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22-Elec-A5 Electronics · Undated paper

Question 4 of 5: NMOS amplifier with n-channel enhancement load

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

Notes on this paper

National Exams — Electronics, 16-Elec-A5. Closed-book; one approved Casio/Sharp calculator. Answer all five questions; each worth 20 marks. Op-amps assumed ideal with ±15 V supplies unless stated; ground and chassis common. Duration 3 hours.

Reference texts. A. S. Sedra & K. C. Smith, Microelectronic Circuits, 8th ed. (diodes ch. 4; MOSFET & BJT amplifiers ch. 5–8; MOS inverters ch. 13–14) — primary; C. K. Alexander & M. Sadiku, Fundamentals of Electric Circuits, 7th ed. (Thévenin equivalents, ch. 4) for the resistive-bridge question.

The question pages identify this paper as the 16-Elec-A5 Electronics, May 2019 sitting.

Question 4: NMOS amplifier with n-channel enhancement load

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 $M_1$ (gate = $v_{IN}$, source grounded) with a diode-connected enhancement load $M_2$ (gate tied to drain at $V_{DD}$); identical $W/L$, $V_{Tn}=0.2V_{DD}$; output is the common node.

Find. The $v_{OUT}$–$v_{IN}$ transfer curve and the mid-band gain expression.

V DD M 2 v OUT M 1 v IN
Enhancement-load inverter: diode-connected $M_2$ acts as the load, driver $M_1$ steers the current; $v_{OUT}$ is taken between them.

Approach. The diode-connected load is always in saturation when it conducts, so trace the driver through cut-off, saturation and triode; equal devices make the middle region especially clean.

  1. Output-high level. $M_2$ can pull $v_{OUT}$ up only until its own $V_{GS2}=V_{DD}-v_{OUT}$ falls to $V_{Tn}$; below that it stops conducting. Hence $v_{OH}=V_{DD}-V_{Tn}=0.8V_{DD}$ — one threshold below the rail, not $V_{DD}$.
  2. Region I — $M_1$ off ($v_{IN}\lt V_{Tn}=0.2V_{DD}$). No current flows, so $v_{OUT}=v_{OH}=0.8V_{DD}$ (flat).
  3. Region II — both saturated ($0.2V_{DD}\lt v_{IN}\lt 0.6V_{DD}$). Equal $k$ and equal currents give $(v_{IN}-V_{Tn})^2=(V_{DD}-v_{OUT}-V_{Tn})^2$, i.e. $$v_{OUT}=V_{DD}-v_{IN}\quad(\text{slope }=-1).$$ This straight segment runs from $(0.2V_{DD},\,0.8V_{DD})$ to $(0.6V_{DD},\,0.4V_{DD})$.
  4. Region III — $M_1$ in triode ($v_{IN}\gt 0.6V_{DD}$). $M_1$ leaves saturation when $v_{DS1}=v_{OUT}$ falls to $v_{IN}-V_{Tn}$, at $v_{IN}=(V_{DD}+V_{Tn})/2=0.6V_{DD}$. Beyond this $v_{OUT}$ drops steeply toward a low $V_{OL}$ while $M_2$ stays saturated.
  5. (b) Mid-band gain. The load is the diode-connected $M_2$, whose small-signal resistance is $1/g_{m2}$ (in parallel with $r_{o1}\Vert r_{o2}$). Thus $$A_v=-g_{m1}\left(\frac{1}{g_{m2}}\,\Big\Vert\,r_{o1}\Vert r_{o2}\right)\approx-\frac{g_{m1}}{g_{m2}}=-\sqrt{\frac{(W/L)_1}{(W/L)_2}}=\boxed{-1}$$ for identical devices — consistent with the exact $-1$ slope of Region II.
1 2 3 4 5 1 2 3 4 5 v_IN v_OUT VTC (illustrative V_DD=5 V, V_Tn=0.2V_DD=1 V)
Transfer characteristic: flat at $0.8V_{DD}$ until $v_{IN}=0.2V_{DD}$, a unity-magnitude slope segment down to $(0.6V_{DD},0.4V_{DD})$, then a steep triode fall. Drawn for $V_{DD}=5\text{ V}$, $V_{Tn}=1\text{ V}$.
Question 4 breakpoints
$v_{OH}$$M_1$ turns onSlope $-1$ region$M_1\to$ triodeMid-band gain
$0.8V_{DD}$$v_{IN}=0.2V_{DD}$$v_{OUT}=V_{DD}-v_{IN}$$v_{IN}=0.6V_{DD}$$A_v=-1$