Karnataka II PUC Electronics (KSEAB) common mistakes
Karnataka II PUC Electronics (KSEAB) · Understand the traps and how to avoid them.
When computing the ripple factor or efficiency of a half-wave rectifier, the candidate plugs in the RMS value of the secondary voltage in place of the peak value, or forgets the √2 conversion entirely, so the ripple factor comes out as 1.21 instead of 1.21 only after correct substitution, and efficiency is reported as 40.6% of the wrong base quantity.
Textbook formulas for ripple factor (1.21 for half-wave, 0.48 for full-wave) and maximum efficiency (40.6% half-wave, 81.2% full-wave) are memorised as bare numbers, but the derivation steps involving Vm, Vrms and Idc are skipped, so the candidate cannot tell which voltage the formula expects.
What to do instead
Always write the defining relations first: Vrms = Vm/√2, ripple factor r = √((Vrms/VDC)² − 1), and η = PDC/PAC. Recompute from these instead of quoting the final constant, and check the answer against the known limiting value (1.21, 0.48, 40.6%, 81.2%) to catch a wrong substitution.
The candidate marks a PN junction diode as forward biased by looking only at whether the battery's positive terminal touches the p-side, ignoring the series resistor and the actual potential difference across the junction, or treats an ideal diode and a practical diode identically when solving a clipper circuit and predicts an output that never exceeds 0.7 V.
Circuit diagrams with multiple sources and resistors are read top-to-bottom rather than by applying Kirchhoff's voltage law around the loop, and the 0.7 V silicon drop is applied inconsistently - sometimes included, sometimes dropped - within a single solution.
What to do instead
For every diode circuit, first assume a state (ON or OFF), then verify: an ON diode must carry forward current and have ~0.7 V (Si) or ~0.3 V (Ge) across it; an OFF diode must have reverse voltage. State the assumed model explicitly at the top of the answer and keep it consistent through all branches, then check the resulting currents have the correct sign.
When asked for the operating point of a BJT, the candidate computes IB from VCC/RB, multiplies by an assumed β of 100 regardless of the value given, and reports IC without checking whether the transistor is actually in active mode or has been driven into saturation, so a circuit that is saturated is analysed as if it were a linear amplifier.
The DC load-line concept and the saturation condition VCE ≈ 0.2 V are treated as separate topics from the current-gain relation IC = βIB, so the candidate never applies the consistency check that IC cannot exceed VCC/RC.
What to do instead
After computing IC = βIB, calculate VCE = VCC − ICRC. If VCE falls below about 0.3 V, declare the device saturated, set VCE ≈ 0.2 V, and recompute IC = (VCC − 0.2)/RC. Always state the region of operation (cut-off, active, saturation) as part of the answer, and for a fixed-bias circuit note that the Q-point is stable only in the active region.
The candidate confuses the transconductance parameter of a JFET with that of a MOSFET, or uses the JFET square-law relation ID = IDSS(1 − VGS/VP)² for a MOSFET without adjusting for the threshold voltage, and mixes up depletion-mode and enhancement-mode device behaviour when predicting whether the channel is on at VGS = 0.
The symbols for JFET, depletion MOSFET and enhancement MOSFET look similar and the transfer characteristics are studied as separate graphs, so the candidate does not connect the device structure to the sign of VGS required to control the channel.
What to do instead
Build a one-page comparison: for a JFET, VGS ranges from VP (cut-off) to 0 with the gate reverse-biased; for a depletion MOSFET, VGS can go positive or negative around VGS(off); for an enhancement MOSFET, the channel exists only for VGS above VGS(th). Write the correct square-law or linear-region equation next to each symbol before solving, and check the sign of VGS against the device type.
While calculating the voltage gain of a CE amplifier with an unbypassed emitter resistor, the candidate uses Av = −RC/re only and ignores the emitter degeneration, or applies the hybrid-π model without converting hfe and hie into the small-signal parameters, producing a gain that is far too large and a phase shift that is stated incorrectly.
