Desperately want to understand impedance.

[bassmannate]

Just making sure I have a good basic understanding of how impedance works and that I'm applying Ohm's Law correctly.

We want input impedance higher than the output of the previous circuit so that our current remains low and the voltage (signal amplitude) stays high. That way, we still have something to work with in terms of a wave form.

If you want maximum VOLTAGE transfer then yes, source Z << load Z. 

However sometimes you want maximum POWER transfer (like power amp into speaker).  In that case you want source Z = load Z.

Rarely (but still good to know) you may want maximum CURRENT transfer: source Z >> load Z.

How does one determine which is best? Obviously, you want as much power going from power amp to speaker but what about other situations?

[R.G.]

Unless you can clearly say what "best" means, there cannot be any best.

In normal audio practice, you want maximum signal voltage transfer. To get that, you want the maximum possible MISmatch in the direction of low impedance at the source and high impedance at the load. This prevents any signal from being lost to the voltage divider nature of the source impedance and the input impedance of the load reducing the voltage.

In current-input situations (there are a few, but they're rare) you want the lowest possible impedance at the load compared to the source, so the load does not reduce the transferred current.

In transferring power, you want matched loads. There is a maximum-power-transfer theorm where you can show that the most possible power transfers from a source  to a load when the load and the source impedances are identical. This is where the "matched impedances" thing comes from. However, it's really only used in practice in RF design.

Contrary to the speaker example, you don't usually want matched amplifier and speaker impedances, even in tube amps. The output impedance of an audio power amp generally should be as low as possible, much lower than the speaker load. This is expressed in the "damping factor" number which is part of audio power amp specifications. In general, the damping factor is the ratio of the load impedance to the amplifier impedance, and this is often several hundred for SS amps, and generally 3-10 for tube amps.

The reason you're not looking for maximum power transfer in audio power amps is that maximum power transfer is often destructive, and always wasteful of power unless there is no other way to do it. For maximum power transfer, the impedances are equal, so equal power is dissipated in both load and source impedances. If the amp has a lower source impedance than the load, it dissipates less than 50% of the power transferred to the load (neglecting the other losses in the amp, which can be significant).  And long experience has found that for best audio reproduction, you want speakers which are controlled very tightly by the amplifier - that is, well damped. Speakers which are driven from high impedances show more of their own resonances.

Maximum power transfer is destructive for things like batteries. A load equal to the internal impedance of a 9V battery or car battery will usually damage the battery from high current flow and high internal dissipation. Batteries can and do explode under such use.

For audio, you will be looking for maximum voltage transfer in almost all cases.

[PRR]

> sometimes you want maximum POWER transfer

That truth usually leads to real problems.

It is true that I can get maximum power (as for heating) in my house by matching the utility company impedance. This will cause 120V to drop to 60V. The power in my house is equal to the power loss in the utility wires and windings; before I can read my meter the lines will melt.

It is nearly true that IF you have a pure resistance audio source NOT near overload, which you can not change, you can get "maximum power" by matching the source. However you get more Power by increasing the gain and level, and loading with 2X source impedance. Rasing the supply also increases power until dissipation is a limit. Raising the load still more raises output without raising dissipation.

> source Z = load Z

This really causes more trouble than good. (And among more educated men than all of us.)

In AMPLIFIER systems, mis-matching is MUCH more useful and more common.

> (like power amp into speaker).  In that case you want source Z = load Z.

No. I won't say "never". However the "source Z" of audio power amps is very tricky.

Measure the Zout of a large transistor amp. Probably below 0.1 ohm. (If it is direct-coupled, you can simply put your ohm-meter to it while on and silent; this gives a close approximation of the low frequency Zout.) Would you load it in 0.1 ohms?

Measure the Zout of an older Fender Twin tube-amp. (Ohm meter will not work; drive to some small level and change the load, plot the voltage.) Probably around 16 ohms on the 8 ohm connection. Do the same with an Ampeg VT-40: more like 50 ohms. Yet these amps do work best close-to their rated impedances. A 50 ohm speaker on the VT-40 sure would be less overwhelming than the same drivers wound and wired for 8 ohms.

The optimum load for a loudspeaker power amp is somewhat higher than the sum of a lot of parasitic series resistances and parasitic shunt conductances which you can NOT see directly. In the Twin we have 200 ohms in the PT, 800 ohms in the 6L6 when bottomed, 33K when linear, 2:1 leverage in the OT..... we load with 4K-8K.

> In transferring power, you want matched loads.

The language is slippery.

If you can NOT modify the source, max-power is matched.

Half the total power is wasted in the source, NOT available to you.

If you have any say in the source, you want MIS-matched.

