So, I've read just about everything I can find on the internet about impedance and impedance matching. I've always had a very basic understanding of high impedance being a very weak, quiet signal (like coming from a guitar pickup) and low impedance being a very strong, loud signal (like coming out of an amp and into a speaker)
Then, I discovered you could make a preamp that lowered the impedance of the signal but was unity gain. I may be getting concepts terribly mixed up at this point as I generally associate high gain to increasing the volume of a signal and thus lowering the impedance.
Everyone here seems to understand (or at least pretend to understand) impedance but I've been trying to understand it since high school. Help!
Quote from: bassmannate on November 22, 2010, 09:53:22 PM
So, I've read just about everything I can find on the internet about impedance and impedance matching. I've always had a very basic understanding of high impedance being a very weak, quiet signal (like coming from a guitar pickup) and low impedance being a very strong, loud signal (like coming out of an amp and into a speaker)
You're fighting some misconceptions. Impedance has nothing to do with strong or weak signals.
Impedance is the generalized AC version of resistance. Resistors have nothing to do with how big the signals through them are. All they do is *resist* current flow. Impedances *impede* current flow, and it's a more general, AC kind of thing than resistance.
One of the really fundamental, awe-striking things you pick up in EE classes is the so-called Thevenin model for voltage sources, named after the man.
Every possible voltage/signal source can be modeled as a perfect voltage source, able to provide its rated voltage into any load, no matter how much current the load needs. If you had a perfect voltage source of, say, 9V, you could put a solid copper bar a foot thick across it, and the voltage would still be 9V. It would not drop at all.
The impedance of a theoretical perfect voltage source is zero. But if we put a resistor in series with that perfect voltage source, we make it into a closer version of a real one. Thevenin's theorm is that all voltage sources can be modeled as a perfect voltage source in series with some impedance (think resistance). Eight AA cells in series makes a 12V battery. So does a car battery. But the internal impedance (think "resistance") of the AA batteries is much larger than the car battery, so it can't put out as much current. 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.
The Thevenin model for a car battery and for the eight AA cells is the same - a 12V voltage source with a resistor in series. The resistor just has a different value for the AA cells than for the car battery. The car battery has a much lower impedance.
How we doing so far? Questions? There is more.
So, to relate R.G.'s (really good) response back to those guitar pickups - a piezo pickup has a very high impedance, so it can only supply a limited amount of current before the voltage starts to drop. (Ohm's law here; voltage = current X impedance; if you can't draw the current you need from the pickup, your voltage drops, and all an audio signal really is, is an AC voltage.)
What your buffer preamp does, then, is keep the voltage the same at the output as the input without drawing more current than your pickup can provide, but it gives you more available current at the output. More current can drive a lower input impedance without a voltage drop. And since your guitar cable has an impedance itself, more current can drive longer cable runs, etc.
The big difference between impedance and resistance is that impedance is frequency dependent, where resistance (be a DC phenomena) doesn't have a frequency component at all. Or, I guess, you might better think of it as; resistance is impedance at 0Hz.
By the way, R.G., that was a better explanation of that than I've seen before. You should write a basic electronics theory book. I'd buy it.
Gabriel
> Everyone here seems to understand (or at least pretend to understand) impedance
Few really do.
That is because few have mastered basic DC circuits.
Even as simple as a battery and a lamp.
Take my truck tail-light: 12v 1.2 Amp.
Do you see that this is 10 ohms?
Take R.G.'s handful of cheap AA-cells with 12V unloaded voltage and 10 ohms internal resistance.
What is the voltage across the lamp?
Take his car battery, 12V 0.001 ohms internal resistance.
What is the voltage across the lamp?
Some folks can solve this very simple problem, but can't get beyond it. If a guitar is 1 volts unloaded and 50K resistance, what is the signal level when loaded with 500K? With 50K? With 5K?
Sometimes we want Power and sometimes we may simplify and only look at voltage. Figure the load Power, Voltage, and Current for 100V 10 ohm source with 100 ohm, 10 ohm, and 1 ohm loads. Which has the most Power, Voltage, Current?
> a preamp that lowered the impedance of the signal
NO. That may be a stage-useful understanding but is NOT what happens.
