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The key question is, what is the best or right way to modify input impedance?

I am a self-teaching novice at electronics. I'm using the LM386 as a learning platform. Signal input comes from a "headphone" type jack (tip).

enter image description here

I have tried to use line out from a small mp3 player as a test source, and it works, using the datasheet example minimal circuit. It was very tinny and distorted at 200x gain, somewhat better at 20x (default) gain, but not great. My goal from here is to learn to improve the circuit. I think I will want higher gain but I want higher fidelity at low gain before I worry about that. Fidelity is not great even at low gain and low output volume and I assume that's my fault and fixable due to the enduring popularity of this chip.

I have noticed when using the mp3 player the circuit seems more stable and less distorted. But when I plug in the guitar instead the circuit seems more distorted, less true to the frequency response across the guitar audio spectrum, less polyphonic, and I even think it drains the battery faster and makes it hotter.

This has led me to try to understand impedance matching. The guitar is said to be a very high impedance device. I have learned that modern impedance "matching" really is about making sure output impedance is as low as possible and input impedance is relatively high, because this is optimal for transference of signal via voltage, rather than power. I think I need to attenuate the input so the high impedance guitar output is handled better, basically so signal is not clipped or otherwise distorted.

Here is my big question: Many diagrams that describe impedance matching issues show "representative" resistors at the output of device "A" and at the input of device "B". Here is a link to such a diagram:

https://www.learnabout-electronics.org/ac_theory/images/Fig-7-2-3a.gif

From containing page:

https://www.learnabout-electronics.org/ac_theory/impedance72.php

What puzzles me is the input representation shows the resistor basically in parallel with the input of device B, while the output representation shows the resistor in series with the output of device A. It seems to me that these are two very different things. Device A representation seems logical, as a resistance in series would increase the total resistance of the output. But the Device B representation is illogical to me. Adding a resistor in parallel should REDUCE the total resistance of this part of the circuit by adding a "new path" for current to follow. So it doesn't seem to represent what it means to. It means to represent the resistance to the input.... but it is not doing that, it's reducing that resistance if anything. I understand it is being compared to a voltage divider, so adding any second load would reduce the available voltage that goes down the first leg since they have to share. So I guess I can see that either approach could reduce the amplitude of the input signal. I'm just not sure why this representation is chosen to explain the theory, we are talking about input impedance so why not conceptualize it in series? I think I am missing something here, probably something important.

For my bench circuit to make my guitar sound better, I would think I'd want to add some resistance in series at the input of the LM386 which is MY Device "B" (the guitar is my Device A). I've read 1M ohm at the input is actually a reasonable estimate to attenuate for incoming electric guitar signal. My practical question is where do I put it? In parallel at the signal input pin as the theory diagram shows? In series at the signal input pin as seems more logical to me? What am I missing?

Please answer the theoretical question with sentences and metaphorical type answers (IE compare to water flow or such) if at all possible. I can look up the math anywhere and don't need to see it again. I'm asking here because I want some expert who understands this perfectly to provide the more metaphorical understanding. Thank you in advance for your assistance. I really want to understand in general how to combine multi-stage components. For example, I have a PT2399 I got in a sort of grab bag, and I want to add some "reverb" to the fun. I've actually tried that and I have gotten some results (not musical ones yet... too much high frequency in the echo). But that portion is put on hold while I get the LM386 to sound decent alone.

The example use of the LM386 is given as reference so you know what I'm attacking (as learning experience), and this question can be answered in direct practical terms. What should I do?. I also welcome comments about the LM386 in general if that's something you want to add. I already have found some good references on that such as...

https://hackaday.com/2016/12/07/you-can-have-my-lm386s-when-you-pry-them-from-my-cold-dead-hands/

but feel free to add your two cents. That's not my main question though.

For reference and clarity:

A) Why is the input impedance conceptualized as a resistor in parallel in the diagram? What is the important concept I'm missing here?

B) Where do I put the 1M resistor I want to experiment with? I know I can try both and I probably will. I just want the experts to give thoughts/guidance also.

