Modeler "Input Impedance" -vs- Real Tube Amp "Input Impedance" ?

[jay mitchell]

I don't quite follow? To clarify, I didn't mean to imply that the value of the input impedance didn't matter. Just that 1Mohm input impedance is 1Mohm input impedance regardless of whether its a tube amp, a solid state pedal, or a digital whatever.

While that is a true statement, I haven't seen an assertion to the contrary.

 

[Boudoir Guitar]

While that is a true statement, I haven't seen an assertion to the contrary.

I just found your statement ambiguous, not contrary, which is why I sought clarity

 

[ucnick]

"real-tube Amps" vs "modeler" doesn't really have a distinction in this case. The "input impedance" on a modeler is defined by the analog circuitry prior to the DAC or anything else, and in a tube amp (or pedal in front of a tube amp) before any gain stage, tube or otherwise. I.e, to the extent anyone is worried that there is a difference between "real" input impedance, "tube" input impedance, "digital" input impedance etc., that conversation doesn't really make much sense.

Actually, not to nit pick, but "prior to the ADC", not "DAC".

 

[Humbug]

Actually, not to nit pick, but "prior to the ADC", not "DAC".

Technically correct, which is the best kind of correct.

 

[Boudoir Guitar]

Technically correct, which is the best kind of correct.

If we wanna get nipicky, it's actually technically incorrect.

The correct statement would be "prior to the ADC and the DAC" not "prior to the ADC, not the DAC".

But the point was made clearly enough for a forum such as this, so who's counting...

 

[jay mitchell]

The correct statement would be "prior to the ADC and the DAC" not "prior to the ADC, not the DAC".

Everything that affects the impedance the guitar sees is "prior to" the ADC, the DSP, the DAC, the output buffering, etc., etc. The only thing that affects the impedance load on the guitar is the first stage of the analog input. It has nothing to do with digitization.
I'm aware that you know this, I'm just trying to keep it simple and unambiguous.

 

[Boudoir Guitar]

The only thing that affects the impedance load on the guitar is the first stage of the analog input. It has nothing to do with digitization.

Thanks for making my original point significantly more clearly

 

[mercifulfuzziness]

I assumed you were getting ready to go split hairs in this vs. that land

I read fractal is better in this

 

[HomespunEffects]

The input impedance on a real-world amp is set at the input jacks. You can check the schematics.

 

[JoeBfstplk]

The input impedance on a real-world amp is set at the input jacks. You can check the schematics.

Yup.

A typical black panel Fender with dual inputs has a pair of 68k resistors to mix the two inputs (and offer at least some isolation from the two guitars' controls interacting) when both inputs are used. These both connect to the input grid and a 1M resistor tying rhe grid's DC potential to ground.

When Input 1 is used alone, the jacks' switching contacts put the two 68k resistors in parallel, reducing them to 34k, for a total of 1034k input impedance (+/- 10%).

When Input 2 is used alone, the same switching instead puts the Input 1 jack's hot to ground, making the three resistors into a voltage divider, with 68k in series with the Input 2 signal and Input 1's 68k in parallel with the 1M grid resistor, giving 63.67k for thelower half of the voltage divider, giving 48.355% of the original signal in (-6.31dB) and approximately 131k input impedance....

When jumpering from Input 1 to another channel's Input 2, you get more compliated math, but end up with the guitar seeing about 592k total load....

 

[HomespunEffects]

Yup.

A typical black panel Fender with dual inputs has a pair of 68k resistors to mix the two inputs (and offer at least some isolation from the two guitars' controls interacting) when both inputs are used. These both connect to the input grid and a 1M resistor tying rhe grid's DC potential to ground.

When Input 1 is used alone, the jacks' switching contacts put the two 68k resistors in parallel, reducing them to 34k, for a total of 1034k input impedance (+/- 10%).

When Input 2 is used alone, the same switching instead puts the Input 1 jack's hot to ground, making the three resistors into a voltage divider, with 68k in series with the Input 2 signal and Input 1's 68k in parallel with the 1M grid resistor, giving 63.67k for thelower half of the voltage divider, giving 48.355% of the original signal in (-6.31dB) and approximately 131k input impedance....

When jumpering from Input 1 to another channel's Input 2, you get more compliated math, but end up with the guitar seeing about 592k total load....

i was told there’d be no math.

 

[Boudoir Guitar]

The input impedance on a real-world amp is set at the input jacks. You can check the schematics.

As is the input impedance on a modeler.

 

[HomespunEffects]

As is the input impedance on a modeler.

The point is you can look at the components on a real amp and know the input impedance.

 

[jay mitchell]

Yup.

When Input 1 is used alone, the jacks' switching contacts put the two 68k resistors in parallel, reducing them to 34k, for a total of 1034k input impedance (+/- 10%).

This is inaccurate. The input device (guitar) to Input 1 sees the 1M resistor to ground. The two 68k resistors are indeed in parallel, but their output sides see the grid of the input tube, which is not at ground potential. In normal bias mode (grid nonconducting) - which is always the case with a guitar signal - the grid-cathode circuit is a reverse-biased diode, which presents an extremely high, mostly capacative, impedance. Ergo, the guitar sees 1M in parallel with a series circuit consisting of 34k resistance and the grid/plate capacitance (multiplied by the Miller effect). IOW, very close to 1M, not 1.034M.

 

[jay mitchell]

The point is you can look at the components on a real amp and know the input impedance.

You can do exactly the same with a modeler. I've had occasion to do just that.

 

[HomespunEffects]

You can do exactly the same with a modeler. I've had occasion to do just that.

Again, the point is that while you can set the input impedance on a modeler with a setting in the input block, on any real amp, you can look at the schematic. I’m just participating in the actual topic of this thread without pedantics.

 

[Boudoir Guitar]

The point is you can look at the components on a real amp and know the input impedance.

The exact same components exist in a modeler input stage? On many you don’t need to crack the case to see the value of resistors connected - just go to the input impedance setting to see.

 

[jay mitchell]

Again, the point is that while you can set the input impedance on a modeler with a setting in the input block,

Not generally. I know of two examples that allow you to do that. I can name quite a few other examples in which the input impedance is fixed.

on any real amp, you can look at the schematic. I’m just participating in the actual topic of this thread without pedantics.

On most modelers, you can also look at the schematic. The ones with user-settable input impedances are in the minority.

 

[HomespunEffects]

Not generally. I know of two examples that allow you to do that. I can name quite a few other examples in which the input impedance is fixed.

On most modelers, you can also look at the schematic. The ones with user-settable input impedances are in the minority.

What I’m saying is none of that answers the questions posed in the OP.

 

[jay mitchell]

What I’m saying is none of that answers the questions posed in the OP.

I answered them in post #15.