New PP amp...

[charliewphelps@gmail.com]

Charlie[/quote]

Class sounds very neat. I'd like to do that as well. Can you post any enrollment info?

A little suggestion: after the usual variac startup, and before a full line voltage startup, consider starting with an incandescent lamp in series, with similar power ratings as the Fisher. one last step to avoid "booms", magic smoke dispersal, exploding fuses and caps.

Slick[/quote]


They even covered "bulb testers" in class. Even had one lying around that one of the students bought for a grand total of $5.

Charlie

[randy warren]

Charlie

Class sounds very neat. I'd like to do that as well. Can you post any enrollment info?

A little suggestion: after the usual variac startup, and before a full line voltage startup, consider starting with an incandescent lamp in series, with similar power ratings as the Fisher. one last step to avoid "booms", magic smoke dispersal, exploding fuses and caps.

Slick[/quote]


They even covered "bulb testers" in class. Even had one lying around that one of the students bought for a grand total of $5.

Charlie[/quote]

those tube tester vary greatly in their capabilities, I do restoration on hi-end tube testers. The dim-bulb trick is an old one, here is an article on a great little portable one...

http://www.jestineyong.com/homemade-power-box/

[TubeDriver]

Transparent Valve covers might not be a bad idea. You can see what is going on, I could spot check to make sure non of my keepers is damaged, you can check to see what your oil looks like!


Guy, no problem about using my phono pre for a while, it has been sitting in my basement for about a decade!

[Jetbat]

Roscoe, question about your schematic. What is the reason that the grid from V1B is tied to the junction between R8 and R9? I see it is a virtual ground. It is some kind of error correction also?

Scott

[Roscoe Primrose]

Roscoe, question about your schematic. What is the reason that the grid from V1B is tied to the junction between R8 and R9? I see it is a virtual ground. It is some kind of error correction also?

Scott

That's the phase splitter for the PP output stage. The junction of R8 & R9 is half way between the plate of V1A & V1B, if V1A's plate voltage increases due to a signal at the grid of V1A, that'll cause the junction of R8 & R9 to increase in voltage, causing V1B to conduct more, which in turn causes V1B's plate voltage to decrease. In theory, this action should cause V1B to always generate an exact inverse AC waveform on it's plate as compared to V1A's plate.

Roscoe

[Guy]

Roscoe: How dare you try to hijack back your own thread! Now I'm flipping between screens to check the schematic and your comments.

Charlie: I've asked the Museum for instructions for the radio-repair class. I'll post whatever I find on this site, but maybe under General Info or something.

[Jetbat]

Roscoe,
Normally the grid to V1B is tied to ground. I can tell the junction between R8 and R9 would be 0 V DC. It looks like the signal from V1A and the inverted signal from V1B would meet at the at that junction and cancel each other out.

Scott

[Roscoe Primrose]

Roscoe,
Normally the grid to V1B is tied to ground. I can tell the junction between R8 and R9 would be 0 V DC. It looks like the signal from V1A and the inverted signal from V1B would meet at the at that junction and cancel each other out.

Scott

R8 & R9 are connected to the plates of V1A and V1B respectively. The plates of V1A and V1B are at about 160Vdc, which, under no signal conditions, is where the junction of R8 & R9 is. The operation of the phase splitter does the very best it can to keep the junction of R8 & R9 at 0V AC.

Let's look at the circuit with no signal input. The plates of V1A & V1B are both at 160v, no current flows through R8 & R9, and the junction of R8 & R9 is also at 160v. If a positive voltage is applied to the grid of V1A, V1A starts to conduct more, causing the plate voltage on V1A to decrease. For ease of discussion, let's the voltage on V1A's plate changes to 150v. As there's now a difference in voltage between the plates of V1A & V1B, current flows through R8 & R9, and the voltage at the junction of R8 & R9 is half way between 160v & 150v, or 155v. Now, from a DC standpoint, nothing else happens because of C2, however, if you look at it from and AC standpoint, the decrease in voltage at the junction of R8 & R9 passes through C2, and causes V1B to conduct less, which causes the voltage on V1B's plate to increase. Of course, this all happens essentially instantaneously in the AC signal condition with the voltage at the plate of V1B following the voltage of V1A, but out of phase....

Roscoe

Roscoe

[Jetbat]

I don't know what I was thinking about the 0V DC. As soon as I started to read your explanation I realized my mistake.

Isn't V1B a common grid stage? Or does that change with the grid tied to the R8-R9 junction?

Scott

[Roscoe Primrose]

Isn't V1B a common grid stage? Or does that change with the grid tied to the R8-R9 junction?

Scott

No, it's a standard common cathode stage. Input is the grid, output from the plate. The connection of R2 to the junction of R4 & R7 just provides a DC feedback to keep the bias current in the output tubes more stable (as does the R1/R3/R6 combo).

Roscoe