Electronics
What this domain tests
Semiconductor diodes and rectification, smoothing, the bipolar transistor as a switch, the operational amplifier in inverting, non-inverting and comparator configurations, and logic gates. Questions ask you to read a circuit, predict output states and calculate gains against supply-rail limits.
Diodes and power supplies
- A silicon diode conducts above about 0.7 V in forward bias and blocks in reverse bias; the I–V curve is exponential forward, near-zero reverse until breakdown.
- Half-wave rectification uses one diode; full-wave uses a centre-tapped transformer with two diodes or a bridge of four. Output ripple frequency doubles in full-wave rectification.
- A smoothing capacitor charges at the peaks and discharges through the load; larger C, larger R and higher frequency all reduce ripple, but ripple is never eliminated.
- A Zener diode in reverse breakdown holds a steady voltage and is used for simple regulation.
Transistors and op-amps
- npn transistor: conduction begins when VBE ≈ 0.7 V; IB is small and IC = βIB; the device saturates as a switch when the base is driven hard.
- Ideal op-amp: infinite open-loop gain and input impedance, zero output impedance. In reality the output saturates near the supply rails.
- Inverting amplifier: gain = −Rf/Rin, with the inverting input held at a virtual earth.
- Non-inverting amplifier: gain = 1 + Rf/Rin.
- Comparator: output swings to the positive or negative rail depending on which input is higher.
Logic gates
AND, OR, NOT, NAND, NOR and XOR are defined by their truth tables; NAND and NOR are universal and can build any other gate. Combine gates by tracking intermediate outputs, not by guessing the final column.
Worked example
An inverting amplifier has Rf = 100 kΩ, Rin = 10 kΩ and rails of ±9 V. Find Vout for Vin = 0.25 V, then for Vin = 1.5 V.
Gain = −Rf/Rin = −100/10 = −10. Vin = 0.25 V → ideal Vout = −10 × 0.25 = −2.5 V, which lies inside the ±9 V rails, so it is achieved. Vin = 1.5 V → ideal Vout = −15 V, but the op-amp cannot exceed its supply, so the output saturates at about −9 V and the signal is clipped.
Traps that cost marks
- Quoting the ideal op-amp output without checking it against the supply rails.
- Ignoring the 0.7 V base–emitter drop when calculating base current.
- Treating β as a fixed constant for all currents and temperatures.
- Swapping the inverting gain (−Rf/Rin) with the non-inverting gain (1 + Rf/Rin).
- Claiming a smoothing capacitor removes ripple completely.
- Leaving logic inputs unconnected, or reading a truth table row by row without combining intermediate outputs.