# Small Diode Emitter Resistance,RE

#### seikoemill89

Joined Sep 21, 2011
3
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

#### t_n_k

Joined Mar 6, 2009
5,455
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

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

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

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

$$\frac{d(V_d)}{dI_d}=\frac{\frac{kT}{\normalsize{q}}}{\normalsize{I_d}}$$

typically at room temp

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

and

$$\frac{d(V_d)}{dI_d}=\frac{0.026}{\normalsize{I_d}}$$

If the term

$$\frac{d(V_d)}{dI_d}$$

is recognized as effectively a small signal dynamic resistance rd

then

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

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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