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#### riccardo manfrin

- Jan 1, 1970

- 0

http://pui.chorwong.com/devry/wp-content/uploads/2009/08/bjtbias1.jpg

Ignore the values reported in the image. I'm interested in understanding the input impedance seen at the base by the AC source, using the \pi model of the BJT.

For AC analysis DC sources are shutdown (become grounds), the capacitor near the emitter shortcircuits the parallel resistor, and all the other caps caps are shortcircuited hence letting AC through.

The impedance is the ratio between voltage and current, so in my case,

r_\pi = v_{BE} / i_B

Unfortunaltely v_BE and i_B are related to the exponential relationship between current and voltage of a diode.

Looking around, I found mention to formulas like

r_\pi = V_T / i_B

where V_T= kT/q, that is the thermal voltage of the BE junction which accounts for the so called "ohmic effects" of the diode model.

Unfortunately, if both the above formulas are correct it looks like V_{BE} = V_T. If this is the case, then my input source signal (V_s in the image) is not varying the V_{BE} voltage which is only driven by thermal noise. Is this a correct guess?

Can you explain me the procedure to derive the input impedance in AC, given the knowledge of the voltage V_s and its series resistance R_s.

Thanks,

RM