High vs. low input impedance

[bassmeister]

I've seen designs for guitar effects with input impedances ranging from lower than 100 kOhm up to several MOhms. High input Z is always preferable to avoid loading down the guitar signal, as far as I know. Why isn't as high input Z as possible always used? High Z = more noise, certainly. But, do there exist designs that could suffer from too high input Z? I'm not very experienced (Are you...?! :wink: ) with Fuzz faces, but I think that this effect interacts a lot with the guitar pickup, a case where the effect would suffer from having a too high input Z.

Could anyone enlighten me or direct me to a source on the net regarding this issue?

Thanks!

[onboard]

Have a read through the "Technology of the Fuzz Face" over at geofex.com - the whole circuit is covered from front to back with a good explanation of the why's and how's, not just the what.

I'm not very experienced either, but *some* things are starting to make sense... :wink:

[niftydog]

Input impedance is generally a trade off between signal loading and noise. Too low an impedance loads a signal, but makes for lower noise entering the circuit. Too high an impedance doesn't load the signal, but allows noise to pass more easily.

Any circuit with a high gain stage could suffer from having too high an input impedance.

The benefit of a 1Mohm impedance vs. a 100kohm impedance is quite small, in fact almost negligable. However, the signal to noise ratio might be drastically affected.

see my post near the bottom of this old thread.

[petemoore]

With even numbers I attempt to type my first equation...hope it works..
 First circuit outputs 1v into second circuit, first circuit is 1k output impedance second circuit is 2k input impedance so:
 2k / [1k] = .5v is what the input to the second circuit would see under these conditions ?
 Correct ?

[niftydog]

think of it as a voltage divider circuit.

The output voltage you read with a CRO is not affected by the output impedance of the effect due to the very high input impedance of the CRO. (you effectively see the source voltage with NO (read; extremely low) output impedance.) Thus, imagine an output as being a perfect voltage source followed by a series resistor representing the output impedance.

Then, at the input of the next effect you can just imagine it as being a resistor to ground. For the sake of completeness, imagine this resistor hangs from an infinitely high input impedance device. (ie; a perfect op amp)

Thus;

Code Select Expand

OUTPUT SIGNAL SOURCE
  |
 .-.
 | | 1k OUTPUT Z
 | | FX 1
 '-'
  |
  o------o Vout effective into infinity ohms input device
  |
 .-.
 | | 2k INPUT Z
 | | FX 2
 '-'
  |
  |
 ===
 GND

Then, to remember the voltage divider equation, I go through this process:

Work out the current through both resistors - then use that current to work out the voltage in the bottom resistor.

The Voltage Divider Equation;
V = (V*R2)/(R1+R2)
Vout effective = (SIGNAL * R2)/(R1+R2)
= (1*2k)/(1k+2k)
= 2k/3k
= 0.66666666666666...

[brett]

So......
In practice most of us use 1M or 2M2 pulldown resistors.  Is anybody suggesting that we might be adding significant noise over a 470k resistor at the input?

(I'm talking about high Z circuits, such as JFET and MOSFET boosters)

Man, I really don't want to go back and re-wire 20 pedals. :cry:

[R.G.]

Why isn't as high input Z as possible always used?

There are a number of reasons.
(1) ignorance; early (in the effects era) designers and guitarists were largely non-technical and simply didn't know the reasons and results of high or low input impedance
(2) intent; some people **like** the treble rolloff of loaded pickups. It's good as one example in front of a distortion that has no bass rolloff and could be prone to getting muddy (See "The Technology of the Tube Screamer", GEO)
(3) cost; it doesn't cost much granted, but in production runs, every penny not spent on a part goes directly to profits.

I guess the more fundamental reason is that not everyone knows or cares what the guitarist playing the pedal sounds like. The commercial world only wants to sell them and the uninformed effects designer only wants the thing to finally work.

High Z = more noise, certainly. But, do there exist designs that could suffer from too high input Z?