My home gets power from a dam 10 miles up the road. Say my house wants 240V 100A, a 2.4 ohm 24,000 Watt load. I could run a 2.4 ohm 10 mile wire to a 480V 100A 48,000 Watt tap on the dam. The dam owner would of course want to be paid for the whole 48,000 Watts, even though I only get 24,000 Watts in my house.

The 48,000W would cost say $5/hour, $40,000 per year. That could pay for a lower-resistance wire. Say we triple the wire, 0.8 ohms. At 100 Amps the line-loss is 80V, 8,000W. Total bill is now on 24KW+8KW= 32KW, 16KW or $13,000 less per year. (If you work out copper-cost, this is a bad deal any way you look at it.....)

Also: in the warm month I turn off the 24,000 Watt heater but may need a hundred watts of light. In fact I may not need all 24,000W all the time. As load drops from 24KW to 1KW, voltage at my end un-sags from 240V to 476V. I can get lamps in any voltage, but I may not wish to change them every time the heat level changes. While the copper economics may not justify fatter wire, the awkwardness of varying voltage with varying load might.

So what means "transferring" power?

If I have power one place and want to use as much as possible in another place, I want LOW losses. The load impedance should be much higher than generator and line impedance.

In the ugly situation where source is precious and load requires all the power it can get, we are really SUCKING power from a hardly-adequate source, and accepting large losses to get what we can. When tubes were expensive or when wires are very long, this may be the best we can do. On short wires with cheap amplifiers, you can get ample power and often best sound withOUT "matching".

> maximum power transfer is often destructive, and always wasteful of power

Well said.

> A load equal to the internal impedance of a 9V battery or car battery will usually damage the battery from high current flow and high internal dissipation. Batteries can and do explode under such use.

Depends how good it is. 6V Lantern batteries sell for $4 or $11, different weights. I have dead-shorted the $4 kind (oops). It gets HOT. If worked into a matched load, it would get very-warm. It does not explode, won't even sizzle spit. The $11 kind worked "matched" might blisted skin or melt plastic. And we know the denser laptop PC batteries do get violent when they develop internal shorts.

Here is a familar "matching" situation. Battery powered drill. Lean on it until the RPM drops to half the no-load RPM. This actually is very-nearly the maximum power to the bit. It also makes BIG heat in the battery. And if you do something like pumping water, so you can count actual work done, running light at 95% full speed eventually moves almost TWICE as much water as working hard at half RPM, because you don't waste half as heat in the battery.

[Paul Marossy]

Everything above was an interesting read.

[mac]

maximum power transfer is often destructive, and always wasteful of power

Remember the bridge over Tacoma River.
http://www.youtube.com/watch?v=3mclp9QmCGs

This a case of matching impedances.
The bridge, or load, had a natural resonance frequency determined by the physical properties of the bridge. In the case of a simple spring this frequency is given by w = sqrt(k/m).
When the velocity of the wind, the external force or driver in electronics, divided by some bridge lenght parameter had a value close to the natural frequency of oscillation of the bridge, the transfer of energy was maximized and the bridge collapsed.

See the Vr/Vin vs Freq graph of the following link,
http://www.youtube.com/watch?v=aDaOgu2CQtI

> Everything I read doesn't really mention current

It is awkward to measure current, especially at audio, and audio techs are lazy.

Funny since charge, mass and spin are the only real things 

mac

[BadIdeas]

And straight into the bookmarks we go.
Is anyone else thinking this thread should be made sticky or linked in the wiki or something?  +  +  = exploding head smily.
I'm wondering what the average impedance of a 9v battery or AC adapter is. I'm guessing I don't read much about it because it only accounts for <10% of whatever. I've read on geofex that it is a big part of the Fuzz Face sound, but I don't remember why.

I like ashcat's post on the pickups; a lot of information crammed in to a little bit of space. I really haven't given much thought to their characteristics just yet, but maybe now those charts on Seymore Duncan's website will make sense now.

[ayayay!]

Is anyone else thinking this thread should be made sticky or linked in the wiki or something?

Not a bad idea!

I'm wondering what the average impedance of a 9v battery or AC adapter is.

R.G. explained that early on in his "12 Volt Battery(ies)" example.  That's DC only, which is easier to comprehend in this case. 

[BadIdeas]

R.G. explained that early on in his "12 Volt Battery(ies)" example.  That's DC only, which is easier to comprehend in this case. 

If we imagine that eight AA batteries had an internal impedance of 10 ohms (I'm making up the numbers) then no matter how we tried, we could not get more than 12V/10 ohms = 1.2A out of them. The car battery probably has an internal impedance of maybe 10 MILLI-ohms. So it would happily put out 1200 Amperes into a dead short.

Maybe I've lost track, is this what you're talking about?

