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BJT as amplifier: gm, rπ, re

Short answer

The BJT amplifies because a small change in vBE moves the operating point along the exponential iC–vBE curve, producing a collector current change ic = gm vbe. gm = IC/VT, rπ = β/gm = β VT/IC, and re = α/gm ≈ VT/IE. Since rπ = β/gm, β = gm rπ, and rπ = (β+1) re because ie/ib = β+1.

Describe functions of BJT as amplifier with the help of transfer characteristics (ic - vBE graph), and find expressions for gm, rπ and re. Also show that β = gm rπ and rπ = (β+1) re.

BJT as an amplifier (transfer characteristic)

The transfer characteristic of the BJT is the plot of iC against vBE (exponential): iC = IS exp(vBE/VT).

 iC
  |            /
  |          _/
  |       _.-  <- slope = gm at Q
  |    _.* Q
  |__.-'
  +--+-----+-------- vBE
     0.5   0.7

The transistor is biased in the active region at Q (VBE, IC). A small signal vbe added to VBE moves the operating point along the curve and gives a collector current change ic = gm vbe. This current through RC gives an output voltage vo = -ic RC = -gm RC vbe, which is larger than the input, so the BJT amplifies. For small signals (vbe < 10 mV) the curve at Q is nearly a straight line, and the distortion is low.

gm, rπ and re

Transconductance:

gm = ∂iC/∂vBE |Q = (IS/VT) exp(VBE/VT) = IC / VT

Input resistance seen at the base: since ib = ic/β = gm vbe/β

rπ = vbe/ib = β/gm = β VT/IC

Resistance seen at the emitter: since ie = ic/α = gm vbe/α

re = vbe/ie = α/gm = VT/IE

Proofs

  1. β = gm rπ: from rπ = β/gm, gm rπ = β. ✓.
  2. rπ = (β+1) re: since ie = (β+1) ib and the same vbe,
rπ/re = (vbe/ib)/(vbe/ie) = ie/ib = β + 1
rπ = (β+1) re

✓ (also α/gm × (β+1) = β/gm since α = β/(β+1)).

✓ Answer checked by Claude Opus 5.5 (high effort)

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