17-Phys-A5, Undated paper, Q1

Question 1 (15 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. $v_{in}$ drives diode $D_1$ (0.7 V forward drop) in series with $R_1=10\,\text{k}\Omega$ into node A, where $v_o$ is measured. From node A, $R_2=10\,\text{k}\Omega$ connects in series with an ideal $V_1=2\,\text{V}$ source (positive terminal toward $R_2$) down to ground — i.e. $R_2$ returns to a fixed 2 V rail, not directly to ground. The output terminal draws no current (ideal open-circuit measurement).

Find. $v_o(v_{in})$ and the sketch of $v_o$ vs. $v_{in}$ over $\pm 8\text{V}$, with $v_o$ marked at $v_{in}=\pm 8\text{V}$.

Fig. 1 — diode $D_1$ + $R_1$ feed node A; $R_2$ returns node A to a fixed $V_1=2\text{V}$ rail (not ground).

Approach. Find the input voltage at which $D_1$ is on the verge of conducting (zero-current boundary), then solve the series loop for $v_o$ above that threshold; below it, no current can flow, so $v_o$ is pinned by $V_1$ alone.

Part (a) — turn-on threshold. If $D_1$ is OFF, no current flows anywhere in the circuit (the only path from node A is through $R_2$ into the ideal $V_1$ source, and the output terminal is open), so $v_o=V_1=2\,\text{V}$ exactly, with zero drop across $R_2$. $D_1$ is on the verge of conducting when $v_{in}-0.7=v_o=2\,\text{V}$, i.e. $$\boxed{v_{in,\text{th}}=V_1+0.7\,\text{V}=2.7\,\text{V}}$$
Below threshold ($v_{in}<2.7\,\text{V}$). $D_1$ stays reverse-biased, so $v_o=V_1=2\,\text{V}$, independent of $v_{in}$ — confirmed at $v_{in}=-8\,\text{V}$: $v_o=2.00\,\text{V}$.
Above threshold ($v_{in}\ge 2.7\,\text{V}$). $D_1$ conducts; the same current $I$ flows through $R_1$ and $R_2$ in series with the 0.7 V drop and the 2 V source: $v_{in}-0.7=v_o+I R_1$ and $v_o=V_1+I R_2=2+I R_2$. Eliminating $I$: $$I=\frac{v_{in}-0.7-V_1}{R_1+R_2}=\frac{v_{in}-2.7}{20\,\text{k}\Omega}$$ $$\boxed{v_o = V_1+\frac{R_2}{R_1+R_2}(v_{in}-2.7) = 0.5\,v_{in}+0.65\ \ (\text{V},\ v_{in}\ge 2.7\text{V})}$$
Endpoint at $v_{in}=+8\,\text{V}$. $$v_o = 0.5(8)+0.65=\boxed{4.65\,\text{V}}$$

$v_o$ vs. $v_{in}$: flat at 2.0 V for $v_{in}<2.7\text{V}$, then rising with slope 0.5 to 4.65 V at $v_{in}=8\text{V}$.

Quantity	Value
Turn-on threshold $v_{in,\text{th}}$	2.70 V
$v_o$ for $v_{in}<2.7\text{V}$	2.00 V (constant)
Slope for $v_{in}\ge 2.7\text{V}$	0.5 V/V
$v_o(v_{in}=-8\text{V})$	2.00 V
$v_o(v_{in}=+8\text{V})$	4.65 V

Concept & Theory

Topic: diode limiter (clipper) referenced to a fixed DC rail.

Key relations: ideal-diode piecewise model ($v_D=0.7\text{V}$ when ON, open when OFF); series KVL/KCL loop analysis.

Why this works: with the output an open circuit, node A can only be held at a voltage other than $V_1$ if current actually flows through $R_2$ — and that only happens once $D_1$ conducts. Below threshold $R_2$ carries zero current, so it drops nothing and $v_o$ sits exactly at the reference rail $V_1$; above threshold the circuit becomes a simple $R_1$–$R_2$ voltage divider referenced to $V_1$ instead of ground.

Common pitfall: assuming $D_1$ turns on at $v_{in}=0.7\text{V}$ (as it would if $R_2$ returned to ground) instead of accounting for the $V_1=2\text{V}$ offset the reference branch adds to the threshold.

Source: Sedra & Smith, Microelectronic Circuits, Ch. 4 (diode limiting and clamping circuits).

Question 1 of 6

Question 1 (15 marks)