Question 1 of 6: Enhancement-Load NMOS Inverter — Voltage Transfer Characteristic

Question 1: Enhancement-Load NMOS Inverter — Voltage Transfer Characteristic (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. An enhancement-load NMOS inverter (Fig. Q1): driver M₁ with its gate driven by $v_{IN}$, and load $M_2$ with its gate tied to $+V_{DD}$. Both transistors are identical (same threshold $V_t$ and same $k=\mu_n C_{ox}(W/L)$). No numeric $V_{DD}$ or $V_t$ is supplied.

Find. The complete VTC with every logic level and noise margin marked, plus the operating mode of each device in every region.

Approach. Because no numbers are given, work the levels symbolically from the drain-current equalities of the two devices, then draw the VTC for representative values.

1. Load is always saturated; it fixes the output-high level. With its gate at $V_{DD}$, the load $M_2$ has $v_{GS2}=V_{DD}-v_{OUT}$ and $v_{DS2}=V_{DD}-v_{OUT}=v_{GS2}$, so $v_{DS2}\ge v_{GS2}-V_t$ — $M_2$ is in saturation whenever it conducts. It stops conducting when $v_{GS2}=V_t$, i.e. when the output reaches $$\boxed{V_{OH}=V_{DD}-V_t}$$ This one-threshold loss below $V_{DD}$ is the hallmark of the saturated enhancement load.
2. Region I — input low. For $v_{IN}\lt V_t$ the driver $M_1$ is cut off. No current flows, the load pulls the output up, and $v_{OUT}=V_{OH}=V_{DD}-V_t$. Mode: M₁ cut-off, M₂ saturated.
3. Region II — both saturated. For $V_t\lt v_{IN}\lt \tfrac{V_{DD}+V_t}{2}$ both devices saturate. Equating currents, $\tfrac{1}{2}k(v_{IN}-V_t)^2=\tfrac{1}{2}k(V_{DD}-v_{OUT}-V_t)^2$; with identical devices this gives $v_{IN}-V_t=V_{DD}-v_{OUT}-V_t$, hence $$\boxed{v_{OUT}=V_{DD}-v_{IN}}$$ a straight line of slope exactly $-1$. Mode: both saturated.
4. Transition into triode. The driver leaves saturation when $v_{DS1}=v_{OUT}=v_{IN}-V_t$. Combined with $v_{OUT}=V_{DD}-v_{IN}$ this occurs at $v_{IN}=\tfrac{V_{DD}+V_t}{2}$, $v_{OUT}=\tfrac{V_{DD}-V_t}{2}$.
5. Region III — driver in triode. For $v_{IN}\gt \tfrac{V_{DD}+V_t}{2}$, equating the triode driver to the saturated load, $$k\!\left[(v_{IN}-V_t)v_{OUT}-\tfrac{1}{2}v_{OUT}^2\right]=\tfrac{1}{2}k\,(V_{DD}-v_{OUT}-V_t)^2 .$$ Evaluated at the largest logic input $v_{IN}=V_{OH}=V_{DD}-V_t$ this yields the output-low level $V_{OL}$ (small but not zero). Mode: M₁ triode, M₂ saturated.
6. Input thresholds and noise margins. Because Region II has slope exactly $-1$ throughout, the two unity-gain points are its endpoints: $V_{IL}=V_t$ and $V_{IH}=\tfrac{V_{DD}+V_t}{2}$. Then $NM_H=V_{OH}-V_{IH}$ and $NM_L=V_{IL}-V_{OL}$.

Drawing the VTC for the common textbook values $V_{DD}=5\ \text{V}$, $V_t=1\ \text{V}$ (chosen only to give the curve a scale) gives $V_{OH}=4\ \text{V}$, the transition at $(3,2)$, and $V_{OL}\approx1.44\ \text{V}$: