Impedance explained?

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Ok, I've been recording as a hobby for nearly 20 years, and have yet to understand impedance completely...

I don't need a definition of impedance (a/c resistance, simple resistance plus effects of reactance I believe) but something I don't understand is OUTPUT impedance!

In a guitar, for example, would this be the impedance of the coil in the pickup? Or the impedance of the transformer secondary in a microphone? Does a 120vac wall outlet have an impedance?

The specs for the Art Tube MP state:
Input impedance: 2k ohms (XLR), 840K ohms (1/4”)
Output impedance: 600 ohms (XLR), 300 ohms (1/4”)

Do guitar pickups have a lot higher output impedance than microphones? Is that why the 1/4 input is so high? I've heard that your input impedance should be at least 4 times as high as the output impedance on the device your plugging into it. Can I measure approximate impedance with a VOM?

My head is spinning - - - I must understand this once and for all!

Thanks! :- )
 
Impedance is resistance in motion. When you measure resistance, it's done by throwing a DC voltage across the thing in question and measuring how the thing fights back. As you get into AC signals, things get a little trickier.

If it's a speaker (for example), the motion of the speaker, the air in the cabinet, the bass port, and the particular frequency all throw in their little contributions, and the impedance can wind up being higher or lower than the DC resistance reading. An 8 ohm speaker can vary from around 6 ohms to several hundred ohms, depending on where it's measured.

With transformers and coils and AC signals, there is no single impedance unless we're talking about a fixed frequency BUT, you can kinda average things out and plug in a number that'll work for most calculations.

Guitar pickups can have a coil impedance of anywhere from 2,500 ohms to about 10,000 ohms. That's why the volume controls on the guitar are usually in the range of 100,000 to 500,000 ohms - about 10 times higher than the pickup impedance - to avoid loading the pickup. The input impedance on the amp is higher still than that, usually in the megohm range, again to avoid loading the pickup which can affect the tone.

The source impedance of the AC coming out of your wall isn't really important till you put a load across it. Even then, it's negligable for applications like lights and toasters.
 
Harvey's right on the money. But let me throw 2 cents in to try and explain it in a more basic way, so that folks can really get a grip on the basics of the problem. I'll use a little bit of EE gobbletygook here, but hopefully it'll help and not hurt...

We'll start out with just DC resistance, and leave impedance for later. In circuit design, there's the concept of what's called an "ideal voltage source". This mythical beastie is a voltage generator that puts out the same voltage no matter what you do to its output load. Let's imagine just for a second an ideal 9V battery, for example. Hang a 1ohm resistor across it, and the voltage stays put at exactly 9 volts. But other fun things happen: if the voltage across that resistor is 9V, then the current passing through it will be 9 amps (according to Ohm's Law: V=IR). Even more fun, the power dissipated by that resistor will be 81 watts: Pd = IsquaredR. Things'll get right toasty, no? Put a dead short across that puppy, and the voltage would still be 9V, and the current would be infinite. And the power would be infinity squared. Now, that'd run the Big Muff Pi for a while...

Needless to say, there _is_ no such beast. Take a real 9V battery, and put a 1ohm resistor across it, and the voltage will drop down to abut 1V, and you'll only get 1amp- and not for long, either.

So where'd the rest go? For the purposes of argument, we can model that good ol' coppertop as an ideal 9V source, that just happens to have a 8ohm resistor in series with it. When we stick our 1ohm load on, 8volts is dropped across the 8 ohm resistor we can't see (because it is an internal part of the battery), and only 1 volt is is left for us to use, and it gets dropped across our 1ohm external load. That's called Kirchoff's Voltage Law, if anybody is still reading now: the sum of all the voltage drops must be equal to the sum of all the voltage sources. Between Ohm and Kirchoff, you can get a handle on just about any circuit problem in the audio frequency range.

Here's the key. If the voltage sags under load, it has to be going _somewhere_. And how much it changes with a known load lets us figure out what the internal resistance of the battery is. Viola: we just determnined that our 9V coppertop is actually a 9V voltage source with an output *resistance* of 8 ohms! No matter what we do, that 8 ohm resistor is always there, inside the battery where we can't see it.

