Impedance explanation for dummies

[lmorse]

I am trying to gain a bit of understanding and intuition about impedance, but I am struggling a little with getting a real good feel for the topic.

I have read that impedance is similar to resistance, only for AC circuits, I have read articles about circuits 'seeing' high or low impedance inputs. I have read that passive guitar pickups are high impedance devices. I know that all circuits have a Thevenin equivalent that is analogous to a single resistor in series with a single voltage source. I understand that input buffers manipulate the input signal to help reduce loss of high-end etc. I think I understand that voltage is 'potential' in terms of electrons/current that is available, current is rate of electron flow, resistance and impedance limit current, and power is the 'output' (e.g. Heat, audio, light etc) of that current flow. But I am still struggling to really understand at a fundamental intuitive level what Input/output impedance really means, how it fits into this picture and how it might impact on pedal circuit design. Why does it matter if a circuit sees a high or low impedance input? How does a circuit 'see' the internal resistance of a previous circuit? Does a circuit not only see a current (rate of electron flow) coming into it?

Can anyone point me to some in-depth articles or reference? Would anyone care to try to educate a creaky old brain?

I tend to find it easy to learn first through analogy, then diagram, and then mathematics.

[GibsonGM]

Good questions, lmorse!  

Some basic stuff to look at:  http://artsites.ucsc.edu/ems/music/tech_background/z/impedance.html

You COULD say that Impedances want to be equal, if they can....like water in 2 tanks, water will not flow back and forth if the tanks are at the same pressure.  We settle for "low output impedance" and "high input impedance", usually, which sort of 'sets up the flow' in the input direction....I can give you some analogy, altho my way of thinking on this may be a little clunky, ha ha:

Picture impedance (abbrev. as "Z") as a FREQUENCY VARIABLE resistance that takes into account the important circuit elements such as what capacitances are involved and the like.   To know impedance, we need to specify a frequency of interest.   For guitar, that is typically oh, 80Hz to 10kHz or so - a HUGE range, obviously.    

Things that have capacitance (like capacitors), and things that have inductance (like guitar pickups) have a DC resistance (our 'normal' resistance we measure on a meter) as well as REACTANCE.   That is the frequency-dependent 'resistance' of these components.   Mostly we deal with capacitive reactance, unless you have a pickup or a coil involved....

The impedance formula for SERIES or PARALLEL impedances works all this stuff together to get us a kind of "Ohm's Law that has been set up for AC".  

So, in a nutshell (a lot to digest!), impedance is frequency-dependent resistance.  Power transfers best if impedances are equal.

A signal voltage has to go from high to low potential, so at some point it leaves one place and must enter another...that is "looking into" the next circuit.     The circuit doesn't 'see' the internal resistance of the last circuit, it simply is more ABLE to draw current, or is less able....impedance mis-match, such as low input impedance, is first experienced as a loss of highs, as these are "more liked" by the low impedance (ok, of course there's more to it, but close for now!).      

High impedance, while not intuitive, is kind of like 'efficiency'...it's GOOD when it takes LESS voltage to drive a circuit!  If it takes very little signal to drive the high Z, then the circuit 'has plenty of signal' and won't DRAG DOWN the circuit before it trying to 'get more' (called "LOADING").  Low impedance "sucks up" power and wastes it, loading our signal down and losing the highs, then the mids, then all of it...So, the circuit sees the current coming into it, yes, but since guitar stuff is a complex signal, it may see SOME of it LESS, depending on the impedance!!    

Read up on this where you can...I learned a lot by reading ARRL ham radio handbooks (!)....Google will help...sorry if my picture is hard to understand, it's the best I can do and be brief, and not take all night to write it!     You'll get it, you just have to immerse yourself in it for a while...takes time.....

Reactances + resistances are the data you need, and the formula gives you the rest.    Sometimes it's just seat of the pants, tho, as in DIY audio - don't need to re-invent the wheel here!!  

