How does impedance matching work on both line and guitar/mic levels? And Headphones?

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Disease8

Disease8

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I am not exactly a newb but I am when it comes down to the technical side of it... Ohms have always confused me. I know it is a way of measuring resistance and impedance. I wonder why a DI box can accept both a line level and a high impedance guitar jack or mic? It is impedance matching so any ohms will go in it and come out the right ohms for a mic input on a mixer (which usually won't like having a line level in its XLR input)?
I know its a complex subject but makes me wonder when people always want their keyboards DI'd and I tell them they don't need to because its a line level and just goes straight into a line input on a mixer, it never caused any problems, I understand if the keys cause a ground loop because the stage power is a different breaker (think that's the term) to the mixer power then I would use a DI box.. Also headphone outputs they are not line ohms right? So why is a headphone output fine as a line source?

I find it hard to understand the descriptions on Wiki as its all using electronics terminology if someone could break it don to simple terms I would much appreciate it! (if that's possible lol)...
Thanks, Dom.
 
An Ohm is an Ohm.

Source/output impedance, roughly, is ability to drive a load.

Load/input impedance is the load the output needs to drive.

So long as the load (in ohms) is greater than the source by a significant multiple, you're OK. Very high multiples are usually OK, with exceptions that I won't go into right now.

For example, a typical headphone is maybe 24 ohms (or 16, or 32. whatever). Therefore, in order to drive that load easily, a headphone amp must be very very low source impedance, ideally well under 1 ohm. That same 1 ohm headphone amp will be perfectly happy driving a 10K ohm line input impedance. Lighter loads are OK (high impedance = light load).

However, a line output with a 100 ohm source impedance cannot successfully drive a 24 ohm pair of headphones.

The load always indicates the amount of current required for a given voltage. This is Ohm's Law: I (current)= V/R. But source and load impedance form a voltage divider, therefore if the source impedance is a significant fraction of the load, then power will be lost in the source, and voltage to the load will drop. That's generally bad.

Next is the concept of operating level. The reason why a headphone amp *might* not be a good line level source is because the voltage isn't very high. For example, 2V into 16 ohm is 0.125A (Ohm's Law again), which is 250mW (V*I). That power into a headphone could melt your face. But a line level input might be looking for a voltage of perhaps 15V--at 10K ohm, that's a mere 0.0015A. So the headphone amp can successfully drive the line input up to 2V, but maybe it doesn't have any more volts to give (because its power supply might be limited to that figure). Another headphone amp can do both high current and/or high voltage. That's a better design, but not all headphone amps are that flexible.
 
mshilarious said:
An Ohm is an Ohm.

Source/output impedance, roughly, is ability to drive a load.

Load/input impedance is the load the output needs to drive.

So long as the load (in ohms) is greater than the source by a significant multiple, you're OK. Very high multiples are usually OK, with exceptions that I won't go into right now.

For example, a typical headphone is maybe 24 ohms (or 16, or 32. whatever). Therefore, in order to drive that load easily, a headphone amp must be very very low source impedance, ideally well under 1 ohm. That same 1 ohm headphone amp will be perfectly happy driving a 10K ohm line input impedance. Lighter loads are OK (high impedance = light load).

However, a line output with a 100 ohm source impedance cannot successfully drive a 24 ohm pair of headphones.

The load always indicates the amount of current required for a given voltage. This is Ohm's Law: I (current)= V/R. But source and load impedance form a voltage divider, therefore if the source impedance is a significant fraction of the load, then power will be lost in the source, and voltage to the load will drop. That's generally bad.

