Where does current enter into all this?

[smoguzbenjamin]

Hey all!

I'm brushing up on my theory and have come to the apalling discovery that I don't fully understand where current is entered into audio. I know that the voltage of a guitar signal is AC. Hold on that stand for Alternating Current.  :?  :shock: I'm completely confuzzled. Is a higher-current signal also louder or is it brighter or am I completely stupid??? Oh well dinnertime.

I have been thinkin for ages tryin to figure it out myself but the genius just doesn't come

[Mike Burgundy]

AC does mean alternating current, but *it's all related and interconnected*
It's all just V=IR.
Lets say point A sits at a different voltage level (potential) than point B.
If there's infinite resistance between the two, no current will flow no matter what antics the voltage level throws.
If there is NO resistance, there will be "massive" infinite current and no voltage difference between the two. All this is V=IR.
Now, if Va (voltage level in A) > Vb, and R has a certain value, then current will flowfrom A to B, while the whole "system" TRIES to equalise both points. If you don't keep feeding current in somewhere or take it away somewhere else, the system will stabilise with Va=Vb.
If you do supply current (so you can compensate for the bleeding away of electrons by the current between A and B), a voltage drop over R will result. Once again, there is no current without voltage and vice-versa.
If Va <Vb, a current will flow from B to A.
If Va alternates, so that sometimes Va>Vb and sometimes Va<Vb then an alternating current will result.

In the world of pedals and electronics, the whole thing boils down to what kind of circuit youre using and the trade-offs/benifits they all have.
A voltage amplifier magnifies the input *voltage* and draws/delivers very little current. You can't draw too much current from it's output, or it won't be able to supply enough to keep the voltage up: so if you draw too much, teh amplifier "sags".
This behaviour usually is different fot different frequencies.
The same thing goes for current amps - the voltage might skyrocket if youre feeding too high an impedance, or the amp stage might fry if youre feeding into too low an imedance.

Long story short:
I'ts all V=IR, and impedances
hih

[smoguzbenjamin]

Read it 5 times. Ha. I'm starting to grasp that. Okido, now to let that run through my head all night and I'll get it by tomorrow :mrgreen:

[smoguzbenjamin]

Forgot to ask: What kind of amplifiers do we use? Voltage right? But what if I want to power a low-impedance reverb tank? I'd need high current to keep the voltage up. So how would I accomplish this? Would an opamp suffice? :?

[Peter Snowberg]

Forgot to ask: What kind of amplifiers do we use? Voltage right? But what if I want to power a low-impedance reverb tank? I'd need high current to keep the voltage up. So how would I accomplish this? Would an opamp suffice? :?

We use both voltage and current amplifiers, but the focus is usually on one or the other. In the end, the voltage is not amplified anywhere near as much as the current is, but both get quite a boost.

In the early stages, the focus is on voltage, but the last stage is almost always about getting maximum current out. There may be voltage gain in the last stage too though.

As Mike said, "*it's all related and interconnected*"

When it comes to the current output of opamps, most of them will not put out very large amounts of it. The NE5532 is a notable exception though. That one will drive a tank or a speaker with ease.

Is that confusing enough?

Take care,
-Peter

[petemoore]

I'm with Ben, that it 'takes a while' sometimes!!!
 I've been reading explanations of this stuff for sometime now...that time it hit me...of course that raises even more curiosities...Kool...I really like it when I get to mull over the 'motions' and the gears mesh!!!
 that's a good one     Thanks to you Mike Burgundy and Peter S.

[Joe Davisson]

This is more general, but might help...

Voltage is proportional to the number of electrons being swept away from their planetary orbits. That void can be filled by more electrons.

Current is the velocity at which those electrons will move towards the voltage.

Resistance is related to both voltage and current. A large resistance reduces the number of atoms whose electrons can be swept away. There is now a "bottleneck". Electrons "clog-up" in the bottleneck, reducing the velocity. In this way, resistance limits both voltage and current.

So there you have it. Voltage is the number of electron "cars" on the highway, current is the MPH, and resistance is the traffic jam.

-Joe

[Rick]

* So there you have it. Voltage is the number of electron "cars" on the highway, current is the MPH, and resistance is the traffic jam.
-Joe *

I'm not lucid tonight but isn't voltage more the pressure or MPH and current the width of the highway. I think it depends on how you look at the whole picture at once. Maybe the power formula says it more usefully.
P = I x E   or  P = I (squared) x R.  Power in watts is volts x amps or amps (squared) x resistance. I just find this way of looking at it more usefull. Am I offbase   ...RG  ...anyone  ..yawn going to bed.

[Peter Snowberg]

I think you got things reversed there Joe.

The basic unit of electrical charge is the Coulomb.

One Volt is equal to one Coulomb per Joule.

A more in depth way to say that is that moving a charge of one Coulomb through an electrical potential difference of one Volt produces a potential energy difference of one Joule.

One Amp is equal to one Coulomb per second of charge transfer.

The Amp measures a quantity of energy that is passing a fixed point per second. It's somewhat of a funny term because we don't have a special word for kilometers per hour or other "per unit of time" types of measurements. Oh well....

So in English: The Volt is more like the MPH (or K/H) and the Amp is more like the number of cars on the road.

I like to analogize to plumbing. Voltage is somewhat like water pressure, and Amperage is like the quantity of water flowing.

Take care,
-Peter

[smoguzbenjamin]

I'm getting all this in physics class But it's nowhere near specialised enough for me. So what has drained into my resistant brain so far:

For a low impedance input (like a speaker or reverb tank) You need a large current gain. For a high impedance input like a pedal you need more voltage.

