Questions about impedance, volts, current, etc.

[ahrenba]

Hey Guys,

I am a total recording newb, but I have been trying to learn the ways of how recording equipment and signals work over the last few days. It has involved A LOT of research and many questions. Get ready for a beast of a post.

Ok. I know that Guitars output a high impedance signal (hi-z), while something like a mic outputs a low impedance signal (lo-z). Hi-z signals have a high voltage, but low current. Lo-z signals have a low voltage, but high current.

I am going to list what I know...let me know if I am right. Then, I have a couple questions.
--------------------------------------------
So, I understand that power equals voltage times current.

Same impedance
If you plug a device with, let's say, 600 ohm's into an input (load) of 600 ohm's, you will match the impedance. This allows for the most power to be passed on, however nothing is boosted.

Lower impedance into a higher impedance
If you plug a device with, let's say, 10k ohm's into an input (load) of 100k ohm's, the voltage will be boosted, but the current will decrease. This is called "bridging" because the load impedance is 10 times that of the output's.

Higher impedance into a lower impedance
Not sure on this: If you plug a device with 10k ohm's into an input (load) of 2k ohm's, the current will be boosted but the voltage will decrease. Could someone verify this?

Am I correct with my logic so far?
--------------------------------------------
Ok, now the questions:

Situation: You have a Hi-z input on a mixer (let's just say it's 100k ohm's). Then you have a guitar that outputs 10k ohm's. When you plug the guitar into the hi-z input, the notion is that the lower impedance of the guitar going into a higher impedance of the input will boost the voltage, but lower the current. This is where I am confused.
Question #1: If the voltage is increased, but the current is lowered, doesn't the overall power output remain the same (not boosted), since one variable is increased, and the other is decreased?

Situation: In an article that I read, they say this:
Nowadays, nearly all devices are connected bridging -- low-Z out to high-Z in -- because we want the most voltage transferred between components.
Question #2:With audio, why do we want the most voltage transferred, and not the most current? Why not both? Why is voltage more important than current? I am not really sure what the difference is in terms of signal.

Question #3: What exactly does a pre-amp boost, the voltage or the current? Aren't these two variables bound? By this, I mean how can you increase one without affecting the other?

Question #4: What exactly happens when you plug a higher Z item into a lower Z input? Was I right in my logic above?

Question #5: This information is from a Shure Mixer (inputs):
Line: Designed for use with - / < 10k ohms, 50k ohms internal, +36 dBV clipping level
Is the "50k ohms internal" the impedance of the load (input)? If so, this unit must not be bridging capable, correct? The reason I ask is because it says it is designed to work with a max of 10k ohm source, but it's input ohm level is 50k, which is only five times that of the output. This is not true bridging, is it?

Question #6:Is power (voltage x current) how "loud" a signal is coming from let's say a guitar into a mixer? By this I mean when someone says the signal is not strong enough, are they referring to the power coming from that output? Is power measured in dBu's?

Question #7: What is the typical ohm (impedance) rating on a standard line level input? Also, decibel level?

Question #8: How much power output is to be expected from a Lo-z mic, a hi-z guitar, line level output, and a speaker?
--------------------------------------------

Wow, that's a lot of stuff. I realize that this stuff isn't really necessary for home recorders, but I am the type of guy who like to know the technical details of things I work with. Thanks guys!

 

[dementedchord]

you've got too much confusion here for me to address it in your form...

1st. why are you so hung up on power??? the only place power becomes a real issue is in driving speakers....ie severely low impedence...
and if you really want to tackle this through power a more effective version would be current (I) squared times resistence (R)


2nd. you've got a big misunderstanding about current.... the current isn't increased as in showing a gain.... it a draw... as in this piece of shit better be able to handle the load.... ie.for any given voltage if the resistence increases then the power supply has less load on it... the current needed to push the load has decreased as a requirement.... conversely if that same voltage is applyied to a lower resistence the the power supply has to be able to handle the increased demand for current..... clearer????

3rd. consider your guitar example.... when you plug the thing into an impedence higherthan it's own(typicly 10X's) everythings copaecetic.... as the impedenc of the input decreases more current is required from the guitar... and since it has no power supply to push it there's a point wher the voltage it puts out sags (decreases) because it just cant drive it anymore...

4th. ohm's law is way more important to understand.... first make sure you understand the ramifications of it in the DC world before trying to apply it to an AC signal.... when the voltage starts to go down we're in trouble!!!!! things are rapidly collapseing around us.... so the ability to output current is the issue... when we run out of current we're done for... so.... for any voltage (E) the supplied current (I) necessary to sustain our house of cards must equall the voltage devided by the resistence.... think about it... i know that's not the way you normally think of ohm's law but there it is... same thing different direction... E/R=I and it's all beacause you cant think of the current as an infinate supply to be drawn from...

 

[drossfile]

get ready, this is a long one. i have the following explanation courtesy of al keltz in a word document on my computer, so i copied and pasted it for you.

first, +1 to dementedchord on ohm's law. very important stuff.

second, i applaud you for pursuing this. too few people are interested in the science of audio, of which a good understanding will answer a lot of newbie questions.

third, i can answer a few of your questions generally. power is measured in dB to the extent that 0 dBm=1 milliwatt. Watts=Volts X Amps, but at signals with such small amounts of power, milliwatts become the more practical measurement in audio signals. all decibel scales are related, and it gets confusing (0 dBu=.775 volts, 0 dBV=1 volt, etc.)

preamps boost gain, which is simply amplitude, which ultimately is watts (in audio, pressure in air as dB SPL), although dBV is usually what the boost is measured in, so i think you can look at preamps as boosting voltage.

now, on to the main event:


High and Low Impedance Signals
by Al Keltz

The output from most electronic devices in an audio system will be of low impedance in nature, usually 150 Ohms or less. However, the output from many passive devices, such as a high impedance microphone or passive guitar pickup can have much greater output impedance. What's the difference and why is it important to know how to deal with these signals in an audio system?

Impedance (Z) is the measure of the total opposition to current flow in an alternating current circuit. It is made up of the sum of two components, resistance (R) and reactance (X).
Z = R + X


Resistance is essentially constant at all frequencies in an audio circuit and is measured in Ohms.

