03 · Intermediate

Transconductance & control variables

Turn each transistor’s nonlinear transfer law into a local control strength at its DC operating point. Define gm as a partial derivative with the output voltage held fixed, derive gm=IC/VT for the exponential BJT model, derive the JFET gm relation from the Shockley transfer curve with VP<0, and derive gm=kVOV=2ID/VOV for the long-channel saturated MOSFET. Compare gm/ID, distinguish intrinsic device gm from the effective transconductance seen through source/emitter degeneration, and quantify when a local linear approximation stops being accurate.

190–245 minIntermediate

In this lesson

Define transconductance dimensionally as gm=(∂Iout/∂Vcontrol)|Q with the other independent terminal voltage held fixed.
Derive BJT gm=ICQ/VT from the forward-active exponential law and explain why β is not needed for this intrinsic derivative.
Derive n-JFET gm=−(2IDSS/VP)(1−VGS/VP) for the declared VP<0 transfer model and rewrite it in terms of ID/IDSS.
Derive long-channel saturated nMOS gm=kVOV=2ID/VOV=sqrt(2kID).
Distinguish MOS triode-region ∂ID/∂VGS=kVDS from saturation-region gm and explain why gm is operating-region dependent.
Compare the bias-point gm values of the BJT, JFET and MOS worked examples from chapter 02 with correct SI units.
Define transconductance efficiency gm/ID and derive 1/VT for the ideal exponential BJT and 2/VOV for the long-channel saturated MOS model.
Distinguish intrinsic device gm from an externally observed current/control-voltage slope after source or emitter degeneration.
Derive gm,ext≈gm/(1+gmRS) for a FET with unbypassed source resistance under the stated small-signal assumptions.
Use second derivatives/Taylor terms to explain why gm is local and why larger signal excursions create distortion even when the DC operating region does not change.

Starting question

Bias fixes where a transistor operates. Transconductance tells us how strongly a small change of the control voltage changes the main current at that particular point.

03.1gm is a partial derivative with units of siemens

Figure 3.1 — gm is the local slope at Q

The one question is: what does the tangent to the transfer curve measure?

The tangent slope depends on the bias point. A transistor family therefore does not have one gm independent of Q.

g_m=[∂I_out/∂V_control]_{Q, other terminal voltage fixed}

[g_m]=A/V=S

For the BJT we normally mean ∂IC/∂VBE at fixed VCE within the chosen active model. For JFET/MOSFET we normally mean ∂ID/∂VGS at fixed VDS within the chosen region.

The “other voltage fixed” clause matters: output-voltage dependence is a different derivative, later represented by output conductance.

03.2BJT gm follows directly from the exponential law

Using the forward-active exponential model

I_C=I_S,BJTexp(V_BE/V_T)

differentiate at Q:

g_m,BJT=∂I_C/∂V_BE=I_CQ/V_T

β is absent from this intrinsic derivative. β matters for base current and input resistance, but the ideal collector-current sensitivity to VBE follows from the exponential transport law.

At 300 K and ICQ≈1.90 mA from chapter 02,

g_m≈1.90 mA/25.85 mV≈73.5 mS

03.3JFET gm from the transfer curve

With the chapter-02 convention VP<0 and

I_D=I_DSS(1−V_GS/V_P)²

differentiate:

g_m,JFET=−(2I_DSS/V_P)(1−V_GS/V_P)

Because VP<0 and the bracket is nonnegative in the valid transfer interval, gm is positive. Define

g_m0=−2I_DSS/V_P=2I_DSS/|V_P|

g_m=g_m0sqrt(I_D/I_DSS)

For IDSS=8 mA, VP=−4 V, VGS=−2 V and ID=2 mA, gm0=4 mS and gm=2 mS.

03.4MOS gm depends on operating region

For long-channel saturation without channel-length modulation,

I_D=(k/2)V_OV², V_OV=V_GS−V_TH

therefore

g_m,sat=kV_OV=2I_D/V_OV=sqrt(2kI_D)

For the chapter-02 point k=1 mA/V² and VOV=2 V, gm=2 mS.

In the long-channel triode equation

I_D=k[V_OVV_DS−V_DS²/2]

holding VDS fixed gives

g_m,triode=kV_DS

So “the MOSFET gm” is incomplete without a bias point and region.