This preamp is an ACTIVE device. It has a "sense" input" which takes very little current (high impedance). It has a Power input, R.G.'s pile of AA cells with 10 ohms. It has a do-hicky which passes a variable amount of boring power from the 10 ohm power source according to what it feels on its "sense" input, and makes that available to its output. Assume the dohicky can show 1 Meg impedance at its sense input and pass the 10 ohm battery to an output. It may be unity voltage gain but the output current and power can be much higher than what it absorbs at it input.
> of high impedance being a very weak, quiet signal.... and low impedance being a very strong, loud signal
Ah, I had not formulated the reason this APPEARS true in common experience.
For reasons I won't type tonight, our everyday electric toys have a W_I_D_E range of Power, a relatively narrow range of voltage, but a huge range of current.
Power levels range from 0.000,000,02 Watts to over 100 Watts, a 1:20,000,000,000 range.
We like to keep voltage-things above 0.1V but under 100V, a 1:1,000 range.
If everything would be the same resistance/impedance, then 1:1,000 of voltage only allows 1:1,000,000 range of power.
Rather than let voltage get out of hand, we lever current over a 0.000,000,1 to 10A range. So when faced with smaller power at smaller voltages, we favor much smaller currents; when working with power which suggests high voltage we instead look at higher currents.
Which is not the whole story. The lowest-voltage common audio source today is the dynamic mike, typically 150 ohm impedance, which is the same as a 100 Watt 120V lamp. And in fact for the same reason: the economics of long copper wire favor impedances around a hundred.
Thanks! This all makes a bit more sense now! Everything I read doesn't really mention current and I guess I just didn't think of it even though I know it's there! I'm pretty good when it comes to calculating voltage/current/resistance in dc circuits because we covered them a decent amount in high school physics. When it comes to ac circuits, it's a whole different ball game.
But it makes sense when you put impedance/current as the focus of understanding impedance vs impedance/voltage.
One more question. Why is it that some circuits you can set the input impedance by putting a resistor across the signal input and ground? How does that work?
Water pump with hose on it analogy, what the hose dumps into or if it's almost sealed at the far end:
Big pump and a firm hose with no leaks, at the far end it's nearly sealed off, only a tiny bit of current is let through = the pressure rising and falling a clear waveshape of +/- pressures '~' identical to the pump end of the hose.
Teeny pump, a floppy hose with leaks which drain off to ground, by the time the wave-pulse gets to the far end of the hose [after all the leaks], there's not much wave left, and the finer bits of shape [high frequencies] are really misshapen when compared to the pump end of the hose.
Quote from: bassmannate on November 23, 2010, 08:16:26 AM
One more question. Why is it that some circuits you can set the input impedance by putting a resistor across the signal input and ground? How does that work?
The "sense input" on something like an opamp has an extremely high impedance - maybe hundreds of millions of ohms - which you can think of like a very large resistor to ground. When you put some smaller resistor to ground before this, it's in parallel with the opamp input. The math works out to where the opamp input, being so much larger than the resistor you've added, has very little impact on the total resistance, and can generally be ignored. The real impedance will be some tiny bit less than the value of your resistor - close enough for government work.
It occurs to me that in order to understand what I said above, it might help to realize that this is a circuit, like a circle or loop. It's not just from the pickup to the input of the device, but from the pickup through the input of the device and back to the pickup. To that end, it might help if we don't call it "ground" but rather "signal return". As far as the pickup is concerned, it's circuit is closed at the sense input of the first active component, whether that's a transistor or opamp or whatever.
Quote from: ashcat_lt on November 23, 2010, 12:58:08 PM
It occurs to me that in order to understand what I said above, it might help to realize that this is a circuit, like a circle or loop. It's not just from the pickup to the input of the device, but from the pickup through the input of the device and back to the pickup. To that end, it might help if we don't call it "ground" but rather "signal return". As far as the pickup is concerned, it's circuit is closed at the sense input of the first active component, whether that's a transistor or opamp or whatever.
But instead of putting a resistor to ground (back to the pickup) wouldn't we want all of the signal to go to the input of the opamp or do we place the resistor to keep too much signal from getting to the input? This is the part I have a hard time with is why do we let some signal (although, very small if it's something like 1meg-ohm) bleed to ground instead of putting it through the active device?