So your question about output impedances really maps right back to DC. How does the output change in the face of a known load? Measure the sag, and you can calculate the output resistance (the internal resistance that exists in all real-world cricuits). Plot the sag with respect to frequency with using an AC signal, and you can kindasorta average the resulting curve and assign an output _impedance_ (a frequency-dependant resistance, if you will). It'll change with the exact frequency, but you can get a handle on it. So the output impedance is actually a measure of how the output will change under load.

Now, why does this matter for mics and guitars? Capacitive loading. The behavior of a piece of cable is a pretty simple thing at audio frequencies. The cable will have some capacitance per foot (for round numbers, let's say 50pF: fifty picofarads per foot). Let's see what happens using a cable made of that wire with a 100ohm mic and a 10Kohm guitar pickup, just for jollies. To make the math less painful (and more familiar!), let's use a 20 foot cable. That's 20*50pf, or 1E-9Farad for the whole cable.

The source (whether it's the mic or the guitar) and the cable form a simple lowpass filter. The source wants to push signal up the cable, the cable wants to load down the high frequencies: at DC, a capacitor looks like an open circuit, but as the frequency goes up more and more current will pass through it. You're gonna have to trust me on this one, but you can calculate the -3dB point (the rolloff) of a simple lowpass like this- it is 1/(2*pi*R*C). With the low impedance mic, that comes out to be 1/(2*pi*100*1E-9), or 1.59E6 Hz. 159kHz, even higher than the stuff that bothers the dogs. That cable doesn't kill the highs from the mic *at all*: the source impedance is low enough to overcome the capacitive loading.

The guitar, however, is a different story. Same equation, but lookee here at what that much higher 10Kohm source impedance does: 1/(2*pi*10000*1E-9), or 15.9E3. That decimal point marches the wrong way 2 places... You're down 3dB at 15.9 kHz! Right in the middle of the "air" range... That's why you don't use StarQuad for your Telecaster.

The capacitance of the wire doesn't exactly eat the highs: but filter that is created by the circuit of guitar and cable *taken together* can't push the highs out the other end of the wire. The source impedance is too high to overcome the capacitive loading. The problem is _not_ the wire, it's the combination of wire and source impedance. Make sense?

If the signal changes with loading, the output impedance is the reason: the signal has to be going _somewhere_. These are grossly simplified, but hopefully they'll make some sense.

If anyone's still reading at this point, you deserve a prize....
 
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Great post. Skippy.

I usually try to make it as simple as possible to explain a concept, but not because it's easier to understand - it's because I always get into trouble when I get too specific - I usually forget the right formula or I make a glaring mistake in my calculations, and then I get EE's jumping my ass all over the place ("He he, you dummy, you said 10 to the minus 23rd power, when it's obviously 10 to the minus 22.9999987th power.") At my age, I have trouble remembering that R does NOT equal MC squared.

Hopefully, my explanation will appeal to all the chowderheads out there (like me), and yours will get all the techies who wanna learn more than my simple version provides.
 
Ok, good stuff here guys. Let me ask a few questions for verification.

I get that I am losing critical frequencies through the cable on a guitar. I assume this can be offset by putting EQ in the signal chain prior to other modifying devices i.e. distortion, phase, and other pedals. Or, should I EQ at the end of the signal chain to rebuild the signal once all the pedals modify the sound? Possibly it does not matter.

With mics this high freq rolloff is not occuring as there is less high frequency stripping going on because of low impedence. Just wondering though, using the same mic, would I get drastic tonality differences depending on the quality of cable I am using on my dynamic and condenser mics?

Last of all, is phantom power shaping the tone at all or just driving a hotter signal?
 
Middleman said:
Ok, good stuff here guys. Let me ask a few questions for verification.

I get that I am losing critical frequencies through the cable on a guitar. I assume this can be offset by putting EQ in the signal chain prior to other modifying devices i.e. distortion, phase, and other pedals. Or, should I EQ at the end of the signal chain to rebuild the signal once all the pedals modify the sound? Possibly it does not matter.

In general, the answer is that (assuming a decent cable that isn't too long), it usually doesn't matter. IF you're gonna eq to boost the top end, do it before the pedals, otherwise you're also boosting all their added pedal noise and hiss as well.

