MOSFET Small-Signal Model for Circuit Analysis

The MOSFET Small-Signal Model

The MOSFET is the most-fabricated device in human history, and almost every analog block built in CMOS rests on the same two-parameter small-signal picture we develop here. In saturation a MOSFET behaves to first order as a square-law voltage-controlled current source: the drain current depends on the square of the gate overdrive. As with the bipolar transistor, that nonlinearity is intractable for hand analysis, so we linearize around a DC operating point and replace the device with a transconductance $g_m$ and an output resistance $r_o$. Unlike the BJT, an ideal long-channel MOSFET draws no DC gate current, so there is no input resistance analogous to $r_\pi$ — the gate is, to first order, an open circuit. This article derives the three equivalent forms of $g_m$, explains the output resistance and channel- length modulation, notes the body effect, and works a concrete numerical example end to end.

The Square-Law Starting Point

In saturation, neglecting channel-length modulation, the drain current of an NMOS device is

where $k = \mu_n C_{ox} (W/L)$ is the transconductance parameter and $V_{ov} = V_{GS} - V_{th}$ is the overdrive (or effective) voltage. Note that conventions vary: some texts (Sedra & Smith) write $k_n = \mu_n C_{ox}$ separately and carry $W/L$ explicitly, while Razavi often lumps them. Whatever the notation, the small-signal transconductance is the derivative of $I_D$ with respect to $V_{GS}$ at the Q-point.

Three Equivalent Faces of g_m

Differentiating the square law gives transconductance. Because the same quantity can be written in terms of overdrive, in terms of the device parameter and current, or as a ratio of current to overdrive, three algebraically identical expressions appear constantly in the literature:

Each form is useful in a different situation. Use $g_m = 2 I_D / V_{ov}$ when you know the bias current and the overdrive. Use $g_m = \sqrt{2 k I_D}$ when you know the device size and current. The middle form exposes a key contrast with the BJT: a MOSFET's transconductance grows only as the square root of bias current, whereas a BJT's grows linearly. At the same bias current a bipolar transistor almost always delivers far more transconductance, which is why BJTs dominate high-gain, low-noise input stages while MOSFETs win on integration density and zero gate current.

Output Resistance and Channel-Length Modulation

Real MOSFETs in saturation show a slight upward slope in $I_D$ versus $V_{DS}$, the analog of the Early effect, called channel-length modulation and parameterized by $\lambda$. The output resistance is

The intrinsic gain of the device is again the product $g_m r_o$. Substituting, $g_m r_o = \frac{2 I_D}{V_{ov}} \cdot \frac{1}{\lambda I_D} = \frac{2}{\lambda V_{ov}}$. Lowering the overdrive raises the intrinsic gain — but at the cost of speed and headroom. Because $\lambda$ shrinks with longer channels, designers chasing gain reach for long devices, while those chasing bandwidth use short ones.

The Body Effect

When the source and body (bulk) terminals are not tied together, the source-to-body voltage modulates the threshold voltage. This introduces a second controlled source in the small-signal model, a back-gate transconductance:

For most introductory analyses the source and body share a node, so $V_{SB} = 0$, the body current vanishes, and you may ignore $g_{mb}$ entirely. It becomes important in cascodes, source followers, and any stage where the source sits at a voltage different from the bulk, where it typically degrades gain or impedance by 10 to 30 percent.

The Complete Small-Signal Picture

Putting the pieces together, the low-frequency small-signal model of a saturated MOSFET is compact: a single dependent current source $g_m v_{gs}$ from drain to source, the output resistance $r_o$ across the same terminals, the optional body-effect source $g_{mb} v_{bs}$, and an open-circuit gate. That is the entire device for hand analysis. Compared with the bipolar hybrid-pi, the absence of a base-current path makes the MOSFET model simpler to apply: there is no $r_\pi$ to worry about, no $\beta$ to track, and no DC input current to bias around. This simplicity, multiplied across millions of devices on a chip, is a large part of why CMOS came to dominate integrated electronics.

It is important to remember the boundaries of the model. It is valid only in saturation, only for small signals, and only at low to moderate frequencies where the device capacitances can be ignored. Push the input amplitude up and the square-law curvature reappears as distortion; the linear $g_m$ is just the first term of a Taylor expansion, and the second-order term generates harmonics. Push the frequency up and $C_{gs}$ and $C_{gd}$ must be added, turning the simple model into the high-frequency model used for bandwidth analysis. And drive $V_{DS}$ below $V_{ov}$ and the device enters triode, where it becomes a controlled resistor with no useful gain. Knowing exactly when each assumption holds is what separates a reliable hand analysis from a misleading one.

One more contrast with the BJT is worth internalizing. Because $g_m$ scales only as the square root of current, you cannot buy transconductance cheaply by raising the bias; doubling $I_D$ only multiplies $g_m$ by about 1.41. To get more gain at fixed current, MOSFET designers instead widen the device (raising $k$) or lower the overdrive, both of which raise $g_m$ for the same $I_D$. This degree of freedom — sizing — has no bipolar analog and is one of the reasons MOSFET design feels different in practice from bipolar design.

A Fully Worked Numerical Example

Consider an NMOS transistor with $k = \mu_n C_{ox} (W/L) = 2\ \text{mA/V}^2$, biased at $I_D = 1\ \text{mA}$, with $\lambda = 0.02\ \text{V}^{-1}$ (so $V_A = 50\ \text{V}$). First find the overdrive from the square law:

Now compute the transconductance two ways to confirm they agree:

Then the output resistance and intrinsic gain:

An intrinsic gain of 100 (40 dB) is typical for a moderate-channel MOSFET — and notice it is about 40 times lower than the BJT's 4000 from the companion article, biased at the same 1 mA. That gap is the quantitative statement of "BJTs have higher gain per device." Loaded into a 10 kΩ drain resistor, this device would give a common-source gain of $-g_m (R_D \parallel r_o) = -(2\ \text{mS})(10\,\text{k} \parallel 50\,\text{k}) = -(2\ \text{mS})(8.33\ \text{k}\Omega) \approx -16.7$.

BJT vs. MOSFET Small-Signal Comparison

Common Mistakes

• Using $g_m = I_D / V_{ov}$ without the factor of 2. The correct relation is $g_m = 2 I_D / V_{ov}$. Dropping the 2 halves your gain.
• Treating the gate as a current node. An ideal MOSFET gate draws no DC current; there is no $r_\pi$. Looking into the gate you see (ideally) an open circuit, with only capacitance at high frequency.
• Forgetting the body effect when the source floats. If $V_{SB} \ne 0$, add the $g_{mb} v_{bs}$ source. It silently reduces gain in followers and cascodes.
• Mixing up $k$ conventions. Some authors fold the $\tfrac{1}{2}$ into $k$; others keep $W/L$ separate. Check your textbook's definition before plugging numbers in.
• Assuming saturation without checking. The model only holds for $V_{DS} \ge V_{ov}$. In triode the device is a voltage-controlled resistor, not an amplifier.

With these parameters in hand, work through the common-source amplifier analysis, contrast the device with the BJT hybrid-pi model, or drop a MOSFET onto the CircuitMath editor to see its small-signal equivalent generated automatically.