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Operating point importance; BJT transconductance

Short answer

The Q-point must be centered in the active region to avoid clipping and ensure faithful amplification. Transconductance gm = IC/VT, derived from iC = IS exp(vBE/VT) giving gm = ∂iC/∂vBE = IC/VT ≈ 40·IC at room temperature.

For the faithful amplification of signal, selection of operating point is utmost importance. Justify the above statement. Derive transconductance of bipolar junction transistor.

Selection of the operating point (Q-point)

The Q-point (ICQ, VCEQ) is the dc condition when no signal is applied. A signal makes the instantaneous operating point move along the load line around Q. For faithful amplification:

  • Q must be in the active region, in the middle of the load line (VCEQ ≈ VCC/2), so that the positive and negative output swings are equal and not clipped.
  • If Q is too near saturation (VCE small), the negative half-cycle of the output is clipped (saturation clipping).
  • If Q is near cut-off (IC small), the positive half-cycle is clipped (cut-off clipping).
  • If the Q-point is unstable (because of changes in β or temperature), it can drift later into a region where clipping or distortion occurs; hence stable bias circuits are used.
  • The Q-point also sets gm = ICQ/VT, rπ, the gain and the input resistance.
 IC
 IC(sat)|\
        | \  Q too high (saturation clipping)
        |  \
   ICQ  |   *Q  (best: centre)
        |    \  Q too low (cut-off clipping)
        +-----\---- VCE
             VCEQ  VCC

Transconductance of the BJT

gm = ∂iC/∂vBE at the Q-point: the small-signal change in collector current per unit change in base-emitter voltage.

In the active region, iC = IS exp(vBE/VT). With vBE = VBE + vbe and vbe ≪ VT:

iC = IC exp(vbe/VT) ≈ IC (1 + vbe/VT)
ic = (IC/VT) vbe

gm = IC / VT

At room temperature VT ≈ 25 mV, so gm = 40 IC. For IC = 1 mA, gm = 40 mA/V. Then rπ = β/gm and re = α/gm.

✓ Answer checked by Claude Opus 5.5 (high effort)

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