Question 1 of 5: Resistor Bridge with a Diode-Coupled Arm

Question 1: Resistor Bridge with a Diode-Coupled Arm (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. A resistive diamond bridge driven top-to-bottom by v_IN; the left divider taps v₁, the right divider taps v₂, and a series arm R₅–D₁ bridges the two taps (diode cathode toward v₁).

Find. Accurate time sketches of v₁(t) and v₂(t) over one period, with the diode’s effect shown.

Approach. With the diode blocked the two taps are independent resistive dividers of v_IN; the bridge arm only carries current when D₁ is forward-biased, which (given its orientation) happens on part of the negative half-cycle. Solve the off state first, find the conduction threshold, then solve the two-node network while the diode conducts.

1. Diode OFF — two independent dividers. With no current in the bridge arm, each tap is a simple divider of the source:
   $$v_1=\frac{R_3}{R_1+R_3}\,v_{IN}=\frac{3}{1+3}\,v_{IN}=0.75\,v_{IN},\qquad v_2=\frac{R_4}{R_2+R_4}\,v_{IN}=\frac{1}{3+1}\,v_{IN}=0.25\,v_{IN}.$$
2. Conduction condition. The diode’s anode faces v₂ and its cathode faces v₁, so it can only conduct when $v_2-v_1\gt V_\gamma$. Using the off-state tap voltages, $v_2-v_1=(0.25-0.75)v_{IN}=-0.5\,v_{IN}$, so the diode turns on when $-0.5\,v_{IN}\gt 0.7$, i.e.
   $$v_{IN}\lt -1.4\ \text{V}.$$
   On the entire positive half-cycle (and for small negative v_IN) the diode is OFF and the clean 0.75/0.25 dividers hold.
3. Diode ON — node equations. For $v_{IN}\lt -1.4$ V let a current I flow from v₂ through R₅ and D₁ into v₁. KCL at the two taps plus the branch relation give
   $$\frac{v_{IN}-v_1}{R_1}+I=\frac{v_1}{R_3},\qquad \frac{v_{IN}-v_2}{R_2}=\frac{v_2}{R_4}+I,\qquad I=\frac{v_2-v_1-V_\gamma}{R_5}.$$
   Adding the first two eliminates I and yields the tidy result $v_1+v_2=v_{IN}$; back-substitution gives the conducting laws
   $$v_1=0.7174\,v_{IN}-0.0457,\qquad v_2=0.2826\,v_{IN}+0.0457.$$
4. Evaluate the peaks. At the positive peak $v_{IN}=+10$ V (diode off): $v_1=+7.5$ V, $v_2=+2.5$ V. At the negative peak $v_{IN}=-10$ V (diode on) the large R₅ limits the correction, so the taps move only slightly from their −7.5 V / −2.5 V projections:
   $$\boxed{v_1\big|_{-10}= -7.22\ \text{V},\qquad v_2\big|_{-10}= -2.78\ \text{V}.}$$
   Both waveforms are therefore scaled triangles — v₁ at 0.75× and v₂ at 0.25× the input — with a small slope-break on the negative peaks where the diode conducts (v₁ pulled up, v₂ pushed down).