Currents in the ideal-diode ladder: 22-Elec-A5 Electronics, May 2018, Q4

Question 4: Currents in the ideal-diode ladder

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.

[Figure not reproduced: Question 4 — diode ladder. Node 1 (below R₁ from +10 V): D₁ feeds R₂ to −10 V (I₂), D₂ feeds node 2. Node 2: D₃ to ground, R₃ to −20 V (I₃). See the official exam paper or the cited reference text.]

Question 4 — diode ladder. Node 1 (below R₁ from +10 V): D₁ feeds R₂ to −10 V (I₂), D₂ feeds node 2. Node 2: D₃ to ground, R₃ to −20 V (I₃).

Given. Two internal nodes. Node 1 connects to +10 V through R₁ (current $I_1$ down), to −10 V through D₁+R₂ (current $I_2$, D₁ anode at node 1), and to node 2 through D₂ (anode at node 1). Node 2 connects to ground through D₃ (anode at node 2) and to −20 V through R₃ (current $I_3$ down). All diodes ideal, $V_\gamma=0.6\ \text{V}$; every R = 10 kΩ.

Find. $I_1$, $I_2$, $I_3$.

Approach. Assume a diode state, solve the resulting linear node equations, then confirm every ON diode carries forward current and every OFF diode is reverse-biased.

Test “all diodes on.” If D₃ conducted it would pin $V_2=+0.6\ \text{V}$ and D₂ would give $V_1=1.2\ \text{V}$; then KCL at node 1 needs $I_{D2}=I_1-I_2=0.88-1.06=-0.18\ \text{mA}$, which is negative. Impossible — so this state is wrong.
Identify the real state. The −20 V rail pulls node 2 strongly negative through R₃, holding D₃ (anode at node 2, cathode at ground) reverse-biased. So D₃ is OFF, while D₁ and D₂ conduct. With D₂ on, $V_2=V_1-0.6\ \text{V}$.
KCL at node 1. Current in from R₁ equals current out through D₁+R₂ and through D₂ (which, with D₃ off, all flows on into R₃):$$\frac{10-V_1}{R_1}=\frac{V_1-0.6+10}{R_2}+\frac{V_2+20}{R_3},\qquad V_2=V_1-0.6.$$
Solve. With all R = 10 kΩ the equation reduces to $10-V_1=2V_1+28.8$, so$$V_1=\boxed{-6.27\ \text{V}},\qquad V_2=V_1-0.6=\boxed{-6.87\ \text{V}}.$$
Branch currents. Substituting back:$$I_1=\frac{10-V_1}{R_1}=\boxed{1.63\ \text{mA}},\quad I_2=\frac{V_1-0.6+10}{R_2}=\boxed{0.313\ \text{mA}},\quad I_3=\frac{V_2+20}{R_3}=\boxed{1.31\ \text{mA}}.$$
Consistency check. $I_2>0$ and $I_{D2}=I_3>0$ (D₁, D₂ forward); $V_2=-6.87\ \text{V}$ is far below 0.6 V, so D₃ is genuinely reverse-biased. KCL closes: $I_1=I_2+I_3=0.313+1.31=1.63\ \text{mA}$. State confirmed.

Question 4 — results
Quantity	Value
$I_1$	1.63 mA
$I_2$	0.313 mA
$I_3$	1.31 mA
Diode states	D₁ on, D₂ on, D₃ off

Topic. Analysis of ideal-diode networks by the assumed-state method.

Key relations. Each ideal diode is either ON (drop $V_\gamma$, any forward current) or OFF (zero current, reverse voltage). A guess is valid only if every ON diode has $I\ge0$ and every OFF diode has $V_{\text{anode}}-V_{\text{cathode}}\le V_\gamma$.

Why this works. Fixing the state linearises the circuit into node equations solvable by inspection; the checks weed out inconsistent guesses (here the −20 V rail forces D₃ off).

Common pitfall. Blindly turning every diode on. A rail that pulls a node hard in one direction often reverse-biases a diode referenced to ground — always re-test the sign of each current and voltage after solving.

Source. Sedra & Smith, Microelectronic Circuits, 8th ed., Ch. 4 (ideal-diode models, analysis of diode circuits).

Question 4 of 5: Currents in the ideal-diode ladder

Question 4: Currents in the ideal-diode ladder