mshilarious
Banned
Warning: this post contains math!!!
From another thread, if you are interested, here are the basics of Ohm's Law, which is the governing principle underlying all electronics. It will unlock the mysteries of audio signal interconnections and "impedance matching".
Ohm's Law:
V = I * R
which can be restated any way you like:
I = V / R
etc.
where V is voltage, measured in Volts (V), I is current (no one knows why), measured in Amps (A), and R is resistance, measured in Ohms (Ω).
Another term we will kick around is impedance, or Z, also measured in Ohms. Some people like to think of impedance as frequency-dependent resistance, or AC resistance. So when we are talking DC (direct current), we are concerned with resistance, and AC (alternating current), we are concerned with impedance.
Let's take a simple circuit of a DC supply (say a 9V battery) and a resistor. What is a resistor? It's a clever little device that ideally has no "reactance". What is reactance? Let's just call it the property of certain devices (such as capacitors and inductors, but don't worry about them for now) that cause them to have different impedance at different frequencies.
OK, so our resistor doesn't have any of that funny reactance stuff, so no matter what frequency we are concerned with, its impedance is always the same as its resistance.
So, we have a resistor, let's say it is 1KΩ. Ohm's Law:
9V / 1000Ω = 0.009A. Let's call that 9mA for convenience.
Alright that is just batteries and bulbs stuff (bulbs are a bit complicated though, but we'll ignore that). What about audio signals, that's what we care about, right?
An audio signal is AC, so we are concerned with impedance. But wait, one more thing, I am a filthy liar about our 9V battery example. Why? Because every source has a source impedance/resistance, which is like a limit on how much current it can supply (talking 9V batteries again, they have internal resistance of about 3Ω, so I should have divided 9V by 1003Ω. Meh, sue me).
Back to audio, that is a critically important concept, because every audio source has a source (or output) impedance. And every audio load--that's where we are sending the signal--has a load (or input) impedance. The load is going to place a demand on the source for some power. How much? Ohm's Law will tell us.
The next question is what are we trying to accomplish? There are two usual goals in audio signal transmission: either a transfer of *signal* (which is voltage), or a transfer of *power*.
What is power? This:
P = V * I
P is power, measured in Watts. We can also get clever and solve for this:
P = V ^ 2 / R
Generally, until we try to drive a speaker (or a headphone, or maybe something like a spring reverb tank), we want to transfer signal and not power.
That brings us to "impedance matching". What it is? Strictly, impedance matching means that source and load impedance are the same. That will maximize transfer of power. Let's investigate that using Ohm's Law. Let's keep our 9V battery with its 3Ω internal resistance. Let's put it in a short circuit:
9V / 3Ω = 3A;
9V * 3A = 27W.
So if you short a 9V battery, it will attempt to dissipate 27W, which is actually a whole lotta power and I really do NOT recommend you try that (you won't actually get that much because the internal resistance increases as the load gets heavier and the battery dies *rapidly*, but never mind that, just don't do it).
27W, great, but that does not do any work at all outside of the battery. That is unhelpful, presumably we bought our 9V battery for a purpose other than starting a fire. So let's load it with a 3Ω resistor (don't try that either, please):
9V / (3Ω + 3Ω) = 1.5A;
9V * 1.5A = 13.5W.
OK, great, only 13.5W, smaller fire now. But wait, how much work is done where? Again, Ohm's Law. Current must be the same everywhere in our simple circuit, so let's solve for voltage in the external resistor:
1.5A * 3Ω = 4.5V;
so 4.5V drops inside the battery, and 4.5V drops in the external resistor, so power:
4.5V * 1.5A = 6.75W
That is the maximum possible power delivered to a load from our imaginary 9V battery (realistically, 2W is about the max for a real-life 9V that you don't want to die in 15 minutes). Don't believe me? Solve for 2Ω and 4Ω for yourself.
When it is important to transfer maximum possible power? When power is a very limited commodity. Tube amps are the #1 application. They are expensive and highly inefficient, so we need to scrape every last possible watt out of them.
That's impedance matching for you. So what do we really want to do in most audio circuits, and why? We want to *bridge* impedance. This means that the source impedance should be much lower, ideally 10x lower or more, than the load attached to it. What does that do? Well, let's go back to our 9V, but let's say it's an audio signal, 9VAC (volts alternating current, an "average" or RMS, root of the mean squares, measurement). And let's say our source impedance is a more typical 100Ω. And let's stick with purely resistive sources and loads.
Into a short circuit:
9VAC / 100Ω = 0.09A, or 90mA
OK, we can get 90mA, which would give us 0.81W, again all dissipated in the source. Not too helpful. Small-signal audio devices, such as microphones, mic amplifiers, line-level devices, etc. can be thought of as voltage input. So if we short the output, we get no signal to the next device. Not very helpful. If we *match* impedance, we get half of the signal into the next device. Not horrible, but suboptimal.
But if we *bridge* impedance with a load of 10x the source:
9VAC / (100Ω + 1000Ω) = ~8.2mA.