[lmorse]

Phew! I've read through two or three times, I'll need another 10 or so passes before I get the subtleties I think . A bit there to take in! Thank you. This is the first time I have heard of impedance as being frequency related resistance.

[lmorse]

Is the concept of 'drawing' a current important? I had always imagined a signal as sort of being 'pushed' or 'driving under its own steam' out out of a circuit, rather than being extracted from the output.

[Seljer]

By "similar to resistance, only for AC circuits" you kind of covered that "frequency dependent resistance". AC is not just the 50 Hz (or 60Hz depending on where you live) coming out the socket on the wall but practically any signal be it 1khz or 1000khz.

Now theres some math with complex numbers that I'll glance over but you the formula for the impedance of a capacitor is:
Z = 1 / (2* pi * f * C)
Where pi is 3.14, f is your frequency, and C is your capacitance. What you should take away form this is that capacitors have low impedance at higher frequencies.

The formula for the impedance of an inductor is
Z = 2*pi*f*L
Here it's basically the opposite. At low frequencies they have low impedance, at high frequencies high impedance.

[to be correct, the above equations are for reactance of these two components, not impedance]


Think of the output and input impedance of two stages as a voltage divider. If the top resistor in a voltage divider is very small you don't lose any voltage over it an get a voltage ratio very close to 1.00 .
Many (most?) pedals are voltage based devices so we want to transfer as much voltage as possible from one stage to the next .

A guitar cable is basically a big long cylindrical capacitor (well theres some inductance too but at audio frequencies it's effect is negligible). The cable's capacitance is basically wired in parallel with the input impedance of next stage. Since capacitors have lower impedance at lower frequencies, it'll basically lower the overall input impedance presented to preceding stage so unless you have a very low output impedance you'll get some voltage loss at higher frequencies. Hence a buffer can drive a long cable without any treble loss.

In theory you get maximum *power* transfer if you match the input and output impedance but that mainly applies to specific applications (mainly high frequency radio circuits and transmission lines). As I mentioned, we're generally going for minimum voltage loss. There are exceptions to this like certain fuzz pedals, where you've got a bipolar junction transistor at the input which has a low impedance (because transistors are current amplifiers, the current needs to flow into the base so it the transistor can even work, opamps and FETs work from voltages).

Is the concept of 'drawing' a current important? I had always imagined a signal as sort of being 'pushed' or 'driving under its own steam' out out of a circuit, rather than being extracted from the output.

It kind of all happens at once. But yes, this is the general concept. If you don't have a load hooked up no current can run and you get the full voltage on the output terminal (regardless of any internal output impedance). As soon as you hook up a load a current flows, that current causes a voltage drop over the internal impedance so the voltage on the terminal is now lower.

[lmorse]

Think of the output and input impedance of two stages as a voltage divider. If the top resistor in a voltage divider is very small you don't lose any voltage over it an get a voltage ratio very close to 1.00 .

So a low impedance means less voltage loss (depending on frequency)? Therefore a low impedance output is preferred?

Since capacitors have lower impedance at lower frequencies

I assume this is a typo and you mean higher impedance, hence blocking DC? Sorry, not being smart, I'm just trying to ensure I understand.

The cable's capacitance is basically wired in parallel

Is this also a typo? How come it's parallel and not series?

So many questions... Every new step I make opens up more! For example... Pi, strange how a constant like that relates to electron flow?

[Seljer]

So a low impedance means less voltage loss (depending on frequency)? Therefore a low impedance output is preferred?

yep


Another issue around this is that guitar pickups are basically whopping big inductors. Their DC resistance already a couple of kiloohms because its a very thin wire. The typical inductance of the coil is easily a couple of henries. So in the treble frequencies the total output impedance of the pickup is easily 100kiloohms. Which is why that first run of cable from guitar->amp or your first pedal is the one that'll have most influence on your tone. Then again, tone is subjective, you might like the "vintage" sound of having your guitar's brittleness softened by a long coily cable.

