22-Elec-A5 Electronics · May 2016

Question 1 of 5: Enhancement-Load NMOS Amplifier

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 1: Enhancement-Load NMOS Amplifier (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. Driver M₁ (gate = v_IN, source grounded) and an enhancement load M₂ whose gate and drain are tied to V_DD (source at v_OUT). Both devices share the same W/L (hence the same transconductance parameter K) and the same threshold $V_T=0.2\,V_{DD}$. Channel-length modulation is neglected for the transfer curve.

Find. (a) the DC transfer characteristic $v_{OUT}(v_{IN})$ and (b) the small-signal mid-band voltage gain.

Approach. Because M₂’s gate is held at V_DD, it is always saturated whenever it conducts; walk the output through three regions as v_IN rises, equating the drain currents of the two devices in each region, then linearise about the operating point for the gain.

1. Load is a saturated device one threshold below the rail. With $v_{G2}=V_{DD}$ and source at $v_{OUT}$, $v_{GS2}=V_{DD}-v_{OUT}$ and $v_{DS2}=V_{DD}-v_{OUT}=v_{GS2}$, so $v_{DS2}\ge v_{GS2}-V_T$ is automatic — M₂ never leaves saturation. It stops conducting when $v_{GS2}=V_T$, i.e. at $$\boxed{V_{OH}=V_{DD}-V_T=0.8\,V_{DD}}$$ which the output reaches for any $v_{IN}\lt V_T$ (driver off).
2. Region II — both saturated (the amplifying region). For $v_{IN}\gt V_T$ the driver turns on saturated. Equating saturation currents (equal K): $\tfrac12K(v_{IN}-V_T)^2=\tfrac12K(V_{DD}-v_{OUT}-V_T)^2$. Taking the physical root, $v_{IN}-V_T=V_{DD}-v_{OUT}-V_T$, i.e. $$\boxed{v_{OUT}=V_{DD}-v_{IN}\quad(\text{slope}=-1)}$$ This straight, unity-slope segment is the heart of the transfer curve.
3. Region-II endpoints (unity-gain points). The segment begins at $v_{IN}=V_T$ (output $=V_{OH}$) and ends when the driver enters the triode region, $v_{OUT}=v_{IN}-V_T$. Substituting $v_{OUT}=V_{DD}-v_{IN}$ gives the upper corner $$V_{IH}=\tfrac12(V_{DD}+V_T),\qquad V_{IL}=V_T.$$
4. Region III — driver triode, load saturated. For $v_{IN}\gt V_{IH}$ set $K\!\left[(v_{IN}-V_T)v_{OUT}-\tfrac12v_{OUT}^2\right] =\tfrac12K(V_{DD}-v_{OUT}-V_T)^2$ and solve for the small $v_{OUT}$. The output does not reach zero: at $v_{IN}=V_{DD}$ (illustrative $V_{DD}=5,\;V_T=1$) it settles at $$\boxed{V_{OL}\approx1.17\ \text{V}}$$ a comparatively high logic-low — the well-known drawback of the enhancement load (a negative low-level noise margin, $NM_L=V_{IL}-V_{OL}=1-1.17=-0.17$ V here).
5. Assemble the sketch. Flat at $V_{OH}=0.8V_{DD}$ up to $v_{IN}=V_T$; a straight slope−1 fall to the corner at $v_{IN}=\tfrac12(V_{DD}+V_T)$; then a concave tail flattening toward $V_{OL}$ (not 0). Figure 1b draws this to scale for $V_{DD}=5$ V.
6. (b) Small-signal mid-band gain. The driver is common-source with $g_{m1}$; the load presents a small-signal resistance $1/g_{m2}$ to AC ground (its gate is at the AC-ground rail). With $\lambda=0$ the gain is $$\boxed{A_v=-\dfrac{g_{m1}}{g_{m2}}=-\sqrt{\dfrac{(W/L)_1}{(W/L)_2}}=-1}$$ since the two devices are identical and, in region II, carry equal overdrive $\Rightarrow g_{m1}=g_{m2}$. The unity magnitude is exactly the slope−1 found in step 2 — the graded insight is that this inverter’s gain is fixed by a device ratio, not by bias current.