With mics this high freq rolloff is not occuring as there is less high frequency stripping going on because of low impedence. Just wondering though, using the same mic, would I get drastic tonality differences depending on the quality of cable I am using on my dynamic and condenser mics?

Usually you have to go out to several 100 feet to hear the effects of a cable on a low impedance mic. Some bad cables can sound bad even when they're pretty short.

Last of all, is phantom power shaping the tone at all or just driving a hotter signal?

Phantom power is used for different reasons in condenser mics; to charge the capsule (if it's not an electret), or to power some of the built-in electronics, or both. If the voltage is lower than the capsule needs to work properly, it can affect the tone, but normally, you do not vary the phantom power to change the tone of a microphone.
Click to expand...
 
Harvey: Boy, do I hear you. That's one of the reasons I quit posting anything on Usenet, back in about 1995. It's no fun playing the moral equivalent of intellectual chicken with folks. Just about the only reason I can screw my courage to the sticking-place and post techie cruft here on HR is that the community is generally pretty supportive, and not just out for blood... Your explanation was great, and not just for chowderheads! I figured that I'd just throw in the background for anyone who cared, because this topic really is a FAQ, and it ain't easy. But believe me, man- I know that you have already forgotten more about this art than I'll ever know!

Middleman: the order really does matter, and dramatically. if you use stomp boxes in series with your guitar signal, everything changes- and *how* it changes depends on the design of the boxes themselves. If the boxes are designed with a buffer amp that is in circuit all the time, your rolloff problem ends right there: the guitar itself just has to drive the cable down to the input of the box. After that, the active electronics in the box do the rest of the driving work, and they will be much lower impedance than your pickups.

But if the box has a bypass mode that cuts *all* the active circuitry out and just passes the guitar signal, then the loading changes (maybe dramatically!) with the box in or out. If you put active EQ before your distortion box, the buffer amps that are part of the EQ will drive the distortion box- you'll get one result. If you reverse the order, the loading will be different, the distortion box will maybe get a lot less HF at its input, and the tone will be *much* different.

Better or worse? It's your call. My personal opinion is that it's better not to let any more HF get sucked up by the loading than you have to, so I like to see the signal get buffered very early on. However, a lot of guitarists absolutely _hate_ that sound- that's why active guitar electronics still get looked at kind of sideways by a lot of folks (all of whom are better guitarists than I'll ever be). For that, all I can say is "Find a tone you like and don't worry about it...".

If you sometimes work live with long cable runs from your pedalboard to the backline amps, you _may_ find it useful to set up your pedal chain with an active buffer early on (like an EQ). That way, the loading change caused by bypassing other effects will be minimized: you'll have a short run from your guitar to the EQ, and then any long run from the output of the pedalboard to the backline will be buffered by the EQ, *not* the guitar itself. It'll give you greater consistency from stage to stage....

One size does not fit all!
 
can you guys give a good link so I can read about this too?
I didn't get it from the above explanations. I asked myself many times the same original questions that schenkerguy did.
I don't understand what is is output impedance, and how does a low impedance "load" a guitaqr pickup.
So if you can point me to somewhere or better, if you can explain this clearer, it'll be great.

Cheers, Andres
 
I've shot my wad on the explanation front- if rereading that article of mine doesn't help, I'm at a little bit of a loss.

Here's another online resource I just found that might help: http://www.tape.com/Bartlett_Articles/impedance.html . See if that lights any bulbs! I judt did a Google search on "impedance" and "FAQ"...

Anybody else want to chime in here? Impedance isn't a trivial subject, but it's not impossible either. I've just been living with it so long that I can't really explain it any better.
 
Thanks for the info!!

I'm going to print out this info and re-read it several times... but I think I'm getting it!

Thanks a ton for this great info, I really appreciate it! It's really great to be able to understand technical concepts fully, and get such expert replies.
 
thanks also for the link.
Is there any logic to follow regarding how different impedances affect tone? Like, if you put a bass into a mic input, does it loose the lows, the highs, etc?
 
Sure. It sometimes may seem a little odd, but there definitely is.

Here's a good hack way to simplify things and get a handle on it. Think of capacitors and inductors as "frequency variable" resistors (which, as it turns out, isn't far from the truth: that's basically half of the story). With a capacitor, the higher the frequency of the input signal, the more current will go through it. It looks like an open circuit at DC, and like a lower and lower resistance at higher and higher frequencies. With an inductor, you have just the opposite case: it looks like a higher resistance at higher frequencies. At DC, it's essentially a dead short, but as you go up in frequency, less current will flow through.