That depends on what you mean by "suffer". Tone is so subjective that it's almost impossible to give a universal rule for it.

I'm not very experienced (Are you...?!  ) with Fuzz faces, but I think that this effect interacts a lot with the guitar pickup, a case where the effect would suffer from having a too high input Z.

In the FF case, it was most likely a happy accident (reason 1) because most distortions profit by having some bass rolloff to control muddiness. Most non-distortions don't need or profit by bass rolloff - unless that's part of the sound.

Then there's the whole genre of treble boosters. These things had their genesis at a time when input impedance effects were not well understood (in the limited effects-designer community) and I believe that they were instituted as a counter to the treble loss that low impedance gives you, as well as a modest distortion device.

[petemoore]

think of it as a voltage divider circuit.

The output voltage you read with a CRO is not affected by the output impedance of the effect due to the very high input impedance of the CRO. (you effectively see the source voltage with NO (read; extremely low) output impedance.) Thus, imagine an output as being a perfect voltage source followed by a series resistor representing the output impedance.
  >>>What is "CRO" ?
Then, at the input of the next effect you can just imagine it as being a resistor to ground. For the sake of completeness, imagine this resistor hangs from an infinitely high input impedance device. (ie; a perfect op amp)

Thus;

Code Select Expand

OUTPUT SIGNAL SOURCE
  |
 .-.
 | | 1k OUTPUT Z
 | | FX 1
 '-'
  |
  o------o Vout effective into infinity ohms input device
  |
 .-.
 | | 2k INPUT Z
 | | FX 2
 '-'
  |
  |
 ===
 GND

Then, to remember the voltage divider equation, I go through this process:

Work out the current through both resistors - then use that current to work out the voltage in the bottom resistor.

The Voltage Divider Equation;
V = (V*R2)/(R1+R2)
Vout effective = (SIGNAL * R2)/(R1+R2)
= (1*2k)/(1k+2k)
= 2k/3k
= 0.66666666666666...

>>>What does * signify in these equations? ... I'll try again .. and certainly appreciate all the assistance !!!

[niftydog]

* =  multiply

nothing sinister! Just didn't want to mix up x's with x's and dots with fullstops etc.

[Satch12879]

Niftydog:

Funny, semi-related story...

I recently used this same methodology on another board to disprove that a Mackie mic pre (Zin=1k5 ohm) presents an increased load to a Shure SM-57 (Zout=310 ohm) than a Neve mic pre (Zin=1k2 ohm).  The cat I was arguing against said that the Mackie's increased load is one of the major reasons that its sound is inferior to the Neve.  I could smell the ignorant BS a mile away and so I went to the Mackie, Neve, and Shure websites, got the specs and did the calculation.  Turned out the discrepancy between the two circuits was about 2%.  My counter arguement was since the loading was insignificant, the timbral difference had to be due to 1) the design of the circuits 2) subjectivity by the operator/listener.

After reading your post, I'm glad I wasn't just flapping my gums again!

[niftydog]

awesome! Glad to be a part of the suppression of blatent ignorance.

[amz-fx]

In my mods to the Rat, one of the things I do is lower the input impedance....  this cuts down on the noise, which in a  high gain device like the Rat is very important.  The clipping circuit generates lots of harmonics so the overall effect on the tone is minimal.

http://www.muzique.com/fx/fat-rat.htm

regards, Jack

[bassmeister]

How do you quote a certain person? I just use copy/paste, like this:

In my mods to the Rat, one of the things I do is lower the input impedance.... this cuts down on the noise, which in a high gain device like the Rat is very important. The clipping circuit generates lots of harmonics so the overall effect on the tone is minimal.

Actually, that was one of the designs I had in mind when asking my questions. I understand more now with the answers you have given. Thanks a lot!

BTW, did anyone get the reference to the Hendrix album title?  :wink:

I'm not very experienced (Are you...?!)

[niftydog]

How do you quote a certain person?

Like that?

Like this;

Code Select Expand

[quote="bassmeister"]How do you quote a certain person?[/quote]