Reactance is the measure of opposition to the flow of alternating current caused by the effects of inductance and capacitance in a circuit. It is also measured in Ohms but it will vary with frequency.

The following formula for inductive reactance illustrates how its opposition to current flow increases as the frequency and/or the amount of inductance increases:


where F = the frequency in Hertz (cycles per second) and L = the inductance in Henrys.

The following formula for capacitive reactance illustrates how its opposition to current flow decreases as the frequency and/or capacitance increases.


where F = the frequency in Hertz and C = the capacitance in Farads.

These formulas also point out the fact that a specified impedance for an audio device is only going to be valid for a single frequency - the actual operating impedance will vary greatly over the audio frequency spectrum.

The Low vs. High Difference
A high impedance microphone or guitar will usually output a greater signal (voltage) than a low impedance microphone. This high impedance signal works fine and even has some advantages in a sound system as the mixer or amplifier doesn't need to boost the signal as much. Therefore, any noise on the line is also not amplified as much and this results in an improved signal to noise ratio.

Keep in mind however, that the impedance of the transmission line (or cable) is affected by the impedances of the devices that are connected to it. A low impedance microphone will lower the impedance of the entire line connected to it. Similarly, if you connect a high impedance microphone, you will have a higher impedance line all the way from the microphone to the mixer. This can become a problem as the length of the cable increases.

High impedance lines are more adversely affected by the inherent capacitance that is present in the cable itself. This capacitance combines with the impedances of the source and destination to set up a filter. As the impedance increases and/or the capacitance per foot increases, the active frequency at which the filter comes into play gets lower. The frequencies above this point actually begin to "short out" across the cable's conductors before they ever get to their intended destination. Keeping impedance low and using quality cables can be important issues for maintaining wide frequency response in long lines.

A high impedance line that is interacting with outside electrical interference will act more like an "antenna" than a low impedance line. This problem can get worse as the cable gets longer. This effect is usually insignificant for a guitar or high-Z microphone plugged into an amp with a 15' cord but it can have a big effect if that same signal is sent 100' down a snake. These are reasons why a high impedance signal is almost always converted to low impedance with the use of a Direct Box (DI) before being sent long distances.

Another reason for the use of a DI is that it takes a two conductor unbalanced line and converts it to a three conductor balanced line. This is a separate issue, not to be confused with impedance. It is a common misconception that all lines that use regular tip/sleeve ¼" guitar cord type connectors or RCA Phono connectors are high impedance. Not so. The output from a guitar that has a battery operated active preamplifier or pickup system will be low impedance in nature and so will the output of an electronic keyboard, guitar preamplifier, guitar effects processor. The same goes for the RCA phono output from CD players, tape decks, etc. The signals are unbalanced but LOW impedance in nature.

Low Feeds High
In order to preserve signal level and frequency response, it's important to drive equipment with a source signal that is lower in impedance than the destination equipment's input impedance. If the input impedance of a device is not significantly higher than the source impedance, the signal will be reduced or "loaded down" and its signal to noise ratio and frequency response will suffer.

Think of this as having a nozzle at the end of a garden hose. The garden hose is a low impedance source (there is little resistance to the flow of water) and the nozzle is the higher impedance of the input being fed by the hose.

When the nozzle valve is closed (open circuit):
• Input impedance is VERY high
• Pressure (voltage) is at maximum
• Flow (current) is zero
Now open the nozzle just a little:
• Input impedance reduces but remains high
• Pressure reduces but remains high
• Flow is small
As you continue to open up the nozzle:
• Input impedance reduces further
• Pressure reduces
• Flow increases


With the nozzle open all the way:
• Input impedance is very low
• Pressure falls dramatically
• Flow is greatest


In the case of a high impedance guitar output (7,000 to 15,000 Ohms or more) driving a relatively low impedance input of a mixer (2,000 to 10,000 Ohms), it's like connecting a garden hose to a fire nozzle. The hose just can't produce enough flow (current) for the size of the opening (impedance) to maintain the pressure (voltage).

Splitting Signals
When a signal needs to be split and sent to more than one destination, the impedances of the destinations provide additional paths for the electrical current. This has the effect of reducing the overall impedance presented to that signal. In terms of our garden hose analogy, we've now added a second open nozzle which provides an additional path for the water (less resistance to flow causes reduced pressure in the entire system).

As a general rule of thumb, it's wise to try and maintain an input impedance of at least 10 times the amount of the source impedance.

For example, if we are going to connect the output of a mixer to several amplifiers, calculate the total load provided by the amplifiers by using the formulas below. If that total is approximately 10 times the output impedance of the mixer, then simple passive, parallel splits (like "Y" connections) will usually work fine. The same general principle applies to splitting microphones too. (There can be other issues involving ground loops and isolation - see "Microphone Splitters" ).

The formula for calculating the total load presented by a number of different parallel impedances is:




If there are only two differing impedances, use the following:


or "the product over the sum".

If there are parallel impedances of the same value, then just divide that value by the number of impedances.
For example:
Two 10,000 Ohm loads = 10000/2 = 5,000 Ohm total impedance.
Three 20,000 Ohm loads = 20000/3 = 6,666.66 Ohm total impedance.

If a microphone has its signal split to two mixers which have a 5,000 Ohm input impedance each, the total load to the microphone is 5,000/2 = 2,500 Ohms.

If a mixer output with an impedance of 100 Ohms is split to 4 amplifiers, each with an input impedance of 20,000 Ohms, the total impedance of the load is 20,000/4 = 5,000 Ohms. This is well within the 10:1 load impedance ratio and illustrates how a mixer's output can be passively split several times to a bank of amplifiers without the need for an active distribution amplifier.

 

[Oancient1]

Hey drossfile - good stuff, especially explaining the concept of impedance as opposed to resistance. To that, with a nod to demented chord on the importance of ohms low I submit the following description, with the disclaimer that it is oversimplified -- using easy numbers and assuming pure resistance to keep thing simple.

To get a feel what is going on, it will help to have an understanding of the math behind series circuits. For that we need to first lets establish a few symbols:

Resistors in series --------////5ohms//// -----////10ohms//// -----------------

Ground -----}}}

Reference points (x) , etc.