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    \$\begingroup\$ Far too lengthy and wordy and lacking in embedded schematics. Please focus and bear in mind that this is a question and answer site. \$\endgroup\$ – Andy aka Jun 20 at 16:28
  • \$\begingroup\$ I am sorry. I'm not much good at speaking tersely, and myself prefer when people speak as a human. This may not be such a human site. I wanted you to understand who you are talking to and what type of answer is sought, to get better answers to the question. I will plan to remove that portion in an edit. The only schematic would be the example from the LM386 datasheet, easy to find and linked. Would I chop that out and post here? Is it not a copyrighted image? \$\endgroup\$ – ChronoLogic Jun 20 at 16:37
  • \$\begingroup\$ Thank you for your consideration and feedback. I have removed most extraneous humanizing information. I believe it is now much more focused and sterile. Please feel free to notify me if you find more useless and unwanted information that should be removed. \$\endgroup\$ – ChronoLogic Jun 20 at 16:46
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    \$\begingroup\$ I've simplified the start and added the relevant pictures - if that is not acceptable please let me know and I'll roll-back the changes (or you might do it. A guitar amp needs a much higher input impedance by the way. \$\endgroup\$ – Andy aka Jun 20 at 17:03
  • \$\begingroup\$ Highly acceptable as you are the expert, I am not. Thank you! \$\endgroup\$ – ChronoLogic Jun 20 at 17:05
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I have been thinking about this further and I think I will attempt to answer my own question but I can't accept my answer as I am not qualified.

I think the answer probably lies in thinking of it as a voltage divider:

enter image description here

When concerned about power, you need to worry about both current and voltage because P=IV, in other words, the power is related to how much voltage is dissipated.
So equal resistors turns out to be the best, since the power is equalized across both legs (Z1/Z2). But when concerned about just voltage you want Z2 to be larger, in order to get more voltage from Vin to Vout. And yes it is "in parallel" to Vout because that is how a voltage divider works, and it is based on relative resistances. And we ARE concerned about voltage (not power as in the telephone "old days") as I understand it. My guess is the theory is best explained that way.

If this is correct, I'd be pleased if someone qualified accepts. If it isn't, I'd be pleased to get corrections. Either way I'd be pleased to see clarifying comments.

EDIT July 30 2020: I've had more time to think about this and read a little more. I now think I could answer my own question even better and I would say I've come to believe that in practice Z1 in this image is the output impedance of a source, which you don't necessarily control. IE, maybe it's a guitar input. It is what it is. Z2 however you do control. And by putting a very large Z2 resistor (like 1M) that goes to ground at the input of your circuit you effectively build a voltage divider, half here that you control (Z2) and half from the circuit you don't (Z1, IE the guitar). In this case, if Z2 is much larger than Z1, you encourage most of the voltage to be distributed to Vout rather than going to ground. Since the audio signal is represented by the voltage wave, this is great, you want the largest voltage wave you can get. This is not really matching impedance per se but it is making it how you want it. I guess for some applications they want to transfer maximum power not maximum voltage. In that case they would want to basically make Z2 close to equal to Z1, which is what impedance matching means in that context. Again if I'm wrong feel free to correct me, I want to know and to share good info. But I feel like I got the hang of it now. If this is right I hope my way of explaining it makes sense to someone else out there that was confused with the way others explain it.

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  • \$\begingroup\$ The solution you propose is often applied to radio-frequency amplifiers where Z1, Z2 are often 50 ohms....amplifier power gain then becomes an important spec. Not so much applied to audio, where the ideal situation has Z1 approaching zero ohms and Z2 approaches infinite ohms. In this case, amplifier voltage gain is the important spec. This is the case for LM386 audio amplifier...it has a voltage gain of 20. \$\endgroup\$ – glen_geek Jun 20 at 19:39
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    \$\begingroup\$ Answering your own question is a very good idea because it guides others in what direction you are thinking. A meaningful answer shows that you already have an opinion on the issue and so respects the respondents. I would like to recommend that you familiarize yourself with a related question, to which I have given the following answer a few days ago -electronics.stackexchange.com/a/506092/61398 \$\endgroup\$ – Circuit fantasist Jun 20 at 20:59
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this question can be answered in direct practical terms. What should I do?. I also welcome comments about the LM386 in general if that's something you want to add.

