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Guitar signal

Discussion in 'Electronic Basics' started by Music Man, Mar 17, 2005.

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  1. Music Man

    Music Man Guest

    Seeing the guitar pickup is a magnet,it produces an A/C current.Yes?

    Does the Preamp input turn it into a DC current?

    How does the input deal with the voltage supplied,and does the impedance
    change
    when the signal declines.

    Also in a guitar preamp why would you earth if the amp had earth
    problem,wouldn't
    it go to the earth plate in the guitar,or is that a new thing.That is ,did
    old electric guitars have no earth?

    Thanks
     
  2. Yes. And AC voltage, also.
    No. It increases the amplitude of the AC voltage.
    The impedances are fixed. The output voltage is proportional to the
    input voltage, just larger. The power amplifier makes the AC voltage
    even bigger and that AC drives the speakers.
    The guitar gets whatever ground from the input jack on the amplifier.
    This is a zero volt reference that the input signal is measured
    against, but it doesn't have to actually be a connection to Earth.
    The main value of having the amplifier actually connected to Earth is
    that this means there is no difference in voltage between the cases of
    the pickups (that act as shields around the coils) and Earth. This
    helps reduce hum pickup because there is less capacitive current
    traveling up and down the guitar cable shield. But having a good
    ground connection also means that you will feel a shock if you are
    touching the metal on the guitar and also touch something else that
    has voltage with respect to Earth on it. So many tube amplifiers have
    a floating guitar common (outside barrel of the cable plug).
     
  3. Yes. The vibrating string in a magnetic field changes the induced current
    in a coil.
    No, that would make it impossible to hear. The guitar preamp has two
    important functions. To amplify the signal, and to convert the high
    impedance to a lower impedance. That simply means making it stronger,
    both voltage-wise and current-wise.
    The impedance of the guitar pickup is always high, no matter how strong
    the signal gets.
    Not sure if I lost you there, but the guitar pickup has two connections,
    two ends of the coil which is wound around one or six permanent magnets.
    One end is usually connected to the metal parts of the pickup and should
    be used as the ground side of the pickup. Connect it to other grounded
    metal parts in the guitar. This reduces noise from outside.

    Sometimes guitarists want to connect one pickup backwards and mix it with
    another pickup, to create more harmonics. Then they ignore this
    precaution and use the ground side of one pickup as the signal side and
    vice versa. So it is not necessary to connect a guitar pickup correctly,
    it is just the preferred way to do it.


    Here is a very popular guitar effects box, the Ibanez Tube Screamer.

    http://www.generalguitargadgets.com/diagrams/ts-808_sc.gif

    Look at the transistor Q1, it is the preamp which converts the guitar
    signal from high to low impedance. In this case the next stage is an op
    amp so no amplification is needed in the first stage.

    Here is another input stage which both fixes the impedance and gives
    amplification, in the Q1 transistor.

    http://www.generalguitargadgets.com/diagrams/bmpsc.gif

    Here is a web page with a lot of information about guitar preamp circuits
    built with transistors of different types:

    http://www.muzique.com/lab/lowvolt.htm

    In all these links you can go up one or more directories and find a lot
    more guitar electronics information and schematics.
     
  4. Active8

    Active8 Guest

    At the risk of sounding pedantic, it's a *voltage* that is induced
    by a time varying mag field.
     
  5. At the risk of sounding even more pedantic.. :)

    The electromagnetic influence causes an electromagnetic force which is
    both a voltage and a current in a conductor. And magnetism around it.

    It is scientifically incorrect to see voltage as a cause of current or
    vise versa, but we often do both to simplify understanding.

    Think of the ignition in a car. A voltage from the battery "causes" a
    current in the ignition coil. The power supply is suddenly cut off, the
    current continues to flow through the coil, that is the nature of coils.
    This current "causes" a very high voltage to appear at the ignition plug
    and results in a spark.

    ...and things of that nature.. (Schwarzenegger)
     
  6. Active8

    Active8 Guest

    I was referring to Faradays law of induction but EMF is the correct
    term, not voltage. It is a simplification. Good thing I reread the
    above, because I read "electromagnetic force" as "electromotive
    force" and thought, "wait!"
    Because it's the electric field that causes the conduction electrons
    to move.
    You wouldn't say that if you knew how much auto crap I've dealt with
    lately ;)
    Yeah, the current can't change from on to off instantly.
    The coil does what it has to do resist the sudden change in current.
    More like the charges keep moving to one end where they're so dense
    in comparison to the other end it effects a great potential.
    Poisson.
    Despite the wording of Faraday's law, that EMF could not occur if
    charge weren't moved (current) to one end of the coil. It's when I
    hear "induced" that the old boy comes to mind.

    Hey. Why is it that I recall hearing of you or some other Johansson
    outside of this group? It's would have been many years ago, but it
    rings a bell. Are you "famous" or is it just that I haven't seen you
    in SED for a while?
     
  7. Not really, but the point is a bit subtle. The ultimate cause of *all*
    of E&M is "charge", sort of. "Charge" directly produces the electric
    field, to which voltage is associated with. Current is the flow of
    charge. It is the electric field by (static) charge that actually causes
    the flow of this charge. That is, the *accelerating* electric field
    produces a force that accelerates charges, i.e. Electric field (voltage)
    is causing current. The *motion* of charges is really secondary.