For example, a 100V series circuit with two 10 ohm resistors in series would look like this:

100V----------////10ohms////-----------////10ohms////-----------}}}

Ohms law says that I = E/R , where
I = Current in amps
E = voltage in volts
R = resistance in ohms


Ignoring the negligible resistance in the wire part of the reference circuit shown above, E= 100V and R = 20 ohms (two 10 ohm resistors in series), ohms law tells us that current through the circuit is:

I = E/R = 100V/20ohms = 5A.

As you noted, power is voltage times current. Expressed mathematically, P (in watts) = I x E. So power dissipated in our total reference circuit is easy to figure:

P = I x E – 100V x 5A = 500W.

In a series circuit, the same value of current flows through all resistors in the circuit, and power dissipation is divided between the individual resistances in the circuit. If the resistances are equal, each resistor dissipates an equal share of the total power of the circuit, or 250W per resistor in the case of our reference circuit.

Before discussing what happens when resistances are not equal, let’s first consider voltage drop within the circuit. For that we must add some measurement points to the reference circuit:

100V—(a)--------////10ohms////-----(b)------////10ohms////---------(z)--}}}

If we measure the voltage between (a) and (z), it will measure 100V, because we are measuring the voltage between the 100V source and ground. e.g. voltage at (a) is 100V, voltage at (z) is 0V. As current passes through a series circuit, each resistor “drops” part of the voltage. Again, because our reference circuit has two equal (10ohm) resistances, it is easy to figure that half of the voltages is dropped by each resistor. e.g:

Voltage across the first resistor (measured between (a) and (b)) is 50V.
Voltage across the second resistor (measured between (b) and (z)) is 50V.

Before moving on, let’s expand the math a little bit. Ohms law in can be expressed three ways:
I = E/R
R= E/I
and
E = I x R

Therefore, in order to express power (P) in terms of voltage and resistance, we can substitute “I x R” in place of the “E” in the original power equation. So instead of :

P = I x E

we have:

P = I x I x R , or the square of the current the resistance.

Likewise, substituting “E/R” in place of the “I” in the power equation we get:

P = I x E = (E/R) x E – (E x E)/R, or the voltage squared divided by the resistance.

If you still follow, lets go back to out reference circuit use the math to prove our assumption of each resistor dissipating half of the power.


100V--(a)--------////10ohms////-----(b)------////10ohms////---------(z)--}}}


Current through each 10 ohm resistor is 5A, so:

P = I x I x R = 5A x 5A x 10ohms = 250W

And voltage across each 10ohm resistor is 50V, so:

P = (E x E)/R = (50V x 50V)/10ohms = 2500/10 = 250W.

All of this just to show that in a series circuit with two equal resistances, half of the power is dissipated by each of the resistances. The key point related to your questions being that in a the first resistance in the series can represent the internal resistance of the output of one device, while the second represents the input impedance of another device.

OK, now let’s examine mathematically what happens when load resistance is too high. For that we will need a new reference circuit:

100V--(a)--------////10ohms////-----(b)------////30ohms////---------(z)--}}}


Current flow:

I = E/R = 100V/40ohms = 2.5A

Voltage drop across the 30ohm load:

E = I x R = 2.5A x 30ohms = 75V

Power dissipated by the 30ohm load:

P = I x I x R = 2.5A x 2.5A x 30ohms = 187.5 W

Or using voltage across the load to calculate:

P = (E x E)/R = (75 x 75)/30 = 187.5 W

As you can see from the above, the voltage across the load is indeed higher, and the current is lower. But the power dissipated is reduced significantly from the maximum value of 250W.

The same power reduction is evident when the load resistance is too low:

100V--(a)--------////10ohms////-----(b)------////5ohms////---------(z)--}}}

Current flow:

I = E/R = 100V/15ohms = 6.66A

Voltage drop across the 5ohm load:

E = I x R = 6.66A x 5ohms = 33.33V

Power dissipated by the 5ohm load:

P = I x I x R = 6.66A x 6.66A x 5ohms = approx 222 W

Or using voltage across the load to calculate:

P = (E x E)/R = (33.33 x 33.33)/5 = approx 222 W

Yes the voltage is lower and the current is higher. But here again, the power disspated by the load is less than the 250W that would be dissipated by a matched 10ohm load.

This of course just illustrates how impedance matching assures max power transfer. I agree that it is most important when dealing with amp outputs into speakers, but impedances should be matched as closely as possible for all connections to assure performance of the gear in the envoronment it was designed for.

Regards,
Tom

 

[ahrenba]

Thank you guys so much for the extremely detailed response. This forum is great! If you don't mind, I have some questions for each of your responses. I divided it up into sections. Thanks!

@dementedchord

Yes, I am confused about the power situation. I thought that power (volts x amps) is what determines how much "signal" gets passed from, let's say, something like a guitar to a mixer input. Isn't that what power does? If not, what actually controls the "level" on a signal? Just the voltage? This is one thing that I remained confused about.

Ok, that makes sense about current. So, let's see if I can apply this logic. You have a small water hose (high pressure [impedance], low current). It is fed into a large pipe, which demands a high current in order to keep that same voltage. Since the small hose can only supply a small amount of current, it cannot supply the demand of the large pipe, therefore decreasing the pressure (impedance) and "loading down" the "power supply" (the small hose). So, in this case would voltage be the amount of water you have to begin with?

I understand now how that works with a guitar into a higher impedance input (10x) works. Question here though. When it reaches a point where the voltage begins to sag because the power supply cannot supply the current it demands, about when is this? Wouldn't the input (load) impedance have to actually be LOWER than the output for this to happen? The reason I ask is because if matching impedance provides maximum power transfer, doesn't this mean that nearly the original voltage and current levels are passed on?

@drossfile
Thanks for the applaud! Thanks for the explanation of power measurement, that makes sense now.

preamps boost gain, which is simply amplitude, which ultimately is watts (in audio, pressure in air as dB SPL), although dBV is usually what the boost is measured in, so i think you can look at preamps as boosting voltage.

So amplitude can also be called gain, or power (voltage x current)? If not, is it a completely different thing?

