Impedance questions

[Colas LeGrippa]

Hi folks,


Though I ve read a lot about impedance, I'm still confused ( my general state hey hey ).
 Lately I built an amplifier line out box according to Jensen's web site using a.hi fi transformer taken from a McCurdy broadcast board . My goal was to take the speaker out from a 12 w amplifier in order to amplify it with a more powerful guitar amp. It works perfectly. Impedance wise, I am confused. A guitar has an impedance of around  20k ( am I wrong ? ) and Jensen say that their line out device has an impedance of...100 R.  So I think I would have to build a 20k device, from speaker out to amp in, in order to respect impedance, no ?  What kinda issue ( is there any ? ) can I face feeding a tube amplifier with a 100 R device ?


Tnanx
.

[pdf64]

Non power / low level signal voltage connections are generally made using impedance bridging, ie the aim is to avoid the connection losing signal voltage. Ideally, to achieve that, the input impedance should be at least 10x the source impedance, but practically, anything above 3x gives reasonably low loss signal transfer.
I think the treble resonant peak impedance of guitar pickups is typically around 50k - 100k. Hence input impedances much lower than the typical 1M tend muffle the tone, due to the resonance being damped.

[Colas LeGrippa]

Ok so how can I build a 8 ohm (speaker out from my little amp ,) to 1 M ( input to power amp)  line   out  device ?  (Guitar level should I say ).
In other words, what typical resistor arrangement will raise up the.impedance to 1M ?
A voltage divider made of 2 resistors, 1M and 100k will cut the signal by 10.Same result with
1k and 100R. How can we calculate the impedance of each circuit ? ( I understand I can use a 1:1 transformer without using a share gnd, and without changing the impedance ).

[jjasilli]

Hi Colas:   there are already circuits that take signal from a speaker output and step it down for an amp input.  As a practical matter, I suggest you plagiarize one of those.


Re theory: a guitar with passive PU's has an output of, say, 250K. Why? The PU's are typically 4K - 13K; they may be wired to function independently; or be placed in series, or parallel.  BUT, the guitars PU's, vol pot(s) and tone pot(s)are all in parallel with one another.  The pots are typically 250K or 500K.  Given the Rule of Tens, the PU impedance values do not much alter the pot R values.  Note that the pots, in turn (no pun intended), are in parallel with the grid leak R of the amp's input stage, typically 1M.  This somewhat reduces the nominal output impedance of the guitar. 



Impedance matching can be done to favor either voltage, or current.  We're only concerned with voltage.  I.e., guitar PU's don't produce much current.  They produce some voltage and we don't want to lose too much of it.  Anyway, the output impedance of the source device should "match" the input impedance of the input device.  "Matching" does not mean "equal" (except for current matching). To avoid voltage loss the input impedance of the input device should be 10X that of the source device's output impedance.  Note that guitars and guitar amps do not fully follow this rule: at the input, or at insertion point of signal into the typical tonestack.  So, some signal voltage is lost.  Because we're talking impedance, some frequencies are affected more than others: it's hi's that are lost.  This turns out to be OK (given proper overall amp design), because every gain stage adds harmonic overtones (hi's) to the original signal.

[Colas LeGrippa]

Hi buddy,


If my understanding is good,  the higher the resistors values in a given line out voltage divider, the higher the losses  'cause there is a higher voltage build up accross higher resistors.


So the impedance of a 1M / 100k voltage divider would be the dc resistance measured at it's output, that is 100k. The impedance of a 2k / 200R voltage divider would be 200R and therefore would leave the signal more intact.


But there is a problem,  100k as source impedance seems just right to feed a 1M guitar amp input following the rule of 1:10 while 200R  is 1:500....


Colas

[jjasilli]

Not exactly.  It's the proportion value, or ratio, of the R's in a voltage divider that counts.  A voltage divider comprised of 2X 1 Ohm R's; of 2X 200 Ohm R's; or 2X 470K Ohm R's would each divide voltage by 50%.  (Though, higher R values will limit current).