03.5Compare gm at the three worked bias points

Device/modelBias pointgm

BJT exponentialIC≈1.90 mA, 300 K≈73.5 mS

JFET transfer lawID=2 mA, IDSS=8 mA, VP=−4 V2 mS

long-channel nMOS sat.ID=2 mA, VOV=2 V2 mS

Equal DC current does not imply equal gm. The BJT exponential law gives a much steeper local current-versus-control-voltage slope at this particular comparison point.

That observation alone does not rank complete amplifiers: input current, noise, headroom, capacitance, output resistance, power and technology constraints also matter.

03.6gm/ID is a transconductance-efficiency metric

Normalize local control strength by DC current:

transconductance efficiency = g_m/I_D/C

For the ideal BJT exponential model,

g_m/I_C=1/V_T

At 300 K this is about 38.7 V⁻¹. For the long-channel MOS saturation square law,

g_m/I_D=2/V_OV

At VOV=2 V the example gives 1 V⁻¹. This is a model-level comparison; modern MOS design spans weak/moderate/strong inversion where the square-law expression does not cover the full gm/ID behavior.

03.7Intrinsic gm is not always the gm seen from the external input node

Figure 3.2 — Source degeneration reduces externally observed transconductance

The one question is: why does source resistance turn current change into negative feedback?

An increase in id raises vs=idRS, which reduces vgs=vg−vs and therefore opposes the original current increase.

For a FET small-signal model with drain-current increment id=gm vgs, negligible gate current and source resistance RS, let the external input be vg:

v_gs=v_g−v_s

v_s≈i_dR_S

Then

i_d=g_m(v_g−i_dR_S)

g_m,ext=i_d/v_g≈g_m/(1+g_mR_S)

The transistor’s intrinsic gm did not change; the circuit added negative feedback.

For a BJT with unbypassed emitter resistance and finite β, an analogous first model gives approximately gm,ext=gm/[1+gm(1+1/β)RE] for collector current versus base-voltage increment under the stated hybrid-π assumptions.

03.8gm is local because nonlinear curvature remains

For the BJT exponential law, the incremental collector current contains

i_c≈g_mv_be+[g_m/(2V_T)]v_be²+...

so the quadratic/linear magnitude ratio is approximately |vbe|/(2VT).

For the MOS square-law saturation model,

i_d≈kV_OVv_gs+(k/2)v_gs²

and the corresponding ratio is |vgs|/(2VOV). Local linearization therefore has an amplitude-dependent error even if the device stays in the same DC region throughout the waveform.

03.9Control variable and output dependence are separate derivatives

A realistic transistor current generally depends on both a control voltage and an output voltage:

ΔI_out≈g_mΔV_control+g_oΔV_out

gm is the control derivative; go is the output derivative. Treating one as the other causes incorrect gain and output-resistance predictions. Chapter 04 will place both terms into complete small-signal equivalent circuits.

03.10Mastery exercises

At 300 K, find BJT gm for IC=0.5 mA.
For IDSS=12 mA, VP=−6 V and VGS=−3 V, find JFET gm.
For long-channel MOS saturation with k=2 mA/V² and VOV=0.5 V, find ID and gm.
For the same MOS bias, find gm/ID.
A FET has gm=5 mS and RS=200 Ω. Estimate gm,ext.
Why is gm=∂ID/∂VGS incomplete unless VDS and region are specified?
For a BJT at 300 K, what |vbe| makes the quadratic Taylor term about 10% of the linear term?
For a MOS square law with VOV=0.2 V, what |vgs| gives the same 10% ratio?

Solutions

1. 0.5 mA/25.85 mV≈19.3 mS.

2. gm0=2IDSS/|VP|=4 mS; at VGS=−3 V the factor is 0.5, so gm=2 mS.

3. ID=(2/2)(0.5²) mA=0.25 mA; gm=kVOV=1 mS.

4. gm/ID=4 V⁻¹.

5. 5 mS/(1+0.005·200)=2.5 mS.

6. Because the local current slope can depend on both output bias and operating-region equation.

7. |vbe|≈0.2VT≈5.17 mV.

8. |vgs|≈0.2VOV=40 mV.

Chapter 04 will turn these local derivatives into complete BJT/JFET/MOS small-signal equivalent circuits, including input and output paths.