Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — December 2017 · 16-Elec-A5, Electronics. Closed-book, 3 hours. Answer all FIVE (5) questions; each worth 20 marks. Unless stated, op-amps are ideal and supply rails are ±15 V.
Reference texts: A. S. Sedra & K. C. Smith, Microelectronic Circuits, 7th ed. (Oxford) — Ch. 4 (diode circuits & rectifiers), Ch. 5–7 (MOSFET and BJT amplifiers, biasing and small-signal models), Ch. 2 & 9 (ideal op-amps, closed-loop bandwidth and slew rate). Cross-reference: R. C. Jaeger & T. N. Blalock, Microelectronic Circuit Design.
Given. The single MOSFET and its bias/coupling network:
Given data — Question 1
Quantity
Value
Quantity
Value
$V_{TH}$
1 V
$V_{CC}$
10 V
$K$ (transconductance param.)
1 mA/V$^2$
$I_{bias}$
2 mA
$\lambda$
0.1 V$^{-1}$
$R_1,\,R_2$
10 kΩ, 5 kΩ
$R_D$
2 kΩ
$C_1,C_2$
$\infty$ (AC short)
Find. The small-signal voltage gain $A_v=v_o/v_{in}$, the input resistance $R_{in}$ seen looking into the source, and the output resistance $R_o$ seen looking into the drain.
[Figure not reproduced: Q1: common-gate MOSFET amplifier (redrawn from exam). Gate AC-grounded by R1/R2, input at source through C1, output at drain through C2. See the official exam paper or the cited reference text.]
Q1: common-gate MOSFET amplifier (redrawn from exam). Gate AC-grounded by R1/R2, input at source through C1, output at drain through C2.
Approach. Recognise the topology (input at the source, output at the drain, gate held at AC ground) as a common-gate stage; fix the operating point from the source current source, evaluate $g_m$ and $r_o$, then apply the common-gate small-signal relations.
Identify the configuration. Although the header says “common base,” $M_1$ is a MOSFET whose gate is tied through the $R_1/R_2$ divider (a DC bias point that is an AC ground), whose source receives $v_{in}$ through $C_1$, and whose drain delivers $v_o$ through $C_2$. This is a common-gate amplifier — non-inverting, with low input resistance and moderate gain. The body is tied to the source, so there is no body effect ($g_{mb}=0$).
Set the operating point. The ideal current source in the source lead fixes the drain current directly: $I_D=I_{bias}=2\text{ mA}$. From the saturation law (neglecting the small $(1+\lambda v_{DS})$ correction in the overdrive), the overdrive is
$$V_{ov}=v_{GS}-V_{TH}=\sqrt{\frac{2I_D}{K}}=\sqrt{\frac{2(2\text{ mA})}{1\text{ mA/V}^2}}=2\text{ V}.$$
A DC check confirms saturation: $V_D=V_{CC}-I_D R_D=10-(2\text{ mA})(2\text{ k}\Omega)=6\text{ V}$ and $V_G=V_{CC}\,R_2/(R_1+R_2)=3.33\text{ V}$, so $V_{DS}\gt 0$ and the device is in saturation.
Compute the small-signal parameters. The transconductance and output resistance at this bias are
$$g_m=\sqrt{2KI_D}=K\,V_{ov}=(1\text{ mA/V}^2)(2\text{ V})=2\text{ mA/V},\qquad r_o=\frac{1}{\lambda I_D}=\frac{1}{(0.1)(2\text{ mA})}=5\text{ k}\Omega .$$
Small-signal gain. With the gate grounded, $v_{gs}=-v_{in}$. Writing KCL at the drain node (load $R_D$ to AC ground, $r_o$ from drain to source) gives the exact common-gate result
$$A_v=\frac{v_o}{v_{in}}=\frac{(1+g_m r_o)\,R_D}{R_D+r_o}
=\frac{\bigl(1+(2\text{ mA/V})(5\text{ k}\Omega)\bigr)(2\text{ k}\Omega)}{2\text{ k}\Omega+5\text{ k}\Omega}
=\frac{11(2\text{ k}\Omega)}{7\text{ k}\Omega}.$$
$$\boxed{A_v=+3.14\ \text{V/V}}$$
The result is positive: a common-gate stage is non-inverting. (Neglecting $r_o$ would give the textbook $g_m R_D=4$; the finite $r_o=5\text{ k}\Omega$, comparable to $R_D$, pulls it down to 3.14.)
Input resistance (looking into the source). The source terminal, loaded at the drain by $R_D$, presents
$$R_{in}=\frac{R_D+r_o}{1+g_m r_o}=\frac{2\text{ k}\Omega+5\text{ k}\Omega}{11}=636\ \Omega\;\approx\;\frac{1}{g_m}=500\ \Omega .$$
The low input resistance ($\approx 1/g_m$) is the defining trait of the common-gate stage.
Output resistance (looking into the drain). With the input driven by an ideal source ($R_{sig}=0$), the source node is AC-grounded, $v_{gs}=0$, the controlled source is off, and only $r_o$ and $R_D$ remain in parallel:
$$R_o=R_D\parallel r_o=\frac{(2\text{ k}\Omega)(5\text{ k}\Omega)}{2\text{ k}\Omega+5\text{ k}\Omega}=1.43\text{ k}\Omega .$$