EDIT: From Wikipedia: The preamplifier provides voltage gain (about: 10millivolts to 1volt) but no significant current gain. The power amplifier provides the higher current necessary to drive loudspeakers.

This is one thing I don't get. If a pre-amp boosts the voltage, doesn't that subsequently lower the current, offsetting the power output? Since P= E x I , if E is raised, and I is lowered, wouldn't that counteract the effects of each other, giving you nearly the same power output? I guess I am just trying to ask why a pre-amp needs to boost voltage, and not current? Is voltage what determines the level of the signal, not current?

I guess the hardest part for me right now is to see what each part contributes to the signal. By this I mean what voltage and current each provide for the overall output of the signal.

Since the dBv and dBu scales are based on the 600 ohm impedance. How do you adjust the scale for a different impedance?

Thanks for that big article. The funny thing is that I actually read that before posting here, but it now makes quite a bit more sense with your guys' help.

@Oancient1

Wow! Thanks for all the mathematical stuff. That was actually pretty useful.

I have a question though. Let's use your circuit:
100V—(a)--------////10ohms////-----(b)------////10ohms////---------(z)--}}}

I am kind of unsure about circuits. I am trying to relate this to a real life situation, hmmm...let's see, I'll try to give an example and let me know if I am right.

Is it safe to make the analogy that the stuff to the left in that circuit (voltage of 100V going through 10 ohms of resistance) is an output source (like from a guitar), while the stuff on the right (10 ohm of resistance and }}}) is like the input of a mixer? And then the line between the two (point b) is a cable connecting the two? If I am on the right track, would }}} be the final voltage that makes it through the circuit?

When you say the voltage "drops", doesn't that mean it's lost? If so, doesn't that mean by the time that the signal hits the mixer, it would be at 0v and unusable? I guess I am just having a hard time grasping the concept of dropping voltage. Also, how does the second resistor drop the same amount of voltage if the voltage coming into it has to be different that 100v, doesn't it?

Also, if we were just looking at one half of that circuit above (let's just say the output from a guitar half, ignore the other half), how can we find the voltage, power, and current levels that is being outputted at point b (before being plugged into a load)? I tried to use your equations for this, here's what I got:

100V—(a)--------////10ohms////-----(b)

I = 100v/10ohms = 10A
P = 100v x 10 ohms = 1000 watts
E = 10A x 10 ohms = 100 volts

Are these right?

Now this is where I am confused, how can a 100v signal go through a 10 ohm resistant output, still contain 100v of power? Am I doing something wrong? I mean, even if you did the calculation for E through the whole circuit, you still get:

E = 5A x 20 ohms = 100 volts

 

[dementedchord]

@dementedchord

Yes, I am confused about the power situation. I thought that power (volts x amps) is what determines how much "signal" gets passed from, let's say, something like a guitar to a mixer input. Isn't that what power does? If not, what actually controls the "level" on a signal? Just the voltage? This is one thing that I remained confused about.

Ok, that makes sense about current. So, let's see if I can apply this logic. You have a small water hose (high pressure [impedance], low current). It is fed into a large pipe, which demands a high current in order to keep that same voltage. Since the small hose can only supply a small amount of current, it cannot supply the demand of the large pipe, therefore decreasing the pressure (impedance) and "loading down" the "power supply" (the small hose). So, in this case would voltage be the amount of water you have to begin with?

I understand now how that works with a guitar into a higher impedance input (10x) works. Question here though. When it reaches a point where the voltage begins to sag because the power supply cannot supply the current it demands, about when is this? Wouldn't the input (load) impedance have to actually be LOWER than the output for this to happen? The reason I ask is because if matching impedance provides maximum power transfer, doesn't this mean that nearly the original voltage and current levels are passed on?

you seem to be catching on... good work...

as to the first... yep it's really a matter of voltage on or AC signal... assuming that there's sufficient current available... and this is important in your guitar example in particular cause there's basiclly NO current available for all intents and puposes... those pick-ups are just little inductive generators....


as to the second... close no cigar.... and your hose analogy works fine in fact i often use it nyself in explaining things... surprised i didn't .... anyhow to clrify... the volume of water is the current... the size of the hose is the resistence... and the pressure behind the water is the voltage.... so when you feed your small hose into the big one your current in this case stays the same... it's all you got... but because the resistence is so low.... the pressure cannot be sustained... and we get system failure... or a reduction of signal/no gain which ever we were trying to do.... clearer???


as to the third... right... as long as there's a 1-1 ratio we dont have any real problems... but we're much happier if the ratio is above it... simply because of the nature of impedence in a AC signal as opposed to pure resistence... impedence can vary wildly with respect to freq... ie. consider speakers... a typical woofer is esentially a short at low freq... and an open at the highs... that is to say short is no resistence and open is infinate... not sure how to keep you from thinking of passing on the current though... the current is disapated as heat..... that's what watts is??? though i admit some times it seems kinda counterintuitive...

 

[Oancient1]

So much in today’s digital world is based on software and logic, but analog circuits is one area where math and the related physics still apply. If you really want to understand this stuff, you will have to study some. Perhaps there are electrical DC and AC fundamentals courses in a community college near you if you have the time and inclination. If not, the link below is an option. You could start with the DC fundamentals and then go to the AC fundamentals.

http://www.ibiblio.org/obp/electricCircuits/

I know this is a copout, but the truth is electronics must be learned a step at a time, and complete answers to some of the questions you ask require a bit of basic knowledge to understand.

That said – I will try to help you understand a little better…

100V---(a)------////R1-10ohms////-----(b)-----////R2-10ohms////-----(z)--}}}

I think I confused you a bit with the above. The circuit shown was meant to illustrate a circuit from a 100V source, through two resistors, to ground. I neglected to show that while one side of the 100V source was connected to R1, the other side was connected to ground. So the current flows from the 100V source, through R1, through R2, to ground ()}}), which is also connected to the other side of the voltage source. Therefore the voltage measured between ground and the three points would be:

(a) 100V
(b) 50V
(z) 0V

Perhaps it easier to understand looking at as shown in the image attached at the end of my post.

