BJT transconductance

The BJT transconductance $g_m$ is how strongly the collector current responds to a small change in base–emitter voltage about the DC bias point:

$g_m={\frac{\partial i_C}{\partial v_{BE}}|}_{\text{bias}}=\frac{I_C}{V_T}$

where $I_C$ is the DC collector current and $V_T\approx 25\text{mV}$ is the Thermal voltage. It is the parameter that turns the BJT into an amplifier: an input voltage becomes an output current.

Derivation by linearisation

Bias the BJT in active mode at $(V_{BE},I_C)$, then add a small signal $v_{be}$ on top: $v_{BE}=V_{BE}+v_{be}$. Substitute into the exponential law:

$i_C=I_S{e}^{(V_{BE}+v_{be})/V_T}=\underset{=I_C}{\underbrace{I_S{e}^{V_{BE}/V_T}}}{e}^{v_{be}/V_T}=I_C{e}^{v_{be}/V_T}$

The DC bias factor collapses neatly to $I_C$. Now expand the remaining exponential for small $v_{be}\ll V_T$ (Taylor series $e^x=1+x+\frac{x^2}{2}+\dots$):

$i_C\approx I_C\left(1+\frac{v_{be}}{V_T}+\frac{1}{2}{\left(\frac{v_{be}}{V_T}\right)}^2+\dots \right)$

• The constant term $I_C$ is just the DC current — it carries no signal.
• The linear term is the small-signal collector current:

$i_c=I_C\frac{v_{be}}{V_T}=\frac{I_C}{V_T}{v}_{be}=g_m{v}_{be}\Rightarrow \boxed{g_m=\frac{I_C}{V_T}}$

• The quadratic and higher terms are distortion; keeping $v_{be}$ small (small-signal operation) keeps them negligible. This is exactly the same linearisation logic used for the MOSFET in Common-source amplifier.

Transconductance of the BJT amplifier: $g_m=I_C/V_T=40\text{mA/V}$ at $I_C=1\text{mA}$ with $V_T=25\text{mV}$ — higher than a MOSFET at the same bias current.

Why it beats the MOSFET

At $I_C=1\text{mA}$ and $V_T=25\text{mV}$:

$g_m=\frac{1\text{mA}}{25\text{mV}}=40\text{mA/V}$

Compare a MOSFET carrying the same $I_D=1\text{mA}$ biased at an overdrive $V_{OV}=0.2\text{V}$. Its MOSFET transconductance is $g_m=2I_D/V_{OV}=2(1\text{mA})/0.2\text{V}=10\text{mA/V}$. The BJT delivers four times the transconductance for the same bias current. The root cause is structural: $g_m=I_C/V_T$ depends only on the thermal voltage (a constant ≈ 25 mV), whereas the MOSFET’s $g_m=2I_D/V_{OV}$ is throttled by an overdrive that is usually hundreds of millivolts. The BJT’s exponential i–v law is simply steeper than the MOSFET square law. This high transconductance-per-current is the single biggest reason BJTs are still chosen for high-gain, low-noise analog front-ends despite CMOS dominating everything digital.

$g_m$ is the central element of the BJT small-signal model; combined with $\beta$ it sets the BJT input resistance $r_{\pi }=\beta /g_m$, and it directly sets amplifier gain — e.g. a Common-emitter amplifier has $A_v=-g_m{R}_C$.