Impedance Matching

[swinginguitar]

So I'm pondering something - in stompbox design, we always shoot for low output impedance/hi input impedance (for obvious reasons)

...but you often hear the term "impedance matching".

in context, does that mean a literal match of impedance, or the aforementioned ideal lo out Z into hi in Z?

what about output transformers in tube amps? is it a literal matching of the tube output to the speaker Z? or other transormer coupled circuits?

then there is te case of a Hi Z instrument pickup, for example, feeding a PA system low Z input. In this case, apparently it's ideal to match the low z?

[teemuk]

...but you often hear the term "impedance matching"

...used incorrectly. 



http://en.wikipedia.org/wiki/Impedance_bridging

http://en.wikipedia.org/wiki/Impedance_matching

[wavley]

...but you often hear the term "impedance matching"

...used incorrectly. 



http://en.wikipedia.org/wiki/Impedance_bridging

http://en.wikipedia.org/wiki/Impedance_matching

Beat me to it.

[gcme93]

The basic contrast is to do with power.

To get the maximum power out of an attached load, you must have the output impedance matching the load impedance:


In this case Zs = ZL

This is because Power is dictated by P=IR and so this layout maximises the power from the load: good if you want to have a speaker playing as loud as possible, or you are trying to get as much power out of a low impedance pickup as possible.

However, this works out as pretty inefficient as you are dissipating the same amount of heat in your amp as in your speaker (or wherever else it is applied).

Stompbox designs deal more with the physical voltage of the signal, and therefore if you imagine the output impedance and input impedance as a voltage divider, we want the most signal voltage possible for the next stage. Therefore we make the input impedance as high as possible, and the output impedance as low as possible

So for the voltage divider above, we are looking to have ZL >> Zs, also drawing much less current along the stretch of cable (less losses from the cable resistance)

Does this clarify it at all?

[swinginguitar]

Stompbox designs deal more with the physical voltage of the signal, and therefore if you imagine the output impedance and input impedance as a voltage divider, we want the most signal voltage possible for the next stage. Therefore we make the input impedance as high as possible, and the output impedance as low as possible

yeh this concept came pretty intuitively for me

However, this works out as pretty inefficient as you are dissipating the same amount of heat in your amp as in your speaker (or wherever else it is applied).

so in the case of a tube amp, you want the speaker dissipating the largest share of power, correct? or do you factor in the xfo to the load?

the OT "shows" the output tubes a <high> load impedance compared to the speaker itself?

So for the voltage divider above, we are looking to have ZL >> Zs, also drawing much less current along the stretch of cable (less losses from the cable resistance)

Does this clarify it at all?

let's take the case of a high impedance pickup being fed to a low z input on a mixer (possibly on a long line). what are the advantages brought by bumping the Z down via a DI box or passive xfo?

[swinginguitar]

...but you often hear the term "impedance matching"

...used incorrectly. 



http://en.wikipedia.org/wiki/Impedance_bridging

http://en.wikipedia.org/wiki/Impedance_matching

ahh..so matching is just that, but in our field bridging refers to setting the appriate in/out Z...

[PRR]

> To get the maximum power out of an attached load

Better said as "To get max power out of an available source, you "match".

Foo on pedals. We can't see any "power" there.

My house is powered by a 240 volt line, which is very long, has 0.4 ohms resistance.

I want to boil a bear, melt steel, something that needs ALL the power I can get.

What is the MAX power I can get? How do I get it?

First, let's use easy numbers. Say I had 100V and 1 ohm of wire resistance.

For several likely load resistances, calculate the load volts, amps, and Power.

Source: 100V 1r

Load:
100 ohms --- 99.0V -- 0.99A === Load: === 98 Watts  -  Line: 1V,0.99A === 0.99 Watts
10 ohms ---- 90.9V -- 9.09A ==  Load: == 826 Watts  -  Line: 9.1V, 9.09A == 83 Watts
1 ohms ----- 50V ---- 50A ====   Load: = 2500 Watts  -  Line: 50V,50A ==== 2500 Watts
0.1 ohms --- 9.09V -- 90.9A ==  Load: == 826 Watts  -  Line: 91V,90.9A == 8265 Watts
0.01 ohms -- 0.99V -- 99A ===== Load: === 98 Watts  -  Line: 99.99V,99A = 9899 Watts

The matched load gets the most Power.

But look at the Line power. When matched, the wires are *wasting* as much power as the load is getting! If over-matched (0.1r load on 1r line) the *waste* power is TEN TIMES the Power I get in my load! OTOH, under-matching (10r load on 1r line) gets me the same Power in my load, but *100 times less* waste in the line, and line-waste is 1/10th of power put usefully into load.

So "matching" is extremely hard on the source.

Another thing. If I assume 100V line then I probably have a 100V boiling-pot or steel-furnace. But the "matching" case delivers only 50V. I will have to get some custom loads.

