When designing BJT transistor amplifiers, one calculates the emitter resistance of BJT:


where the voltage is the thermal voltage at room temperature and \$I_c\$ is the bias current.

This is how it's usually done, but why do we calculate it using only the bias (DC) current? If the resistance depends on collector current, why doesn't the \$r_e\$ resistance swing up and down as the collector current swings with the input (base) current?

  • \$\begingroup\$ The emitter diode, as with other diodes, can be accurately modeled using a Taylor Series. This (bit of math) uses progressively-higher-order terms in a series summation. For very tiny changes in voltage, only the first couple of terms matter. For larger and larger voltages, approaching and exceeding that 0.026 volts, the math remains accurate because the distortion is being predicted by the size of the higher-order (higher powers of exponentiation) terms. To start your visit in the interesting room of distortion, read about "intercept points." For diodes, "intercepts" are about 1/3 volt. \$\endgroup\$ Commented Feb 20, 2019 at 5:16
  • \$\begingroup\$ Large signal excursions in active mode are approximated using the Shockley equation (roughly exponential in nature.) Such curves have slopes (\$R=\frac{\text{d}V}{\text{d}I}\$) that look linear if you zoom in close. That slope won't be the same if you are zooming in at a different place on the curve. The collector current tells you "where to look" on the exponential curve in order to find the slope at that point. The slope is \$r_e\$. So yes, the DC quiescent current gives you a starting point. If you move too far away, though, it no longer applies and needs to be recalculated again. \$\endgroup\$
    – jonk
    Commented Feb 20, 2019 at 5:52

2 Answers 2


Because the notion of emitter resistance is only meaningful in the context of the small-signal approximation, where we assume the AC component is too small to cause nonlinear effects.

In other words, because that's how the small-signal emitter resistance is defined.

  • 1
    \$\begingroup\$ Actually the small signal impedance does swing with large excursions of Ic such that when Ic~0 \$r_e\$ is very high with very low Vbe so input signals do get attenuated by the modulated input impedance unless there is Re added \$\endgroup\$ Commented Feb 19, 2019 at 22:47

Considering a Common Emitter configured transistor amplifier.

re = 25mV/Ic

For a fully bypassed emitter voltage gain, Av = Rc/re and the variation of re with collector current causes distortion because Ic varies as the output amplitude varies which varies re, varying the gain. As the output amplitude swings positive, Ic reduces increasing re and reducing the gain. As the output amplitude swings negative, Ic increases, reducing re and increasing the gain the gain. Therefore positive going output excursions are reduced in amplitude and negative going output excursions are increased in amplitude.

For a partially bypassed emitter:

Voltage gain, Av = Rc/(re + RE1) where RE1 = the unbypassed emitter resistance.

So, in this configuration the variation in re with collector current becomes less significant as re is in series with RE1.

So, the advantage of a fully bypassed emitter is higher gain with the drawback that distortion is increased.

Typical Av (No emitter bypassing) = 2 to 10. Typical Av (partially bypassed emitter) = 10 to 50. Typical Av (Fully bypassed emitter) = 50 to 500 but with increased distortion.


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