Help me really understand volts and watts...

Thanks guys for all your explinations! I feel like I'm walking in a whole new world...

Not to intrude on a thread or anything, but im reading the same book and what i cannot grasp is impedance. I dont understand loads and how impedance matching and all this stuff ties to what the above threads explain.

Thanks for all the good info guys, it has helped me understand a bit better too.

good question! If you don't get your answer here, start a new thread on that specific question.

In most audio work, any time you see "impedance" you can pencil-in "resistance", or "resustance to audio frequency signals", or "resistance that is different at different audio frequencies".

Yes, the ideal resistance is the same resistance at any frequency.

There are no ideal resistances, though a common resistor is pretty close at audio frequencies.

Chokes, capacitors, loudspeakers, and other gizmos have a "resistance" that changes with frequency, though we often try to minimize the change across the audio band, or give a worst-case resistance (a speaker may be 6 ohms for DC, 50 ohms for 40Hz, 7 or 8 ohms around 400Hz and 15 ohms above 3KHz: we don't put DC in speakers and the 50 ohms does no harm to the amp, it is that 7 or 8 ohms around 400Hz that matters most).

There is a LOT more to impedance, and outside the practical audio field it can be critical. But we are simple folk, and most "impedances" in audio can be considered "worst-case audio frequency resistance".

Now, having written "resistance everywhere, you go to Ohms Law and Watts law.

Ohm's Law says resistance is the voltage/current ratio.

Watts law says that power is XXXvoltage times resistanceXXX EDIT Voltage times Current.

Neither of these say ANYTHING about the signal level. Unfortunately, we simple practical audio folks tend to associate certain impedance classes with certain general signal levels.

Also, if we think at all, we think in terms of voltage. We touch upon power when we use dBm, but this is out of fashion. We almost never think about current: quick, what is the peak current in an IHF standard line-input when the signal is a home hi-fi CD player at 0dBfs? It is about 0.28milliAmps, but who thinks about it?

Before straining yourself trying to understand impedance matching and different loads, make sure you fully understand the concept of a voltage divider, ie. an ideal voltage source connected to two resistors to ground. Internalize the relationship between the two resistors, and the proportional voltage drop across each of them.

From there, start thinking about not-so-ideal sources. Ie., a voltage source always has some output resistance in series with it, and a current source always has some resistance in parallel with it. Then, using your intuition about voltage dividers, you'll start to see the relationship between a given source and any sort of load you decide to hook up to it.

xx

I guess I get confused about the voltage divider. For example, I was thinking about building a passive mixer, just to help me learn more about designing circuits. I got thinking about it, and it seems like the output of line level devices ranges from 500 ohms to some around 70k ohms. Maybe even higher. I know my alesis HD24xr outputs 220 ohms. How can one accomodate for all these input impedances? Also as a side note, how would I calculate decibel loss and makeup gain?

> it seems like the output of line level devices ranges from 500 ohms to some around 70k ohms. Maybe even higher.

Hmmm. Except the eraly versions of the Dynaco tube hi-fi preamp, and some old DIN standard, and many guitar FX pedals, everything is "low" output impedance. ARP modular synths used a 1K output, most hi-fi and "consumer studio" is 47-470, most broadcast is under 60 unless it is true 600.

Don't confuse output impedance with drive ability. A hi-fi box may be 47 ohms output impedance, but be unable to drive full level into anything less than 2K. The miracle of negative feedback.

Mostly we make inputs 10K or higher because that way we just don't have to care if the source is near-zero or 600 (or even 1K). And because most sources found in a studio chain will drive 10K easily.

> how would I calculate decibel loss and makeup gain?

You could just avoid it. Don't try to do Voltage Divider and dB lesson at the same time.

The passive mixer is a voltage divider with a common "lower" resistor for all channels. In the simplest lowest-loss form, the "lower" resistor for one channel is all the other mix-resistors (and their sources; prove that this can be negligible) in parallel.

So for the made-up case of an 11-input mixer with 10K mix-resistors, with all inputs connected, one input sees a 10K "top" resistor and ten other 10K resistors "in parallel to ground". This makes a 10K/10= 1K "bottom" resistor.

The voltage divider is then 10K and 1K. The output is, at a glance, a little less than 1/10th of the input. It is really 1/11 of the input. And so we see that the "gain" of a mix network with N inputs, expressed as a fraction, is 1/N. To bring the overall gain back to unity (for one input), we need a gain of N. A 64-input mix network needs a make-up amp with a gain of 64.

The output impedance, which the make-up amp input sees, of that 11-input 10K mixer is evidently all 11 of the 10K resistors in parallel. This is 10K/11 or 909 ohms. In general, it is R/N.

This will lead you to a spec for a purpose-made make-up amp.

