How can a "signal" have a high or a low impedance?

[Dolmetscher007]

For the most part, when I don't perfectly understand something, I've learned over time to just not over-think it, and to move on; sooner or later it will "click". However, the terms "high-impedance signal" or "low-Impedance signal" just aren't clicking for me.

It is my understanding that "Impedance" is, essentially, a word that describes how much "resistance" to the flow of an Alternative Current signal within a circuit. I've seen it usually described as, "With a DC signal, the lowering of strength due to dissipation as it flows through a circuit is called 'resistance.' Whereas, 'Impedance' describes the same thing, but in reference to an AC signal." Please don't take me apart if I missed the mark on that definition. I think you get what I mean; people tend to say, "Impedance is like resistance, but for AC." My point with this question is... how can a signal, AC or DC be described as "high" or "low" impedance or resistance?

In my mind, a signal is... well... a signal. Signals don't "have" an impedance. Signals have a strength (Voltage) and a volume (Current), but what impedes or resists that signal refers to the obstacles that signal faces. For example: An AC signal with 120VAC has X amps of current as it passes through a 10k resistor. As it exits the resistor, the current has been lowered by Y amps due to the 10k Ohms of resistance. So, the signal doesn't have the resistance, the conductors within the circuit are what has the resistance. So when someone says that this or that piece of equipment puts out a "High Impedance Signal..." what do they mean? Why is it not, "The signal was connected to a high Impedance INPUT?"

So confused.

[Helmholtz]

In my mind, a signal is... well... a signal. Signals don't "have" an impedance.

You're absolutely right here. "High impedance signal" doesn't make sense.

I think what is meant is a signal provided by a high impedance source (e.g. an output), which can deliver only little current without significantly loading down the voltage.
So a high impedance source should always be connected to an even higher input (load) to preserve the signal voltage level.

[eschertron]

To add: for a signal-carrying conductor, the impedance is in reference to ground, not some other part of the signal.

As Helmholtz indicates, the important thing to note is that an impedance value in a circuit is referring to where it came from (output impedance) or to where it is going (input impedance). It is always* desirable to have any output impedance feeding a larger input impedance, to avoid the circuit being loaded. The impedance is the measure of how little current can flow while preserving signal voltage. A lower output impedance -> more current can flow to prevent the signal from being loaded. A higher input impedance -> less current is needed to provide the circuit with a good signal.

*always when looking to preserve the voltage of the signal.

[Dolmetscher007]

You're absolutely right here. "High impedance signal" doesn't make sense.

I think what is meant is a signal provided by a high impedance source (e.g. an output), which can deliver only little current without significantly loading down the voltage.
So a high impedance source should always be connected to an even higher input (load) to preserve the signal voltage level.

Thank you! Helmholtz
I think that last sentence is the most important thing, right? If you have a device that puts out a "high-impedence output," it should be plugged into a "high-impedence input" of another device. Correct?

My only real-world frame of reference is, when I was a teenager, I had several SM-57 and SM-58 microphones. They all had XLR connections. Then one day, my bass player showed up with some microphone that he had bought that he kept calling a "High - Z" microphone. And it has a 1/4" jack connector rather than the XLR. He did now know what "high-z" meant, and neither did we. Someone had told him at the store that a high-z microphone was "good." This was 25+ years ago. Ha ha ha!!! I still do not fully understand what a high-Z microphone is, but I believe I know that "Z" delineates "impedance," and I guess the 1/4" jack connector was meant to prevent the user from plugging it into the XLR connector of a typical microphone pre-amp. Instead, we were supposed to stick it into a "line in" input as if it were an electronic keyboard, sampler, or whatever.

To add: for a signal-carrying conductor, the impedance is in reference to ground, not some other part of the signal.

As Helmholtz indicates, the important thing to note is that an impedance value in a circuit is referring to where it came from (output impedance) or to where it is going (input impedance). It is always* desirable to have any output impedance feeding a larger input impedance, to avoid the circuit being loaded. The impedance is the measure of how little current can flow while preserving signal voltage. A lower output impedance -> more current can flow to prevent the signal from being loaded. A higher input impedance -> less current is needed to provide the circuit with a good signal.

*always when looking to preserve the voltage of the signal.

Thanks eschertron
I really hate to reveal how confused I really must be about electronics, but... I STILL cannot seem to wrap my head around the concept of "Ground." I honestly think that my main hang-up, both with the above "impedance" question, and with "ground", the disconnect for me might have to do with me confusing DC and AC terms. Here is what I DO totally understand...

The physical stuff...

