We offer three ways to understand the notion of "intrinsic emitter resistance":

• One way is to rely on a graphic definition:
  
  Look at r_e as just the Ic vs Vbe curve, turned on its side:
  
  
  
• Another way is to rely on a bit of calculus, to calculate d(Vbe)/d(Ie);
  
  The transistor's gain is dI_C/dV_BE. But if we write Ebers-Moll the way we did above, as I_C = (approx.) I_S e^V_BE/25mV then d (I_C) / d(V_BE) = (1/25 mV) (I_S e^V_BE/25mV), or, more simply,(1/ 25mV) (I_C). And r_e, the reciprocal of gain, is 25 mV / I_C A, which we prefer to write as 25ohms / I_C (in mA).
  
  
• ...and a third way is to try a wild metaphor:
  
  Alternative `derivation:' Tess o' Bipolarville
  
  Here is an alternative argument to the same result: imagine a lovely milkmaid seated (in the summer twilight) on a stool, tugging dreamily at the emitter of a transistor whose base is fixed. She has pulled gently, until about 1 mA flows. What delta-current falls into her milkpail, for an additional tug?