Jetbat wrote:
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