Quote from: bassmannate on November 23, 2010, 01:17:10 PM
But instead of putting a resistor to ground (back to the pickup) wouldn't we want all of the signal to go to the input of the opamp or do we place the resistor to keep too much signal from getting to the input? This is the part I have a hard time with is why do we let some signal (although, very small if it's something like 1meg-ohm) bleed to ground instead of putting it through the active device?
This is where we get into the difference between resistance and impedance.
First, ashcat is correct, the resistor can only REDUCE the impedance lower than the device that is being driven. You can always reduce the overall impedance which a driver/source "sees" as a load by adding more loading - more resistors in parallel with the thing being driven.
And that does lead directly to your question: if we want all the signal to go to the input, why load it down more?
It's because we want to force the DC level on the signal wire coming from the guitar to be at 0.000000000V, and we want to do that even in the face of disconnecting and reconnecting the effect by bypass switching. There is nearly always a capacitor at the input of an effect. This serves to block the DC from inside the effect from leaking out and causing pops and clicks when we switch the effect. (as an aside, capacitors have an infinitely high DC resistance, but their IMPEDANCE drops with frequency; difference between DC and AC impedances again)
But capacitors are not perfect, and they leak a little DC through them. Not much, but enough to cause clicks. We can drain this leakage to ground with a resistor, and that's what that 1M (typically) resistor is for - to drain the DC and keep clicks and pops from happening when we switch the signal into/out of the effect.
And your intuition is right - it is a step in the wrong direction, loading down the guitar. We have to make a compromise here, though, because we really don't want clicks. So we pick a resistor that's as big a value as possible to not load the guitar too much, but still drain away the capacitor leakage. This is, like so many things, a compromise.
Now it's all coming together! Thank you so much, everyone, for your patience!
I certainly don't understand this stuff as well as some of the other folks around here, but I do what I can, and at least know that "it's out there". I know the limits of my knowledge and try not to overstep them.
One thing I think is worth mentioning now that the basic questions have been answered is that the Thevenin model mentioned above is actually not really a very good model of a passive guitar pickup. To properly model a pickup you need the voltage source in series with a fairly large resistance and a hefty inductance, both of which are parallel to a relatively small capacitance. Moving outward from there you generally have a couple of parallel resistances in the form of the V and T controls. Then you've got to plug it into something, which introduces the generally negligible series resistance of the cable conductors and the considerable amount of parallel capacitance between the two. Then you get to whatever components are at the front end of the first active device.
All of this adds up to a system which impedes the higher frequencies quite a lot more than it does the lower. Specifically, it creates low-pass filter which generally cuts off somewhere between 4K-8KHz. (Personally, I don't call that treble, but rather high-mids) There is generally a resonant peak right before the cutoff frequency. The height of this peak is mostly determined by the overall resistance between hot and ground of the circuit to which the pickup is connected. That is, the total resistance derived from the V and T in parallel with the input/pulldown resistor and whatever else.
Many folks don't like the way it sounds when the amplitude of that peak gets too high. I'd suggest that this might be at least part of the reason why we try to set that "input impedance" around 500K-1M. It's all well and good to think that in theory we'd like to get everything we can out of a given circuit, but to a very large extent the things we tend to like about some of the more popular designs ("vintage" tube amps for example) are the ways in which they fail to perform their specific task in a theoretical ideal way.
> Everything I read doesn't really mention current
It is awkward to measure current, especially at audio, and audio techs are lazy.
> and I guess I just didn't think of it
Then you have not fully ABSORBED resistances. R=V/I. V=I*R. I=V/R. All three are always true and often useful. Get in the habit of computing all three aspects.
Perfect resistance is the same at any frequency or voltage.
Impedance is resistance that isn't the same all the time.
Constantly changing impedance is a royal pain in the head.
IN AUDIO, we hope to either have "mild" impedance changes, OR/AND a situation where wild impedance swing does NOT matter.
We almost NEVER "match" audio impedances.
The convention is: "Low Z output, High Z input".
There WAS a convention (and a reason-why) pro-gear was 600 Ohms in and out. However as soon as tubes got cheap output impedances dropped toward 100 ohms and inputs became 2K to 20K. Today any "600" port is >100 out into >10K in.