I assume this is a typo and you mean higher impedance, hence blocking DC? Sorry, not being smart, I'm just trying to ensure I understand.

Doh, yep, precisely that. As you go lower and lower in frequency the impedance converges towards infinity at DC ("0Hz").

Is this also a typo? How come it's parallel and not series?

Code Select Expand

     Zoutput
    -------
----       -----------------+
    -------        |        |
                   |       | |
                 -----     | |  Zinput
         Ccable  -----     | |
                   |       | |
                   |        |
                 -----    -----
                  ---      ---
                   -        -




So many questions... Every new step I make opens up more! For example... Pi, strange how a constant like that relates to electron flow?

The mathematics of this universe are uncanny  
When you're doing the equations, you have to use 2*pi to convert frequency into angular frequency so your trigonometry functions (like for a sine wave) match up appropriately.

[lmorse]

Another issue around this is that guitar pickups are basically whopping big inductors

So I am guessing that humbuckers have a higher impedance than single coils - that's why single coils sound brighter? More windings = more wire = higher impedance?

[Seljer]

Yep, thats a part of physics behind it all. But there is also a bunch of stuff like inter winding capacitances and resulting resonances other phenomena that influence what the total end result sounds like.

[lmorse]

I think I have just figured why the cable capacitance is in parallel... is it because the shielding etc is part of the long thin capacitor which is in parallel to the signal travelling through the cable? I originally thought you were saying the capacitance of the cable is in parallel to the device it is going in to 

[PRR]

Leave "impedance" for a later time.

You seem to be struggling with Resistive Voltage Dividers.

Forget audio/guitar.... you know too much and not enough.

Come over to my house where the power wire is too long.

I get 250V (125V+125V) through 1 Ohm of too-long wire.

> Is the concept of 'drawing' a current important?

I turn off all the loads in the house. Infinite resistance. What is the current in my power wire? 250V/infinity is Zero Amps. What is the voltage at my house? What is the voltage-drop in 1 Ohm at Zero Amps? 1r*zero is Zero Volts. I get full 250V at the house.

I connect a 9 Ohm load. The total resistance in the circuit is now 1+9 or 10 Ohms. What is the current? 250V/10r is 25 Amps. What is the voltage-drop? 1r*25A is 25 Volts dropped. 250V-25V= 225V at the house. Lamps dim a bit.

What about a 4 Ohm load? Total is 1+4= 5 Ohms. Current must be 250V/5r= 50 Amps. 50Amps in 1 Ohm is 50 Volts drop. I have just 200V at the house.

Rather than mess around with explicit Volts and Amps, voltage-dividers can be computed from the resistance values.


____________________________________________
In modern audio, 95% of the time "impedance" is pretty near Resistance.

The output of my CD player is rated 100 Ohms output "impedance". If I measure it I might find 200 Ohms at 20Hz, 100 Ohms at 100Hz, 110 Ohms at 20KHz.

The input of my cassette recorder is "22K" and pretty sure 20K-25K all across the audio band.

*Say* the CD outputs 1.000 Volts.

If I connect CD Out to Cass In, the cassette deck gets 0.990V at 20Hz to 0.995V above 100Hz. The frequency response has an "error" of 0.05dB, which is inaudible (and also far less than the response error of tape).

You can use different numbers, but generally systems are designed so interface errors are very small.

However, this is a guitar forum. When full-up, a passive guitar with its cable *does* have an impedance which is very far from a Pure Resistance. Full analysis is messy. But assume something like:

5K Ohms 82Hz to 160Hz
10K at 200Hz
50K at 1KHz
100K at 2KHz
200K around 4KHz
100K near 8KHz

For illustration, *assume* the guitar can deliver any frequency at 1.00 Volts.