And

I will try a 1M ohm input resistor to attenuate for guitar

I'm not sure that putting a 1 Mohm is going to cut the mustard for the LM386. Ideally you would need a high impedance buffer amplifier and feed its output to the potentiometer in one of the circuits I edited into your question. Google is your friend and it revealed this when I searched: guitar amp circuit using LM386 and an input buffer.

enter image description here

Here's a link to the design that should help. Note that it only uses a 9 volt supply voltage. Here's a independent review of said amplifier so it's clearly attracting attention in several places. It pretty much deconstructs it technically in a rather nice way so it should also be quite useful. Here's yet another spin off so this design has cred with guitarists I estimate (being one myself).

There is also a search engine on this site to find any possibilities of other amplifier designs.

For audio and pretty much up to about a MHz, output impedances tend to be no more than one-tenth the input impedance of the device it feeds. Typcally a poweramp will have an output impedance sub 0.5 ohm and, as a comparison a speaker is 4 ohms or more.

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  • \$\begingroup\$ Thank you, I have seen that but I don't have a JFET (yet) to try this and I've seen schematics (for example, in the datasheet) without them so I thought I'd start without. It seems I need to add to my inventory. However, I now also suspect that Justme is correct and I'm asking too much of a single 9 volt battery as well. I will move to a wall wart and either attempt to flatten it a bit more or maybe just use an L7809 regulator (which I do have) to flatten it. \$\endgroup\$ – ChronoLogic Jun 21 at 16:07
  • \$\begingroup\$ You can use a 12 volt battery of course. See one of the links I left in my answer. \$\endgroup\$ – Andy aka Jun 21 at 16:34
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You say

What puzzles me is the input representation shows the resistor basically in parallel with the input of device B, while the output representation shows the resistor in series with the output of device A. It seems to me that these are two very different things. Device A representation seems logical, as a resistance in series would increase the total resistance of the output. But the Device B representation is illogical to me.

There is a fundamental problem here: you should see output and input resistance as parts of the devices A and B, not as something you add in front or at the end of. I suggest you try to see your stages as two-ports. Let's also talk about resistance to simplify things.

So, for the output resistance of a stage: it depends on how you model it. If you model it as a voltage source you will have a voltage generator with a (possibly small, ideally zero) series Rout; if you model it as a current source you will have a current generator with a (possibly high, ideally infinite) parallel Rout. In your schematic above, the resistance should be draw inside the triangle that represents the amplifier. Something like this:

enter image description here

Image taken from here: https://i.stack.imgur.com/enMZ0.gif

Note that you can go from one output representation to the other by applying Norton or Thevenin's theorems. Keep in mind that a good voltage source will make a lousy current source, and vice versa.

As for the input resistance, this is the resistance you see 'looking' into the... input port. If your stage is designed to receive voltage as an input, you will be better off with a high input impedance because when you attach a voltage generator to its input you create a voltage divider with Rin and the internal resistance Rs of the voltage source (ie the output resistance of the preceding stage). But if your stage is there to accept a current, then you will be better off with the lowest input impedance you can get. Because if you model the preceding stage as a current generator with an internal conductance Gp - corresponding to a parallel resistor Rp, then you are having a current divider and you will get most of the current if your Rin is much lower than Rp.

Redraw your circuits as two ports and it will be clear where those topologies come from.

When you combine your stages, depending which goes after which, you will get either a voltage divider or a current divider. Good or bad depends on the relative values of Rout and Rin. And no, the best solution is not necessarily Rout stage(N) = Rin stage(N+1). Not even if you want to get the maximum power into the last stage. I wrote an extensive answer to this other question about impedance matching (How voltage and power transfer in cascaded amplifiers work?)

Basically, the maximum power transfer theorem is about how to extract maximum power from a given source, and NOT about how to get a load absorb its maximum (shall we call it 'nominal'?) power. If you want your load to absorb the maximum power allowed, then you should seek totally unmatched solutions, corresponding to making the stage that powers it as similar as possibile to an ideal voltage source (Rout=0, with Rout in series with your RL - forming a voltage divider) or and ideal current source (Gout=0, or Rout=infinity with Rout in parallel with your RL - forming a current divider).

When you cannot act on the output resistance of the powering stage, you mighr end up stuck with a bad voltage divider (or a bad current divider). In that case it makes sense to add an intermediate stage whose purpose is to show to the powering stage and impedance that will make the voltage divider look better, and at the same time allows you to select the impedance it shows to your load. This is what the FET stage in your modified circuit does.