• In the United States, our high-voltage lines travel for miles and electricity enters our homes after being reduced in voltage down to 240VAC through a center tapped transformer (120VAC - 0 - 120VAC) on the utility pole.
• This means that we essentially have two 120VAC circuits coming into each home. Circuit A is the first 120VAC wire as the "Hot wire" and the 0VAC center tap wire is the "Neutral wire".
• The center tap wire is also connected to another wire that is connected to an actual metal spike that is driven into the actual ground at the base of the utility pole.
• This "ground wire" splits off from the center tap, going on one end to the spike in the ground, and to other end goes into our house and shows up as the "third-prong" of a wall outlet.

The theoretical stuff...

• The way a transformer works is that the high voltage electricity pulses through the primary winding of the transformer, and induces a voltage/current into the secondary windings.
• With DC, such as with a 9V battery, electricity can only flow when you have a conductive wire connecting a terminal with a higher number of electrons (negatively charged) to a terminal with fewer electrons (positive charge).
• When connected, the electrons pass through the wire carries the surplus electrons from the negative terminal to fill the electron "holes" found at the positive terminal until they equalize... i.e. battery is dead.
• If you complete this connection without some form of load to absorb some of that energy, it is called a "short circuit," because the energy released by the flow of electrons dissipates causing more heat than the physical wire material can support.
• If you "cut" that conductive wire in half, but leave it connected to the power source, you now have 1 wire connected to the negative terminal, and another wire connected to the positive.
• If you connect the other ends of those two wires to the terminals on some load (light bulb), the circuit is closed, and... assuming the light bulb can handle the voltage and current... you don't have a short circuit, just a closed circuit.
• I also know that electrons always try to get from the negative terminal to the positive terminal through the path of least resistance. So, in the light bulb scenario, if you strip some of the insulation aware from the middle of the wire that connected to the negative battery terminal, and you solder another wire to the exposed copper wire, and then connect the other end of that wire to the positive battery terminal, the electrons leaving the negative terminal will not bother going through the resistant light bulb, but will, instead take an immediate short-cut through the new wire, straight to the positive, i.e. another "short circuit."

Where I get weird, is... with a US wall outlet, you don't have a "positive" and a "negative" terminal like you do with a battery. You have a "Hot" wire, a "Neutral" wire, and a "Ground" wire.
If you connect the Hot wire to a light bulb, and the neutral wire to the other terminal of the light bulb, the electrons pulse back and fourth, 60 times per second, through the light bulb.
I cannot find a YouTube video or an analogy that explains to me in a way that sticks with me the following things...

Why does a light bulb that is connected to a DC battery, with electrons flowing straight through its filament light up, but then... the exact same light bulb can be connected to a wire that passes electrons through it's filament back and fourth really fast... and it lights up just the same. I guess... in my mind I can say... if you take a belt sander, turn it on, and have someone hold it to your forearm... as the grit pushes over your skin, it will scratch and injure you and it whips past in one direction. However, if the same belt sander could reverse direction 60 times per second, that doesn't change the fact that it's scratching and injuring you. It's just doing it back and forth rather than all in one direction.
But I really cannot see how having a hot wire that connects to a neutral wire to complete this back-and-forth loop of electrons, needs a third wire, to... what...? carry electrons back and forth between the hot wire an the metal spike in the ground outside, underneath the utility pole?

I've heard so many terms like... "provides a reference... provides a clear path to ground... as a safety measure... to prevent the charge from flowing through you..." but... it just doesn't stick. Supremely frustrating.

Do you have any tips or clarifying phrases that I may be missing. Or even a recommendation like... "Dude... you sound like you have total ADHD and possibly a learning disability. If I were you I'd watch [THIS VIDEO]." or whatever?

I really thank you both for your help. You have no idea how frustrating it gets when you have no one to ask and you just have to worm your way through text and videos that are just eeeever soooo slightly out of reach, because you seem to be missing some super fundamental concepts.

Michael

[mhuss]

The discussions above (low into high Z) are relative to signal *voltage* levels, which is how we usually view low level signals.

The most *power* is transferred when source and sink match, e.g., the "16 ohm" tap on a tube amp into a 16 ohm speaker load.

On the second question, "ground" is just the reference point used to measure voltages. Grounds could be (virtually) the same in a given system, say, guitar into pedal into amp - all three "grounds" will normally show continuity between each other, and if the amp is grounded properly, there would also be continuity to earth ground (the third hole in on the AC outlet). However, they don't have to be. A balanced circuit (e.g., most studio equipment) with the ground lift switch engaged *could* have no continuity between the source and destination grounds - but may anyway due to internal connections to mains power ground.

[Helmholtz]

The most *power* is transferred when source and sink match, e.g., the "16 ohm" tap on a tube amp into a 16 ohm speaker load.

I agree with most of your post, but (audio) amps are not "power matched" to the speakers.
Tube amps can have a source impedances between 1 Ohm and 100 Ohm at the 8 Ohm output, depending on tubes and amount of NFB.
SS amps are typically below 1 Ohm.

