In short: Linearise the MOSFET around its DC bias point and it becomes a voltage-controlled current source gm·vgs (gm = 2·ID/VOV) in parallel with an output resistance ro = 1/(λ·ID), and the T-model is the same device redrawn with a 1/gm resistor on the source side.
Why we need a small-signal model at all
A saturated MOSFET obeys the square law, ID = ½·μnCox·(W/L)·(VGS − Vth)²·(1 + λVDS). That is a nonlinear equation, and nobody solves nonlinear equations by hand while sizing a differential pair at 2 a.m. The trick every analog designer uses is to split every voltage and current into a fixed bias part and a tiny AC part, then keep only the first-order term of the Taylor expansion. The result is a linear circuit made of resistors and controlled sources, which you can analyse with plain Kirchhoff's laws. If you have not yet seen how the bias point is chosen, read our walkthrough MOSFET as an amplifier first.
Key fact: "Small" means the AC swing is small enough that the quadratic term is negligible. A common rule of thumb is vgs below about 0.1·VOV, which keeps second-harmonic distortion near 5 percent or lower.
Deriving gm from the square law
Write vGS = VGS + vgs and iD = ID + id. The transconductance is the slope of the transfer curve at the bias point:
gm = ∂iD/∂vGS = μnCox·(W/L)·VOV = 2·ID/VOV = √(2·μnCox·(W/L)·ID)
The three forms are all useful. The first tells you gm grows linearly with overdrive at fixed W/L. The second is the one you quote in interviews: for a given current, a lower overdrive buys more gm, which is why we call gm/ID the efficiency metric. The third shows that gm only grows as the square root of current, so doubling the bias current buys you just 41 percent more gain.
ro and the channel-length modulation story
In saturation the drain current is not perfectly flat. A higher VDS shortens the effective channel and iD creeps up. Differentiating iD with respect to vDS gives the output conductance gds = λ·ID/(1 + λVDS) ≈ λ·ID, so ro = 1/(λ·ID) = VA/ID, where VA = 1/λ is the Early voltage. With VA = 10 V and ID = 100 µA you get ro = 100 kΩ.
Note the scaling: λ is roughly inversely proportional to channel length. A 180 nm device drawn at minimum L has a VA of maybe 5 V, while the same process at 4×Lmin might reach 20 V. That is the whole reason analog designers spend extra area on long channels for current mirrors and gain stages.
Key fact: The intrinsic gain gm·ro = 2·VA/VOV is the most you can ever get from a single transistor. With VA = 10 V and VOV = 0.22 V, that is about 90 V/V, or 39 dB.
Assembling the hybrid-pi model
At low frequency the gate is an open circuit (oxide insulates it), so the gate-source branch has infinite resistance. The drain-source branch has the controlled current source gm·vgs in parallel with ro. That is the entire hybrid-pi model, and it fits on a napkin.
Body effect: the third parameter
If the source is not tied to the bulk, the bulk acts as a second gate. Add a second current source gmb·vbs in parallel with gm·vgs, where gmb = χ·gm and χ = γ/(2·√(2φF + VSB)) is typically 0.1 to 0.3. In a source-follower or a cascode, ignoring gmb is a classic reason hand calculations miss simulation by 15 to 20 percent.
The T-model: same device, different bookkeeping
The T-model moves the controlled source so it feeds a resistor of value 1/gm that sits between gate node and source. Check KCL: the source delivers gm·vgs from drain into the gate node, the resistor 1/gm carries exactly that same current to the source, and the true gate terminal draws nothing. The two models are electrically identical at the terminals, but the T form makes one calculation almost trivial: the resistance looking into the source of a MOSFET is 1/gm (when the gate is held at AC ground and ro is large).
Worked example with real numbers
Take an NMOS with μnCox = 200 µA/V², W/L = 20/1, VA = 10 V, biased at ID = 100 µA.
- VOV = √(2·ID / (μnCox·W/L)) = √(200 µ / 4 m) = 0.224 V
- gm = 2·ID/VOV = 0.89 mS, so 1/gm ≈ 1.12 kΩ
- ro = VA/ID = 100 kΩ, giving gm·ro ≈ 89
- Common-source stage with RD = 20 kΩ: Av = −gm·(RD ∥ ro) = −0.89 m × 16.7 k ≈ −14.9 V/V
| Parameter | Formula | Value above |
|---|---|---|
| gm | 2·ID/VOV | 0.89 mS |
| ro | VA/ID | 100 kΩ |
| 1/gm | VOV/(2·ID) | 1.12 kΩ |
| gm·ro | 2·VA/VOV | ≈ 89 |
In Cadence Spectre or Synopsys HSPICE, run a .op analysis and read gm, gds and gmb straight from the operating-point printout. They should match your hand numbers within the accuracy of the level-1 assumptions. Larger gaps point to velocity saturation or short-channel effects, which is when designers move from square-law hand analysis to gm/ID lookup tables. Our teams at Avecas design services use exactly this hand-versus-simulator check on every analog block.
Where the model breaks, and what to add
At high frequency, add Cgs (roughly ⅔·W·L·Cox in saturation), Cgd and the junction capacitances Cdb and Csb. The unity-gain frequency becomes fT ≈ gm/(2π·(Cgs + Cgd)). Gate thermal and drain thermal noise appear as a current 4kTγ·gm in parallel with the source. For more on how amplifier topologies reuse this model, see common-source amplifier gain.
Key fact: Interviewers love the trap "what is gm of a MOSFET in triode?" It is μnCox·(W/L)·VDS, which is independent of VGS, a very different answer from saturation.
Common interview traps
- Forgetting that gm changes with bias: a transistor at 10 µA and 100 µA has gm differing by 3.2×, not 10×.
- Using VDD as VA. Early voltage is a device parameter, not a supply.
- Ignoring ro when the load is a current source. Then the gain is simply gm·ro.
- Forgetting that a PMOS has the same small-signal model; only the DC polarity differs.
If you want to practise these derivations under guided exercises and real Spectre labs, our CourseTron analog courses take you through them. For hands-on training with live instructors, ChipXpert VLSI training is our sister brand.
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FAQ
What is the difference between the hybrid-pi and T-model?
They are equivalent at the terminals. Hybrid-pi uses a controlled source from drain to source with an open gate; the T-model puts a 1/gm resistor between gate node and source, which simplifies calculations of resistance seen at the source.
Why does gm increase with the square root of current?
Because gm = √(2·μnCox·(W/L)·ID) in strong inversion. Deep in weak inversion it becomes gm = ID/(n·VT), which is linear in current.
Is ro always large enough to ignore?
No. In a gain stage loaded by an active current source, ro sets the gain. It can only be ignored when a much smaller resistor shunts it.
Does the small-signal model include DC values?
No. All DC supplies become AC ground and all DC current sources become open circuits. DC only sets the numerical values of gm and ro.