Channel Length Modulation

What do you need to know to understand this topic?

• The MOSFET operation
• The MOSFET model

What is channel length modulation?

Pinch-off

It was previously said that a MOS transistor pinches off its channel when the drain-source voltage reaches the override voltage ($V_{DS} = V_{GS} - V_{TH}$). But what happens as we keep increasing the drain voltage? As we know, when the drain reaches pinch-off, we have $V_{GD} = V_{TH}$. More generally, the pinch-off occurs in the channel when the difference in potential between gate and that point in the channel is $V_{TH}$. If $V_{DS}$ keeps increasing, the pinch-off must occur at a position $x$ within the channel, somewhere between the drain and source. $$V_{Gx} = V_{TH}$$ $$V_G - V_x = V_{TH}$$ $$V_x = V_G - V_{TH}$$ $$V_{xS} = V_{GS} - V_{TH}$$ and $$V_{DS} > V_{GS} - V_{TH}.$$ There is a voltage drop between the pinch-off point $x$ and the drain. Whatever distance $\Delta L$ remains between these two is a depletion region. Since the channel is being modulated by $V_{DS}$, we call this effect channel-length modulation.

Channel Length modulation

What effect has channel length modulation?

To see the effect of channel length modulation on transistor operation, we replace the length of the transistor by the effective length $L - \Delta L$ in the $I_{DS}$ equation in saturation: $$I_{DS} = \frac{1}{2}\mu C_{ox}\frac{W}{L - \Delta L}(V_{GS} - V_{TH})^2$$ $$I_{DS} = \frac{1}{2}\mu C_{ox}\frac{W}{L} \frac{1}{1 - \Delta L/L}(V_{GS} - V_{TH})^2.$$ Assuming that $\Delta L \ll L$, we can linearize $\frac{1}{1 - \Delta L/L}$ around $\Delta L/L = 0$ to get $\frac{1}{1 - \Delta L/L} \approx 1 + \frac{\Delta L}{L}$. Then: $$\begin{equation}I_{DS} = \frac{1}{2}\mu C_{ox}\frac{W}{L} (1 + \frac{\Delta L}{L})(V_{GS} - V_{TH})^2.\label{eq:a}\end{equation}$$ Next we state that $\Delta L$ is influenced linearly by $V_{DS}$ by a factor of $\lambda'$ (this is more or less true), which is a process parameter: $$\Delta L = \lambda' V_{DS}.$$ But usually, a parameter $\lambda = \lambda'/L$ is used instead, so: $$\begin{equation}\frac{\Delta L}{L} = \frac{\lambda' V_{DS}}{L} = \lambda V_{DS}\label{eq:b}\end{equation}.$$ We can replace $\eqref{eq:b}$ in $\eqref{eq:a}$ to get $$I_{DS} = \frac{1}{2}\mu C_{ox}\frac{W}{L}(V_{GS} - V_{TH})^2(1 + \lambda V_{DS}).$$ By the above equation it is easy to see that increasing $V_{DS}$ increases $I_{DS}$. The following plots show the $I_{DS}-V_{DS}$ curve with channel-length modulation.

The $I_{DS}-V_{DS}$ curve

Now the plateaus are not flat, but have some slope. The meaning and value of the slope is discussed in the small-signal model.