ETA CETa common mistakes
ETA CETa · Understand the traps and how to avoid them.
Applying Ohm's law (V = IR) directly to a junction diode or Zener diode and treating it as a fixed resistor when calculating current in a bias network.
Candidates memorize V = IR as a universal rule and forget that a forward-biased diode has an approximately constant voltage drop (about 0.7 V for silicon, 0.3 V for germanium) that must be subtracted from the supply before dividing the remaining voltage by the series resistance.
What to do instead
Before any calculation, redraw the circuit and label each element as linear (resistor) or non-linear (diode, transistor junction). For a diode branch, write I = (V_supply - V_diode) / R_series and state the assumed V_diode explicitly. Practise mixed networks where one branch has a diode and another does not, so the distinction becomes automatic.
Mixing up the three transistor configurations (common emitter, common base, common collector) when asked for current gain, voltage gain, or input/output impedance.
The names describe which terminal is shared between input and output, but candidates often associate the name with the terminal that is grounded or with the terminal where the signal enters, leading to inverted answers about phase and gain.
What to do instead
Build a one-page comparison table with rows for input impedance, output impedance, voltage gain, current gain, and phase inversion, and columns for the three configurations. Reconstruct the table from memory before each study session, then verify against a textbook. Anchor common emitter as the only configuration with 180-degree phase inversion and moderate input/output impedance.
Computing the total resistance of a series-parallel network by adding all resistor values, or by adding parallel branches as if they were in series.
Under time pressure candidates scan for the largest and smallest values and jump to a formula without tracing the actual current paths, and they confuse the reciprocal rule for parallel resistors with simple addition.
What to do instead
Redraw the network in stages, collapsing one series group or one parallel group at a time and writing the intermediate value beside each collapsed block. Use the product-over-sum shortcut only for exactly two parallel resistors, and use the reciprocal-of-reciprocals form for three or more. Check the answer against bounds: total resistance must be lower than the smallest parallel resistor and higher than the largest series resistor.
Treating AC and DC quantities as interchangeable, for example adding a DC bias voltage directly to an RMS AC value or ignoring phase when summing voltages across a series RLC circuit.
Candidates learn that RMS values behave like DC for power calculations and then overgeneralize, forgetting that instantaneous voltages across reactive elements are out of phase and that RMS values add in quadrature only for certain combinations.
What to do instead
Classify every quantity as DC, peak, peak-to-peak, RMS, or instantaneous before combining. For series RLC, compute the phasor sum using reactances (X_L = 2πfL, X_C = 1/(2πfC)) and the impedance magnitude Z = sqrt(R² + (X_L - X_C)²). For power, use P = I²R with RMS current and keep DC bias separate from AC signal calculations.
Reading a resistor colour code from the wrong end, or misreading the multiplier band as a tolerance band.
The tolerance band is set slightly apart and is often gold or silver, but candidates count bands from left to right without checking spacing, and they treat the third band as tolerance when it is the multiplier.
What to do instead
Always locate the tolerance band first (the band with the widest gap from the others) and read from the opposite end. Memorize the sequence black through white with a mnemonic, and drill ten random four-band and five-band resistors per session, writing the value and tolerance in ohms before checking.
Assuming a capacitor blocks DC completely in every context, or assuming it passes AC without limit at all frequencies.
The rule 'capacitors block DC and pass AC' is taught as a slogan, so candidates forget that capacitive reactance X_C = 1/(2πfC) is finite and frequency-dependent, and that a capacitor needs time to charge through a resistor.
What to do instead
Work numerical examples: compute X_C at 50 Hz, 1 kHz, and 1 MHz for the same capacitor and note how the value changes. For DC, analyse the charging curve with the time constant τ = RC and state the steady-state condition (capacitor acts as an open circuit). For AC, treat the capacitor as an impedance and combine it with resistance using phasor addition.
Applying Kirchhoff's current law at a node without assigning consistent current directions, then getting a sign error.
Candidates write equations node by node but change the assumed direction of a current midway through the analysis, so some currents are counted as entering and others as leaving without a consistent convention.
What to do instead
Before writing any equation, mark an arrow on every branch and define entering as positive and leaving as negative (or the reverse) for the whole problem. Write the node equation once from the marked arrows, then solve. If a computed current is negative, interpret it as flowing opposite to the assumed arrow rather than redoing the equation.
Confusing the definitions of peak, peak-to-peak, RMS, and average value of a sine wave, and using the wrong conversion factor.
The factors 0.707 and 1.414 look similar and are easy to swap, and the average value of a full sine cycle is zero, which contradicts the intuition that an average should be a positive number.
What to do instead
Write the definitions side by side: V_peak = V_rms × 1.414, V_rms = V_peak × 0.707, V_peak-to-peak = 2 × V_peak, and V_average over a full cycle = 0 (over a half cycle = 0.637 × V_peak). Drill conversions in both directions and include one question where the answer is zero to test whether the full-cycle average rule is understood rather than memorized.
Ignoring the forward voltage drop of an LED when sizing its series resistor, or using the supply voltage as the voltage across the resistor.
Candidates treat the LED like a resistor and compute R = V_supply / I_desired, which produces a resistor that allows far too much current and, in a real circuit, destroys the LED.
What to do instead
Always subtract the LED forward voltage first: R = (V_supply - V_LED) / I_LED. Practise with several supply voltages and LED colours (different forward voltages) and verify the power dissipated in the resistor with P = I²R. State the assumed forward voltage in the answer.
Misidentifying the terminals of a bipolar transistor or the polarity of a junction in a schematic symbol, especially for PNP devices.
The arrow on the emitter shows conventional current direction, but candidates memorize 'arrow points out' for NPN without checking whether the device is PNP, and they confuse the collector and emitter positions on the symbol.
What to do instead
For every transistor problem, first state the device type and the arrow direction (NPN: arrow points out of the emitter; PNP: arrow points in). Then check the bias conditions: for NPN, the base-emitter junction is forward biased when V_B > V_E by about 0.7 V. Redraw the symbol from memory with labelled terminals before solving, and include at least one PNP example in every practice set.