OK, but wait, I said we didn't really care about current, but, continuing through with Mr. Ohm, we learn that the voltage dropped across the load is:
8.2mA * 1000Ω = 8.2VAC.
Hey, not bad, we only lost 0.8V of signal. For the decibel-minded (decibels are a handy way to express ratios using a logarithmic scale), you can convert:
dB = 20* log (8.2 / 9) = -0.8dB.
We can live with that and be happy.
When is impedance "mismatched"? When the load impedance is less than the source. Then we will lose most of our signal. Here's 10x the wrong way:
9VAC / (100Ω + 10Ω) = ~82mA
82mA * 10Ω = 0.8VAC
dB = 20 * log (0.8 / 9) = -21dB
That kinda sucks. But unfortunately, impedance mismatches get even worse than a simple pad. This is because of that guy in the corner we've ignored, reactance. Let's say we have some, so source impedance varies by frequency. Let's assume a variance of 2 over our range of interest (it is often much worse). Let's analyze with bridged impedance at nominal 10x, matched impedance, and mismatched impedance at 10x:
Bridged:
9VAC / (100Ω + 1000Ω) = ~8.2mA
8.2mA * 1000Ω = 8.2VAC
dB = 20 * log (8.2 / 9) = -0.8dB
9VAC / (200Ω + 1000Ω) = 7.5mA
7.5mA * 1000Ω = 7.5VAC
dB = 20 * log (7.5 / 9) = -1.5dB
Matched:
9VAC / (100Ω + 100Ω) = 45mA
45mA * 100Ω = 4.5VAC
dB = 20 * log (4.5 / 9) = -6dB
9VAC / (200Ω + 100Ω) = 30mA
30mA * 100Ω = 3VAC
dB = 20 * log (3 / 9) = -9.5dB
Mismatched:
9VAC / (100Ω + 10Ω) = ~82mA
82mA * 10Ω = 0.8VAC
dB = 20 * log (0.8 / 9) = -21dB
9VAC / (200Ω + 10Ω) = ~43mA
43mA * 10Ω = 0.4VAC
dB = 20 * log (0.4 / 9) = -27dB
So the frequency response varies as follows:
Bridged: 0.7dB
Matched: 3.5dB
Mismatched: 6dB
Since real-world impedance can vary a lot more than 2x, this can really screw with our happiness. So where complex source or load impedances exist, we need to bridge impedance *if* fidelity to source signal is our primary goal. It isn't always, but that is a post for another day.
From another thread, if you are interested, here are the basics of Ohm's Law, which is the governing principle underlying all electronics. It will unlock the mysteries of audio signal interconnections and "impedance matching".
Ohm's Law:
V = I * R
which can be restated any way you like:
I = V / R
etc.
where V is voltage, measured in Volts (V), I is current (no one knows why), measured in Amps (A), and R is resistance, measured in Ohms (Ω).
Another term we will kick around is impedance, or Z, also measured in Ohms. Some people like to think of impedance as frequency-dependent resistance, or AC resistance. So when we are talking DC (direct current), we are concerned with resistance, and AC (alternating current), we are concerned with impedance.
Let's take a simple circuit of a DC supply (say a 9V battery) and a resistor. What is a resistor? It's a clever little device that ideally has no "reactance". What is reactance? Let's just call it the property of certain devices (such as capacitors and inductors, but don't worry about them for now) that cause them to have different impedance at different frequencies.
OK, so our resistor doesn't have any of that funny reactance stuff, so no matter what frequency we are concerned with, its impedance is always the same as its resistance.
So, we have a resistor, let's say it is 1KΩ. Ohm's Law:
9V / 1000Ω = 0.009A. Let's call that 9mA for convenience.
Alright that is just batteries and bulbs stuff (bulbs are a bit complicated though, but we'll ignore that). What about audio signals, that's what we care about, right?
An audio signal is AC, so we are concerned with impedance. But wait, one more thing, I am a filthy liar about our 9V battery example. Why? Because every source has a source impedance/resistance, which is like a limit on how much current it can supply (talking 9V batteries again, they have internal resistance of about 3Ω, so I should have divided 9V by 1003Ω. Meh, sue me).
Back to audio, that is a critically important concept, because every audio source has a source (or output) impedance. And every audio load--that's where we are sending the signal--has a load (or input) impedance. The load is going to place a demand on the source for some power. How much? Ohm's Law will tell us.
The next question is what are we trying to accomplish? There are two usual goals in audio signal transmission: either a transfer of *signal* (which is voltage), or a transfer of *power*.
What is power? This:
P = V * I
P is power, measured in Watts. We can also get clever and solve for this:
P = V ^ 2 / R
Generally, until we try to drive a speaker (or a headphone, or maybe something like a spring reverb tank), we want to transfer signal and not power.
