output impedance ?

[chunks717]

Hay guys..................................................................
    so, i understand that impedance is the sum of reactance(cap/or induct)...and resistance.  in the case of a 1M resistance, the reactance is comparatively minuscule, and can for practical purposes be ignored.  that takes care of matching the pick-ups Z.  If the next pedal in the chain offers 1M input z also, than for maximum power transfer, it reasons to have the outputs be 1M as well, as i have frequently seen,  tho I'm not sure if that is as simple as just adding the res. to ground, as the output of some trans. set ups seems to be much lower z.  Now we come to the amplifier........which tend to have lower input impedance...right?
   any thoughts on matching Z from stage to stage?....can this be ignored as long as the end result matches the input of the next circuit?
   also, what would be an average input Z of a tube amp?  are solid state amps much different?(by design?)
   If my assumption is correct, than one would do well to offer the amp a lower Z input signal..............thank you all.

[R.G.]

so, i understand that impedance is the sum of reactance(cap/or induct)...and resistance.

Correct.

in the case of a 1M resistance, the reactance is comparatively minuscule, and can for practical purposes be ignored.

Not necessarily correct. The reactance of a cap goes down as frequency goes up, and the reactance of an inductor goes up as frequency goes down. So at frequency extremes, the parasitic capacitance and inductance MAY be a problem. In fact, at high frequencies, the parasitic capacitance across the leads of a 1M resistor becomes a factor quickly. Inductance is generally less of a problem with high resistances, but can get to you with low resistances.

For a 1M resistor, parasitic capacitance of C = 1/(2*pi*20,000*1M) = 8pF causes an effect at 20kHz audio signals. Resistors much above 1M get to being more capacitive than resistive at high audio frequencies. You have to compute the frequency. But for guitar signals that mostly lie under 7kHz, yes, you can start by ignoring the reactances. You gotta know the numbers.

that takes care of matching the pick-ups Z.  If the next pedal in the chain offers 1M input z also, than for maximum power transfer, it reasons to have the outputs be 1M as well,

And here we have a problem. Maximum power transfer is NOT what you want for audio signals. You want maximum voltage transfer, so that means you want the driven side of the transfer to be more than 10X the driving side impedance, preferably more than 100 or 1000 times. Guitar pickups have an impedance that is complex (composed of winding resistance, self capacitance, and the inductance of all those turns of wire) that is perhaps as low as a few K at DC and goes up to maybe 100K or before self capacitance starts making fall off again. That happens out at 5K-10kHz. It depends on the pickup, whether single coil or humbucking, and how it's wound.

So - maximum power transfer is NOT what you want.

Now we come to the amplifier........which tend to have lower input impedance...right?

No. Most tube amps have a 68K resistor or two to mix two inputs, then a 1M to ground; this is followed by a triode grid, which is very much like a few pF cap to ground.
Transistor amps sometimes have lower input resistances. This makes the guitar sound much duller as it loads down the high frequencies where the pickup impedance gets quite high.

any thoughts on matching Z from stage to stage?

Yes. Don't. MISMATCH to keep all the signal voltage you can.

also, what would be an average input Z of a tube amp?  are solid state amps much different?(by design?)

See above.

If my assumption is correct, than one would do well to offer the amp a lower Z input signal..............thank you all.

No, it's actually better to make it as high as you can get it.

[chunks717]

sir, thank you, sir

still.......... hi in, low out, hi in, low out..stage to stage, circuit to circuit (by a magnitude of 100-1000x).........up until the amplifier and then "as high as you can get it" into the amp?!?
pardon my simplification............................

(Aron, thanks for the spell check, that I may come off as less the buffoon!)

[chunks717]

I suppose that my next question would be....
If I were to build a buffer circuit to condition my signal post pedal chain, and pre amp, what characteristics would this signal posses?
(assuming all other preferences being equal?.....or perhaps a few cases?)

[Transmogrifox]

Matching input impedance for maximum power is applicable to RF receivers and power amplifiers.  You have a radio wave, which is electromagnetic energy in the air.  On the input of a high frequency receiver, you don't want to amplify the wideband signal sensed on the antenna before you filter or you amplify the noise (generally out-of-band signals), causing input stage to overload and distort at low signal levels, thus you demodulate the noise and have a real mess to sort out.

