Interstage Impedance Question

[DylanBenton]

Hi,

Not the first person to be stumped by impedance, but I was wondering when setting the values between stages, where do the components that increase and decrease each respective impedance start and end?

For example, If I had a Guitar (Hi) going into an opamp buffer, I would expect a Low impedance output on the buffer, but I'm wondering how a volume control or resistor after it changes the output Z (if at all) and where components after the buffer start to influence the input Z of the next stage (Or what separates the 2).

I hope I've articulated that correctly, but here's a basic block diagram of what I am trying to ask

[FiveseveN]

Output Z is a feature of a source and input Z is a feature of a load. Z in series makes it bigger, parallel makes it smaller.
Does it help if you stick to DC and only look at the resistance?

[FSFX]

Hi,

Not the first person to be stumped by impedance, but I was wondering when setting the values between stages, where do the components that increase and decrease each respective impedance start and end?

For example, If I had a Guitar (Hi) going into an opamp buffer, I would expect a Low impedance output on the buffer, but I'm wondering how a volume control or resistor after it changes the output Z (if at all) and where components after the buffer start to influence the input Z of the next stage (Or what separates the 2).

I hope I've articulated that correctly, but here's a basic block diagram of what I am trying to ask

The answer to this is to calculate the Thevenin equivalent for the circuit. That is something that is one of the first thing to learn along with Ohms Law and Norton equivalence.
It may be a bit much for beginners or amateur pedal builders to grasp initially but is the first principles of electronics that EEs get taught.

Simulation programs like TINA-TI or LTSpice allow you to determine the impedance without resorting to any maths, but to really understand what is happening does require using the application of Thevenin's Theorem. to the circuit.

The effect of the position of the potentiometer may not be quite intuitive, but if you work it out then if it is driven from a low impedance, the potentiometer has its lowest impedance of one quarter of its resistance value when it is at its centre position. 

[antonis]

Graphic illustration of what Fiveseven said..

[FSFX]

Graphic illustration of what Fiveseven said..

With respect, I don't think you are applying Thevenin correctly to the impedance that you state out of the potentiometer.

[antonis]

With respect, I don't think you are applying Thevenin correctly to the impedance that you state out of the potentiometer.

I presume that (Z3 + 10k - X)*X / (Z3 + 10k) should puzzle OP more..

Let's make it more comprehensible..


@DylanBenton: You can sit at any point you like and look around yourself to estimate particular point impedance..
(what's behind you is considered previous stage Zout and what's ahead is considered next stage Zin..)

[ElectricDruid]

The "simplified" way I usually think about it is to consider Zout as a resistor in series with the output and to consider Zin as a resistor to ground. Seen like that, it's clear that series resistance after an output is increasing the impedance, and that resistors to ground in front of an input are decreasing it since they're in parallel with Zin.

It also makes clear why any of this matters, since Zout and Zin make your classic voltage divider, so if Zout is large and Zin is small, you just turned down your signal!

[antonis]

Seen like that, it's clear that series resistance after an output is increasing the impedance,

That might lead into eroneous induction, Tom..
(although, you're absolutely right..)

A classic example is the resistor in series with the Base of an BJT CE amp (like in Big Muff) ..
Many people think it raises stage's impedance..

[FSFX]

The output impedance of op amps and the effect of negative feedback on the resultant output impedance is explained fully in this TI video tutorial.

https://training.ti.com/modern-op-amps-output-impedance

   

[DylanBenton]

Thanks guys,

Sorry for the late reply. Been running around like crazy.

Ok, i think ive got it, Assuming the input impedance of an opamp ~ 10M and the output Impedance is roughly 200 ohms, does this look correct? (Pot at 50% Linear taper).



The

@DylanBenton: You can sit at any point you like and look around yourself to estimate particular point impedance..
(what's behind you is considered the previous stage Zout and what's ahead is considered the next stage Zin..)

One issue I'm facing is, where do I determine where the values for input and output begin and end? for example, what stops me from using the 1M in parallel for the opamps output calculation (Even though it doesn't change the resistance being parallel) or taking the influence of the pot into the next stages input Z Calculation? I know in this example it won't actually change it, but there must be more complex situations where it does.

I feel im almost there, just need clarification

[antonis]

Ok, i think ive got it, Assuming the input impedance of an opamp ~ 10M and the output Impedance is roughly 200 ohms, does this look correct? (Pot at 50% Linear taper).

Wrong for Zout..
(which is 5k200)
Lower 5k should be considered in parallel with 1M//Next stage input impedance..

One issue I'm facing is, where do I determine where the values for input and output begin and end?

On the very same point..
(which point is where a voltage dividing effect happens.. e.g. the joint of 5k resistors above..)

[DylanBenton]

Oh! i think it's clicking!

Graphic illustration of what Fiveseven said..

I think I had overcomplicated it in my head

All I need to remember is the divider is the "splitting point" between out + in z.

[PRR]

> 200 ohms, does this look correct? (Pot at 50% Linear taper).

Assuming Audio, so the capacitor may be "low low impedance", and likewise for opamp out, and opamp-in  verylarge, then yes. The halfway-up wiper is 1/4 of the total pot.

But the fact you have to ask (and onlookers are disputing), IMHO means you should do a month of Boot Camp. Do a hundred Ohms Law problems before breakfast. Just resistors!! Master that before you even admit there are amplifiers.

Example (and not the best): https://physics.info/circuits-r/problems.shtml

[DylanBenton]

Yeah. I do.

Trying to multitask a 5 year old, work and love life leaves me little time, but luckily im making some.

I'll go back to the study desk and try again

[antonis]

> 200 ohms, does this look correct? (Pot at 50% Linear taper).
Assuming Audio, so the capacitor may be "low low impedance", and likewise for opamp out, and opamp-in  verylarge, then yes. The halfway-up wiper is 1/4 of the total pot.

I think I'm confused now, Paul..

Are we talking about 10k or 1k audio pot..??