[PRR]

>> I'm wondering what the average impedance of a 9v battery
> R.G. explained that early on

No. He pulled "10 ohms" out of his hat, a nice round number for a simple example.

From 1985 data on Radio Shed's $0.59 9V batt, I get a value near 50 oums.

And in fact his "eight AA-cells at 12V and 10 ohms" is very nearly the value I get using RS's $0.27 (cheapest) AA batts.

> I'm wondering......

Do you know any electricity? Hint: measure a 9V without and with a 100 ohm load. Assume the drop is due to a voltage-divider, the 100 against some resistor hidden inside the battery.

BTW: for high-current loads, a low-price battery "gets weak" not so much because its unloaded voltage drops, but because that internal resistance rises. (The zinc gets covered with gas bubbles, hardly any path left for electricity to get through.)

[Krallum]

Definite sticky/wiki content.

[phector2004]

What a great thread. Have you guys considered collaborating to make a book? 

This was very helpful... Gotta re-read it to digest it some more, but here's a question:

If you overwind a pickup, you're increasing its resistance (more wire)/inductance (more coils). Is there therefore LESS of a Z mismatch, meaning more power is transferred at the cost of signal quality?

I'm guessing there's more current flow, but the variance of the voltage will increase, depending on the frequency?

[R.G.]

If you overwind a pickup, you're increasing its resistance (more wire)/inductance (more coils). Is there therefore LESS of a Z mismatch, meaning more power is transferred at the cost of signal quality?

I'm guessing there's more current flow, but the variance of the voltage will increase, depending on the frequency?

Pickups are complicated devices. If you overwind, you increase resistance, a bit. A single coil is often about 4K ohms resistive and 1/2 to 1 HENRY of inductance. Inductance increases as the square of the number of turns, so increasing turns by 41% doubles inductance. The distributed self capacitance of the winding comes to dominate things above about 7kHz, and there may be a resonant peak or two from the capacitance and either the full inductance or sub-sections. This last is very, very dependent on exactly where every turn in the coil is compared to every other turn and the magnets, metal holder/fittings, strings, etc.

You have turned over the next rock and are staring into a new and different set of ugly squirming things: AC impedance issues. It's not as simple as the power-transfer or impedance matching ideas from earlier in this thread.

[ashcat_lt]

In running some quick sims it looks like changing the DC resistance of the pickup over a reasonable range (ie - through values typical in real pickups from light SC to hot HB) makes no real difference into a 1M input.  I'd guess this is because we're so far past the 10:1 ratio that these changes are minor in comparison. 

Sweeping the inductance through a similarly reasonable range has a noticeable effect in moving the center of the resonant peak downward in frequency and also damping it a bit in amplitude.  Sweeping the self-capacitance does something similar, though I just used arbitrary numbers around the "typical" value I got from somebody else.

[PRR]

> here's a question:

Arg. Pickup impedance is NOT simple.

Don't forget cable capacitance (~30pFd/foot) (VERY significant) and any on-guitar volume pot (moderates all other impedances).

Here's a tip: let the pickup winder worry about it.

The typical design path is to use more turns of finer wire, giving more bass-mids but ultimately a treble-peak/fall lower in frequency. We need that top-drop to take the inharmonicity (end-effect) off the strings. If we manage to go too far all the zing is gone. It is easy to plot ideal parts, tough to estimate actual windings, and the "musicality" has to judged by ear. I suppose most winders skip the theory and just try various things.

And in fact pickups evolved within context of the HIGH input impedance of a tube grid.

Which goes back to low output impedance HIGH input impedance. In this case we don't control the pickup+cable impedance, but we can make sure our input impedance is high like a tube guitar amp.

FWIW, I can't find a plausible set of L-R-C values giving significant peak over 250K. And that peak is so narrow that intuition and experience says it may be blunted a bit without loss of musicality. The customary 1 Meg is probably "HIGH enough". Values 470K to 2Meg7 seem to interchange without comment. Values toward 68K cause significant top-droop, a useful cheap-trick.

[ashcat_lt]

This is maybe only tangentially related to the point of this thread, but some might find it interesting:
GuitarFreq2.2

It's an Excell spreadsheet which simulates a guitar circuit and show the frequency response as various values are changed.  It was put together by one of the mods on the GuitarNutz forum, and I trust his macros.  He's included a number of "presets" for common pickup types and cable lengths and whatnot.  He'd also incorporated several common mods - such as "treble bleed" cap across the V control.

I personally took some of the values from this and built my own sim in 5spice so I can get deeper into messing with the circuit.

[Resynthesis]

Anybody seen this? It might be of help to some

http://online.physics.uiuc.edu/courses/phys498pom/Lab_Handouts/Electric_Guitar_Pickup_Measurements.pdf