Current in a series circuit flows in a loop from the one side of the voltage source, through resistance in the loop, to the other side of the voltage source. In the attached figure, Rs (R1 in my figure) would be the theoretical internal impedance of the output device and RL (R2 in my figure) would be the loading input impedance of the connected device. Thus point (b) in my figure would be both the output of the first device and the input to the next. The analogy to the guitar-mixer connection is OK – with the understanding that the guitar is capable of generating only microvolts, and a typical input impedance of a dedicated guitar input is 600 k-ohms.

As dementedchord said earlier, you really do not need to be all that concerned with impedance matching except connections to speakers. You can see this by simply comparing specifications of devices that are routinely connected together. As an example, the preamps in my cheapy digital recorder are usually ok for guitar, but lousy for microphones, so I use a single channel preamp between the mic and the line input of the recorder. The output impedance of the preamp is listed as 350 ohms and the line input impedance of the recorder is 12 k-ohms, but the preamp provides plenty of gain so who cares!

Tom

 

[ahrenba]

@dementedchord

Thanks for the response. Ok, I understand why plugging a higher impedance into a lower impedance doesn't work too well, since it will load down the source.

One quick question. When you plug a lower impedance into a higher impedance (let's say the load is 10 times the output), does the voltage actually become higher than it originally was when it left the source? Would current decrease or stay the same?

The reason I ask because if you take the hose analogy: plug a small hose into an even smaller one, wouldn't the pressure (voltage) actually increase?

One clarification: So, power (voltage x current) is what determines how much signal there is? Or is it just voltage? If it is voltage, what exactly does power do? For some reason, I can't grasp this concept.

Also, for the other two who posted above, it would be great if you could take a look at the questions I posed to you in the post a bit higher than this one. Thanks!
----------------------------------------------------
EDIT: Just saw Oancient1's response.

In this circuit:
100V---(a)------////R1-10ohms////-----(b)-----////R2-10ohms////-----(z)--}}}

Even though a guitar pickup produces no where near 100 v, let's use that as the example for the source. So, would I be right with this: The pickup's produce 100v which goes through the internal impedance of 10 ohms, and by the time it outputs, the signal is now only 50 volts? Then only 50 volts would be available to go into the mixer?

The output impedance of the preamp is listed as 350 ohms and the line input impedance of the recorder is 12 k-ohms, but the preamp provides plenty of gain so who cares!

Ok, in this situation, voltage transfer would be at a max, but the current flow would be extremely decreased, correct? When you say it doesn't matter because the pre-amp provides plenty of gain, does gain mean that the signal's power is boosted (voltage x current), voltage is boosted, or current is boosted?

 

[drossfile]

Thanks for the applaud! Thanks for the explanation of power measurement, that makes sense now.


Quote:
preamps boost gain, which is simply amplitude, which ultimately is watts (in audio, pressure in air as dB SPL), although dBV is usually what the boost is measured in, so i think you can look at preamps as boosting voltage.

So amplitude can also be called gain, or power (voltage x current)? If not, is it a completely different thing?

EDIT: From Wikipedia: The preamplifier provides voltage gain (about: 10millivolts to 1volt) but no significant current gain. The power amplifier provides the higher current necessary to drive loudspeakers.

This is one thing I don't get. If a pre-amp boosts the voltage, doesn't that subsequently lower the current, offsetting the power output? Since P= E x I , if E is raised, and I is lowered, wouldn't that counteract the effects of each other, giving you nearly the same power output? I guess I am just trying to ask why a pre-amp needs to boost voltage, and not current? Is voltage what determines the level of the signal, not current?

I guess the hardest part for me right now is to see what each part contributes to the signal. By this I mean what voltage and current each provide for the overall output of the signal.

Since the dBv and dBu scales are based on the 600 ohm impedance. How do you adjust the scale for a different impedance?

Thanks for that big article. The funny thing is that I actually read that before posting here, but it now makes quite a bit more sense with your guys' help.

ok, "gain" is just a synonym for "increase," so my original answer was a bit unclear. the "gain," or "increase" that takes place is indeed in voltage through a preamp, but remember that you have 4 factors here: voltage, amperage, resistance, and reactance (the latter 2 of which comprise impedance). if all a preamp did was increase voltage, you would have a simple step-up transformer that had a poteniometer in series with its secondary winding, and it would result in a decrease in either amperage or resistance. in a preamp, however, you will notice that the electronics are far more advanced than that. now, i'm not going to pretend to have a full understanding of the inner workings of capacitors, resistors, transistors, tubes, and all the other goodies under the hood of a preamp, but the moral of the story is: voltage and current are only inversely related when resistance remains constant. by manipulating all factors, our overall amplitude can be increased (gained). ideally this would occur equally across the frequency spectrum (which would give a "flat" and "uncolored" response). however, due to the nature of electronics, this is usually not the case. varying grades of semiconductors, insulators, conductors, etc are available which make the whole thing as much an art as it is a science, and this is why we have frequency response charts, useless as they may be.

dBu/dBv/dBA/dBm etc are not "scalable" according to impedance, at least as far as i understand it. decibels in general were created as a way to translate the linear numbers of psychoacoustics (the human ear has a trillion to one ratio between highest level tolerable and lowest level percieved) into exponents, which is simply more practical when we're talking about numbers so large. look at it this way: 1 volt is expressed as 10 to the zero power (sorry no superscript here) in scientific notation. we can further take this to a logarithm of 0 (the correlated exponent of ten in this instance), which equals 0 bels (named after alexander graham bell) and then to 0 decibels (10 times 0). if you increase in a linear fashion for 10, 100, 1000, 10000, etc. this goes to 10 to the 1, 2, 3, 4, then to logarithm 1, 2, 3, 4, then 1, 2, 3, 4 bels, which will equal 10, 20, 30, 40 decibels. you also have to understand that decibels relate to several different things: physical pressure in the air, electrical signals in an audio system, the human ear's reaction to it, and everything in between. so, dBv and dBu are already "scaled" based on electronic signals.

holy cow, now i've confused myself. what was the original question? anyway, all i can really suggest to you is that if you're as fascinated with this drivel as i am, spend the $90 on "total recording" by david moulton at www.kiqproductions.com it's worth 10 times that (or log1 )

one other thing: as far as i can tell, discerning the difference between what voltage and current separately provide regarding the signal is irrelevant. they both will work in concert throughout the course of the signal's route. you can apply voltage to any device, but until that device begins to make some sort of energy conversion, no current is present. a load device is something that converts electricity into some other form of energy. voltage is simply the excitation of electrons. you could have a wire carrying 480 VAC, and if an open switch was standing between it and a motor, there would be no work being done and as a result no current (amperage) being drawn. once you close that switch, the motor goes to work as an inductive (electromagnetic) device, and suddenly you still have 480 VAC supplied to that motor, but the work being generated is drawing a current, as well as counter-emf, harmonic distortion, and lots of other fun electrical peripherals.

i did it again. i have no idea where all of that was going, but there it is. it probably isn't going to help immensely, but good luck anyway. buy moulton's book.