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  • \$\begingroup\$ Thank you. Yes I understand there is an internal impedance, internal to the device, no matter how simple. The drawing models it as parallel. I was wondering why. I guess to me it seems useful for two things: 1) to model what it actually does (IE it does create a voltage divider, it seems) and 2) if I need to adjust the input impedance externally to a discreet component (like an IC) then I think the diagram suggests I do that via adding an external parallel impedance (not series). But I may be wrong on that? This is the meat of my question and uncertainty. \$\endgroup\$ – ChronoLogic Jun 21 at 16:27
  • \$\begingroup\$ The problem with 'adding' impedance is that it will not ameliorate the power transfer where you need it. You will just add a dissipating element that will draw more power from the source, but won't necessarily steer power inside the rest of your amplifier. You should work on reducing the output impedance of the source, instead. But that's not always possible. See my answer to this other question: electronics.stackexchange.com/questions/505799/… \$\endgroup\$ – Sredni Vashtar Jun 21 at 16:53
  • \$\begingroup\$ Ok that makes sense. So I think it suggests that you really need a "translator" device to basically speak the right language (in terms of signal transmission) on both ends of the connection. That is probably why the solution of adding a JFET or op-amp solves the issue. My attempt at finding an alternative was misguided it seems, in this case. \$\endgroup\$ – ChronoLogic Jun 21 at 18:56
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Impedance is an issue, but not where you think.

An 9V battery has very large output impedance compared to a power supply or other batteries, so it can't provide much current and the output voltage will droop.

Most likely this is your main source of distortion. Try with a real power source.

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  • \$\begingroup\$ Thank you I will try that! This is new information. \$\endgroup\$ – ChronoLogic Jun 21 at 15:56
  • \$\begingroup\$ What are your thoughts regarding the ruby amp: robrobinette.com/How_the_Ruby_Amp_Works.htm It does use a standard 9V battery. I had seen that and this was part of why I thought I was fine with the standard 9V. Is the addition of the JFET instrumental in overcoming the battery issue? I would not think so... in fact, it was this circuit that involved a JFET that made me think the guitar impedance was a big issue. I don't have a JFET (yet) so I was trying to attenuate a different way in the meantime (and trying to LEARN which is the main point here). I appreciate your time. \$\endgroup\$ – ChronoLogic Jun 21 at 16:42
  • \$\begingroup\$ My thoughts? The website never says anything about powering it with 9V battery. However it specifically mentions powering it with a 9V wall wart power supply. And LM386 input is not directly suitable for guitar, it has too low impedance for that., about 50 kohm. Most guitar amps have about 1 Mohm input impedance. The 10k potentiometer will make the input impedance even worse, to about 8 kohms. The impedance is off by a factor of 120. That is why there is a JFET buffer in the schematic, but you could use an op-amp buffer. \$\endgroup\$ – Justme Jun 21 at 17:13
  • \$\begingroup\$ Ok I had wondered about using an op-amp instead of the JFET. I may try that because I have one on hand. As far as the battery, it doesn't mention it in text but the image does show a battery connector, and other sites I read that talk about the ruby do mention using this connector. I believe one of them said the wall-wart connector would disconnect the battery (so both are not connected). That's why there are two red wires depicted at the power connector. This is the specific image: robrobinette.com/images/Guitar/Ruby/… Thank you, have a great day. \$\endgroup\$ – ChronoLogic Jun 21 at 18:52
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About your wondering...

"why the input representation shows the resistor basically in parallel with the input of device B, while the output representation shows the resistor in series with the output of device A."

... you have correctly guessed that this is the conceptual circuit of the ubiquitous voltage divider. Let's first say some words about it...

This humble circuit of two resistors in series is the most common electrical circuit. Why?

For some reason (it would be interesting to see what... but this is another topic), we prefer to use voltage rather than current as a data carrier in the low power electronics. The voltage divider is a device needed to convert (proportionally decrease) the voltage. From this point of view, it is a useful device. An example of such a desired application is the 10 k potentiometer in your circuit above.

