Electronics 1: practice problems and exam questions

Diodes, bipolar transistors, op-amp circuits and RC networks: the first electronics course after circuit analysis.

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Cram pack (PDF)Formula sheets, method steps and practice problems with full solutions, ready to print.

Topics in this course

  1. Diodes & transistorsdiode circuits · BJT bias point · BJT as a switchcalculations
  2. Op-amp circuitsinverting amplifier · non-inverting amplifier · summing amplifiercalculations
  3. RC & RL transientsRC charging · RL circuit · capacitor with initial chargecalculations
  4. Series & parallel resistors3 resistors · 4 resistors · ladder of 6 resistorscalculations
  5. DC-DC convertersideal conversion ratios · inductor ripple and currents · output ripple and the CCM boundarycalculations
  6. Wheatstone bridge & strain gaugesbalance and node voltages · unbalanced bridges · strain gauges and load cellscalculations
  7. Digital logic: gates, K-maps & flip-flopsgates, truth tables and delays · Boolean algebra, K-maps and multiplexers · don't-cares, flip-flops and counterscalculations

Exam tips for Electronics 1

  • Assume a state (diode on or off, transistor active or saturated), solve, then check the assumption.
  • With an ideal op-amp and negative feedback, the two inputs sit at the same voltage.
  • Use V_BE = 0.7 V for silicon transistors unless the question says otherwise.

Worked examples

Diodes & transistorsEasy
+−18 V470 Ω3.3 VIR​ = ?

A 3.3 V Zener diode is fed from 18 V through a 470 Ω resistor, with no load connected. Find the current IR in the resistor.

Vin = 18 VVZ = 3.3 VR = 470 Ω

Find IR (mA)

Show hints
  1. The Zener holds the output at VZ.
  2. VR = 18 − 3.3 = 14.7 V.
  3. I = V/R.
Show the answer and the step-by-step solution

Zener regulator: a reverse-biased Zener diode holds its voltage VZ while its current stays above a minimum IZ,min. The series resistor R takes the difference between input and output: IR = (Vin − VZ)/R. That current splits between the load, IL = VZ/RL, and the Zener: IZ = IR − IL. Design for the worst cases: the lowest input with the largest load must still leave IZ,min, and the highest input with no load gives the largest Zener power PZ = VZ·IZ.

Resistor voltage

VR = 18 − 3.3 = 14.7 V

Current

IR = 14.7/470 = 31.28 mA

Answer

IR = 31.28 mA
Op-amp circuitsMedium
−+Vout​Rf​ = 47 kΩR1​ = 1 kΩVin​

The op-amp is ideal and runs from ±15 V supplies; its output saturates at ±12 V. The non-inverting amplifier is driven by a sine wave. Find the largest input amplitude Vin,max (peak value) for which the output does not clip.

R1 = 1 kΩRf = 47 kΩVsat = ±12 V

Find Vin,max (V)

Show hints
  1. For a non-inverting amplifier the gain is 1 + Rf/R₁.
  2. The output peak |gain|·Vin must stay below Vsat.
  3. |Av| = 48.
Show the answer and the step-by-step solution

Ideal op-amp rules: with negative feedback, the op-amp drives its output so that its two inputs sit at the same voltage (V₋ = V₊), and no current flows into either input. Writing KCL at the inverting input with these two facts gives the output.

Gain

Av = 1 + Rf/R1 = 1 + 47/1 = 48

Largest input before clipping

Vin,max = Vsat/|Av| = 12/48 = 0.25 V

Answer

Vin,max = 0.25 V

How practising works

  • “Practise this course” mixes problems from all its topics, like a real exam.
  • Every problem has new values, so you learn the method instead of an answer.
  • Stuck? Open a small hint first, and the full explanation only when you need it.
  • Mock exams are timed and have no hints, and you get your score and a full review at the end.

Calculators for this course

Series and parallel resistor calculatorVoltage divider calculatorRC time constant calculatorOp-amp gain calculatorLED resistor calculator555 timer calculator (astable and monostable)Ohm’s law calculator (voltage, current, resistance and power)Resistor colour code calculator (4, 5 and 6 bands)Capacitor calculator (series, parallel, charge and energy)

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Practise this course for freePractise this courseTake a mock exam5 free problems every day. No card needed.