LTSpice and MXR Dist+ Impedance question

[bushidov]

Hi All,

Sorry to keep bothering everyone, but I keep getting curious about things and have questions. So, this time, its the MXR Distortion+. Using this as my schematic:


Per ElectroSmash, the input impedance is calculated as:
Zin = R1 + ( R2 // ZinOp-Amp ) + ZVirtual Ground
Zin = 10KΩ + ( 1MΩ // 2MΩ ) + 500KΩ = 1176KΩ or around 1M

I know ElectroSmash is something to take with a grain of salt, but this also matches with the official Jim Dunlop site for the officially declared Input Impedance of the pedal. However, because of C1 and it's reactance, shouldn't it be much much lower, like under 200K at 1kHz?

What is weirder is that LTSpice seems to agree.



At 1kHz, it is giving me around 150K of impedance.

Am I measuring it wrong?
Is the "declared impedance" called out for a different frequency (and I was wrong to assume 1kHz)?
Is this just something LTSpice doesn't do out of the box?

[antonis]

At 1 kHz, C6 capacitive reactance is 159 Ohms so you can safely ignore Zin virtual ground (500k)

C2 capacitive reactance is 15k900 added in series to R1..

So, you have 26k + (1M//2M) -> 692k at 1kHz..

No comes into game V1 output impedance toghether with C1 reactance at 1kHz (159k)..

[bushidov]

Doesn't the reactance of C1 play a role?

Also, when a pedal company is quoting "Input Impedance", at what frequency are they quoting it at, as by definition, impedance is frequency dependent?

[antonis]

Sorry..!!! (typo)

"Now" instead of "No"..

[PRR]

It may be clearer if you limit the frequency range and plot on log paper. But my Idiot agrees with yours.