Help me count the buffer In and Out impedance

[Agung Kurniawan]

Hi guys. its been a long time since my last post.
just want to confirm, can you guys please count the in and out Z for me. Im using this kind of buffer on every pedal I have make.

[ashcat_lt]

Assuming that's an FET opamp, you can basically consider the input to be an open circuit and the output to be a short to ground.  So for a "close enough" estimate of impedance, imagine the input of the device doesn't even connect to the opamp, and the output connected to ground where the opamp is.  Now just add up the resistance between each jack and ground.

[key-bored]

Had someone asked me, I would have said the input impedance was 1K.
I'd have also said the output impedance was 1K.
Now, if it's not, I got me some learnin' to do.  But *at* the input, and *at* the output, the impedance is 1K, idnit?

KB

[samhay]

^ no (input) and yes (output).

[GibsonGM]

I'd also say output would be 1k due to the series resistor...changeable based on what follows it.

Now, for input...isn't that 970k in parallel with the opamp's input impedance?  And isn't that massively high if this is FET based?  So it'd be safe to just use 1Meg as a reasonable assumption?   I have never had adverse results from making this sort of assumption, anyway.  This is very much what Ashcat said, I know.

Anyone else?

[Rixen]

Input impedence of an op-amp is transformed by the op-amps gain/feedback. In the case of a non-inverting op-amp circuit it becomes very high, and you only really need to consider the 910k resistor to ground:

HyperPhysics "Practical Benefits:Negative Feedback"

Z_in=(1+A₀B)⋅Z_ino

where

A₀ is the gain without feedback (the open loop gain)
B is the fraction of the output which feeds back as a negative voltage at the input
Z_ino is input impedance without negative feedback

In the case of a TL071 used as a buffer A₀ is 35000 (minimum), B is 1, and Z_ino 10¹² giving an overall of 3.5x10¹⁷ ohms. So the 910k is really the only factor. Even a bipolar op-amp in this configuration will show an extremely high Z_in


In the case of an inverting amplifier the opposite is the case, and the impedence is low, which is why a summing amplifier can work.

[Agung Kurniawan]

ok well, what I have to do to make the in Z to be 1M and the out Z to be 10k?
its a 1/2 of TL072

[Rixen]

Change the 910k at the input to 1M, change the 1k at the output to 10k.

You could get by with a smaller input capacitor, as currently the 100 n at the input would give you a lower frequency cutoff of 1.59 Hz

[Agung Kurniawan]

did you guys see the bias resistor as a 910k? its a 470k
sorry, my bad.

You could get by with a smaller input capacitor, as currently the 100 n at the input would give you a lower frequency cutoff of 1.59 Hz

Im using 100n-470n for in or out cap to get all the frequencys. honestly I dont want to cut those low freq. they bring the power of my sound.

[Agung Kurniawan]

is that anything wrong with that buffer design Im working in?

[Rixen]

no, looks fine

[PRR]

> I would have said the input impedance was 1K.

The far end of the input 1K does not go to ground, or anything very groundy.

Stack a Corgi on an Elephant on the ground. Is this 12 inches high? Or 12 feet high?

It goes to 910K/470K to ground. (Now said to be 470K.) That's the elephant.

And an opamp, but even a '709 amp connected this way will be many-Megs, "negligible".

So it is a hair under 471K input over most of the audio band.

[Phoenix]

You'll also want a DC reference resistor ("anti-pop" resistor), on the other side of the input cap. So that value would be in parallel with the other resistances. So if you used another 470k there, it would be (in the passband) 470k||470k+1k||opamp input impedance ≈ 235k if the op amp is a FET input one, lower if it's a common bipolar type.

I'd also suggest you move the position of the 1k resistor on the input, from before the cap, to after the cap and between the non-inverting input of the opamp and the 470k resistor. The way you have them arranged is a voltage divider, there's no need to attenuate the signal like that at the input, even if it is only a very small amount of attenuation, the other way keeps better signal-to-noise ratio, it's better practice.

Don't know why you'd want the output impedance to be 10k (unless you were making some filter or something), but yes, just replace the 1k with a 10k.

[Agung Kurniawan]

well, thank you everyone

[antonis]

Just have in mind to take into account C1 & R2 for HPF in Phoenix schematic..
(R3 should be only considered in Input Impendance calculation..)

[duck_arse]

like gibson, I read 970k. just saying, for completeness.

[GibsonGM]

Amazingly, my instinct was just about correct on this one. Or more likely, I had read that somewhere re. input Z and it just stuck with me.  I didn't derive it using my math, for sure.

Agung, your "4" really does look like a "9"! 

[Kipper4]

Agung, your "4" really does look like a "9"! 

And his six is a nine

[Agung Kurniawan]

ahahah, sory guys. I am bad on writing