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Amplification in vacuum tubes

  1. Dec 7, 2008 #1
    hi, we say amplification µ=plate resis.(rp) x gm = δVp/δVg=plate voltage/grid voltage
    but im confused where is the amplification taking place it is just that to maintain the same plate current we have to EXTERNALLY increase the plate voltage,δVp, in response to an decrease in grid voltage,δVg. Also i have one more doubt that, since we do not make the grid voltage positive w.r.t. cathode so amplification cannot take place as the presence of -ve voltage on grid will only restrict the no. of electrons reaching the plate.
     
  2. jcsd
  3. Dec 8, 2008 #2

    dlgoff

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    Triode tubes amplify a signal placed on the grid that is biased above the cathode. (so you are correct that -v on the grid will restrict electron flow) The plate potential should be above the grid bias. Hence you vary the plate current by varying the grid potential.

    Welcome to PF
     
  4. May 14, 2011 #3
    Given amplification µ=plate resis.(rp) x gm = δVp/δVg,
    You say that in order to maintain a constant plate current, a decrease in δVg should be responded with an externally driven increase in δVp.

    In order to maintain [tex]\mu[/tex] (amplification) one would have to maintain the ratio δVp/δVg. I think the physical idea is that to maintain the current for a given initial δVp/δVg, the difference in voltage at the plate and grid must stay the same. An increase in one would result in an increase in the other, while affecting other physical properties of the circuit.

    As for the cathode and grid relationship, I'm not 100% sure, but I think a decrease in δVg increase flow from plate to grid, but restricts flow from the grid to the cathode. If δVg < δVc, then is the "cathode" even functioning as a cathode anymore?

    Also,
    If anyone wants to talk about the physics of music recording electronics, I would have fun with that.

    -Nate
     
  5. May 15, 2011 #4

    dlgoff

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  6. May 15, 2011 #5
    Thanks! The video is cool.
     
  7. May 15, 2011 #6
    here is a review of vacuum tube operation, including some real equations:

    http://www.john-a-harper.com/tubes201/

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

    Electrons are emitted from the hot cathode via thermionic emission (Richardson's equation, Dushmanns law)

    A large positive potential on the plate will draw all the emission electrons to the plate. The grid itself does not itself intercept the electrons, but creates a space charge density (very slow electrons) that limits the plate current (Child's law). Modulating the grid voltage modulates the plate current.

    The plate voltage signal is produced by the plate current flowing through (across) the external plate circuit resistance (impedance). This can be a resistor or an LC circuit, or an output transformer primary.

    Bob S
     
  8. May 15, 2011 #7

    dlgoff

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