If I fed it to my 22K cassette input:
82Hz -- 5K guitar -- 22K load -- 0.8V
200Hz -- 10K guitar -- 22K load -- 0.68V
1KHz -- 50K guitar -- 22K load -- 0.3V
2KHz -- 100K guitar -- 22K load -- 0.18V
4KHz -- 200K guitar -- 22K load -- 0.1V
8KHz -- 100K guitar -- 22K load -- 0.18V

It sucks all over, with a 8:1 suck-out in the scream/zing overtone area. (Try it. Tack a 22K resistor across that busted cord you need to fix anyway, and listen.)

If I fed it to a typical 1meg (1,000K) guitar amp input:
82Hz -- 5K guitar --   1Meg load -- 0.995V
200Hz -- 10K guitar -- 1Meg load -- 0.990V
1KHz -- 50K guitar --  1Meg load -- 0.95V
2KHz -- 100K guitar -- 1Meg load -- 0.91V
4KHz -- 200K guitar -- 1Meg load -- 0.83V
8KHz -- 100K guitar -- 1Meg load -- 0.91V

The suck-out at 4KHz is quite small. For music RE-production, I would fret a bit (though larger errors are often tolerated). For music PROduction, we tweak the amp knobs, speaker choice, pickup balance, to get the pleasing tone.

If you work the numbers for some other likely amp-input resistances, you will see that 470K and 2Meg are hardly-different from 1Meg, nothing like a guitar driving say 22K. We can use various guitar-amps without concern for impedance; but non-guitar inputs may "load" the expected guitar tone in unhappy ways.

[PRR]

> Pi, strange how a constant like that relates to electron flow?

It doesn't.

Sine waves can be described as the path of a point moving around a circle, like a crank.

If you stayed awake in Trig class you did circle problems with Radians. It is like you used the radius as a yard-stick to measure around the outside of the circle.

The total circumference of the circle is Two * Pi.

The basic units of reactance are defined with radian math.

There's actually some logic to that. The peak amplitude of the wave is the radius of the circle. If we want to know how the amplitude changes, we would use the radius as our yard-stick.

If F and H were re-defined differently, we could take 2Pi out of many everyday formulas. However it makes more complicated formulas more messy. So 6.28 has to be at your fingertips.

[GibsonGM]

Nice, guys.



To be explicit, the cable's capacitance (a phenomenon that occurs between 2 conductors separated by a dielectric) is in parallel, as shown in that ASCII drawing, because it is happening between the center 'signal carrying conductor' and the grounded shield of the cable..it's not a property of the circuit, but of the physical item used (the cable).     

It's just letting high frequencies jump from the conductor to the grounded shield.

This can happen basically anywhere...between signal wires in an effect or amplifier, it's even part of how the elements of transistors and tubes  interact! (Miller Effect)    It's considered at design time.


----------------------------------------------------------------------------------------------------------------------------------------------------------------------
Caps pass higher frequencies easier, yes.  A GIVEN capacitor will have a LOWER REACTANCE ("AC resistance") at a higher frequency.  Accordingly, it will have a higher reactance as you lower the frequency.     

[lmorse]

it is happening between the center 'signal carrying conductor' and the grounded shield of the cable..it's not a property of the circuit, but of the physical item used (the cable).

Cool, I do understand this correctly now then.

PRR - Thank you, great explanation. I now understand that signal loss at various frequencies is a function of both the signal generation circuit impedance and the load circuit impedance:

Signal Loss for a certain frequency = (V-out / (R-out (for that freq) + R-in (for that freq)) ) * R-out (for that freq)

[DrAlx]

On the subject of analogies, you may have come across something like this...
http://en.wikipedia.org/wiki/Hydraulic_analogy
It doesn't help with the complex variable side of things but does give some physical examples
of how flow can be impeded in a frequency dependent way (i.e. slow changes impeded differently to fast changes).

[Seljer]

^
Or if you've ever been somewhere with a weak water mains and for example, if someone flushes the toilet while you're in the shower. The extra water flow for refilling the toilet's reservoir and the "resistance" of the main cold supply pipe prior to where it splits between shower/toilet cause the cold water's pressure to drop in the shower and you get left scalding hot water in the shower.