Small Diode Emitter Resistance,RE

Thread Starter


Joined 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


Joined 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

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

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

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

\(r_e=\frac{26}{\normalsize{I_E[mA]}} \ ohms\)
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Thread Starter


Joined 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!