The mid-band gain formula is memorised for the bypassed-emitter case, and the effect of RE in reducing gain and improving linearity is treated as a qualitative remark rather than a quantitative term in the gain expression.
What to do instead
Write the general expression Av = −RC/(re + RE) for the unbypassed case and Av = −RC/re when RE is fully bypassed, and note the 180° phase inversion explicitly. For hybrid parameters, use Av = −hfe·RL/hie with RL the parallel combination of RC and the load. Sanity-check the magnitude: a CE stage typically gives gains of 50–300, not 5000.
The candidate states that negative feedback increases the bandwidth and input impedance without specifying whether the feedback is series or shunt at the input and whether it is voltage or current at the output, so the answer gives the wrong direction of change for input and output resistance in a given topology.
The four feedback topologies are memorised as a list of names rather than as a table of effects, and the rule that series feedback raises input resistance while shunt feedback lowers it is not linked to the sampling method at the output (voltage sampling lowers output resistance, current sampling raises it).
What to do instead
Draw the four-topology table and fill it in from first principles: series mixing at the input always increases Rin, shunt mixing decreases it; voltage sampling at the output decreases Rout, current sampling increases it. For each problem, identify the mixing and sampling nodes before writing any impedance change, and remember that gain is reduced by the factor (1 + Aβ) in all cases.
In an oscillator problem, the candidate applies the Barkhausen criterion as |Aβ| = 1 and a phase shift of 0° or 360° but then designs a phase-shift oscillator with three RC sections each giving 60° instead of the required 60° per section at the frequency where the total is 180°, or states the frequency of a Wien bridge oscillator as 1/(2πRC) without checking which R and C are in the frequency-selective network.
The phase-shift and Wien-bridge circuits are memorised as formulas for frequency only, and the condition that the amplifier must supply the remaining 180° (or 0°) of loop phase is not applied when the RC network is analysed.
What to do instead
For every oscillator, derive the loop phase first: a three-section RC ladder gives 60° per section only at f = 1/(2πRC√6), and the amplifier must then provide 180°. For the Wien bridge, the lead-lag network gives 0° at f = 1/(2πRC) and the amplifier must give 0° (non-inverting) with gain ≥ 3. Always state both the frequency condition and the gain condition together.
The candidate analyses an astable multivibrator by assuming both transistors are simultaneously ON or simultaneously OFF, or computes the time period of a 555 astable circuit as T = 1.1(R1 + R2)C instead of T = 0.693(R1 + 2R2)C, so the duty cycle is reported as 50% when it cannot be.
The cross-coupled nature of the multivibrator is not traced through the regenerative feedback loop, and the 555 charging path (through R1 + R2) is confused with the discharging path (through R2 only), so the two half-periods are treated as equal.
What to do instead
Trace the multivibrator state by state: in an astable, Q1 ON forces Q2 OFF, and the capacitor voltage determines when the state flips. For the 555, write tHIGH = 0.693(R1 + R2)C and tLOW = 0.693·R2·C separately, then T = tHIGH + tLOW and duty cycle = (R1 + R2)/(R1 + 2R2). Note that a duty cycle below 50% requires a diode across R2.
The candidate applies the ideal op-amp assumptions (zero input current, virtual short between the inputs) to a circuit where the op-amp is operating in saturation, or computes the output of an inverting amplifier as −(Rf/Rin)·Vin without checking that the result stays within the supply rails, and reports a gain of 100 for a circuit that would clip at ±12 V.
Virtual short is memorised as a universal rule rather than a consequence of negative feedback, so the candidate does not check that feedback is present and that the output is not railed before applying it.
What to do instead
Before using virtual short, confirm negative feedback exists (a resistor or direct connection from output to the inverting input) and that the required output lies between the supply rails. If the computed output exceeds the rails, state that the op-amp saturates at approximately ±VCC and the virtual-short condition no longer holds. For comparator circuits, do not apply virtual short at all.