 

[drossfile]

oh, and amplitude simply refers to the power level at any given point. so gain isn't amplitude, but a gain IN amplitude will be an increase in power. amplitude sort of refers to the level of loudness, but this is a dynamic scale due to psychoacoustics which depends on frequencies, overtones, transients (peak) vs. rms, etc...ugh i have to stop now...

 

[ahrenba]

Ah! So, basically, voltage really only directly impacts current (and visa verca) when output and iniput impedances are equal? That seems to make sense!

And so..... power is actually how "loud" or strong the signal is? And Voltage, impedance, and current are factors that control how strong this power is? That would make sense also.

I guess where I got confused about power was when dementedchord asked why I was concerned with power. So, like I asked above, isn't power (Voltage x amperage or Amperage squared x resistance) what determines how strong the signal is?

Your explanations and analogies were great! Thanks for those.
--------------------------------
Here are my questions that remain from the post earlier:

One quick question. When you plug a lower impedance into a higher impedance (let's say the load is 10 times the output), does the voltage actually become higher than it originally was when it left the source? Would current decrease or stay the same?

The reason I ask because if you take the hose analogy: plug a small hose into an even smaller one, wouldn't the pressure (voltage) actually increase?

In this circuit:
100V---(a)------////R1-10ohms////-----(b)-----////R2-10ohms////-----(z)--}}}

Even though a guitar pickup produces no where near 100 v, let's use that as the example for the source. So, would I be right with this: The pickup's produce 100v which goes through the internal impedance of 10 ohms, and by the time it outputs, the signal is now only 50 volts? Then only 50 volts would be available to go into the mixer?

The output impedance of the preamp is listed as 350 ohms and the line input impedance of the recorder is 12 k-ohms, but the preamp provides plenty of gain so who cares!

Ok, in this situation, voltage transfer would be at a max, but the current flow would be extremely decreased, correct? When you say it doesn't matter because the pre-amp provides plenty of gain, does gain mean that the signal's power is boosted (voltage x current), voltage is boosted, or current is boosted?

 

[drossfile]

Ah! So, basically, voltage really only directly impacts current (and visa verca) when output and iniput impedances are equal? That seems to make sense!

well, voltage and current will share a relationship on both "sides" of any circuit component, whether the signal is passing through connections that are of matched or unmatched impedance. i was basically making a reference to ohm's law when i said that current and voltage will exhibit an inverse relationship when resistance (R) remains constant. this is not the same as impedance (Z, or resistance + reactance). resistance is simply one component of impedance (and the most easily calculatable one at that).

voltage by itself doesn't effect anything. it is simply electron movement through a conductor. sound waves are just excited air molecules, which vibrate a microphone capsule, which transduces the modulating air pressure into an electrical signal. we can manipulate those signals 6 ways to sunday.

but volts (E), current (I), and resistance (R) all have an indivisible relationship, just like pressure, temperature, and volume. these are laws of physics that cannot be bent. look at ohm's law as a pie chart:

you can see that these things are inseparable, and we can depend on two known quantities to calculate the third, and it ALWAYS WORKS!

but once you introduce reactance to the equation, everything goes nuts (still provable, but a hell of a lot more complex). remember that reactance includes induction, which is fairly unpredictable from one system to the next due to variations in quality of power and components (think of a square wave going through a high vs. low end transformer). further, you have to add in capacitance, which will vary for the same reasons.

damn it, you drew me in again. i really have too much time on my hands, as do you. i'm done for tonight. believe it or not, i'm enjoying this.

 

[ahrenba]

damn it, you drew me in again. i really have too much time on my hands, as do you. i'm done for tonight. believe it or not, i'm enjoying this.

Lol! You have been awesome with my questions so far! Thanks!

Ok, I think your explanation in the last post makes sense. The variables (voltage, current, impedance) are bound?

So, with a pre-amp...it would boost voltage and current, correct? Since the variables are connected, what would be adversely affected, the impedance? If it does boost both voltage and current, this is thing I am slightly confused about. If both are these are connected, wouldn't it mean you can't raise one with out lowering the other?

One thing I just need confirmed is this (from my other post):
And so..... power is actually how "loud" or strong the signal is? And Voltage, impedance, and current are factors that control how strong this power is? That would make sense also.

I guess where I got confused about power was when dementedchord asked why I was concerned with power. So, like I asked above, isn't power (Voltage x amperage or Amperage squared x resistance) what determines how strong the signal is?
------------------------------------------
My questions for Oancient1 (you can answer them if you feel inclined) :

One quick question. When you plug a lower impedance into a higher impedance (let's say the load is 10 times the output), does the voltage actually become higher than it originally was when it left the source? Would current decrease or stay the same?

The reason I ask because if you take the hose analogy: plug a small hose into an even smaller one, wouldn't the pressure (voltage) actually increase?

In this circuit:
100V---(a)------////R1-10ohms////-----(b)-----////R2-10ohms////-----(z)--}}}

Even though a guitar pickup produces no where near 100 v, let's use that as the example for the source. So, would I be right with this: The pickup's produce 100v which goes through the internal impedance of 10 ohms, and by the time it outputs, the signal is now only 50 volts? Then only 50 volts would be available to go into the mixer?

