Channel Length Modulation

In the ideal MOSFET model, the drain current in saturation is completely independent of VDS. However, in real devices, increasing VDS beyond pinch-off slightly reduces the effective channel length, causing the drain current to increase gradually. This effect is called channel length modulation and it introduces a finite output resistance in the transistor, which is critical for analog circuit performance.

Core Concept: Why Channel Length Modulation Occurs

In the ideal saturation model, once the channel pinches off at the drain end, the current saturates. But the pinch-off point is not fixed. As VDS increases beyond VDSsat = VGS - Vt, the depletion region at the drain side of the channel widens slightly, effectively shortening the physical channel length L to an effective length L_eff = L - deltaL. Since the saturation current depends on 1/L (through the kn term), a shorter L_eff means higher ID.

This results in a gradual upward slope in the ID vs VDS curve in the saturation region, instead of the perfectly flat characteristic predicted by the ideal model. The slope is parameterized by the channel length modulation coefficient lambda (units: V^-1). A longer channel transistor has smaller lambda because the relative change in L_eff is smaller for the same VDS increase.

Mathematical Expression

The modified drain current equation in saturation incorporating channel length modulation is ID = (kn/2)(VGS - Vt)^2 * (1 + lambda * VDS). The term (1 + lambda * VDS) captures the gradual increase in current with VDS. For short channel devices, lambda is large and the output resistance is low. For long channel devices, lambda is small (closer to ideal behavior).

The output resistance ro is defined as ro = dVDS/dID, which from the modified equation evaluates to ro = 1 / (lambda * ID). This output resistance appears in parallel with the drain in small-signal models and directly limits the voltage gain of amplifiers. The intrinsic gain of a MOSFET is Av = -gm * ro = -gm / (lambda * ID).

Practical Understanding

In analog circuit design, high output resistance is desirable for large voltage gain and better current mirror accuracy. Cascode topologies are commonly used to boost the effective output resistance by placing transistors in series, multiplying ro and greatly improving gain. Channel length modulation is the primary reason why analog designers use longer channel lengths (2L or 4L minimum) even in advanced nodes.

In current mirrors, a mismatch in VDS between reference and output transistors causes a current error due to lambda. This is called the finite output impedance error. Using cascode mirrors or Wilson mirrors reduces this error by keeping the drain voltages equal.

Given:
VGS = 1.5 V, Vt = 0.6 V, VDS = 2.0 V
kn = 150 uA/V^2, lambda = 0.05 V^-1

Why this formula applies:
VDS = 2.0 V > VGS - Vt = 0.9 V, so device is in saturation. Channel length modulation is present.

Formula:
ID = (kn/2)(VGS - Vt)^2 * (1 + lambda*VDS)

Substitution:
ID = (150e-6/2)(1.5-0.6)^2 * (1 + 0.05*2.0)

Calculation:
ID = 75e-6 * 0.81 * 1.1
ID = 75e-6 * 0.891

Final Answer:
ID = 66.8 uA

Output resistance:
ro = 1/(lambda*ID) = 1/(0.05 * 66.8e-6) = 299 kOhm

Exam Tip: For GATE, ro = 1/(lambda*ID). If lambda is not given, assume ideal (ro = infinity). Also remember: lambda is inversely proportional to channel length L. Doubling L roughly halves lambda and doubles ro.

• Channel length modulation is caused by drain depletion region expansion shortening L_eff as VDS increases in saturation.
• Modified saturation current: ID = (kn/2)(VGS-Vt)^2 * (1 + lambda * VDS).
• Output resistance ro = 1/(lambda * ID). Larger lambda means lower ro and lower gain.
• Lambda is inversely related to channel length. Longer channels give smaller lambda and higher ro.
• Intrinsic voltage gain: |Av| = gm * ro. Maximized by increasing gm and ro simultaneously.

Quick Revision

• Channel length modulation: L_eff = L - deltaL as VDS increases beyond saturation onset.
• Saturation current with CLM: ID = (kn/2)(VGS-Vt)^2 * (1 + lambda*VDS).
• ro = 1/(lambda * ID). This is the small-signal output resistance at the drain.
• Intrinsic gain = gm * ro = gm / (lambda * ID) = sqrt(2*kn*ID) / (lambda*ID) = sqrt(2*kn/ID) / lambda.
• Lambda is small for long-channel devices and large for short-channel devices.
• Trap: In GATE, if ID is given along with lambda, compute ro directly. Do not confuse ro with RD (external load).
• Cascode amplifiers multiply ro by approximately gm*ro, greatly increasing gain.