the voltages seem fine. my .02

[catnine]

If the pin 8 is 56.1 volts then the pin 7 grid should be about 2 volts more negative at 54 volts, not 34.2 volts more negative like 21.9 would imply. If thats the case then the tube is in cut off and would explain the high plate voltage. The rest of the voltages seem fine. my .02

You're right and I checked it twice and that's the split load phase inverter . I'll have to find out what's going on there. I know I checked the resisters and the ground and all was fine yet when I look at other fender princetons and all use the same split load PI pin 7 and pin 8 are only about 2 volts difference. Thank you for pointing that out . I wonder if it's the 12ax7 gone bad.

[loudthud]

If the pin 8 is 56.1 volts then the pin 7 grid should be about 2 volts more negative at 54 volts, not 34.2 volts more negative like 21.9 would imply. If thats the case then the tube is in cut off and would explain the high plate voltage. The rest of the voltages seem fine. my .02

Measurements at the grid of a concertina (or an LTP) are unreliable because the meter's input resistance pulls the grid towards ground. That changes the bias on the tube so the cathode current drops. You'll hear a big thump from the speaker if the amp is working.

[catnine]

I know what I did now. I was measuring on the V2 pins ,well stupid me. On a split load PI like mine has you can measure the cathode pin 8 on the tube socket pin because the cathode is in series with a 1.5K ohm and 56K ohm resister to ground BUT to read pin 7 there is a 1 meg ohm resister between pin 7 on the tube socket and the junction between the 1.5K ohm and 56K ohm to ground so if you take the voltage reading off pin 7 of the tube socket you have 1 meg ohm + 56K ohm in series= 1.056 meg . So looking at the fender schematic they show the reading for pin 8 at the tube and pin 7 after the 1 meg ohm . On pin 8 there is 56K + 1.5K = 57.5K .Now on V2 pin#8 I read 56.5 VDC and after the 1 meg resister at the junction I read for pin# 7 I read 54.5VDC .

[Enzo]

Yes, and that 1 meg resistor combines with the resistance of your meter to form a voltage divider. That is why you cannot reliably measure pin 2 or 7 to ground. The only way to measure what is on the grid is to measure if from the cathode. That way your meter is in parallel with the 1 meg, and since no current flows, no voltage drops. If your cathode measures 90 volts, and you get 1.2v between cathode and grid, you can infer that the grid sits at 88.8v. In fact, that 1.2 or whatever will be what is dropped across that 1.5k, since it is the only resistor there with current flowing through it. There should be no voltage drop acros the 1 meg itself.

[catnine]

Yes, and that 1 meg resistor combines with the resistance of your meter to form a voltage divider. That is why you cannot reliably measure pin 2 or 7 to ground. The only way to measure what is on the grid is to measure if from the cathode. That way your meter is in parallel with the 1 meg, and since no current flows, no voltage drops. If your cathode measures 90 volts, and you get 1.2v between cathode and grid, you can infer that the grid sits at 88.8v. In fact, that 1.2 or whatever will be what is dropped across that 1.5k, since it is the only resistor there with current flowing through it. There should be no voltage drop acros the 1 meg itself.

Yes, and that 1 meg resistor combines with the resistance of your meter to form a voltage divider. That is why you cannot reliably measure pin 2 or 7 to ground. The only way to measure what is on the grid is to measure if from the cathode. That way your meter is in parallel with the 1 meg, and since no current flows, no voltage drops. If your cathode measures 90 volts, and you get 1.2v between cathode and grid, you can infer that the grid sits at 88.8v. In fact, that 1.2 or whatever will be what is dropped across that 1.5k, since it is the only resistor there with current flowing through it. There should be no voltage drop acros the 1 meg itself.

Are we talking about the same thing? Looking at the Phase inverter only and mine is just like a tweed 5E3 uses by fender. V2 uses one triode of the 12ax7 as the second gain stage . The plate of the second gain stage is pin 1 grid pin 2 and cathode pin 3 . Pin 3 is 151.5 VDC then there is a coupling cap . 022uf then to the grid of the PI pin 7 . Pin 8 the cathode goes through a 1.5K in series with a 56K to ground . Pin 7 grid goes through a 1 meg resister which is series with the same 56K to ground. The part I don't understand is how there is no voltage drop on pin 7 if read at the socket. Pin 6 plate gets it's B+ after the second dropping resister through a 56K resister the through a coupling cap to the grid of one 6v6 . the cathode pin 8 goes through another coupling cap before 1.5k and 56k to ground then to the grid of the other 6v6 . The two 6v6 grids have the two 220k ohm resisters to form the Y for the fixed bias circuit then the 1.5K grid stoppers the the 6v6 grids.