The output impedance of the preamp is listed as 350 ohms and the line input impedance of the recorder is 12 k-ohms, but the preamp provides plenty of gain so who cares!

Ok, in this situation, voltage transfer would be at a max, but the current flow would be extremely decreased, correct? When you say it doesn't matter because the pre-amp provides plenty of gain, does gain mean that the signal's power is boosted (voltage x current), voltage is boosted, or current is boosted?

 

[Oancient1]

In this circuit:
100V---(a)------////R1-10ohms////-----(b)-----////R2-10ohms////-----(z)--}}}

Even though a guitar pickup produces no where near 100 v, let's use that as the example for the source. So, would I be right with this: The pickup's produce 100v which goes through the internal impedance of 10 ohms, and by the time it outputs, the signal is now only 50 volts? Then only 50 volts would be available to go into the mixer?

Ok, in this situation, voltage transfer would be at a max, but the current flow would be extremely decreased, correct? When you say it doesn't matter because the pre-amp provides plenty of gain, does gain mean that the signal's power is boosted (voltage x current), voltage is boosted, or current is boosted?

Re the first paragraph -- yes in that inaccurate (for a guitar) connection, if the output impedance of the guitar and input impedance of the mixer were equal, then the voltage at the input of the mixer would be 50V. However, after having a look some typical specs, I must take back part of what I said previously and say that there appear to be cases where impedances are deliberately mismatched. For example, using typical pickup output voltage (100 millivolts (.1V)) and internal resistance values (10 ohms) found on the internet, and the listed guitar input impedance value for my recorder (600 kohms):

100mV---(a)---////R1-10kohms////-----(b)---////R2-600kohms////--(z)--}}}

I = V/R = .1V/610,000ohms = approx .00000016A (only .16 microamps!)

V drop across R1 (voltage drop across R1):

V = IR = .00000016A x 10,000ohms = .0016V (equals 1.6 millivolts)

V drop across R2 (voltage drop across R2):

V = IR = .00000016A x 600,000ohms = .0984V (equals 98.4mV)

Power disspated by R2:

P - IE = .00000016A x .0984V = approx. .000000016W (equals only .016 microwatts).

If the typical values I found are accurate, it seems obvious that the impedance mismatch is intentional in order to ensure that the voltage drop across the guitar pickups internal resistance is insignificant, thus providing the highest possible voltage level at the input to the recorder . Likewise, the fact the the current is so small (and thus the power dissapted by R2) would indicate that power is indeed a less important factor than voltage in this connection.

Regarding the second quoted paragraph, the word gain is used to describe power gain, voltage gain, or current gain, and you have to be careful to understand in which context the word is used, and often the context is not clear in specifications, etc. Again having a look at the specs of some of my gear it would appear that amplifying devices provide both voltage and power gain, although the amounts of gain are not equal because the input impedance is typically much higher than the specified load impedance.

Its been kind of cool remembering all this stuff from back in my days as a tech. But my head is swimming, and I look forward to spending some time just connecting the stuff together and doing a little recording.

Tom

 

[dementedchord]

@dementedchord

Thanks for the response. Ok, I understand why plugging a higher impedance into a lower impedance doesn't work too well, since it will load down the source.

One quick question. When you plug a lower impedance into a higher impedance (let's say the load is 10 times the output), does the voltage actually become higher than it originally was when it left the source? Would current decrease or stay the same?

The reason I ask because if you take the hose analogy: plug a small hose into an even smaller one, wouldn't the pressure (voltage) actually increase?

One clarification: So, power (voltage x current) is what determines how much signal there is? Or is it just voltage? If it is voltage, what exactly does power do? For some reason, I can't grasp this concept.

after reading the other responces and coming back to ansewr this i have to admit it's starting to get hard to remember what alls been said so we may need to take a day to digest and perhaps start another thread with whatever concerns you still have... and cant say absolutely but i suspect your problem really boils down to this need to understand visa vie the power equations... on a practcal level power plays very little part in this up too the primary amplification stages... units are not in the watts... but rather milliwatts untill then with the possible exception of tubes but even then on the per tube level where usually dealing with the likes of 1/2 wt...


as to the rest....
no increase in voltage... a marked decrease in current... perhaps this will help... voltage is a potential... it's not a sliding scale commodity where if we dont use it we collect more... and again think of it as the pressure in your hose...


again with the hose... dont know if i can sustain the thought as i have really thought of fluid dynamics since high school physics '68... but no it doesnt increas in either case IIRC... the pressure is distributed evenly... isn't that what makes hydrolic jacks work???? cant remeber is that the 2nd law of fluid dynamics????


ok maybe now we're circleing in on the problem... the voltage determines the signal... the current is just what's been used in getting there and the work it does is expressed as power... which ultimately is just given of as heat... seems i remeber a formulae for converting to BTU's even...

clearer grasshopper????

 

[ahrenba]

@Oancient1

Ah! Very nice, that is the kind of example I was looking for (deliberately mismatched). The only thing that I don't really understand is what "voltage drop" is? Wouldn't "drop" insinuate that voltage is lost each time it hits a resistor? If so, does this mean once that 100mV signal hits the first resistor, it loses 1.6mV, which then allows it to only pass on 98.4mV to the next resistor?

Maybe, you could briefly explain what "voltage drop" is? Thanks! One definition that I found was "That portion of the voltage used or "consumed" by each device in a circuit." If this is true, wouldn't all the voltage be consumed by the time it hits the input?

Its been kind of cool remembering all this stuff from back in my days as a tech. But my head is swimming, and I look forward to spending some time just connecting the stuff together and doing a little recording.

-----------------------------------------
@dementedchord

ok maybe now we're circleing in on the problem... the voltage determines the signal... the current is just what's been used in getting there and the work it does is expressed as power... which ultimately is just given of as heat... seems i remeber a formulae for converting to BTU's even...

Very nice. That is what I was looking for. I was basically confused what actually determines the signal, but now I know. So, now that we have that established, just real quickly....what does power (volt. x amps) provide (or do?)

clearer grasshopper????

Ah, yes, Sensei, I have learned much.

