Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams — 07-Elec-A5 Electronics, December 2013. Closed book; non-communicating calculator permitted. Six questions of 20 marks; any five constitute a complete paper — all six are solved here as a study resource. Op-amps ideal with ±15 V supplies unless stated; diode forward drop as given per question.
Reference texts. A. Sedra & K. Smith, Microelectronic Circuits, 7th ed. (Oxford) — MOS/BJT amplifiers (ch. 7–9), diode wave-shaping (ch. 4), op-amp circuits (ch. 2), CMOS logic (ch. 14). Cross-checks from C. Sadiku, Fundamentals of Electric Circuits.
Check / source notes. All 16 schematics follow the printed drawings; diode orientations are as drawn (Q5 a/b/d, Q6 b). Q2's data block is internally inconsistent (it labels the BJT NPN yet lists VEB(on), names an RL that is not in the drawing, and gives RE = 1 kΩ while part (a) asks you to sizeRE); assumptions used to resolve these are stated in Q2 per the exam's "state any assumptions" instruction.
Given. First stage = the classic instrumentation-amp input pair (here with the "gain-set" resistor equal to R, i.e. unity bridge); A3 = unity difference amplifier. Inputs: v1 a ±1 V triangle (period 2 ms), v2 a ±1 V square wave (period 1 ms).
Find. The closed-form vo(v1,v2) and its time waveform over one 2 ms frame.
Figure 3. Two-stage difference network (feedback and cross-coupling resistors all equal R).
Approach. Use the virtual-short at each op-amp: the − input equals the + input. Solve the two first-stage nodes, then apply the unity difference amplifier.
First-stage outputs.A1's − input is held at v1; KCL there (from v2 through R and feedback through R) gives $v_{A1}=2v_1-v_2$. By symmetry $v_{A2}=2v_2-v_1$.
Difference stage. The unity difference amplifier outputs the difference of its two driving voltages:
$$v_o=v_{A2}-v_{A1}=(2v_2-v_1)-(2v_1-v_2).$$
Result (part a). $$\boxed{v_o=3\,(v_2-v_1)}.$$ The C at the output is driven directly by the ideal (zero-impedance) op-amp A3, so it does not alter vo — it is only an output load.
Waveform (part b). Substitute the two inputs. The square wave v2 switches every 0.5 ms and the triangle v1 ramps at ±2 V/ms, so vo is a set of straight ramps of slope $-3\times(\pm2)=\mp6$ V/ms with a ±6 V jump at every square-wave edge. Key values (V): $t{=}0{:}\,+3$; $0.5^-{:}\,0$; $0.5^+{:}\,-6$; $1.0^-{:}\,-3$; $1.0^+{:}\,+3$; $1.5^-{:}\,+6$; $1.5^+{:}\,0$; $2.0^-{:}\,-3$. The peak excursion is ±6 V, comfortably inside the ±15 V rails (no clipping).
Figure 4. Output vo=3(v2−v1) over one 2 ms frame: piecewise ramps with ±6 V steps at each 0.5 ms square-wave edge.