Given 1 Volt from 100 ohms, what is the received voltage for 10K load? 0.990V. What if it were 100K? 0.9990V. How about five 10K loads? 0.952V. Note that the maximum variation from 100K to 2K is less than 5%, less than 0.5dB, and thus inaudible.
We do the SAME thing in house wiring. Small load, large load, lots of loads, same (near-enuff) wall voltage. The guide is 2% sag per section, 5% sag for the whole house. You have 120.0V with everything off, you should have 114V with EVERYthing on.
Same in household water supply. You have 50psi every tap closed, maybe 40psi with a few major taps open. Pipes cost more than wires and (unless you are taking a shower) pressure is not THAT critical, we accept some loss.
> wouldn't we want all of the signal to go to the input
Signal Voltage? Current? Power?
And why do we care how much signal we get? Gain is CHEAP. As long as what we get/take is a FAITHFUL sample, we can easily bump it up (within some limits).
If we are designing an Input, we may have no control of the signal we get, except by general convention. The best way to deal with that situation is to "sip" the source, then gain-up as needed.
Vacuum tube grids, at audio, need nearly "no" current and nearly "no" power. We may approximate a donkey triode grid as >200Meg||30pFd. Modern opamps are in this general range. (BJTs have more design concerns.)
We must put two "signals" into it. The audio, of course. But also the DC bias.
We may think of this as a "loudspeaker crossover". For frequencies 20Hz and up, the circuit should be through the audio source. For frequencies 20Hz and down (to DC) the circuit should be through the bias source.
Loudspeakers require BIG current, costly Power, so we do not waste. We use a coil to the woofer and a cap to the tweeter.
But grids need "no" power. And it turns out that high-impedance audio coils are not real possible, or at least not cheap. However since we need "no" power, we may use a simple large resistor to complete the DC circuit, and a capacitor to overwhelm the DC bias for inputs 20Hz and up.
There is a small leakage. Exact computation is dubious and leakage varies a lot. The tube-book simplifies this by telling us "1 Meg max grid circuit resistance". So put 1 Meg grid to bias.
Sidebar: it also happens that tube grid DC bias is often signal and power common, "ground". You can just connect a guitar to a grid, no other parts. Until you unplug: then the open grid and leakage forces the infinite impedance to large unknown voltages. In preamps, that's usually safe, until you plug-in again and force grid back to ground instantly: POP. So even if you have a grounded source with <1Meg DC resistance, you throw a Meg on there anyway.
What is the new impedance at the grid? Ignoring the 30pFd for now, we have 200Megs || 1Meg. 0.995Meg and we will call this "1Meg" cuz most audio is 10% at best. Say our source has some DC on it, not what we want for bias. We block it with a cap. How big a cap do we need to overwhelm 1Meg at 20Hz? About 0.01uFd.
Back to that 30pFd. What is its impedance? 265Meg at 20Hz, ignore that. 2.6Meg at 2KHz, which slightly matters next to the 1Meg we have. 265K at 20KHz, which does impose additional loading on the source.
So the working input impedance of a R-C coupled tube amplifier is infinite at DC, approaches 1Meg for 20Hz to 5KHz (MOST of the audio band), sagging to 200K by 20KHz and 2K at 2MHz.
In an Audio application, we need constant response 20Hz-20KHz, so we must absorb 1Meg-200K variation without any big sag. For 10% or 1dB sag, a 20K source will do. For -3dB sag we can get away with a 80K source. If the source is an opamp, so what? But often the source is a bunch of resistors doing audio things, or maybe a tube with large output impedance.
This way we have "mild" impedance changes AND a situation where this impedance swing does NOT matter.
Most audio "impedances", we pretend they are "constant resistance, sorta", and jiggle things so we DON'T have to know the messy details.
This usually leads to lo-Z out, hi-Z in.
Gitar amp inputs are significantly higher than pickup impedance. Speaker amp outputs are lower than speaker impedances. There are important exceptions: some fuzz have 68K input, tube guitar amps show 2X to 10X impedance to the speaker. The reason this works are beyond the reach of my typing fingers tonight; just don't be shocked by deliberate exceptions.
If we had a classic lossless Passive L-C filter which needed EXactly 75 ohm termination, we could use a 75 ohm resistor instead of the 1Meg resistor. But such situations are rare. There is usually another way to do it.
