Maharashtra HSC Electronics Technology (Vocational) common mistakes
Maharashtra HSC Electronics Technology (Vocational) · Understand the traps and how to avoid them.
In a BJT bias numerical, the candidate finds I_B from the base circuit, writes I_C = βI_B, and then uses I_E = I_C when calculating the emitter resistor voltage or current.
The three transistor currents are memorized as separate formulas, and because β is large the candidate assumes I_C and I_E are effectively equal.
What to do instead
Always write the node equation I_E = I_B + I_C first, then substitute I_C = βI_B to get I_E = (β + 1)I_B. Use (β + 1) whenever the emitter current or emitter voltage is asked for, and check that I_E is slightly larger than I_C.
For two capacitors in series, the candidate calculates C_eq by adding the capacitances, or uses the parallel-capacitor formula because the capacitors are drawn one after another.
The resistor series formula is overgeneralized to capacitors, and the reciprocal relationship for series capacitors is not practised enough.
What to do instead
Write the defining relation before substituting: 1/C_eq = 1/C1 + 1/C2 for series, C_eq = C1 + C2 for parallel. Sanity-check that a series combination is smaller than the smallest capacitor, while a parallel combination is larger than the largest.
For a 50 Hz full-wave bridge rectifier, the candidate states the ripple frequency as 50 Hz, treating the output ripple as if it were a half-wave rectifier.
Half-wave and full-wave output waveforms are memorized as labels rather than counted pulses per input cycle, and the bridge waveform is not sketched.
What to do instead
Sketch the input sine wave and mark the conducting diode pairs. Count the output pulses in one input cycle: half-wave gives one pulse, full-wave and bridge give two. Therefore ripple frequency = 2 × 50 Hz = 100 Hz for a 50 Hz full-wave supply.
When asked for the Barkhausen criterion, the candidate writes only Aβ = 1 and omits the loop phase-shift condition, or states that Aβ must be greater than 1 for sustained oscillation.
The criterion is memorized as a single magnitude equation, and the positive-feedback phase requirement is not linked to the oscillator circuit.
What to do instead
State both conditions: at the oscillation frequency the loop gain magnitude must be 1, and the total loop phase shift must be 0° or an integer multiple of 360°. Explain that Aβ slightly greater than 1 starts oscillation, but amplitude limiting brings the effective loop gain back to 1.
The candidate applies De Morgan's theorem as (A + B)' = A' + B', distributing the complement over the OR operation without changing the operator.
The negation sign is treated like a multiplier that can be moved inside brackets, and the rule is recalled without a truth-table check.
What to do instead
Use the memory rule 'break the bar and change the sign': (A + B)' = A' · B' and (A · B)' = A' + B'. Verify any simplified expression with a two- or three-input truth table before finalizing the answer.
For an AM signal with a 5 kHz sinusoidal modulating tone, the candidate gives the bandwidth as 5 kHz instead of 10 kHz, or gives 10 kHz for an SSB signal.
The modulating frequency is confused with the occupied bandwidth, and the sideband spectrum is not drawn.
What to do instead
Draw the carrier at f_c and the two sidebands at f_c ± f_m. For AM and DSB, bandwidth = 2f_m; for SSB, bandwidth = f_m. Substitute the modulating frequency only after identifying the modulation type.
In an 8051 delay-loop calculation, the candidate counts one oscillator period per DJNZ instruction instead of 12 oscillator periods per machine cycle, or ignores the machine-cycle count of the other loop instructions.
The crystal frequency and the machine-cycle frequency are treated as the same quantity, and the instruction timing table is not used.
What to do instead
Convert the crystal frequency to machine-cycle frequency by dividing by 12 for the classic 8051, list the machine cycles for each instruction in the loop, multiply by the loop count, and add overhead before converting to time.
When reading a CRO display, the candidate counts vertical divisions, multiplies only by volts/div, and states the result as RMS voltage, ignoring the 10× probe attenuation and whether the reading is peak-to-peak.
The volts/div scale, probe factor, and peak-to-RMS conversion are treated as independent facts rather than one measurement chain.
What to do instead
Compute V_pp = vertical divisions × volts/div × probe attenuation factor. Convert to RMS only for a sine wave using V_rms = V_pp/(2√2), and state the waveform assumption. Check whether the display is peak-to-peak before converting.
In a single-sided PCB layout question, the candidate routes two copper tracks so that they cross, or places a via to jump over another track.
Schematic connectivity is prioritized over fabrication constraints, and vias are assumed to be available on a single-layer board.
What to do instead
Reroute one track around the obstacle, or use a zero-ohm resistor or insulated jumper wire as a documented crossover. Check clearance and pad access, and remember that a single-sided board has no plated through-hole via for track jumping.
The candidate describes a good hand-soldered joint as dull, grainy, and ball-shaped, or says the iron should be held only on the pad until the solder melts.
A cold joint's appearance is mistaken for a normal joint, and the purpose of heating both the pad and the lead is not understood.
What to do instead
Heat the pad and component lead together, then feed solder into the joint rather than onto the iron tip. A good joint is shiny, smooth, and forms a concave fillet; a dull, grainy, ball-shaped joint is a cold joint and should be reheated with flux.