[Enzo]

Don't confuse neutral and ground, though they are connected together at your electrical panel. The transformer on the pole out back sends the center tapped 240, or 120-0-120 if you like. The zero in the middle there is not ground, it is neutral. The two 120vAC lines are opposite polarity - also called out of phase. In your service panel, the earth stake outside is wired to your neutral bar.

Asd to why does AC light the bulb, well current is current. If we put 9vDC on a bulb, it lights up. On AC, 60 Hz power changes polarity 120 times a second, And during that small interval, the current lights the bulb. The electrons heat the filament by passing through it. They don't have to go anywhere else. Either direction will light the bulb, just as the bulb lights on DC regardless of which end is positive. Current does flow at essentially the speed of light. SO our bulb gets current one way, then current the other way at this quick rate. Instead of a belt sander, thing of a two-man lumber jack saw. The two guys take turns pulling the blade across the tree. CUts either direction, they don't have to only saw one direction.. SO just imagine the bulb has a battery, and someone switches battery polarity back and forth a LOT.

You don't NEED a ground wire - an earth connection - for a circuit to work, after all amps for decades didn't have three wire power cords, hell wall outlets only had two hole, not three.The third ground wire is for safety, not circuit operation.

[bbsailor]

This is a complex question because it depends on if you want to transmit the maximum voltage to have a good signal to noise ratio or want to transmit the maximum power to have the loudest output from a speaker or want to minimize any impact on the tonal quality of the source device.

When dealing with input voltages and impedances you want the source impedance to be about one tenth the load impedance at the highest desired frequencies to have minimal losses. The big issue is what is the impedance at the highest frequency you want to pass into the device you are feeding. With guitar pickups, I use the rule of 40 that says that the value of the volume pot should be about 40 times higher than the DC resistance of the pickup to minimize any high frequency roll off due to loading.

If the output impedance of a device is the same as the input impedance of what it is feeding into, the voltage will be reduced to one half of the unloaded source output voltage. This is why the input impedance of most guitar amps is 1 Meg ohm to minimize the load on guitar pickups. There is however one more issue to understand. In high impedance inputs, the coax cable connecting the guitar to the amp adds capacitance which had a more audible effect at higher impedance connections. A typical guitar cable adds about 30 of per foot of cable. This capacitance tends of affect the transmission of higher frequencies, like adjusting the tone control slightly lower.

Low impedance microphones are rated at about 150 to 250 ohms to feed a mic input that has about a real 2400 ohm input impedance. With this lower impedance, cable length has a much more minimal impact on the tonal alterations compared to high impedance connections.

Bottom line; device output impedance, cable length along with cable capacitance, and device input impedance all combine to affect the final sound you hear.

I hope this helps put this issue into a better perspective.

Joseph J. Rogowski

[J M Fahey]

Referring strictly to *signal*.

True, in principle by Signal we mean an Audio Voltage 99.98% of the time.

How can a *Voltage* have "impedance"?

Well, doubts rise because we use an incomplete description of said Signal, giving the "name" but not the "why".

Truth is that Signal IS voltage (so far so good) but it does not come from out of the blue, .... it comes from a *generator*. (A pickup, microphone, tape head, tube plate, Op Amp, and many more) and it is THE GENERATOR which has an internal impedance.

When we say "high impedance signal" we "should" actually say "Signal coming from a high/low/whatever impedance generator".

In the real World that Signal Voltage source will be loaded by ... ummm ... the load (duh).

If generator internal impedance is low (Op Amp , tube cathode, line amp transformer out, etc.) , it will easily drive most anything with little or no attenuation, after all we always have an (unseen) voltage divider or attenuator at work: generator internal impedance and load impedance.

If ratio is low (say 600 ohm driving 10k) attenuation is nil; if ratio is high (100k pentode plate resistor driving a 10k input impedance SS amp) loss will be HUGE.

Same with, say, a guitar pickup.
DCR is not that bad, a few k, but Inductance is HUGE, so impedance at mid/high frequencies will be very high, and cause treble loss (heard as loss of sparkle or even muddyness) so it will be happier with a High Impedance input (typically 500k or even 1M).

[pdf64]

...The most *power* is transferred when source and sink match, e.g., the "16 ohm" tap on a tube amp into a 16 ohm speaker load...

This is a complex question because it depends on if you want to transmit ... the maximum power to have the loudest output from a speaker...

Just to expand on post #6, the max power transfer theorem finds little application outside RF / transmission lines. Even from the perspective of valve amps, it's awfully wasteful and inefficient
Zollner explains valve amp output impedance rather nicely https://www.gitec-forum-eng.de/wp-co...e-matching.pdf