22-Elec-A5 Electronics · May 2016

Question 3 of 5: Ideal-Diode Resistor Network

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

Notes on this paper

National Exam — 07-Elec-A5 Electronics — May 2016. Closed-book; non-communicating calculator permitted; three hours. Five questions, 20 marks each — all five constitute a complete paper, so every question is solved below. Op-amps are ideal with supply rails of ±15 V unless a question states otherwise.

Reference texts (22-Elec-A5 Electronics): A. S. Sedra & K. C. Smith, Microelectronic Circuits, 7th/8th ed. (Oxford) — diodes (Ch. 4), MOSFETs and biasing (Ch. 5/7), op-amp circuits (Ch. 2), signal generators / wave-shaping (Ch. 17), MOS logic inverters (Ch. 14); R. C. Jaeger & T. N. Blalock, Microelectronic Circuit Design (McGraw-Hill).

Check / figure note: every schematic below was redrawn from the printed figures of the paper. Where the paper omits a numeric supply (Q1), the transfer curve is drawn to scale for an illustrative V_DD = 5 V and the symbolic result is stated alongside.

Question 3: Ideal-Diode Resistor Network (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.

Given data (Question 3)
Element	Value	Connection
R₁	10 kΩ	+10 V to node A (carries I₁)
R₂	10 kΩ	node V_L to −20 V (I₂)
R₃	10 kΩ	node V_B to −10 V (I₃)
D₁	0.6 V	anode A, cathode V_L
D₂	0.6 V	anode A, cathode V_B
D₃	0.6 V	anode ground, cathode V_B

Find. the branch currents $I_1,\,I_2,\,I_3$.

Figure 3. Diode network: node A sits below R₁; D₁/D₂ steer from A, D₃ connects ground to node V_B.

Approach. Ideal diodes are switches with a fixed 0.6 V drop when on; assume an on/off state, solve the resulting linear node equations, then confirm every “on” diode carries forward current and every “off” diode is reverse-biased.

Trial state. Try D₁ on, D₃ on, D₂ off. Then D₃ pins $v_B=0-0.6=-0.6$ V and D₁ gives $v_L=v_A-0.6$.
Node A (with D₂ off). All of I₁ leaves A through D₁ into V_L and down R₂, so $I_1=I_2$. Writing both currents: $\dfrac{10-v_A}{R}=\dfrac{v_L+20}{R}$ with $v_L=v_A-0.6$ $$\Rightarrow 10-v_A=v_A-0.6+20\Rightarrow \boxed{v_A=-4.7\ \text{V}},\ v_L=-5.3\ \text{V}.$$
Currents I₁, I₂. $$I_1=\frac{10-(-4.7)}{10\text{ k}}=1.47\ \text{mA},\qquad I_2=\frac{-5.3+20}{10\text{ k}}=1.47\ \text{mA}.$$ As required $I_1=I_2$ (series through the conducting D₁).
Node V_B and I₃. With $v_B=-0.6$ V, the current down R₃ is $$I_3=\frac{-0.6-(-10)}{10\text{ k}}=\boxed{0.94\ \text{mA}}$$ supplied entirely by D₃ from ground ($I_{D3}=0.94$ mA > 0, so D₃ is legitimately on).
Consistency of D₂ (off). $v_A-v_B=-4.7-(-0.6) =-4.1$ V < 0.6 V, so D₂ is reverse-biased — the assumed state is self-consistent, and $I_1=1.47,\;I_2=1.47,\;I_3=0.94$ mA is the unique solution.

Question 3 — branch currents
Current	Path	Value
$I_1$	+10 V → R₁ → A	1.47 mA
$I_2$	V_L → R₂ → −20 V	1.47 mA
$I_3$	V_B → R₃ → −10 V	0.94 mA

Topic. Piecewise-linear (ideal-diode) DC network analysis.

Key relations. Ideal diode: on ⇒ $v=0.6$ V, $i\ge0$; off ⇒ $i=0$, $v\le0.6$. Ohm’s law + KCL at each node.

Why this works. With three diodes there are eight on/off combinations, but only one satisfies both diode inequalities. Guiding the guess by inspection — the +10 V rail forward-biases D₁, and the very negative V_B node lets ground push current in through D₃ while starving D₂ — lands the correct state in one pass; the final inequality checks make the answer rigorous.

Common pitfall. Assuming all three diodes conduct: forcing D₂ on demands a negative (impossible) current through it. Always test every conducting diode for $i\ge0$ and every blocked diode for $v\le V_\gamma$.

Source. Sedra & Smith, Microelectronic Circuits, §4.3–4.4 (ideal-diode / constant-voltage-drop models).

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