Small Diode Emitter Resistance,RE

Discussion in 'General Electronics Chat' started by seikoemill89, Sep 21, 2011.

  1. seikoemill89

    Thread Starter New Member

    Sep 21, 2011
    hi everyone,here i would like to ask on how to prove this below equation
    IE=Is(qVBE/e kT -1) until we get this equation:re=26mV/IE..
    this is from the topic of BJT Biasing...hopefully someone can help me to solve this...TQ
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    For a diode with voltage drop Vd & current Id.

    By Shockley's equation

    I_d=I_s(e^{\frac{\normalsize{q} V_d}{kT}}-1)

    (e^{\frac{\normalsize{q} V_d}{kT}}-1)=\frac{I_d}{I_s}

    e^{\frac{\normalsize{q} V_d}{kT}}=\frac{I_d}{I_s}+1

    Taking the natural logarithm of both sides gives

    \frac{\normalsize{q} V_d}{kT}=ln(\frac{I_d}{I_s}+1)


    Differentiating wrt Id gives

    \frac{d(V_d)}{dI_d}=\frac{kT}{\normalsize{q}} \frac{\frac{1}{\normalsize{I_s}}}{(\frac{ \normalsize {I_d}}{ \normalsize {I_s}}+1)}


    Normally Id>>Is

    So to a good approximation


    typically at room temp

    \frac{kT}{\normalsize{q}}=0.026 \ volt



    If the term


    is recognized as effectively a small signal dynamic resistance rd


    r_d=\frac{0.026}{\normalsize{I_d}} \ ohms

    For the BJT emitter region diode junction we can state (by a similar process) the equivalent dynamic emitter resistance for a quiescent emitter current IE

    r_e=\frac{0.026}{\normalsize{I_E}} \ ohms

    Or if IE is expressed in mA

    r_e=\frac{26}{\normalsize{I_E[mA]}} \ ohms
    Last edited: Sep 21, 2011
    anhnha likes this.
  3. seikoemill89

    Thread Starter New Member

    Sep 21, 2011
    thank you so much for t_n_k...the answer is so meaningful for me because i am a fresh degree student,so i have forgotten what i have study before in diploma...a lot of thanks for you...peace!