Also, voltage dividers affect frequency response.  See, e.g.:  https://www.analog.com/en/analog-dialogue/studentzone/studentzone-november-2018.html



For the output impedance of a voltage divider, articles are here:  https://www.google.com/search?q=voltage+divider+output+impedance&ei=Y1DCYbaFNKaV0PEP2raq2AM&oq=voltage+divider+out&gs_lcp=Cgdnd3Mtd2l6EAEYATIFCAAQgAQyBQgAEIAEMgUIABCABDIFCAAQgAQyBggAEBYQHjoFCAAQkQI6DgguEIAEELEDEMcBEKMCOgsILhCABBCxAxCDAToLCC4QgAQQxwEQowI6BAgAEEM6CAguEIAEELEDOg4ILhCABBCxAxDHARDRAzoHCAAQsQMQQzoICAAQgAQQsQM6CAgAELEDEJECOggIABAWEAoQHjoECAAQDToICAAQCBANEB5KBAhBGABKBAhGGABQAFjoI2CbNmgDcAF4AYABlgKIAcEikgEHMC4xMi4xMJgBAKABAcABAQ&sclient=gws-wiz

[pdf64]

…
Impedance matching can be done to favor either voltage, or current.  We're only concerned with voltage.  I.e., guitar PU's don't produce much current.  They produce some voltage and we don't want to lose too much of it.  Anyway, the output impedance of the source device should "match" the input impedance of the input device.  "Matching" does not mean "equal" (except for current matching). To avoid voltage loss the input impedance of the input device should be 10X that of the source device's output impedance.  …

Impedance matching means that the output and input impedances are pretty similar.
The ‘voltage matching’ scenario you are describing is usually referred to as ‘impedance bridging’, see https://en.m.wikipedia.org/wiki/Impedance_bridging

[Colas LeGrippa]

I miss something. A voltage divider with an impedace of 5k could.drive a 50k load. But a guitar amp has a 1M load...where taking about 200 times, not 10......

[pdf64]

That’s immaterial.
Provided that the bridging connection impedances have a ratio of AT LEAST 10, then the coupling will be near perfect.
200 x will be even closer to perfection.

[Colas LeGrippa]

So source impedance has to be much lower than the load impedance, at least 10 times , but even much more is better.
When we're talking about line level or guitar level, what are we referring to ?  Signal voltage ?


Higher source impedance cuts the signal voltage , hence the necessity of low impedance source for min. signal loss, right ?


Thanks

[pdf64]

So source impedance has to be much lower than the load impedance, at least 10 times , but even much more is better.

The ‘x 10’ ratio is a guideline, not a mandate.
As the input impedance increases, the efficiency of signal voltage transfer improves.
If the input impedance = the source impedance, then half the signal voltage will be lost, -6dB.
If the input impedance is 3 x the source impedance, then a quarter will be lost, -2.5dB.
10 x, then 9% loss, -0.83dB
100 x, then 1% loss, -0.083dB.
And so on.
Nothing’s perfect 

When we're talking about line level or guitar level, what are we referring to ?  Signal voltage ?

It applies universally, in any scenario where voltage loss needs to be minimised. From a tiny sensor right up to a power station.

The main applications where we’re wanting max power transfer, and hence matching impedances (and the -6dB signal voltage loss is not a concern), are RF and transmission lines.

The speaker outputs of our valve amps don’t use voltage bridging or impedance matching. The goal is for the impedance that gives the desired loadline.

Higher source impedance cuts the signal voltage , hence the necessity of low impedance source for min. signal loss, right ?

And if we’re stuck with a high source impedance, eg piezo pickups, then we need to look to use an input with a very high impedance, eg 10M rather than 1M.

[Colas LeGrippa]

Now that you touch the subject, I have an acoustic guitar with a piezo p/u that I hate because it has a sterile tone. Question: can I replace the 1M input resistor on one of my tube amps by a 10M ? How wpuld react the input 12ax7 ?


Thanxxxx

[pdf64]

With grid leaks much higher than 2M, grid current will create a grid voltage that messes up cathode bias. You could add an input blocking cap and move to grid bias.
eg https://el34world.com/charts/Schematics/files/Fender/Fender_deluxe_5a3_schem.pdf
Or use a FET buffered input.

[sluckey]

Now that you touch the subject, I have an acoustic guitar with a piezo p/u that I hate because it has a sterile tone. Question: can I replace the 1M input resistor on one of my tube amps by a 10M ? How wpuld react the input 12ax7 ?

Does that guitar have a built-in battery operated preamp? If so, nothing needs to be done to the amp.

[Colas LeGrippa]

I will read again all your info and  links. It is much clearer now than it was. Thanks to this forum and all  of your replies.


I wish you a Merry Christmas. Keep off from the virus with all the means possible.


I love you all.


Colas