So what does this mean? Guitar and bass pickups are nothing more than a big coil of wire wrapped around a magnet- in other words, they're really nothing more than a big inductor that just happens to have the strings wiggling around in its magnetic field. The value of the inductance, and the value of the DC resistance of the coil, are very carefully designed to offer maximum output when loaded with the right load: a very high impedance guitar amp, for example. Add too much capacitance in parallel with the pickup (from the cable, for example), and the highs will go away. In fact, the tone control in your guitar is usually nothing much more than a big capacitor in series with a variable resistor that they bridge right across the pickup: turn the knob up all the way to the max resistance, and you keep all your highs. Turn the knob down to a lower resistance, and more and more of your highs will go away. They really ought to call that a lowpass filter, not a tone control, but who's counting?

I once built a Telecaster for a collaborator that had a *bass cut* tone control. All it was was a big ferrite-core inductor in series with the tone pot, right where the usual capacitor would be. Turn the knob up to the max resistance, and you'd keep all the lows. Turn it down towards lower resistance, and more and more of the *lows* would get shunted away through the inductor. So this worked exactly the opposite of a highpass tone control like you'd usually have, and they just loved it. It was hard for me to imagine that anybody'd want a Telecaster to be _more_ twangy, but there you go. There's no accounting for taste.

Anyway, run a highly-inductive bass pickup into a mic preamp with a low input impedance, and you start to roll off the lows. The interaction isn't quite a simple to undertand as the lowpass filter you create with the cable, but it is analogous: an RC filter gives you a lowpass effect (the cable seems to eat your highs), an RL filter gives you a _highpass_ effect (the low load impedance seems to eat your lows, because of the intrinsic inductance of the pickup).

Every pickup has an ideal load. The bottom line is that the higher the input impedance (and thus the lower the load on the pickup), the better your results will *probably* be.

But ony "probably": it's not quite cast in stone. Don't let that stop you: try different types of loadings (like that diseased Telecaster!), and you may just find a tone you really like... Frankly, funny loadings are probably at least half of what these "boutique" preamps with DI instrument inputs actually offer.

Hey, Harvey- you want to kick anything in here?
 
Nope, Skippy.

You nailed it better and simpler than I ever could. That's one of the best definitions of impedence and loading I've ever seen.
 
thanks, that's a great explanation. I'm gonna take time to grasp it cause I've been in the dark for a long time and I have to unlearn a lot. But sounds great to me.
Cheers
 
just a couple of questions.

The bottom line is that you need a high input impedance so you don't load the pickup. And when you load the pickup is when you alter the tone. Right? I don't get the image of "loading the pickup", as I think of it as the generator of the sound/electricity, hence the first link in the chain and as such can't be affected by subsequent links. Maybe that's a wrong assumption?

And what does RL filter and RC filter mean?

Again, I am sorry if my questions are too obvious. Cheers, Andrés
 
It is the first link in the chain. And if it was an ideal source, nothing that happens after *would* affect it. However, it isn't: it's a real-world source, so it has an output impedance associated with it: essentially, a resistor in series.

Let's take a look at that 10Kohm guitar pickup. We can describe that as an ideal source that puts out say 1V (just for argument) that has a 10Kohm resistor in series with it. This is *inside the pickup* (a parasitic resistance, if you will). It is not a separate component that we can see, or somehow remove to try and make things work better. It is just an intrinsic part of the pickup design itself, essentially distributed along the many turns of fine wire that make up the pickup coil.

So with no load at all, the pickup puts out 1V. If we drive an amplifier with an input impedance of 1Megohm (a decently high impedance), we'll have a little bit of signal loss- but it is so small we can ignore it. If you want to know exactly how much, here it is: Vload = Vsource*(Rload/(Rsource+Rload)). So that comes out to be 1*(1,000,000/1,010,000), or .9901V. Less than 1%. Trivial. No problem.

But if we run the 10Kohm pickup into an amp with an input impedance of 10Kohms, look what happens to our no-load 1V: the actual output is only 1*(10,000/20,000), or 1/2 volt. 500mV. That's 50% loss!

