MOSFET I-V

The MOSFET I-V characteristics describe the quantitative relationship between drain current and the terminal voltages VGS and VDS. These equations are the foundation of both digital switching and analog amplifier analysis, and they are among the most frequently tested topics in GATE ECE and ESE.

Core Concept: Two-Region I-V Equations

The MOSFET drain current is described by two separate equations, one for each active operating region. The boundary condition is VDS = VGS - VT, the pinch-off condition. For any operating point, you must first determine which region the device is in before applying the correct formula.

In the linear (triode) region where VDS < VGS - VT: the drain current is ID = kn * [(VGS - VT)*VDS - VDS^2/2]. The quadratic VDS^2 term arises because the gate-to-channel voltage is not uniform along the channel. Near the source end, the effective gate overdrive is VGS - VT, but near the drain end it is VGS - VT - VDS. The average drives the parabolic current response.

In the saturation region where VDS >= VGS - VT: the drain current is ID = (kn/2) * (VGS - VT)^2. Note that VDS no longer appears in the expression (ignoring channel-length modulation). The factor of 1/2 comes from evaluating the linear-region formula at VDS = VGS - VT (the saturation onset point) and substituting, which gives (kn/2)*(VGS-VT)^2 as the maximum current in the pinch-off limited condition.

Mathematical Expression and Derivation Insight

The transconductance parameter kn = mu_n * Cox * (W/L) combines carrier mobility (mu_n), oxide capacitance per area (Cox), and device geometry (W/L). Higher W/L gives a transistor that can carry more current for the same gate overdrive. In saturation, the transconductance gm = dID/dVGS = kn * (VGS - VT). This is the small-signal parameter that determines voltage gain in common-source amplifiers: Av = -gm * RD.

Channel-length modulation is captured by multiplying the saturation current by (1 + lambda * VDS): ID = (kn/2) * (VGS - VT)^2 * (1 + lambda * VDS). In this model, the output resistance ro = 1/(lambda * ID), which appears in the small-signal equivalent circuit of the MOSFET. A large ro (small lambda) means the device is a better current source.

A useful special case: at VDS = VGS - VT, the linear and saturation formulas give the same ID. Substituting VDS = VGS - VT into the linear formula: ID = kn*[(VGS-VT)^2 - (VGS-VT)^2/2] = (kn/2)*(VGS-VT)^2. This confirms the formulas are consistent at the boundary.

Practical Understanding

In digital circuits, the MOSFET is used as a switch. The on-resistance (R_on) when the device is deep in the linear region is approximately 1/(kn*(VGS-VT)). Smaller R_on improves switching speed and reduces voltage drop across the transistor. This is why digital transistors are designed with large W/L ratios.

In analog circuits, the common-source amplifier biases the MOSFET in saturation. The small-signal voltage gain is Av = -gm * (RD || ro). Since gm = sqrt(2 * kn * ID) (derived by expressing gm in terms of ID instead of VGS-VT), increasing the bias current ID increases gm and thus the gain. This relationship between bias current and gain is a key design tradeoff in analog IC design.

Given:
kn' = 200 uA/V^2
W = 20 um, L = 2 um → W/L = 10
VT = 0.8 V
VGS = 2.0 V
VDS = 0.5 V

Why this formula applies:
First check region: VGS - VT = 2.0 - 0.8 = 1.2 V
VDS = 0.5 V < 1.2 V → device is in LINEAR region.

Formula:
ID = kn * [(VGS - VT)*VDS - VDS^2 / 2]
kn = kn' * W/L

Substitution:
kn = 200 uA/V^2 * 10 = 2000 uA/V^2 = 2 mA/V^2
ID = 2 mA/V^2 * [(1.2)(0.5) - (0.5)^2/2]

Calculation:
ID = 2 * [0.6 - 0.125]
ID = 2 * 0.475
ID = 0.95 mA

Final Answer:
ID = 0.95 mA in the linear region.

Exam Tip: When VDS is very small (VDS << 2*(VGS-VT)), the VDS^2 term is negligible, so ID = kn*(VGS-VT)*VDS, giving a linear resistance R = 1/(kn*(VGS-VT)). This is the deep-triode approximation commonly used in switch resistance calculations in GATE problems.

• Cut-off: VGS < VT, ID = 0.
• Linear: VDS < VGS - VT, ID = kn*[(VGS-VT)*VDS - VDS^2/2].
• Saturation: VDS >= VGS - VT, ID = (kn/2)*(VGS-VT)^2.
• kn = mu_n * Cox * (W/L); increasing W/L proportionally increases ID.
• With channel-length modulation: ID_sat = (kn/2)*(VGS-VT)^2*(1+lambda*VDS).

Quick Revision

• Linear region formula: ID = kn*[(VGS-VT)*VDS - VDS^2/2] valid when VDS < VGS-VT.
• Saturation formula: ID = (kn/2)*(VGS-VT)^2 valid when VDS >= VGS-VT.
• Both formulas give the same value at the pinch-off boundary VDS = VGS-VT.
• Transconductance: gm = kn*(VGS-VT) = sqrt(2*kn*ID) in saturation.
• Deep-triode approximation: R_on = 1/(kn*(VGS-VT)) when VDS << VGS-VT.
• Channel-length modulation: multiply saturation current by (1+lambda*VDS); output resistance ro=1/(lambda*ID).
• Exam trap: do NOT use the saturation formula in the linear region; overdrive voltage VGS-VT must be computed before solving any MOSFET circuit.