There is a fashion to use, instead of a special amp, a mike preamp. Such things like to see a source impedance of 100-200 ohms, and usually like to run a gain of around 40dB which is, in fractions, a gain of 100/1. Therefore we really want a mix-network gain of 1/100 and an output impedance of say 150 ohms. Using the 1/N and R/N formulas from above, and working backward, we want a network with 100 inputs of 15K each. Gain is 1/N or 1/100, which when made-up with a 40dB or 100/1 amp comes to unity. Mix network output impedance is R/N or 15K/100 or 150 ohms. The only "problem" is that we may not want 100 inputs. But we can build the 100-input version on paper, assign the 16 inputs we need, take the other 100-16= 84 resistors and compute their parallel resistance, then trade them in on one "swamping" resistor of that value. 15K/84= 178.6 ohms, 1dB errors won't kill us, in simple dividers 10% error is less than 1dB, use 168 or 175 ohms.

Prove to yourself, with simple fractions, that a 16-in 15K mix network with 178 ohms swamping the bus gives gain of 1/100 for any input and an output impedance near 150 ohms.

All that dB do is turn all this gain/loss multiply/divide into add/subtract. When you need to compute the gain/loss of a phone line from here to Podunk, 13 lossy segments with 12 booster amps between, and do not have a calculator, add/subtract dB is much easier than multiply/divide fractions. But for simple work, it is no big deal. And for voltage dividers, fractions are the easy logical way to solve the problem and the dB number is just for the spec-sheet.

Maybe reading from another point will help solidify your understanding of impedance. I know it helps me to see different angles in understanding the whole.

http://www.tape.com/Bartlett_Articles/impedance.html

here is an excerpt:

IMPEDANCE FAQ
by Bruce Bartlett

Impedance is one of audio's more confusing concepts. To
clarify this topic, I'll present a few questions and answers
about impedance.

WHAT IS IMPEDANCE?
Impedance (Z) is the resistance of a circuit to alternating
current, such as an audio signal. Technically, impedance is the
total opposition (including resistance and reactance) that a
circuit has to passing alternating current.
A high impedance circuit tends to have high voltage and low
current. A low impedance circuit tends to have relatively low
voltage and high current.

I'M CONNECTING TWO AUDIO DEVICES. IS IT IMPORTANT TO MATCH THEIR IMPEDANCES? WHAT HAPPENS IF I DON'T?
First some definitions. When you connect two devices, one is
the source and one is the load. The source is the device that
puts out a signal. The load is the device you are feeding the
signal into. The source has a certain output impedance, and the
load has a certain input impedance.
A few decades ago in the vacuum tube era, it was important to
match the output impedance of the source to the input impedance
of the load. Usually the source and load impedances were both
600 ohms. If the source impedance equals the load impedance,
this is called "matching" impedances. It results in maximum
POWER transfer from the source to the load.
In contrast, suppose the source is low Z and the load is high
Z. If the load impedance is 10 times or more the source
impedance, it is called a "bridging" impedance. Bridging results
in maximum VOLTAGE transfer from the source to the load.
Nowadays, nearly all devices are connected bridging -- low-Z out
to high-Z in -- because we want the most voltage transferred
between components.
If you connect a low-Z source to a high-Z load, there is no
distortion or frequency-response change caused by this
connection. But if you connect a high-Z source to a low-Z load,
you might get distortion or altered response. For example,
suppose you connect an electric bass guitar (a high-Z device)
into an XLR-type mic input (a low-Z load). The low frequencies
in the signal will roll off, so the bass will sound thin.
We want the bass guitar to be loaded by a high impedance, and
we want the mic input to be fed by a low-impedance signal. A
direct box or impedance-matching adapter does this. Such
adapters are available from Radio Shack.
The adapter is a tube with a phone jack on one end and a male
XLR connector on the other. Inside the tube is a transformer.
Its primary winding is high Z, wired to the phone jack. The
transformer's secondary winding is low Z, wired to the XLR. You
plug the guitar cord into the phone jack, and plug the XLR into a
mic input in a snake or mixer. Use it with a bass guitar,
electric guitar, or synth.
This impedance-matching adapter works, but is not ideal. The
load it presents to the bass guitar might be 12 kilohms, which
will slighly load down the high-Z guitar pickup, causing thin
bass.
An active direct box solves this problem. In place of a
transformer, the active DI usually has an FET (Field Effect
Transistor). The FET has a very high input impedance that does
not load down the bass guitar.

sometimes word origin can help. my old boss could take any word apart. big Latin buff. something that they don't require nowdays unless you are inprisoned by mean nuns with five oclock shadows.

impedance- impede

etc

Wow. those last 3 responses or so did it, I get it now. Thanks PantinaCreme and PRR, those responses were excellent.

Thought I would bump this - as it always needs to be near the top of the list and never lost
PRRs farm is going down in history

look at the drift speed section


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