LONG signal lines echo like a sewer pipe. Short lines, the echo dies out too quick to matter. If you don't like echo in your long pipe, you stuff wool batting or foam in the ends, same as you do in your concrete cellar so the drums don't ring-on the next day. In wires we put a resistor on the end, and there is a "matching" value. The definition of "long" depends on frequency. For TV signals it may be a few feet. Audio signals hardly show trouble at thousands of feet. We never "match" cables 1 foot to 300 feet, and few of us work further.
Agreed I understand it is there from studying speaker crossover networks, but I don't understand most of it. I work at it in reverse sometimes. For example the input impedance of a distortion circuit. Messing with the CMOS stuff confused me a lot. But to estimate the input impedance, I tried different value input caps until I got in the range that started rolling off the bass. Then, focused on the smallest input cap that seemed to stop rolling off the bass, figuring that the frequency was around 160hz. Then I calculated backwards to find the input impedance. Not very accurate, probably making some of you cringe, but gets in the ball park. You can do the same with treble roll off. I guess it is the quasi-SWAG method. Scientific Wild A** Guess. ;)
Quote from: G. Hoffman on November 22, 2010, 11:52:14 PM
By the way, R.G., that was a better explanation of that than I've seen before. You should write a basic electronics theory book. I'd buy it.
Gabriel
I second that. Starting to study Engineering, haven't taken an EE course yet. A book by R.G. would help make the connection between distant theory and something real that I understand qualitatively.
for Paul R.: fantastic, thank you!
Yearghggggghghghghgh! Blast you, 30 minute time out...! :icon_evil:
Okay big long explanation gone and not typing it out again. Google "reactance" and see if that helps. That's the word that helped me understand impedance 12 years ago. ashcat_lt was eluding to it above. Re-read his post.
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.
Quote from: bassmannate on November 24, 2010, 03:55:28 PM
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.
> Okay big long explanation gone and not typing it out again.
NotePad(*). Type there, THEN click the forum Reply button, paste, post.
That avoids the forum's short attention span problem.
It lets you type in a full window, not a little box.
You can jump between editor and browser and pick-up fragments of forum posts easier (IMHO) than in the forum's Reply view.
You can SAVE your words to your own machine for safe-keeping, handy if the forum or the whole internet goes down. Or if you are drafting a book.
(http://i55.tinypic.com/2pt4c35.jpg)
(*) NotePad is just the Windows freebie. Mac has TextEdit (http://support.apple.com/kb/TA20406?viewlocale=en_US), similar but better, unix usually has a proper plain-text editor, you can get excellent plain-text editors (I like UltraEdit).
Be SURE your editor will copy/paste PLAIN ASCII. Pasting from MS Word usually leaves odd codes which mess with the forum's tiny brain, and using a PLAIN text tool is easier than arguing with Word.
Quote from: earthtonesaudio on November 24, 2010, 04:11:08 PM
Quote from: bassmannate on November 24, 2010, 03:55:28 PM
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?
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.
> 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.
Everything above was an interesting read.
Quotemaximum 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
Quote> 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 ;D
mac
And straight into the bookmarks we go.
Is anyone else thinking this thread should be made sticky or linked in the wiki or something? :icon_redface: + :icon_evil: + ??? = 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.
QuoteIs anyone else thinking this thread should be made sticky or linked in the wiki or something?
Not a bad idea!
QuoteI'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.
QuoteR.G. explained that early on in his "12 Volt Battery(ies)" example. That's DC only, which is easier to comprehend in this case.
QuoteIf 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?
>> 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.)
Definite sticky/wiki content.
What a great thread. Have you guys considered collaborating to make a book? :icon_biggrin:
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?
Quote from: phector2004 on November 28, 2010, 12:24:24 PM
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.
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.
> 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.
This is maybe only tangentially related to the point of this thread, but some might find it interesting:
GuitarFreq2.2 (http://people.smartchat.net.au/~l_jhewitt/circuits/GuitarFreak_2_2.xls)
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.
Anybody seen this? It might be of help to some :icon_biggrin:
http://online.physics.uiuc.edu/courses/phys498pom/Lab_Handouts/Electric_Guitar_Pickup_Measurements.pdf (http://online.physics.uiuc.edu/courses/phys498pom/Lab_Handouts/Electric_Guitar_Pickup_Measurements.pdf)