The candidate simplifies a Boolean expression by cancelling terms that are not adjacent in a Karnaugh map, or applies De Morgan's theorem incorrectly by changing the operator but not the individual terms, and so produces a minimised expression that is not logically equivalent to the original.
Karnaugh-map grouping rules (groups of 1, 2, 4, 8; wrap-around; don't-care cells) are practised on simple examples only, and De Morgan transformations are done in one step without verifying by truth table.
What to do instead
After every simplification, verify by substituting two or three input combinations into both the original and the simplified expression. For De Morgan, apply the rule in two explicit steps: complement each literal, then swap AND for OR (or vice versa). For Karnaugh maps, mark the group boundaries on the map and confirm each group size is a power of two before reading off the term.
The candidate computes the modulation index of an AM wave from the peak and trough amplitudes but then reports the bandwidth of the AM signal as fm instead of 2fm, or states the bandwidth of an FM signal using Carson's rule as 2(Δf + fm) while substituting the modulating frequency for the deviation, so the sideband count and bandwidth are wrong.
The sideband structure of AM (carrier plus upper and lower sidebands) is memorised as a diagram rather than as a frequency-domain consequence of multiplication, and Carson's rule is recalled as a formula without understanding that Δf is the peak frequency deviation, a separate quantity from fm.
What to do instead
For AM, write the spectrum explicitly: fc, fc + fm, fc − fm, so bandwidth = 2fm. For FM, identify Δf (deviation, proportional to modulating amplitude) and fm (modulating frequency) as distinct quantities, then apply Carson's rule BW ≈ 2(Δf + fm) and state the number of significant sidebands from the Bessel table if required. Check units and magnitudes against typical broadcast values.
The candidate draws the output waveform of a full-wave bridge rectifier with only two diodes conducting per half cycle but places the conducting diodes on the wrong pair, or draws the output of a centre-tap rectifier with the same diode count as a bridge, and so predicts the wrong peak inverse voltage across each diode.
The bridge and centre-tap configurations are studied as similar circuits, and the PIV (2Vm for centre-tap, Vm for bridge) is memorised as a number without tracing which diodes are reverse-biased during each half cycle.
What to do instead
For each configuration, redraw the circuit for the positive half cycle and for the negative half cycle separately, mark the conducting diodes and the reverse-biased diodes, and derive PIV from the loop that includes the reverse-biased diode and the secondary winding. Confirm the result: centre-tap PIV = 2Vm, bridge PIV = Vm. Also check the diode count: 2 for centre-tap, 4 for bridge.
When reading a meter, the candidate quotes the measured value to more decimal places than the instrument's resolution allows, or combines a reading and its uncertainty by simple addition instead of root-sum-square, and reports a percentage error calculated against the wrong base (measured value instead of true value).
Significant figures and error propagation are treated as a separate mathematics topic and not applied to the practical readings taken in the electronics laboratory, and the definition of percentage error is recalled loosely.
What to do instead
Record every reading to the least count of the instrument (for example, a 0-1 mA meter with 50 divisions has a resolution of 0.02 mA). Compute percentage error as (measured − true)/true × 100, using the true or accepted value as the denominator. For combined uncertainties, use the root-sum-square rule for independent contributions, and round the final uncertainty to one significant figure.
The candidate describes the working of a CRO or a digital multimeter in generic terms without naming the specific block - for example, omitting the trigger circuit and time-base generator when explaining how a waveform is displayed, or failing to distinguish the input attenuator from the vertical amplifier - and so cannot answer a question about why the trace is unstable or why the reading is wrong on a particular range.
Instrumentation is revised from a block diagram caption rather than by following the signal path from probe to display, so the function of each block and the effect of a fault in that block are not connected.
What to do instead
Learn the CRO signal path in order: probe → input attenuator → vertical amplifier → deflection plates → screen, with the time-base generator driving the horizontal deflection and the trigger circuit synchronising the sweep. For each common fault (unstable trace, no trace, wrong amplitude), map it to a specific block. Do the same for the DMM: input divider, ADC, display, and the effect of a blown current fuse on the current ranges.