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Emitter current gain?

  1. Mar 16, 2012 #1
    I am going through some notes on a very simple BJT circuit setup as a voltage follower. It is simply an NPN with its collector connected to a voltage source, its base connected to a signal, and its emitter connected to a load resistor to ground.

    A schematic is attached.

    Now he setup a list of equations for this circuit:

    1. [itex]i_{b} = \frac{V_{b}-V_{e}}{h_{ie}}[/itex]

    2. [itex]i_{c} = h_{fe}i_{b}[/itex]

    3. [itex]i_{e} = i_{c} + i_{b}[/itex]

    4. [itex]V_{e} = i_{e}R_{L}[/itex]

    I am familiar with equations 2-4. I am not so sure about equation 1. I am not familiar with the term [itex]h_{ie}[/itex] nor the relationship.

    I can infer that the [itex]h_{ie}[/itex] term must have units of ohms, so it is different from the beta gain term I am used to seeing.

    Can anyone explain what this term is and where the relationship comes from?
     

    Attached Files:

  2. jcsd
  3. Mar 16, 2012 #2
    Ok, I worked the transfer function out and then looking in my electronics book, I see that the term matches up with the [itex]r_{\pi}[/itex] term in the Hybrid-pi model of a transistor, which is equal to [itex]\frac{h_{fe}}{g_{m}}[/itex] which is the transconductance.

    Now this is something that has always confused me as a disconnect between my textbooks and transistor datasheets . . along with other paremeters used extensively in my books, there is no term called transconductance in datasheets, and according to my textbook, this can vary as the collector current varies . . so how could I ever design a real gain that I want with a real transitor that doesn't give me this parameter?
     
    Last edited: Mar 16, 2012
  4. Mar 16, 2012 #3

    vk6kro

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    hie is the resistance of the base emitter junction and it is normally given in data sheets.

    I have a data sheet for a 2N2222 and this value depends on the base current chosen, but it is in the range of 250 ohms to 8000 ohms. This variation is due to the B-E junction being a diode which is being turned on by a DC current.
    For calculations if you don't know this value, you can assume about 500 ohms.

    If there is no feedback, the gain of a common emitter amplifier depends on this value so it is important. If a more accurate gain is required, then you would use feedback.
    One easy way to apply feedback is to have an unbypassed emitter resistor in a CE amplifier. Gain is then approximately equal to (collector resistor / emitter resistor)

    The gain on a voltage follower is only slightly dependent on hie and it will always be slightly less than 1.
     
  5. Mar 17, 2012 #4
    Thank you for the thorough answer.

    Now, I understand that the resistor at the emitter is called "emitter degeneration" and actually creates an inherent negative feedback in the first place. When you say add feedback to get an accurate gain, do you mean other than this inherent feedback?
     
  6. Mar 17, 2012 #5

    vk6kro

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    No, that is one method of applying feedback and it is effective at establishing a predictable gain even though the transistors used may have different current gains.
     
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