Transconductance Chart

Discussion in 'General Electronics Chat' started by KCHARROIS, Apr 7, 2013.

  1. KCHARROIS

    Thread Starter Member

    Jun 29, 2012
    292
    1
    Hello,

    I cant make sense of this chart below about transconductance, any clue?

    Thanks
     
    Last edited: Apr 7, 2013
  2. patricktoday

    Member

    Feb 12, 2013
    157
    42
    gm is transconductance and the mho unit tells you what to multiply by the voltage in column 1 to get the resultant output current. Capital "G", Gm, represents "large signal" and lowercase "g", gm, represents "small signal." I'm not sure about column 4. It appears to show the percentage mho that will occur at an arbitrary Icq (collector quiescent current)... I notice that each value in column 3 below the 1st cell could be derived from multiplying the 1st cell (0.038) times the "percentage" in column 4.
     
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  3. KCHARROIS

    Thread Starter Member

    Jun 29, 2012
    292
    1
    Ok but I don't get how they got results in column number 3 with just the signal peak voltage?
     
  4. patricktoday

    Member

    Feb 12, 2013
    157
    42
    What type of device or circuit is this for? You didn't provide any context.
     
  5. KCHARROIS

    Thread Starter Member

    Jun 29, 2012
    292
    1
    Its for a colpitts oscillator, I'm reading a book called build your own transistor radio and I;m reading the oscillator chapter formulas like Ic=Is*e^Vbe/0.026 were given.
     
  6. KCHARROIS

    Thread Starter Member

    Jun 29, 2012
    292
    1
    This might help...

    If mho equals (below) Ic =

    0.037mho 0.962mA
    0.034mho 0.884mA
    0.036mho 0.676mA

    because Ic = mho * 0.026mV
     
  7. patricktoday

    Member

    Feb 12, 2013
    157
    42
    The mho calculation is the inverse of the emitter resistance, so, yes, this sounds correct. To get emitter resistance, we usually calculate 0.026/Ie if Ie is already known... so the mho calculation is 1/(0.026/Ie) (and Ic will be very close to Ie).
     
    Last edited: Apr 7, 2013
  8. KCHARROIS

    Thread Starter Member

    Jun 29, 2012
    292
    1
    Ok so I think I figured it out...

    If mho for 0.013Vp is 0.037mho then with the formula gm = Icq/0.026

    Icq = 0.962 mA//

    Now since the graph is referring to 1mA = 0.0384615mho

    (0.962mA/1mA)*100 = 96.2

    At a signal of 0.013, (0.0384615/100)*96.2 = 0.037//

    I guess the graph is referring to 1mA for everything because current changes as input signal increases.
     
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