Short-channel Effects in MOS transistors

Benefits: Reducing dimensions increases the number of devices and circuits that can be processed at one time on a given wafer. The frequency capability of the active devices continues to increase, as intrinsic fT values increase with smaller dimensions while parasitic capacitances decrease. Significant short-channel effects become important in MOS transistors at channel lengths of about 1 µm or less. The primary effect is to modify the classical MOS square-law transfer characteristic in the saturation or active region to make the device voltage-to-current transfer characteristic more linear.

Velocity Saturation from the Horizontal Field When an MOS transistor operates in the triode region, the average horizontal electric field along the channel is VDS/L. When VDS is small and /or L is large, the horizontal field is low, and the linear relation between carrier velocity and field is valid.

Typical measured electron drift velocity vd versus horizontal electric field ξ in an MOS surface channel. At high fields, however, the carrier velocities approach the thermal velocities, and subsequently the slope of the carrier velocity decreases with increasing field. While the velocity at low field values is proportional to the field, the velocity at high field values approaches a constant called the scattering-limited velocity vscl.

A first-order analytical approximation to this curve is whereξc ~1.5 X l06 V/m and µn ~ 0.07 m2/v-s is the low-field mobility close to the gate. From eqn., as ξc →∞, vd → vscl = µn ξc . In a device with a channel length L = 0.5 µm, we need a voltage drop of only 0.75 V along the channel to produce an average field equal to ξc , and this condition is readily achieved in short-channel MOS transistors. Substituting Vd and ξ(y) in ID eqn.,

Integrating along the channel, we obtain and thus Let VDS(act) represent the maximum value of VDS for which the transistor operates in the triode region. This Equation is valid in the triode region. The minimum value of VDS for which the transistor operates in the active region. In this region current should be independent of VDS because channel-length modulation is not included here. Therefore, VDS(act)  value of VDS that sets ∂ID/∂ VDS = 0.

Where To set, Rearranging, Solving the quadratic equation gives, Since the drain-source voltage must be greater than zero,

To determine VDS(act) without velocity-saturation effects, let ξc →∞ so that the drift velocity is proportional to the electric field, let x = (VGS - Vt) / (ξc L). Then x→0, and a Taylor series can be used to show that Using this eqn., To find the drain current in the active region with velocity saturation, substitute VDS(act) in ID eqn., After rearranging,

when the carrier velocity saturates. when the velocity is completely saturated, let ξc →0. Then the drift velocity approaches the scattering-limited velocity, Vd →Vscl = µnξc then when the carrier velocity saturates. ID linear function of the overdrive (VGS - Vt). ID independent of the channel length. ID in the active region is proportional to Vscl = µnξc. When velocity saturation is not significant, ID inversely proportional to L.

Substituting Vds(act) in ID, where x = (VGS - Vt) / (ξc L) If x << 1 , (1 - x ) ~ 1/(1 + x), and eq.1

Let V’GS gate-source voltage of the ideal square-law transistor. Then, Let VGS be the sum of V’GS and the voltage drop on RSX . Then This sum models the gate-source voltage of a real MOS transistor with velocity saturation.

Rearranging while ignoring the (ID RSX )2 term gives  eq.2 Analysing eq.1 & 2: And,

Thus the influence of velocity saturation on the large-signal characteristics of an MOS transistor can be modeled to first order by a resistor RSX in series with the source of an ideal square-law device. Fig:Model of velocity saturation in an MOSFET by addition of series source resistance to an ideal square-law device.

Transconductance and Transition Frequency The values of all small-signal parameters can change significantly in the presence of short-channel effects. One of the most important changes is to the transconductance. Calculating gives where vscl = µnξc To determine gm without velocity saturation, let Ec→∞ and x = (VGS - Vt) / (ξc L).

The transconductance increases when the overdrive increases or the channel length decreases. On the other hand, letting ξc → 0 to determine gm when the velocity is saturated gives Ratio of the transconductance to the current can be calculated As ξc → 0 , the velocity saturates and When x = (VGS - Vt) / (ξc L)<<1,

Therefore, as ξc → ∞, x → 0, and the above equation collapses to The change in transconductance caused by velocity saturation is important is because it affects the transition frequency fT.

Mobility Degradation from the Vertical Field A vertical field originating from the gate voltage influences carrier velocity. The reason is that increasing the vertical electric field forces the carriers in the channel closer to the surface of the silicon, where surface imperfections impede their movement from the source to the drain, reducing mobility. Since the gate-channel voltage is not constant from the source to the drain, the effect of the vertical field on mobility should be included within the integration . Effective mobility where µn is the mobility with zero vertical field, and θ is inversely proportional to the oxide thickness