Next is the concept of operating level. The reason why a headphone amp *might* not be a good line level source is because the voltage isn't very high. For example, 2V into 16 ohm is 0.125A (Ohm's Law again), which is 250mW (V*I). That power into a headphone could melt your face. But a line level input might be looking for a voltage of perhaps 15V--at 10K ohm, that's a mere 0.0015A. So the headphone amp can successfully drive the line input up to 2V, but maybe it doesn't have any more volts to give (because its power supply might be limited to that figure). Another headphone amp can do both high current and/or high voltage. That's a better design, but not all headphone amps are that flexible.
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Gosh mshilarious your ohming out all over the place today and this explanation helps me out also!!
 
mshilarious said:
...So long as the load (in ohms) is greater than the source by a significant multiple, you're OK. Very high multiples are usually OK, with exceptions that I won't go into right now...
Click to expand...

Just to put this into perspective here are a few examples.

Many low impedance mics = 50 - 200 ohms
Many mic preamp inputs = 500 - 2000 ohms. (should be around 10 times greater than the mic)

Line Level device = 1000 - 10,000 ohms
Line level inputs = 10,000 - 100,000 ohms (should be around 10 times greater than the source)

It's not a perfect world. Specs vary greatly from one manufacturer to the next. Loads are not always 10X the source, but anywhere in the ballpark should work nicely. This is why some devices offer input gain and/or output gain controls...to adjust for those occasional mismatches.
 
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RawDepth said:
Line Level device = 1000 - 10,000 ohms
Line level inputs = 10,000 - 100,000 ohms (should be around 10 times greater than the source)

It's not a perfect world. Specs vary greatly from one manufacturer to the next. Loads are not always 10X the source, but anywhere in the ballpark should work nicely. This is why some devices offer input gain and/or output gain controls...to adjust for those occasional mismatches.
Click to expand...

Those would be poor line level source impedances; 100-300 ohm is much more typical (or even lower). Again, it's not a problem to have a higher ratio than 10x.

The reasons why those devices would be poor are:

- You don't know what load to expect, or how many loads. 100K is an atypical load; 10K is much more common. So a 10K source impedance is likely 1:1, that's not good.

- The source has to drive not only the load, but the cable to get there. At 10K source impedance, the cable length you can drive before you get measurable audio band high frequency loss (let's say 100kHz bandwidth to avoid that fate, assume cable capacitance of 30pF/ft):

ft = 1 / (2 * pi * 0.00000000003F * 10000R * 100000Hz) = 5 ft!

At about 25 ft, we've lost 3dB at 20kHz.

Not very impressive . . . this is why guitar cables are trouble, the source impedance of passive guitar pickups is high.

So we need much lower source impedance than that; at 100 ohm, we get a maximum cable length of 500 ft. And in fact 100 ohm into 10K ohm is the typical line source/load, which works fine everywhere in the world every day.

Finally, from the cable calculation you can see that gain controls cannot adjust for complex impedances, load or source. You could add some simple to complicated EQ to compensate, but it's easier to avoid the problem in the first place.
 
mshilarious said:
An Ohm is an Ohm.

Source/output impedance, roughly, is ability to drive a load.

Load/input impedance is the load the output needs to drive.

So long as the load (in ohms) is greater than the source by a significant multiple, you're OK. Very high multiples are usually OK, with exceptions that I won't go into right now.

For example, a typical headphone is maybe 24 ohms (or 16, or 32. whatever). Therefore, in order to drive that load easily, a headphone amp must be very very low source impedance, ideally well under 1 ohm. That same 1 ohm headphone amp will be perfectly happy driving a 10K ohm line input impedance. Lighter loads are OK (high impedance = light load).

However, a line output with a 100 ohm source impedance cannot successfully drive a 24 ohm pair of headphones.

The load always indicates the amount of current required for a given voltage. This is Ohm's Law: I (current)= V/R. But source and load impedance form a voltage divider, therefore if the source impedance is a significant fraction of the load, then power will be lost in the source, and voltage to the load will drop. That's generally bad.