But what I'm wondering is, say I have an amp, and I can set the voltage and current gain seperately and independantly. If I raised the current but leave the voltage as-is, would the volume increase? If yes, why? The same story with voltage. :?

What I'm getting at is, what do current and voltage changes do to a signal? Hmm that's not the best way to ask that but I don't know how to explain what I want to know :lol:

[Transmogrifox]

But what I'm wondering is, say I have an amp, and I can set the voltage and current gain seperately and independantly. If I raised the current but leave the voltage as-is, would the volume increase? If yes, why? The same story with voltage.  

"Volume" refers to the power delivered to a speaker, therefore, volume depends on current AND voltage.  The deal here is that a speaker has a set impedance, or "resistance" of 4 to 16 ohms.

Let's suppose we want to drive a 4 ohm speaker with our amplifier.

Power, P = V^2/R = I^2*R = IV

If you consider V = IR, you can algebraically show all of the above power identities algebraically equivalent.

Ok, back to the 4 ohm speaker.  Suppose we want to drive it at 10 Vrms.  With V=IR, the RMS current required to maintain 10 Volts RMS on a 4 ohm impedance is 10/4 = 2.5 Amps (RMS).  

What is the "volume" of this?  Well the POWER, P = IV = 25 Watts RMS.

This is like having a 100 Watt guitar amplifier operating at a quarter of its maximum volume.

What if we want  it louder?  We can amplify the current with a voltage controlled current source (Vacuum Tube).  So, suppose we want to get the full 100 Watts RMS out of our 100 watt amp.  We cannot change the impedance of the speaker, so when we amplify the current, a higher voltage results on the speaker....so here we go:

100 W = IV  (full volume)
V= IR, and R=4 ohms
==> 100 W = I *(I*R) = I^2 *R
I = squrt(100/4) = squrt(25) = 5 Amps RMS

So if we are driving a 4 ohm speaker at 5 amps RMS, what does the voltage need to be?
V = IR  (I = 5, R=4)
V = 4*5 = 20 Volts.

Let's check now.  We want 100W RMS = V*I , and V=20 Volts, I = 5 Amps, so

5*20 = 100

Is it starting to get more clear now?

So more volume is more current AND voltage.  For guitar amps, the total current and voltage relationship for more power depends on the impedance of a speaker.  

A 16 ohm speaker requires larger Voltage and smaller Current to deliver 100 W.  A 4 ohm speaker requires a lot of current, but less voltage.

If I raised the current but leave the voltage as-is, would the volume increase? If yes, why? The same story with voltage.  

So the short answer to this question is yes>>>but you must lower the impedance to get more current with the same voltage....And vice versa is true >>Raising voltage, keeping the current the same increases volume.

In your guitar amp, you effectively keep the impedance the same, and change the voltage AND current together.

You can think of it as "I need "X" watts of power for my amp to be as loud as I want, I have a "Z" impedance speaker, so I need "I" amps of current to get this power.  "V" voltage will result if I put "I" current into "Z" impedance.  ~Therefore, my current amplifier must not clip (encounter limiting) when the voltage is "V" Volts.~

Or you can think of it as "I need "X" watts of power for my amp to be as loud as I want, I have a "Z" impedance speaker, so I need "V" voltage on this "Z" impedance to get this much power.  "I" current will result if I apply a "V" voltage to "Z" impedance. ~Therefore, my Voltage amplifier must be able to supply at least "I" current.~


Usually, the guitar amp is a combination of a voltage amplifier (preamp) and a current amplifier (power amp).  If you need "V" volts to get "X" watts of power, you pre-amplify the guitar signal to "V" volts, then apply this to a current amplifier with a voltage gain of 1, or you may preamplify it to a voltage that results in the desired current on a voltage-controlled current source, which is determined by the Current/voltage gain of the amplifier.

Think about these four types of amplifiers, and see if they make sense when given the above:

Amplifier Topology                            Units of gain

1. Current controlled current amplifier   Units [Amps/Amp]
2. Voltage controlled current amplifier   Units [Amps/Volt]

3. Voltage controlled voltage amplifier   Units [Volts/Volt]
4. Current controlled voltage amplifier   Units [Vots/Amp]

What would you use each of these for?
A distortion pedal is generally #3
A power amplifier stage is generally # 2

[smoguzbenjamin]

Hey that sure clears it up! I was thinking along those lines but couldn't make my brain straighten it out like that. Thanks! Hey that's pretty cool. :mrgreen: Can't wait!

[Joe Davisson]

Current is electron velocity. Not quanity, or force, or anything else. I'll try to explain it better...

The velocity of sand through an hourglass is dependent on two factors: gravity, and the cross-section of the pipe.

If the hourglass only has one granule of sand, it won't get stuck in the pipe. It will fall directly down, since the pipe is much larger. (It might hit the side, but...)

When more granules enter the top than can fit in the cross-section of the pipe, they clog up and slow down. Only two things can increase the rate of flow: more gravity, or a wider pipe.

In electronics, the amount of sand is virtually infinite, because the metal wires are pre-filled with electrons. The voltage acts like gravity, pulling electrons through the resistance. More voltage increases the rate of flow, as does less resistance. It's that simple, really!

The idea that current is some sort of quanity is the product of confusion. Voltage is force. Current is velocity. And the mass to be concerned with is how many columbs fit in the "cross-section of the pipe".

This shows the squared attraction force of voltage:
Watts = voltage^2 / (seconds / coulombs)

So, 10^2 volts / (1 second / .5 coulombs) = 50 watts (joules per second)
(The resistance and velocity are "hidden", and the relationships between attraction force, time, and mass that produce energy are shown. For example, increasing the time bracket to 2 seconds nets 25 watts total.

-Joe