[Enzo]

Sorry, not my day. Loudthud mentioned a long tail pair PI, so I had that in my head, but a split load will have that same effect on a grid reading. SO what I said stands, just ignore the part about the other grid in the tube. Stick with pins 6,7,8.

Look at the typical first stage in a Fender, 1.5k from cathode to ground, and a 1 meg from grid to ground. Stated another way, the two resistors are joined together at ground. Now the split load, we take that exact circuit and elevate it from ground by 56k. The relation between grid and cathode has not changed, only the distance from ground has changed. In the input stage, we have about a volt on the cathode and about zero on the grid, so th tube is biased at about a volt. In your split load, that relationship remains intact, just that now both those tube elements are higher above ground. It is the same 1.5k biasing the tube. The same 1 volt DIFFERENCE remains.


The key to all this is Ohm's Law - my vote for the single most important formula in electronics. Voltage is not dropped simply because it flows through a resistor. It only drops when CURRENT flows through a resistor. For a voltage drop to form across the 1 meg resistor, current must flow through it. Except for extreme conditions, no current flows through that grid. The 1 meg resistor might as well be sticking out into space. The current through the tube flows between cathode and plate. Out of the cathode, it flows throuigh the resistors we are looking at, the 1.5k and th 56k. For sake of discussion and convenience, let us say the tube is conducting 1ma of current. Then by Ohm's Law, 1.5v wwill drop across the 1.5k resistor. Current is the same throughout a series circuit, so the 56k will drop 56v across itself from the same current. SO the point where the resistors met will be 1.5v lower (to ground) than the cathode. 56v there, and 57.5v at the cathode in this example

Now we wire the 1 meg from that point of 56v, and run it over to the grid. NO current flows, because the grid is just a wire in a vacuum. Ohm's Law says zero amps times 1 million ohms is zero volts. NO voltage drop through that resistor. If you measure just end to end on that resistor, you should get zero volts. If you measure from grid to cathode, you will get the same voltage then as is across the 1.5k, 1.5v.


Now look at your meter. Let us say it has an internal resistance of 1 meg. When you measure 400 volts of B+, that 1 meg resistance doesn't interfere. But in the case of this gid resistor, one end of the resistor is connected to that 56v point, when you ground a meter and probe with the other lead, when you probe that grid resistor, now you have the 1 meg resistor and your 1 meg meter in series between the 56v point and ground. They will form a voltage divider, and you will get a reading of about 28v.

Obviously the currents in your amp may differ from the example, but the same physics holds.

ANother example of zero current, zero voltage drop is in your B+. Your power tube plates have one voltage, a resistor drops some for teh screen, then another resistor drops the voltage for the PI, and maybe another for the pramp tubes. DO teh B+ starts high, and by the last tube in the row it is much lower. Now remove ALL the tube and take the same readings. Now the B+ at the power tube sockets will be the same at all the othe places. No drops, because the tube currents are not gone.

[catnine]

I had one other question since we are on this subject . On the 6G2 princetons and even later princetons the PI one triode is the PI and the other triode is part of the tremlo this would be a 12ax7 . My build has no tremlo so V2 is the second gain stage and PI . V1 on my build is just the preamp . I built mine like the 5E3 in that respect. I noticed the princetons off the PI cathode use a 1k ohm resister is series with the 56K to ground . In mine I used a 1.5K in series with the 56K to ground . Does this really matter ? I could never figure out why the princetons used a 1K instead of the 1.5K unles it has something to do with the tremlo yet the PI triode is not connected to the tremlo in any way . The main difference between a 5E3 and 6G2 besides the 5E3 has a completely different tone stack is the 5E3 is cathode biased where my build and the 6G2 are fixed bias. The 6G2 also uses a NFB loop and a 12ax7 preamp and my build I have no NFB loop and the preamp is a 12ay7 because I like less gain. That was the main reason V1 on my build is just the preamp since on a 6G2 the V1 uses both triodes one as the preamp and one as the second gain stage and V2 is the PI and the other triode works the tremlo off the bias .

I never changed the 1.5K to a 1K I just wondered why fender changed the value. I can't imagine it has anything to do with the difference between cathode biased output tubes or fixed bias output tubes.

[Enzo]

The tremolo has nothing to do with the PI, they are separate circuits.

The 1k or 1.5k resistors in your PIs are really the same element as the cathode resistor in your input stage. They could use 1k instead of 1.5k there if they wanted. The resistor sets the bias for the tube. That sets the operating point for the tube. I never gave any thought to what might be the reason for th different value. I might consider it gave a more desirable performance in that circuit. Maybe better headroom, so the PI doesn;t clip. Hard to say, and not really critical anyway.