Another thing. Say I end up with *two* loads, perhaps 2 ohms each, so that both together make the 1 ohm "match". One to boil the bear and one to boil the turnips. Both designed for 50V 1250W. The turnips get done first, I unplug that pot. The other 2r pot now gets *66.6V*. It was designed for 50V, so now it is making 1.78 times as much power as it was designed for. It may burn-up. Certainly it will boil much more vigorously than I expect, and ruin the bear.

Similar things happen in audio. A matched load is typically "too heavy". In some cases the source runs hot. The line voltage sags. If we have different numbers of loads, we get a different line voltage for every combination.

"Bridging" actually means we over-design the source to be plenty-strong, and the load to be extra-easy. Now the line voltage is consistent for any reasonable number of loads.

Matching was done, sorta, back in the EARLY days of telephone, when there were no amplifiers except the carbon microphone. Voice power was *precious*. Loads were few, usually just one. Matching gets the maximum Power to the listener. However as sets improved the telephone system moved toward 200 ohm source and 900 ohm load. Not great Bridging, but less line loss, and you could tap a second listener on the line without a huge drop of volume. Broadcast interfaces follow a similar trend: matching when tubes were VERY expensive, moving toward mis-match for more consistency, and finally to 50 ohm source 1Meg loads.

[PRR]

> the OT "shows" the output tubes a <high> load impedance compared to the speaker itself?

Electrons do NOT like to travel through empty space.

We may approximate this as a resistance.

Resistance of a small tube is 10,000 ohms. A large (more expensive) tube may be 1,000 ohms.

For 99% power transfer, we could use a 1K tube and a 100K load. However at any reasonable Voltage, a 100K load takes very little power. (For just 1 Watt, we need 316Vrms 447V peak; 40 Watts needs *2000Vrms*.) We could use more/bigger tubes for 100r source and 10K load, but tubes are expensive. We usually work tubes into 2X to 5X their internal resistance, 4K to 8K loading.

We can wind the secondary to any reasonable impedance which suits our load.

We also mis-match the load to the speaker wire. While a combo-amp may have just a foot of wire, speakers are also used 10 feet or 100 feet from their amps.

The simplest voice-coil might be one turn. But that would be under 1 ohm. If the line is more than a few dozen feet long, it is also near an ohm, so we lose about as much *precious* amplifier Watts in the line as are delivered to the speaker. 5 or 10 ohms is more suitable for getting power to the speaker instead of in line-loss. 15 ohms for theaters (long line and large budget), 3 ohms for car radios (short line and cheaper winding), which somehow morphed to 4/8/16 nominal ratings.

> a high impedance pickup being fed to a low z input

Easy experiment. Solder 5K across your guitar cord. The bass will come down to half, which may be acceptable in this day of cheap gain. However a pickup coil IS a "coil", has inductance, which is a rising impedance. In treble the pickup impedance may be more like 50K, output more like 1/10th. This change, from 1/2 in bass to 1/10th in treble, really fouls-up the tonal balance. A balance which was arrived at over years of experimentation (and assumes tube-input and very-high Z loading).

> bumping the Z down via a DI box

That can work, obviously. Passive transformers are also coils, with inductance. Pickup impedances evolved for higher output, which is higher inductance. And higher than many transformers. Putting two coils together does mess with the pickup designer's assumption. For various reasons, it isn't always bad, but you usually want a rather special transformer.

Why do we run long lines at around 100 ohms? That's the natural impedance of the universe, or at least of long skinny cheap conductors. There is capacitance everywhere, and it sucks worse on high impedance. Every conductor has some inductance, and every affordable conductor has all the resistance we can get away with, so low-low impedance is bad. It works out that over a wide range of sizes, coax is 50-100 ohms, pair-cable is 100-200 ohms (300-900 ohms for wide-space pair like open telco wire on glass pins). That's why low-Z mikes are run at 150 ohms, Ethernet at 85-110 ohms, Cable-TV at 75 ohm, and we do not run long wires at either 4 ohms or 40,000 ohms.

[gcme93]

We'll I've definitely learnt a lot there! I guess I've been looking at too many equations in my course and not applying them as fully as they need

Thanks a lot PRR, very clear and very informative as always!

[Thecomedian]

So, is this impedance business why putting in a larger Ohm volume control "saves" higher frequencies?
Would putting in too high of a volume control Ohm resist (1meg+) create an impedance value in the guitar that weakens the output signal of the stompbox?

Output is what is coming into the stompbox, while input is what's going out of the stompbox? or
Output goes out of stomp, input is what comes into stomp?

[gcme93]

So, is this impedance business why putting in a larger Ohm volume control "saves" higher frequencies?
Would putting in too high of a volume control Ohm resist (1meg+) create an impedance value in the guitar that weakens the output signal of the stompbox?

Output is what is coming into the stompbox, while input is what's going out of the stompbox? or
Output goes out of stomp, input is what comes into stomp?

I'm loving paint at the moment...

[Thecomedian]

saved. thank god.

[bluebunny]

Nice picture, George.  Simple and memorable.  Thanks.