Therefore, you really want to do passive filtering on the input to select a very narrow band of frequencies on the front end of the receiver.  Because it's passive, you need to retain as much power in the signal as possible, and generally the components are almost strictly reactive.  Reactive elements store energy, resistors waste energy (unless you're making a heater, then the energy burned off in a resistor wouldn't be considered "wasted").  This is where you need impedance matching.  If you have a well designed receiver, then each filter-->amplifier-->filter-->mixer.... stage is input/ouput impedance matched for a narrow band of frequencies, retaining any precious power you can, since over-use of amplifier stages adds noise, and increases the challenge of keeping things from interacting in undesirable ways..

In audio, you want to amplify a very wide bandwidth (proportionally speaking, though 20kHz is a pretty narrow bandwidth in the 900MHz range).  This brings us back to what RG stated.  You are looking at voltage only.  There is little consideration for signal power until you get to the output stage of your amplifier.  For the amplifier, you want the output impedance matched to the input impedance.  For the preamplifier, it doesn't matter if you're burning off power on input resistors since they're all such low-power circuits anyway.

As a side note, you could theoretically make a series of power amplifiers from the guitar to a speaker with matched input/output impedances with almost strictly reactive components.  The added complexity would probably result in higher noise, higher distortion and lots of expensive parts.  There's a good reason the audio world has taken to high input impedance/low output impedance amplifiers.

In response to your second question:
Everything should be high in-->low out

Your amplifier input is not a special case where it needs to be higher input impedance than your stompboxes.  It should be just as high (like 300k to 1M)

[Transmogrifox]

I suppose that my next question would be....
If I were to build a buffer circuit to condition my signal post pedal chain, and pre amp, what characteristics would this signal posses?
(assuming all other preferences being equal?.....or perhaps a few cases?)

You snuck that one in. 

The buffer is what allows you to insure that you have a low output impedance to drive your amp, so you don't get much signal loss in the cable and amplifier input.  You can drive low input impedance amplifiers without getting as muddy with a good buffer.

If you're using a guitar preamp last before your amp, you probably don't need a buffer because the preamp more than likely has a very low output impedance and the juice to drive as low of input impedance you'll ever see on a guitar amp.

[chunks717]

trans..........
   sorry for the poor choice in phrasing, by "post ped, pre-amp- i simply meant between the last effect and the amplifier, but i think that answers my question, in that a pre-amp is a good way to end a pedal chain.....buffer may be just as well, (ish) yeah?

[Transmogrifox]

Buffer would be equivalent.  A preamp is a buffer with features such as increased signal level, eq, sometimes distortion and compression.   It doesn't hurt to have a buffer at the end of the pedal chain because you may have times where all of your FX and your preamp are bypassed, thus leaving your guitar to drive a long cable to the amp on its own.  The buffer would supply the guitar with a high input impedance, then drive the cable and amplifier with low output impedance.

If you have built all of your own pedals, and you know they all have a low output impedance, then it would be even better to put your buffer in front of the pedal chain so it's the first thing your guitar sees.  When your pedals are on, they're driving the long section of cable.  When bypassed, the buffer is still driving the long cable.

[chunks717]

aha................back to what has long been stated (buffer at the beginning.........)
thanks for the information, both of you.

I'm sure that R.G. had a reason for stating...
-If my assumption is correct, than one would do well to offer the amp a lower Z input signal..............?
"No, it's actually better to make it as high as you can get it."

curious to hear that perspective................also I have appreciated the rapid fire "dialog" style responses....y'all are fantastic!

[Transmogrifox]

I'm sure that R.G. had a reason for stating...
-If my assumption is correct, than one would do well to offer the amp a lower Z input signal..............?
"No, it's actually better to make it as high as you can get it."

curious to hear that perspective................also I have appreciated the rapid fire "dialog" style responses....y'all are fantastic!

I must have misunderstood your question.  You want to OFFER the amp a very low input impedance, but if you were to modify your amp's input impedance, you would want to make it larger.  That was the direction of my statement "you want it [amplifier input impedance] as high as you can get it".  To clear any misunderstanding, it follows the same pattern: High Z in, Low Z out.  Even between preamp stages within the amplifier, this principle holds true.

[chunks717]

.........contrary, you read the q's perfectly................thanks again
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