Next is the concept of operating level. The reason why a headphone amp *might* not be a good line level source is because the voltage isn't very high. For example, 2V into 16 ohm is 0.125A (Ohm's Law again), which is 250mW (V*I). That power into a headphone could melt your face. But a line level input might be looking for a voltage of perhaps 15V--at 10K ohm, that's a mere 0.0015A. So the headphone amp can successfully drive the line input up to 2V, but maybe it doesn't have any more volts to give (because its power supply might be limited to that figure). Another headphone amp can do both high current and/or high voltage. That's a better design, but not all headphone amps are that flexible.
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doesn't the input drive the load because that where the electricity is coming from? what is the load? the input impedance or the output impedance? or the actual ohms that end up going through the cable after input and output have worked themselves out or combined or whatever the technical term is?
 
Disease8 said:
doesn't the input drive the load because that where the electricity is coming from? what is the load? the input impedance or the output impedance? or the actual ohms that end up going through the cable after input and output have worked themselves out or combined or whatever the technical term is?
Click to expand...

Inputs don't drive a load; they are the load! You will hear people saying things like "This preamp doesn't have enough gain to drive this microphone." That is fundamentally incorrect, the microphone must drive the input impedance of the preamp. What they are trying to say is that the voltage output of the microphone is low, so a lot of voltage gain is required in the preamp. But the microphone with the weak voltage output could have a very low output impedance, such that there is little difference between its unloaded output (open circuit, not connected to any preamp) and its output into a 1K input impedance preamp.

The device that is sending signal has an output (or source) impedance. That's like a limit on how much current that device can supply.

The device that is receiving the signal has the input (or load) impedance. The load decides how much current it wants to draw based upon that impedance and the signal level (voltage) of the output. If the source cannot supply enough current due to its source impedance being too high, the voltage will drop excessively--this is a loss in signal level.

Ohms don't go anywhere; impedance is a physical property of the components in a device. Electrons travel from place to place (really just back and forth a lot); this is current. Voltage is the electrical pressure that determines how much current flows, given the resistance (or impedance) of the load.

Ohm's Law: current = voltage / resistance. It's the fundamental principle of electronics, meditate on it until all becomes clear . . . Ohm . . . . Ohm . . . . Ohm . . . .
 
Hi thank you man I am going to meditate

mshilarious said:
Inputs don't drive a load; they are the load! You will hear people saying things like "This preamp doesn't have enough gain to drive this microphone." That is fundamentally incorrect, the microphone must drive the input impedance of the preamp. What they are trying to say is that the voltage output of the microphone is low, so a lot of voltage gain is required in the preamp. But the microphone with the weak voltage output could have a very low output impedance, such that there is little difference between its unloaded output (open circuit, not connected to any preamp) and its output into a 1K input impedance preamp.

The device that is sending signal has an output (or source) impedance. That's like a limit on how much current that device can supply.

The device that is receiving the signal has the input (or load) impedance. The load decides how much current it wants to draw based upon that impedance and the signal level (voltage) of the output. If the source cannot supply enough current due to its source impedance being too high, the voltage will drop excessively--this is a loss in signal level.

Ohms don't go anywhere; impedance is a physical property of the components in a device. Electrons travel from place to place (really just back and forth a lot); this is current. Voltage is the electrical pressure that determines how much current flows, given the resistance (or impedance) of the load.

Ohm's Law: current = voltage / resistance. It's the fundamental principle of electronics, meditate on it until all becomes clear . . . Ohm . . . . Ohm . . . . Ohm . . . .
Click to expand...

Hey thanks bro I'm just going to have to let it sink in... Forgive me if I'm wrong but...
a guitar is extremely high ohms (very high resistance to electrical current current) it needs a DI box to get it to a level so it can drive the preamp?
 
How can a microphone have an electrical signal out of it when its not pugged into a pre? So a load is the signal a preamp is trying to give the source right? And the source drives the preamp with whatever current is left after the voltage has been divided by the resistance?
 
Disease8 said:
Hey thanks bro I'm just going to have to let it sink in... Forgive me if I'm wrong but...
a guitar is extremely high ohms (very high resistance to electrical current current) it needs a DI box to get it to a level so it can drive the preamp?
Click to expand...

