The 1M resistor doesn't mean anything to it. It could be 1000M or 100 ohms. The unbalanced source resistance is still 47K and thus, the input impedance is still 47K.

Do you understand how to calculate the resistance of two resistors in series?

 

Do you understand how to calculate the resistance of two resistors in series?

I guess you don't understand DC coupling resistors do not establish a current path for operation of the next stage, but instead incorporate with the voltage source as the source resistance. Thus there is no current flow like you are trying to imagine. Because the input signal + is after the 47K and not the input jack.

 

The amplifier input resistance is 1047 k ohms assuming the grid does not swing positive with respect to the cathode. In normal operation the grid can be assumed to be an infinite resistance in parallel with its grid-to-cathode and heater capacitance plus its grid-to-plate capacitance X stage gain. In the case of a 12AX7 stage having gain of 60 this works out to be 1.6 pF + 60 x 1.7 pF = 104 pF. The impedance of this capacitance is -j77k ohms at 20 kHz. The tube input capacitance dominates the input impedance above ~1.5 kHz.






12AX7 datasheet https://www.drtube.com/datasheets/12ax7-rca1962.pdf

 

The amplifier input resistance is 1047 k ohms assuming the grid does not swing positive with respect to the cathode. In normal operation the grid can be assumed to be an infinite resistance in parallel with its grid-to-cathode and heater capacitance plus its grid-to-plate capacitance X stage gain. In the case of a 12AX7 stage having gain of 60 this works out to be 1.6 pF + 60 x 1.7 pF = 104 pF. The impedance of this capacitance is -j77k ohms at 20 kHz. The tube input capacitance dominates the input impedance above ~1.5 kHz.




View attachment 322334

12AX7 datasheet https://www.drtube.com/datasheets/12ax7-rca1962.pdf

I wasn't going to introduce the concept of input capacitance (especially Miller capacitance) until our friend @LadySpark had correctly understood the resistance.

 

The amplifier input resistance is 1047 k ohms assuming the grid does not swing positive with respect to the cathode. In normal operation the grid can be assumed to be an infinite resistance in parallel with its grid-to-cathode and heater capacitance plus its grid-to-plate capacitance X stage gain. In the case of a 12AX7 stage having gain of 60 this works out to be 1.6 pF + 60 x 1.7 pF = 104 pF. The impedance of this capacitance is -j77k ohms at 20 kHz. The tube input capacitance dominates the input impedance above ~1.5 kHz.




View attachment 322334

12AX7 datasheet https://www.drtube.com/datasheets/12ax7-rca1962.pdf

The amplifier input resistance is 1047 k ohms assuming the grid does not swing positive with respect to the cathode. In normal operation the grid can be assumed to be an infinite resistance in parallel with its grid-to-cathode and heater capacitance plus its grid-to-plate capacitance X stage gain. In the case of a 12AX7 stage having gain of 60 this works out to be 1.6 pF + 60 x 1.7 pF = 104 pF. The impedance of this capacitance is -j77k ohms at 20 kHz. The tube input capacitance dominates the input impedance above ~1.5 kHz.




View attachment 322334

12AX7 datasheet https://www.drtube.com/datasheets/12ax7-rca1962.pdf

That is analysis of the 1st amplification stage without a voltage source. The input resistance of the stage with a voltage source connected is 1M.

 

To continue the input impedance thing here is the scalar value of input impedance from 20 Hz to 20 kHz. The Y-axis is in megaohms.

 

That is analysis of the 1st amplification stage without a voltage source. The input resistance of the stage with a voltage source connected is 1M.

Hurrah! (almost).
Actually it's 1.047M, because the 47k is in series;
and the input resistance of the amplifier is the same regardless as to whether there is a source connected.

 

To continue the input impedance thing here is the scalar value of input impedance from 20 Hz to 20 kHz. The Y-axis is in megaohms.

View attachment 322337

If you want to model the entire system, then a good model of a guitar pickup is 6H in series with 6kΩ, and then you need some cable capacitance.
(and if you want the whole system to look like a complete mess, add in the "tone stack"!)

 

Hurrah! (almost).
Actually it's 1.047M, because the 47k is in series;
and the input resistance of the amplifier is the same regardless as to whether there is a source connected.

Its pretty obvious you guys want to argue about it instead of learning about something you don't know.
Bye.

 

The 1M grid resistor connects the grid to 0V.
Current in the 1k cathode resistor self-biases the grid/cathode so that the cathode has a positive voltage reducing the plate/cathode current.

 

Its pretty obvious you guys want to argue about it instead of learning about something you don't know.
Bye.

Read up on resistors in series before you come back!

 

It has high gain and low gain inputs. Low gain giving an attenuation of 6dB.

Look again. Both inputs are identical.

 

Look again. Both inputs are identical.

 

View attachment 322338

Ah, very perceptive! Let's see if I understand this:
Guitar Input 2 only plugged in the guitar sees 92k ohms and the input signal is attenuated by ~50%.
Guitar Input 1 only plugged in the guitar sees 1047k ohms and the input signal is attenuated by ~5%.

 

When 2 is plugged in, 1 still sees a path to ground through input 2. Assuming it is low impedance (compared to 47K) , there is little change to input 1.

 

When 2 is plugged in, 1 still sees a path to ground through input 2. Assuming it is low impedance (compared to 47K) , there is little change to input 1.

But it's not intended for using with two guitars simultaneously. Whenever have you seen two guitarists share an amplifier? That would imply that two guitarists could agree on which amplifier to use!

 

Ah, very perceptive! Let's see if I understand this:
Guitar Input 2 only plugged in the guitar sees 92k ohms and the input signal is attenuated by ~50%.
Guitar Input 1 only plugged in the guitar sees 1047k ohms and the input signal is attenuated by ~5%.

ALmost.
The guitar sees 47K reguardless of input, but plugging into input 2 keeps the amp in default low gain mode as the grid bias resistance is 44.9K. Loading of the guitar pickups are the same and thus, no change in tone.

 

Ah, very perceptive! Let's see if I understand this:
Guitar Input 2 only plugged in the guitar sees 92k ohms and the input signal is attenuated by ~50%.
Guitar Input 1 only plugged in the guitar sees 1047k ohms and the input signal is attenuated by ~5%.

You are perfectly correct.
(We are assuming that there is only one guitar plugged in at any time, and that the choice
of inputs exists in order for the guitarist to choose high gain or low gain, not for the purposes of plugging in two guitars)
There is a good explanation here
https://guitar.com/features/opinion...ifference-between-high-low-inputs-amplifiers/
Note that the low level input has a lower input impedance.

 

I wasn't going to introduce the concept of input capacitance (especially Miller capacitance) until our friend @LadySpark had correctly understood the resistance.

I pull that info from a text book that is out of print for a few decades and is not scanned into the internet. I say that was a good test of your lack of knowledge on the subject and reaffirms what I already know.

 

I pull that info from a text book that is out of print for a few decades and is not scanned into the internet. I say that was a good test of your lack of knowledge on the subject and reaffirms what I already know.

If that’s what you learned from, it’s a good job it is out of print.