Fender input resistance / V1 grid grounding resistance

[sbirkenstock]

Hi,

Fender used in their standard BF/SF (did not check the tweeds) a 1 meg resistor and two 68k resistors.
I thought that the 1 meg is the grounding of the grid in V1.

If I put in a guitar in input 1 (either Normal or Vibrato Chanel)
the guitar "bypasses" the 1 meg. (I use a Strat with 5,7K, all pots on 10)
The resistance between the grid of V1 and ground goes down to about 40K with the guitar plugged in.

In input 2 the 1 meg is shortcut by the jack of input 1, so it does not matter at all.

So what is the 1 meg resistor good for?

Best regards,

Stephan

[tubeswell]

With a jack in the 'hi' socket only, you have a high input impedance (1M) and 34k of series resistance (68k||68k). The input impedance acts together with the output impedance from the guitar (together with the series resistance of the 34k) as a voltage divider that preserves more of the signal than if you have the jack inserted into the 'lo' input jack. The 34k series resistance forms a R/C filter together with the inter-electrode capacitance of the 1st triode, to roll off unwanted HF.

With a jack in the 'lo' socket only, you have a lower input impedance (68k) and 68k of series resistance forming a 1:1 voltage divider. THis amount of voltage division means the signal will be attenuated more (than if you have the jack inserted into the 'hi' input jack). The 68k series resistance forms a R/C filter together with the inter-electrode capacitance of the 1st triode, to roll off more HF than in the above scenario. So you get a weaker signal with more HF reduction.

[sbirkenstock]

Thank you very much for the explanation!
I installed a little switch to put the 1meg in and out and it does make an audible difference.
On the value of the grid stopper there is a nice article on the web.
http://www.valvewizard.co.uk/gridstopper.html
They do come to 40K with a rolloff above 20K Hz.
Which is very close to 34k.

What I did not understand is, why they take the complete Miller capitance into the calculation.
Complete = Cathode to Grid + Grid to Plate
Shouldn´t it only be the Cathode to Grid part?

[JazzGuitarGimp]

The output impedance of a power supply is very low. For purposes of calculating miller capacitance effects, we actually consider that impedance to be zero. So the miller capacitance from the grid to the plate can also be thought of as capacitance from the grid to ground. Now it's easier to see that the two miller capacitances are actually in parallel, so Cm1 + Cm2.

[sbirkenstock]

If the output impedance of the power supply is 0 this makes sense.
But then the signal would be lost as well?
So we usually have a 100K plate load resistor.
Or 50K plus 50k on the cathode.
Doesn´t the plate load resistor have an impact on the grid to plate capacitance?

[Malcolm Irving]

I went back to the original paper by Miller:

http://web.mit.edu/klund/www/papers/jmiller.pdf

and found that the usual formula that we use for the input capacitance at the grid:

Cin = Cgk + Cga(1 + A)

is actually an approximation. The accurate formula does take into account the plate load resistance Ra and the internal plate resistance ra of the tube, as follows:

Cin = Cgk + Cga(1 + A.Ra/(ra+Ra))

If Ra >> ra then the approximate formula is good.

[Malcolm Irving]

Correction to the above:

It depends on what you take 'A' to be. If it is the actual voltage gain of the stage then the usual formula is OK. In the Miller paper formula the 'A' is the mu of the tube.

So, in other words, the plate load and internal plate resistance are already being taken into account in the usual formula, according to the way we calculate the voltage gain 'A'.

[martin manning]

Nice! Thanks for posting this, Malcolm.

[JazzGuitarGimp]

Good stuff. Thanks Malcolm...