That brings us to "impedance matching". What it is? Strictly, impedance matching means that source and load impedance are the same. That will maximize transfer of power. Let's investigate that using Ohm's Law. Let's keep our 9V battery with its 3Ω internal resistance. Let's put it in a short circuit:
9V / 3Ω = 3A;
9V * 3A = 27W.
So if you short a 9V battery, it will attempt to dissipate 27W, which is actually a whole lotta power and I really do NOT recommend you try that (you won't actually get that much because the internal resistance increases as the load gets heavier and the battery dies *rapidly*, but never mind that, just don't do it).
27W, great, but that does not do any work at all outside of the battery. That is unhelpful, presumably we bought our 9V battery for a purpose other than starting a fire. So let's load it with a 3Ω resistor (don't try that either, please):
9V / (3Ω + 3Ω) = 1.5A;
9V * 1.5A = 13.5W.
OK, great, only 13.5W, smaller fire now. But wait, how much work is done where? Again, Ohm's Law. Current must be the same everywhere in our simple circuit, so let's solve for voltage in the external resistor:
1.5A * 3Ω = 4.5V;
so 4.5V drops inside the battery, and 4.5V drops in the external resistor, so power:
4.5V * 1.5A = 6.75W
That is the maximum possible power delivered to a load from our imaginary 9V battery (realistically, 2W is about the max for a real-life 9V that you don't want to die in 15 minutes). Don't believe me? Solve for 2Ω and 4Ω for yourself.
When it is important to transfer maximum possible power? When power is a very limited commodity. Tube amps are the #1 application. They are expensive and highly inefficient, so we need to scrape every last possible watt out of them.
That's impedance matching for you. So what do we really want to do in most audio circuits, and why? We want to *bridge* impedance. This means that the source impedance should be much lower, ideally 10x lower or more, than the load attached to it. What does that do? Well, let's go back to our 9V, but let's say it's an audio signal, 9VAC (volts alternating current, an "average" or RMS, root of the mean squares, measurement). And let's say our source impedance is a more typical 100Ω. And let's stick with purely resistive sources and loads.
Into a short circuit:
9VAC / 100Ω = 0.09A, or 90mA
OK, we can get 90mA, which would give us 0.81W, again all dissipated in the source. Not too helpful. Small-signal audio devices, such as microphones, mic amplifiers, line-level devices, etc. can be thought of as voltage input. So if we short the output, we get no signal to the next device. Not very helpful. If we *match* impedance, we get half of the signal into the next device. Not horrible, but suboptimal.
But if we *bridge* impedance with a load of 10x the source:
9VAC / (100Ω + 1000Ω) = ~8.2mA.
OK, but wait, I said we didn't really care about current, but, continuing through with Mr. Ohm, we learn that the voltage dropped across the load is:
8.2mA * 1000Ω = 8.2VAC.
Hey, not bad, we only lost 0.8V of signal. For the decibel-minded (decibels are a handy way to express ratios using a logarithmic scale), you can convert:
dB = 20* log (8.2 / 9) = -0.8dB.
We can live with that and be happy.
When is impedance "mismatched"? When the load impedance is less than the source. Then we will lose most of our signal. Here's 10x the wrong way:
9VAC / (100Ω + 10Ω) = ~82mA
82mA * 10Ω = 0.8VAC
dB = 20 * log (0.8 / 9) = -21dB
That kinda sucks. But unfortunately, impedance mismatches get even worse than a simple pad. This is because of that guy in the corner we've ignored, reactance. Let's say we have some, so source impedance varies by frequency. Let's assume a variance of 2 over our range of interest (it is often much worse). Let's analyze with bridged impedance at nominal 10x, matched impedance, and mismatched impedance at 10x:
Bridged:
9VAC / (100Ω + 1000Ω) = ~8.2mA
8.2mA * 1000Ω = 8.2VAC
dB = 20 * log (8.2 / 9) = -0.8dB
9VAC / (200Ω + 1000Ω) = 7.5mA
7.5mA * 1000Ω = 7.5VAC
dB = 20 * log (7.5 / 9) = -1.5dB
Matched:
9VAC / (100Ω + 100Ω) = 45mA
45mA * 100Ω = 4.5VAC
dB = 20 * log (4.5 / 9) = -6dB
9VAC / (200Ω + 100Ω) = 30mA
30mA * 100Ω = 3VAC
dB = 20 * log (3 / 9) = -9.5dB
Mismatched:
9VAC / (100Ω + 10Ω) = ~82mA
82mA * 10Ω = 0.8VAC
dB = 20 * log (0.8 / 9) = -21dB
9VAC / (200Ω + 10Ω) = ~43mA
43mA * 10Ω = 0.4VAC
dB = 20 * log (0.4 / 9) = -27dB
So the frequency response varies as follows:
Bridged: 0.7dB
Matched: 3.5dB
Mismatched: 6dB
Since real-world impedance can vary a lot more than 2x, this can really screw with our happiness. So where complex source or load impedances exist, we need to bridge impedance *if* fidelity to source signal is our primary goal. It isn't always, but that is a post for another day.
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