Where did the other 50% go? It was dropped across the intrinsic resistance of the pickup: the output impedance. It went away before we ever got to see it, lost inside the pickup. The pickup is not a perfect generator, so it gets very lossy when loaded heavily. Half our signal _never makes it out of the pickup_, because of the loading.

That's the easy part: that's just a DC voltage divider. But worse yet for us, the frequency response of the pickup will also be affected, and much *more* than just a straight loss. That is a pretty nonlinear and complicated thing, and I'm not going to dig into it here- but the low frequencies will get rolled off more than the highs. What we've built is an RL highpass filter, between the inductance of the source, and the input impedance of the load. So the overall signal is attenuated by 50% (3dB), but in reality the lows will get hammered much worse than the highs because of the inductance of the pickup.

An RC filter is a resistor in series with a capacitor (like our pickup driving the cable). An RL filter is a resistor in series with an inductor, L being the universal shortand for inductance.

Anyway, good questions: nobody was born knowing this stuff, and getting a handle on it make a lot of things easier to understand, work with, and control.
 
thanks, that is a great explanarion.

So the rule is low output Z and high input Z. You know, that looks odd to me, since I thought that if the input Z is high then less signal will pass thru. Then I read an old article by Craig Anderton that says "an input impedance is equivalent to putting a shunting resistor between the input of the amplifier and the ground. An output impedance is equivalent to having a resistance in between the output of the amp and the next stage".
So joining your explanation and the article together, I understand it a little more. The input keeps the signal from going to the ground and sends more of it to the next stage. Is that correct?

the article is at http://guitareffectprojects.tripod.com/buffer.jpg

Cheers, Andrés
 
Well, you can certainly see why Anderton is a writer of tech articles, and I'm not: his description is exactly correct, quite succinct, and only took him 1/10 of 1% of the words I used. Why didn't I think of that? (;-)

The input impedance of an amplifier can be thought of as exactly that: a resistance between signal and ground. As long as it is more than about 10 times as great as the output impedance of the source, it can be throught of as having essentially no loading effect on the source. That's the classical rule of thumb, and it even has a technical name: nerds like me would call that a "bridging" connection. What your amplifier "sees" at its input is essentially what is left over, *after* the signal is dropped across the input impedance (EEs, don't shoot: I know that I'm playing fast and loose with the terminology, just for simplicity). A bridging connection just hangs across the output and doesn't affect it enough to worry about.

Guitars and basses are high-impedance sources, and you always have to have an *extremely* high input impedance to keep from loading them and changing their behaviors. The grid on a vacuum tube, an op-amp input buffer optimized for high-Z use, the primary winding of a high-ratio transformer: good examples of proper loads. Guitars and basses are just flat out wierd, but we have to tolerate them anyway. (;-) And sometimes, a bit of extra loading can make them sound better.

Studio microphones are a the same sort of thing- just the range of their impedances (and their preferred loadings) change. Their output impedance is usually pretty low, in the 100ohm range. Mic preamps are usually designed with input impedances around 10Kohms, so that they also can be considered bridging loads. However, there's a fly in the ointment there: many microphones have transformer outputs, and may actually sound significantly *better* with a lower impedance input (or a higher loading). Once again, the "boutique preamp" phenomenon rears its ugly head: different pres will have different input impedances, and you'll get different sounds _from the exact same mic_ because of that, regardless of any "color" that gets added in the pre itself. The Groove Tubes Vipre preamp takes that to the next level: you can control the input impedance with a knob on the front panel. Twist the knob until you like the sound! Yes, it will make a difference. On some transformer-output mics, like ribbons, it can make a _striking_ difference.
 
ALMOST all guitars and basses are high impedance.
My guitar has an EMG 707 pickup which has a pre-amp built in. It runs on a 9v battery and the signal is low impedance. It makes for a much hotter signal and because it is low impedance, it is not affected by interference from lights or other sorces as much either.

Active rules.
 
Cudos, Skippy and Harvey. My wife thinks I'm nutz walking around all day saying "great post, great post". Suppose I am. Hope all reading realize many people pay LARGE sums of $$$ and invest many hours of study for this kind of education that you can receive on this board. "great board, great board"......... :>})
 
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