Unfortunately, in most cases it is a harmful device that unnecessarily reduces the voltage and overall gain of multistage voltage amplifiers. From this point of view, it is a harmful device. Whether we want it or not, in each case, when we connect (cascade) a voltage source (amplifier output) to a load (amplifier input), such an undesired voltage divider (attenuator) is formed. But what is Rout and what Rin in this configuration?

The picture in the audio site above is misleading since it shows these resistances as external resistors… but they are inside the stage. They are equivalent resistances and in most cases they cannot be seen.

Rout expresses the fact that when we start consuming current from the output of the stage, its voltage drops (dRout = dV/dI). In some cases, Rout can be a visible external resistor. For example, we can intentionally connect such a resistor in series to a perfect ("ideal*) voltage source (amp output) to make it imperfect… e.g. with an educational purpose. I suggest to my students carry out such experiments in the introductory labs to see the difference between a perfect and imperfect voltage source. Or Rout can be line resistance if it is too long… or a protecting resistor in the emitters of the output buffers. Typically, we want Rout to be small (preferably zero) to convey the whole voltage.

Rin expresses the fact that when we apply voltage to the stage input, it begins consuming current from the output of the previous stage and its voltage drops. In BJT Rin is primarily determined by the transistor input resistance… but also there are bias resistances in parallel. Rin can be a visible external resistor as well. As above, we can intentionally connect such a resistor in parallel to a perfect voltage load (amp input) with extremely high resistance to make it imperfect. For example, my students load an intentionally worsened voltage source (with added Rout) to see the impact of Rin on Vout. Or, we can connect a resistor in parallel to the FET input to protect it. Typically, we want Rin to be as high as possible (preferably "open circuit") to convey the whole voltage from the previous stage.

So, to transfer maximum voltage, Rout has to be as much low as possible and Rin as much high as possible. To transfer maximum power in the output stage, we have to make the load resistance RL equal to Rout. The opposite (to make Rout = RL for maximum power) is not true.

Besides these straightforward solutions most typical for electric circuits, in electronics we use more clever techniques based on the ubiquitous negative feedback.

Rout can be decreased almost up to zero by closing the negative feedback loop after it. As a result, the amplifier increases its output voltage with VRout thus compensating it. The amp output can be thought of as a "negative resistor" with resistance -Rout that destroys the "positive resistance" Rout.

Rin can be (theoretically) increased up to infinity by disconnecting its lower end from ground and shifting with Vin. The name of this weird circuit trick is "bootstrapping".

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    \$\begingroup\$ Thank you. Testing today I found that about a 270 resistor helped clean the guitar output a lot. This resistor was run from the "tip" (IE via TRS connector) of the input connector to ground, prior to the volume pot. This is the opposite of what I initially thought I needed (high 1M changed to low 270). I could then add more gain via the control between pins 1 and 8. At first I was confused, but I now suspect that my REAL issue is insufficient supply current as pointed out by Justme, which is why I have to make the signal weaker so it does not clip. I have a lot to learn. \$\endgroup\$ – ChronoLogic Jun 21 at 16:20
  • \$\begingroup\$ It makes a good impression that you answered everyone here... many of the questioners don't. I have no idea what the guitar output is and why a 270 (k?)ohm resistor improves it. The power supply really needs to be good. Also try connecting a large electrolytic capacitor (> 1000 μf) in parallel with the battery. Success! \$\endgroup\$ – Circuit fantasist Jun 21 at 16:59
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    \$\begingroup\$ You said: "Do not remove any "humanizing information". It is not "extraneous"... on the contrary, it is vital here on this site to make it more human" -> Because of the clearly stated rules and intentions of this site (by the site owners) this is extremely bad advice to a newcomer . Newcomers will suffer substantially if they follow your advice, There are many places on the internet where such encouragement is appropriate and welcome. This is not one of them. The clearly stated aim of the site is to provide highly focused Q&A sets that are of long term technical value. \$\endgroup\$ – Russell McMahon Jun 21 at 22:09
  • \$\begingroup\$ Thanks Russell I accept that. I guess you win some and lose some. Sometimes it seems humanity is slipping away regardless, but I understand that this site feels it's better to simply sidestep that entire issue to avoid the bad sides. \$\endgroup\$ – ChronoLogic Jun 22 at 5:01

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