Blues Jr has a neg voltage turning off V1A.

[SoulFetish]

If the cathode is at 0V and there in no voltage drop across the plate load resistor, the tube is not conducting. But I don’t think it’s because it’s in cutoff.
check the heater voltages, or can you see if the tube is lit?
If its not lit, swap in another tube (swapping out the tube is the easiest, and first thing you should do anyways).
I suspect it might be heater related, either in the tube or amp

[Helmholtz]

Without heating there are no free electrons to charge the grid negative.

....of course the meter could be defective.

[vintagekiki]

....of course the meter could be defective.

meter is correct

V1A is cut off it has 265V on both ends of plate resistor and 0V on cathode .

[SoulFetish]

If the cathode is reading 0V, and the grid is at -.5V, it isn’t in cutoff, not even close.

[Helmholtz]

If the cathode is reading 0V, and the grid is at -.5V, it isn’t in cutoff, not even close.

True, but I'd expect the static voltage of the open grid to be considerably more negative than -0.5V.
Connecting the meter provides a current (grid leak) path and lowers the magnitude of the voltage.

It should be easy for the OP to find out if the grid is open.

[Enzo]

Something very basic is happening here. I don't think this is the place to look for subtleties. Just my opinion

[SoulFetish]

I think so too. Have we eliminated that the tube as the cause yet?

[Leo_Gnardo]

I think so too. Have we eliminated that the tube as the cause yet?

I vote for an open filament. Why that would send the grid reading to negative, I dunno. But with no voltage drop across the plate resistor, and zero voltage on the cathode. I don't see any other reason for a failure to amplify, with the evidence presented. .

Yes I know I suggested possibly a shorted cathode bypass cap or other cathode-to-ground short, but that would ordinarily result in a weak distorted signal. Here we have none. So it's open filament time, QED. Slap in a known working 12AX7, shoulda done that way back in the beginning.

[Helmholtz]

Something very basic is happening here.

What could be more basic than a disconnected grid?

[Helmholtz]

Why that would send the grid reading to negative, I dunno.

It's impossible.

[Helmholtz]

The -0.5V is not always present on V1A G1 and today it went much higher as I turned up the volume pot with oscillator into i/p .

PLease don't measure DC voltages with signal present.

[vintagekiki]

If V1 (12AX7) is new (not as new) the measurements indicate

" V1A is cut off it has 265V on both ends of plate resistor and 0V on cathode "
For whatever reason (bad tube, loose solder joint, faulty resistor} the tube is not conducting.

https://www.thetubestore.com/lib/thetubestore/schematics/Fender/Fender-Blues-Junior-Schematic.pdf

1) Correction
If on both sides R10 (100k) has voltage Uz = 265V, and voltage on R11 (1.5k) is 0V, first check that C4 (22u) accidentally is not in short-circuited.
If R11 and C4 are correct, resolder all pins on tube socket V1 there may be somewhere cold solder.
If that doesn't help, replace tube socket V1 most likely is defective (pin2 or pin3 is interrupted)

[SoulFetish]

How would the bypass cap shorting put the tube in cutoff? It would put the tube at ~0Vg to cathode and quite the opposite would happen.

[vintagekiki]

post #27

underline says all

[SoulFetish]

Okay, we need to define “cutoff” as the condition where the grid voltage drops negative to the cathode to the point where the tube no longer conducts.