Essentially, yes. But a guitar has a fairly healthy voltage output. Passive DIs can use that to their advantage--transformers can swap voltage for current, or vice versa, but total power cannot increase (it will decrease slightly due to losses in the transformer, which I will ignore in this post).

Remember power = voltage * current;

So a guitar might be say 100mV (millivolts, thousandth of a volt, so this is one tenth of a volt, not too shabby) output with an output impedance of 10K. Therefore, the most current it can supply is 100mV / 10K = 10uA (microamp), into a short circuit (which isn't too useful). A mic preamp has an input impedance of 1K. That won't work very well, we need to maintain that 1:10 ratio for source/load impedance if we don't want signal (voltage) to drop.

So a transformer might say, hey, let's drop that voltage to 10mV (turns ratio of 10:1--usually passive DIs will be 16:1 or so, but I'm keeping the math easy), and in exchange the output impedance after the transformer is only 100 ohm (law of transformers: voltage changes by the turns ratio; impedance changes by the square of the turns ratio. Turns ratio is the number of windings on the primary vs. the secondary coils). The transformer will also magically make the preamp appear as a 100K load to the guitar. I love transformers!

So power into the transformer is 100mV / 100K = 1uA (microamp, a millionth of an amp) --> 100mV * 1uA = 100nW (nanowatt, a billionth of a watt)

Power out is 10mV / 1K = 10uA --> 10mV * 10uA = 100nW

An active DI uses a transistor instead, which does a different magic trick: it can add power. To do that, of course, it needs a power source, like phantom power or a battery.

A transistor DI might decide it wants to have an input impedance of 1M. That's a very high impedance that will draw very little current from the guitar pickup, so that practically no signal is lost

100mV signal / 1M = 100nA current! That's low. Power = 100mV * 100nA = 10nW.

The maximum possible output depends on the type of transistor, power supply, etc. but for a typical phantom-powered DI you'd have an output impedance of worst case 300 ohm. So let's use 100 ohm instead, that is probably typical. Transistors can have voltage gain, current gain (which is lowered output impedance), or both, but most active DIs don't do any voltage gain since there is already a healthy voltage from the guitar.

OK, so the 100 ohm source impedance of the DI is very good into our 1K mic preamp, that's 1:10, power is:

100mV / 1K = 100uA --> 100mV * 100uA = 10uW (microwatts, millionth of a watt), or 1,000 times the input power.

I like transistors too :)
 
Disease8 said:
How can a microphone have an electrical signal out of it when its not pugged into a pre? So a load is the signal a preamp is trying to give the source right? And the source drives the preamp with whatever current is left after the voltage has been divided by the resistance?
Click to expand...

There's no current, because that requires a circuit. There is voltage, which is potential energy. You will see a measurement of a mic's sensitivity into an open circuit. How can we know what that is, when there is no electricity flowing? Strictly speaking, we can't, BUT, if we hook up a very very light load to the microphone (say greater than 1M), and we know the microphone has a low output impedance (less than 600 ohm, usually much less), then we know the voltage we measure across that 1M resistor is essentially the same as open-circuit because there is very little voltage drop.

You should think of the source and load impedances as a voltage divider. What's that? Simple, any two resistors will divide a voltage across them, like this:

Let's say we have a 11V signal with an output impedance of 100 and an input impedance of 1K. This forms a complete circuit with total impedance of 1100.

Current = voltage / resistance

11 / 1100 = 10mA. OK, current is 10mA. It is 10mA everywhere in the circuit; that is, through both resistors. All of the electrons that leave the starting point of the circuit must return to where they started (they have 11V of potential energy when they start, but 0V when they return--they are all used up :(), so current is always the same in any given circuit.


Flipping around Ohm's Law, what is the voltage drop across each resistor?

voltage = current * resistance

10mA * 1K = 10V
10mA * 100 = 1V

So we actually only have a 10V signal across our load--we lost a volt to the source impedance of the circuit.

So the "open circuit" voltage of our source is 11V, but into a 1K ohm load we only get a 10V signal.
 
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