 

[dementedchord]

mostly what it does is give us an index of what we use so that we can insure that the power supply is designed accordingly.... or that when we are assembling a room or system that we dont excede the amperage of the fuse box... (the sum of the individual wattages)

 

[ahrenba]

Thanks dementedchord for the response.
-------------------
Remaining question for Oancient1

Ah! Very nice, that is the kind of example I was looking for (deliberately mismatched). The only thing that I don't really understand is what "voltage drop" is? Wouldn't "drop" insinuate that voltage is lost each time it hits a resistor? If so, does this mean once that 100mV signal hits the first resistor, it dissipates 1.6mV? I assume this is right. So, wouldn't this allow the remaining 98.4mV to transfer on to the next resistor, which in turn, dissipates this remaining voltage (98.4mV) due to the extremely high resistance? I hope this is right!
-------------------
New question:

I am looking at getting the Lexicon Omega computer recording interface. The impedance for the line inputs are at 20 kOhm unbalanced and 10 kOhm unbalanced. So, this presents a problem with trying to plug in an electric guitar, which from what I have heard, is much higher) like at least 20 kOhm's.

1. If I plug the guitar into it, will I still get a signal (faint probably)?

2. Most dedicated instrument inputs (on a preamp) have an impedance of around 1 MoHm's from what I have found. Let's use an arbitrary number here. Let's say a guitar has an output impedance of 12 koHm's. Since the ratio between the two is extremely large, maximum voltage would be transferred, but current would be severely low, right? This will help me understand a preamp better: From here, what does the preamp do to the various components of the signal (voltage, current, overall power). Obviously it has to boost something by providing gain, but to what?
I sort of asked this before, but am still somewhat unsure.

3. You guys informed me that voltage is what determines the "signal". Is the "signal" that is determined by this voltage, also called level? By level I mean like the ranges that include mic level, instrument level, line level, and speaker level? These levels are in dB's right?

Also, why wouldn't manufacturers just make all inputs on mixers, pre-amps, etc. a very high impedance (something like 5 moHm's)? That way it'll give you the 10x impedance rating that is recommended. What is the disadvantage of this? Current would be low, but voltage would be almost at a maximum? What's the downside? This is probably a very stupid question.

Is this thread getting too bogged down? Should I start a new one with my remaining questions?

 

[ahrenba]

Ok guys. I made a new thread that is more condensed and easy to follow. I have transferred my remaining questions there.

If you could, please take a look: https://homerecording.com/bbs/showthread.php?t=256933

Thanks!

 

[Oancient1]

Thanks dementedchord for the response.
-------------------
Remaining question for Oancient1

Ah! Very nice, that is the kind of example I was looking for (deliberately mismatched). The only thing that I don't really understand is what "voltage drop" is? Wouldn't "drop" insinuate that voltage is lost each time it hits a resistor? If so, does this mean once that 100mV signal hits the first resistor, it dissipates 1.6mV? I assume this is right. So, wouldn't this allow the remaining 98.4mV to transfer on to the next resistor, which in turn, dissipates this remaining voltage (98.4mV) due to the extremely high resistance? I hope this is right!
-------------------
New question:

I am looking at getting the Lexicon Omega computer recording interface. The impedance for the line inputs are at 20 kOhm unbalanced and 10 kOhm unbalanced. So, this presents a problem with trying to plug in an electric guitar, which from what I have heard, is much higher) like at least 20 kOhm's.

1. If I plug the guitar into it, will I still get a signal (faint probably)?

2. Most dedicated instrument inputs (on a preamp) have an impedance of around 1 MoHm's from what I have found. Let's use an arbitrary number here. Let's say a guitar has an output impedance of 12 koHm's. Since the ratio between the two is extremely large, maximum voltage would be transferred, but current would be severely low, right? This will help me understand a preamp better: From here, what does the preamp do to the various components of the signal (voltage, current, overall power). Obviously it has to boost something by providing gain, but to what?
I sort of asked this before, but am still somewhat unsure.

3. You guys informed me that voltage is what determines the "signal". Is the "signal" that is determined by this voltage, also called level? By level I mean like the ranges that include mic level, instrument level, line level, and speaker level? These levels are in dB's right?

Also, why wouldn't manufacturers just make all inputs on mixers, pre-amps, etc. a very high impedance (something like 5 moHm's)? That way it'll give you the 10x impedance rating that is recommended. What is the disadvantage of this? Current would be low, but voltage would be almost at a maximum? What's the downside? This is probably a very stupid question.

Is this thread getting too bogged down? Should I start a new one with my remaining questions?

You are understanding voltage drop correctly. However, "dissapate" is a term generally associated with power. The resistor dissapates the power, and as a result part of the voltage is "dropped" across the resistor.

You should certainly get a faint signal plugging a guitar directly into a line input, but if you are talking about recording only one guitar at a time, then no worries --you can use the Omega's "instrument" input, which has a published input impedance of 1 Mohm. If recording two guitars simultaneously, you will need a preamp for the second guitar. For this I use a single channel cheapy Behringer preamp, on which the TRS input impedance is 1 Mohm to specifically meet this application. The Behringer also has an XLR mic input, with a published impedance around 2 Kohms, optimized for microphones.

Regarding dB -- no -- signal levels are not expressed as pure dB. "dB" (decibels), is logaritmic way of expressing big ratios-- gain ratio in the electronics world. +3dB is a gain of 2 (2x). 10 dB = 10x, 20 dB = 100X, 30db - 1000x, and so on. The ratios so expressed assume that the smaller number and larger number are in the same units: voltage or power generally. It is only when a reference unit is attached to dB that it can be used to specify a certain level of power or voltage. dBV uses 1 volt as reference; dBm uses one milliwatt (mw) as the reference; . So 0 dBm = 1 mw, 10 dB = 10 mw, 20 dBm = 100 mw, and so on. Likewise, -10 dBm = .1mw, -20 dBM = .01mw, etc. Wikipedia.org has a decent description of decibels, including the underlying math.

Your last question isn't stupid -- just oversimplification. You are assuming that the only thing that matters is having enough voltage at the input. But what about having too much? All amplifiers (including preamps) are designed to work within a range of input levels (and frequencies BTW) to produce a range of output levels, and input and output impedances are deliberately selected to optimize performance.

Tom