Q about Impedance Measurement technique

[fuzzy645]

I have been using the following methods to measure both in & out impedance:

http://www.zen22142.zen.co.uk/Theory/inzoz.htm

These methods are working out well for me.  My question is this: When measuring output impedance, I have noticed that my results vary significantly depending upon the pot rotation, particularly volume and tone pots.   If that is the case, is there some kind of  "standard" to follow with regards to pot rotation position when determining the output impedance.    Should pots be on 100% rotation, or perhaps at 50% rotation.  To me, common sense says the pots should be at 50% rotation as a kind of middle ground.

Input impedance seemed to be totally independent of the pot rotation, so that is a non-issue, just the output impedance varied.

Thanks.

Paul

[CynicalMan]

In most cases it's easier to calculate impedances based on the schematic. What effects are you measuring?

Knowing effect companies, the de facto standard is probably the highest input impedance they can get and the lowest output impedance they can get. That being said, in commercial effects, there should be output buffers unless it's a vintage reproduction pedal.

[brett]

Hi
you can easily measure output impedance by feeding a fixed output signal into a 1 Mohm pot. Turn the pot to 100% and measure the AC voltage. Then turn the pot until the signal falls to half. At this point the pot resistance = the output impedance.
Similarly, a pot between a buffer and the input of an effect can be used to measure the input impedance. At the point of 1/2 voltage loss, the resistance of the pot equals the impedance of the input.
cheers
NB the volume pot of many effects will change the output z. This is not usually an issue if the input z of the next circuit is low (ie 1/4 or less of the output z).

[PRR]

If you have a potentiometer-connected pot as the output:

You are not too concerned about exact impedance

Impedance is zero at "0" and generally (not always) low at "10", but R/4 at half-output ("5" on linear pot, "7" on audio-taper)

[fuzzy645]

If you have a potentiometer-connected pot as the output:

You are not too concerned about exact impedance

Impedance is zero at "0" and generally (not always) low at "10", but R/4 at half-output ("5" on linear pot, "7" on audio-taper)

Thanks PRR.

Are any advantages/disadvantages of relocating the aforementioned volume & tone on the input vs. the output of an overdrive?  It seems to me if I switched them over to the input, then my output impedance will be super low (a desirable trait).   

In a nutshell, here are my measurements when I take the pots out of the picture entirely:
input Z = 100K  (I realize that most prefer this # to be 1M)
output Z = 1K  (nice and low )

Now, when I add the pots on the output, the measurements are:
input Z = 100K (same)
output Z varies depending on pot value from as low as 1K to as high as 35K when the vol/tone pots are maxed out (FYI - these pots are both 100K pots).

It seems to me that if I relocated the vol/tone pots to the input side I can guarantee a low output impedance, but the new questions then become:
1. How will this affect the input impedance (good or bad). Hopefully it won't lower it, as 100K is already getting on the low side.
2. Does the tone change/suffer in anyway by moving the pots to the input side, or is that a non-issue. 

I understand that buffers are the best solution to this type dilemma, but I would rather not "over engineer" this thing when I'm making it just for my own personal use. 

[CynicalMan]

Putting them on the input will make the controls affect the amount of distortion, so it's better to keep them at the end. A 35k output impedance isn't really anything worth worrying about with modern equipment, so I wouldn't sweat it.

[PRR]

> output Z = 1K  (nice and low)

How low is low?

For driving loudspeakers, 1K is absurdly "high impedance".

In most tube design, 1K is absurdly low, means you put too much money where it hardly matters.

In porch design, a similar parameter is "flexibility". How much does it bounce/sag under load?

To figure "how low is low flexibility?" you must know the load. One sleeping dog? All 39 members of the dancing club? Maybe I need to bring my 5-ton plow-truck out of the snow to work on it.

So the amount of lumber you stick under a porch (or the amount of gain/loss you put at in/out jacks) is relative to loading.

For the sleeping dog, anything more than a couple scantlings is a waste of my lumber-dollar. The dancing club may need railroad timbers.


Some typical cases have evolved through design and everyday use.

Guitar pickup is 5K in lows and 100K at top resonance. For minimal loading everywhere, we like guitar inputs to be 500K-1Meg. (Your 100K hardly loads the lows but cuts the highs down to half.)

It is not unreasonable for a "guitar cord device" to have output impedance similar to guitar. Full emulation of guitar impedance is rare; 50K is a happy zone. Higher output impedance, if flat, just leads to more loss... since our boxes have ample gain, loss may be utterly tolerable even at 500K output impedance.

30 feet of guitar cord is 10K at 16KHz, 100K at 1.6KHz, >1Meg below 160Hz. In guitar work we don't demand 20KHz. We do like our highs. But in a box with an amplifier we can usually contrive some high-boost. Indeed an overdriven amp spews highs and cord-loading is one possible way to tame the shrills.

Hi-Fi interfaces today are often 100-1K out and 10K-100K in. This comes "free" with the types of circuits usually used. <1K output preserves >20KHz response despite long cables. However a popular Dyna had hi-Z out and adjusted the power amp input impedance to 50K, and that worked fine with 3-foot cables.

Loudspeakers are around 10 ohms and have large kick-back. The mainstream loudspeaker amp drives with a lower impedance. 1/3 of load from days when it was hard to do anything else. In the 1950s cheap amplification made 1/1000th of load possible and became a "feature". Conversely in music creation the speaker kick-back may be useful. Many Fenders have Zout very similar to load, and other designs have Zout about 10X load.

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> relocated the vol/tone pots to the input side

As Alex says: a universally useful "overdrive" needs two gain/loss controls.

At the input, signal of whatever level must be adjsted so that overdrive begins at a musically useful point in the performer's effort range.

At the output, the level must be adjusted to a useful point in the next stage's signal zone.

Example. Clip a 9V opamp. Output is about 2Vrms. Take this to a typical guitar amp, which overloads by 1Vrms, and the guitar amp will overload before the "overdrive" does. Even the 0.4V of diode clippers is hotter than most normal guitar levels.

Output impedance is a consideration but NOT all that important.