    "Charge" is in quotes because, charge itself is nothing more than a
    number describing the exchange of photon momentum between "charges".
    Ultimately, *any* motion can only be instigated by other motion, i.e.
    exchange of motion from one entity to another, so in this sense it is
    always motion causing other motion, i.e. Newton's conservation of
    momentum. However, the motion that is being exchanged in E&M to do this,
    is not the motion associated with the "current" itself, so in this sense
    current itself doesn't cause anything.



    Kevin Aylward

    http://www.anasoft.co.uk
    SuperSpice, a very affordable Mixed-Mode
    Windows Simulator with Schematic Capture,
    Waveform Display, FFT's and Filter Design.
     
  8. I have been around here and in SED off and on for a few years, since
    2001. And elsewhere in usenet since 1996/97, in alt.philosophy, for
    example. Before that I was active in the global amateur computer network
    Fido from 1990-91.

    The first time I came to SED I recognized several author names from books
    and articles in electronics magazines like Wireless World, Popular
    Electronics, Everyday Electronics, Elektor, etc..which I had read since
    the middle of the 60-ies when I started studying electronics.

    Nobody takes me seriously, but there aren't much people around with no
    bodies.
     
  9. The Ultimate Cause of Everything There Is is Desire:
    http://www.godchannel.com/spirit.html

    Cheers!
    Rich

    for further information, please visit http://www.godchannel.com
     
  10. Active8

    Active8 Guest

    I like it, but when you say "accelerating E field", what do you
    mean? IOW, if a stationary or static charge causes another charge
    (charge 2) to accelerate toward it, I can understand that the field
    associated with charge 2 as accelerating, but not the field of the
    original static charge. Maybe you're referring to the net E field?
     
  11. All I am doing here is trying to highlight that the conventional
    terminology is to use the word "accelerating". That is, its is
    recognised that a electric field generates a *force* and a force will
    *accelerate* an *object* as per Newton.
    I am referring to the actual *object* that is moving, not its field. The
    object is accelerating, hence the name accelerating voltage or
    accelerating electric field. It isn't referring to the field, only that
    fields accelerate objects.


    Kevin Aylward

    http://www.anasoft.co.uk
    SuperSpice, a very affordable Mixed-Mode
    Windows Simulator with Schematic Capture,
    Waveform Display, FFT's and Filter Design.
     
  12. Jon Danniken

    Jon Danniken Guest

    I always thought that was due to the collapse of the magnetic field back
    onto the coil, creating a potential between the two ends of the coil.

    Jon
     
  13. Maybe that is a way to see it, but the cause is rather that the magnetic
    field is _not_ collapsing immediately when the power supply is cut off.
    Instead the magnetic field continues to exist, generating more current to
    flow, which results in high voltage in the spark plug.

    A coil is like a pump without a motor, it is started by pushing current
    through it, and it continues to pump current automatically when the power
    supply has been removed. It slows down because of resistance, like a
    motor slows down because of friction.
     
  14. Active8

    Active8 Guest

    doh!<blush>
     
  15. Active8

    Active8 Guest

    I used to think of it as inertia since the mag field lines or flux
    is drawn as circles. Like a wheel that has to decelerate, thus
    giving rise to reactance or resistance to change. But the flaw is
    that you'd then have to think of the current also having to
    decelerate as if the charge carriers (electrons) really zipped
    along, but they don't. It's the charge that propagates quickly while
    the electron drift velocity is snail paced.
     
  16. Fred Abse

    Fred Abse Guest

    No, it doesn't, the current stops (more or less) instantly. In this
    example, there's no capacitor across the contact breaker, to simplify
    things.
    No, it works like this, in inductors, the relationship between voltage and
    current is:

    V = -L di/dt

    If current is cut off suddenly, di/dt is very high. in a perfect world,
    infinite, but we're not in a perfect world, so there is finite decay time.

    L is fixed, hence V is also very high.
     
  17. Active8

    Active8 Guest

    Sorry, V = L di/dt di/dt is negative, decreasing and sets up a
    voltage opposite to that which was present before.

    You can also look at it as a collapsing mag field (changing flux)
    which induces an EMF. But is d_phi/dt dependant on di/dt or is there
    really some inertial thing going on with one or the other or both?
    Either one is just a buncha photon spin action, eh?

    Kevin!!!

    And the current very badly wants to *not* turn off instantly. It's
    one of the premises used to determine initial conditions for a
    transient analysis. And a cap doesn't want to change voltage
    instantly.
    Hey this is cool! We're OT and still talking electronics.
     
  18. Our general terminology in E&M is very useful but not how things
    "really" "are". Of course we all use concepts such as, changing magnetic
    fields "causing" electric fields an other such heuristic descriptions,
    but one should always keep in mind that it is just macro descriptions of
    things we don't even know the "true" operation of.



    Kevin Aylward

    http://www.anasoft.co.uk
    SuperSpice, a very affordable Mixed-Mode
    Windows Simulator with Schematic Capture,
    Waveform Display, FFT's and Filter Design.
     
  19. Fred Abse

    Fred Abse Guest

    No, the direction of the induced EMF is such as to oppose the change of
    current that produced it, hence:

    V = -L di/dt

    As you say, di/dt is negative, hence the induced voltage is positive,
     
  20. Fred Abse

    Fred Abse Guest

    Does it matter?

    If it matters, does it matter that it matters?

    :)
     
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