Weber "Lenora"

[floyd]

Webers" Brown Deluxe layout shows 68K resistors on the input jacks. Since there is only one input per channel , shouldn't these be 33K ?

[sluckey]

Webers" Brown Deluxe layout shows 68K resistors on the input jacks. Since there is only one input per channel , shouldn't these be 33K ?

The value is not critical. Use 10K to 100K.

[ac427v]

In other amps, I think I hear a little more punch with 33k. Hard to tell for sure though because the volume is a little lower.

[HotBluePlates]

In other amps, I think I hear a little more punch with 33k.

Assume you're playing into a 12AX7 1st gain stage. The data sheet says "input capacitance" is 1.8pF, and grid-to-plate capacitance is 1.7pF. Typical stage gain is 60.

Total input capacitance of the stage = grid-to-plate capacitance * (Gain + 1) + input capacitance = 1.7pF * (60+1) + 1.8pF = 105.5pF.

The 33kΩ (or 68kΩ) resistor forms a low-pass filter with the total input capacitance of the 1st stage.
F_-3dB = 1/(2*Π*R*C)

F_-3dB(33kΩ) = 1/(2*Π*33k*105.5pF) = 45.7kHz
F_-3dB(68kΩ) = 1/(2*Π*68k*105.5pF) = 22.2kHz

Both are above the range of hearing and of the speaker's ability to reproduce, so there shouldn't be a sonic effect to using either.

I did not add in any allowance for stray capacitance due to wiring length. The placement of the Daisy Chain switch on the Lenora at the back of the chassis adds an awful lot of extra lead length, and may increase wiring capacitance just enough to make the switch between resistors just-barely audible.

If it were my amp, I'd be more concerned with minimizing that lead length (by putting that switch on the front panel) and putting the resistors right at the tube socket. But that's just me...

[floyd]

My question arises from this.. when most builders of Fender type amps choose to go with only one input jack on each channel , I always see 33K instead of 68k. So , if there is no difference , why not stay with the 68k ?

[sluckey]

The value is not critical. Use 10K to 100K.

[floyd]

The value is not critical. Use 10K to 100K.

So.. you seem to be saying , the value is not critical. Use 10K to 100K.  Is that what you're saying , I just want to be sure ?

[HotBluePlates]

... most builders of Fender type amps choose to go with only one input jack on each channel , I always see 33K instead of 68k. So , if there is no difference , why not stay with the 68k ?

Many amp builders copy whatever came before without really understanding what is going on.

Others have read what's going on and found out that when plugging into the "High" jack, the 68kΩ resistors are in parallel. So they use 33kΩ, because 34kΩ is not a standard value. In their own way, they are also just copying what came before.

The value is not critical. Use 10K to 100K.

So.. you seem to be saying , the value is not critical. Use 10K to 100K.

  No, that's exactly what he said.

If you only have 1 of these resistors per channel, then the only function of the resistor is to stop V1 from becoming an a.m. radio. The grid stopper does that by cutting high frequency response so that the radio issue never has a chance to happen. You can use the smallest value needed to accomplish that goal (maybe 10kΩ, like Sluckey suggested).

Or the resistor can be as large as you want, as long as it doesn't roll off highs you wanted from your guitar (the 100kΩ upper limit Sluckey suggested).

Your guitar cable may also be rolling off highs (due to cable capacitance), which could magnify how much impact the grid stopper seems to have. On the other hand, most guitar speakers are rolling off by 4-6kHz, and are well down by 10kHz. Add it all up, there's a lot of leeway in an acceptable value, which will likely be determined by your guitar cable and ears.