UPMSP Radio & Colour TV Technique common mistakes
UPMSP Radio & Colour TV Technique · Understand the traps and how to avoid them.
In power and rectifier calculations the candidate drops 230 V straight into formulas that need a peak value, or swaps the form factor 1.11 with √2 = 1.414, so the computed diode PIV and capacitor voltage rating come out around 230 V instead of about 325 V (or 650 V for a centre-tap rectifier).
Mains voltage is quoted as an RMS value everywhere in the syllabus and in the lab, and the RMS-to-peak conversion is treated as an afterthought rather than the first step of the calculation.
What to do instead
Start every AC-to-DC numerical by writing Vm = √2 × Vrms, so 230 V mains gives Vm ≈ 325 V. Memorise the two PIV results as a pair: bridge rectifier PIV = Vm ≈ 325 V, centre-tap PIV = 2Vm ≈ 650 V, then choose the nearest standard rating above that figure.
While measuring, the candidate plugs the probes into the wrong multimeter sockets (10 A/current jacks for a voltage test) or selects the resistance range across a live circuit, and reports a reading taken on the wrong range or with the meter loading the circuit.
Bench meters of different makes have the jack layout in different positions, and exam practicals are done under time pressure with a meter the candidate has not used before.
What to do instead
Before touching a circuit, check the jack labels and the rotary switch position, and prove the meter on a known source (a 1.5 V cell or a 230 V socket) first. Start on the highest range and step down; always select the function and range before connecting probes. For resistance tests, isolate the component and confirm zero reading by shorting the probes.
The candidate designs or explains bias networks by treating the transistor as an ideal switch, assuming it is ON the moment a resistor connects the base to the supply, and drawing a Zener regulator with no series resistor or calculating IZ without load current.
Textbook circuit diagrams show the transistor already conducting, so the operating conditions (VBE ≈ 0.7 V for silicon, VCE > VCE(sat), Zener current above the knee) never get written down.
What to do instead
Always state the conditions in the answer: VBE ≈ 0.7 V silicon, IC ≈ β × IB only in the active region, and the Zener carries the difference between the series-resistor current and the load current, so IS = IZ + IL. Show the calculation that proves the assumed region, otherwise treat the device as cut off or saturated.
For a capacitor-input full-wave rectifier the candidate treats ripple as fixed by the capacitor alone, ignores load current, or quotes the diode PIV and a filter-capacitor value without linking them to the computed Vm and output current.
The ripple formula is remembered as a bare constant rather than as a relationship, and the 78XX regulator stage is drawn with only its output capacitor.
What to do instead
Learn the governing relations together: Vr(pp) ≈ I/(2fC) for full-wave and I/(fC) for half-wave, so ripple rises with load current and falls with C or f. Add the regulator decoupling capacitors (typically 0.33 µF input, 0.1 µF output for 78XX series), check input-output voltage headroom and heatsink dissipation PD ≈ (Vin − Vout) × IL.
The candidate uses 10.7 MHz as the intermediate frequency for AM and 455 kHz for FM, or fails to calculate the image frequency and explains AGC and mixer roles in generic terms without giving the actual IF values.
Both figures sit in one chapter and are rehearsed side by side, so they swap under exam pressure; image frequency needs a short calculation that is skipped when time runs short.
What to do instead
Anchor the pair firmly: AM broadcast IF = 455 kHz (some receivers 465 kHz), FM broadcast IF = 10.7 MHz. For image frequency write fIMAGE = fs + 2 × IF = fLO + IF. Be prepared to explain that AGC varies IF-stage gain to hold the demodulator's output roughly constant as signal strength changes.
The candidate computes the length of a half-wave dipole with a bare λ/2 using f in Hz, or mixes feet and metres, producing an antenna length out by a factor of 3.28 or a factor of 10.
The shortcut constant (468/f in feet) and the metric form (143/f) are both remembered, but not which unit goes with which.
What to do instead
Write the units in the working: λ (m) = 300 / f (MHz) and half-wave dipole length ≈ 143 / f (MHz) metres = 468 / f (MHz) feet. The factor reduction accounts for end effects and velocity factor. State the units of the answer, and if a transmission-line question follows, give characteristic impedance as Z0 = √(L/C).
In scanning and sync answers, the candidate describes a 625-line frame being traced in one field, gives 25 fields per second as well as 25 frames per second, or places the colour burst on the front porch of blanking instead of the back porch.
Interlace and the composite-signal timing diagram are learned as a picture to copy rather than as a timeline, so field/frame counts and porch positions get shuffled.
What to do instead
Recite the timeline as a chain: 25 frames/s, each split into two fields, so 50 fields/s, 15,625 lines/s, 64 µs per line, 20 ms per field, 40 ms per frame; horizontal sync meets on the front porch and the colour burst on the back porch. Equalising pulses sit in the vertical blanking interval, and the burst is gated by the horizontal flyback.
The candidate swaps the chroma subcarrier frequencies between systems (quoting 3.58 MHz for PAL or 4.43 MHz for NTSC) or says both NTSC and SECAM carry the colour information in the same way, and confuses hue with saturation.
The three system names and their numbers are similar, and the system diagrams differ only in the phase and modulation blocks, so the answers merge together.
What to do instead
Fix one anchor fact per system: NTSC has a 3.579545 MHz subcarrier and phase-modulated chroma; PAL has 4.43361875 MHz with the V component alternating in phase line by line to cancel hue errors; SECAM sends chroma as FM on subcarriers of 4.40625 MHz and 4.25 MHz, one per line. Keep the definitions apart: hue is the colour itself, saturation is its amount as opposed to white light, and brightness is the luminance level.
For DTH and set-top box questions, the candidate says the dish feeds the Ku-band frequency directly into the STB's tuner, or cannot say what the LNB does with its local oscillator.
The block diagram is memorised as a chain of boxes, with the downconversion arithmetic and the band switching left out.
What to do instead
Learn the frequency conversion, not just the blocks: the satellite's Ku-band 10.7-12.75 GHz signal is downconverted by the LNB, with a local oscillator of 9750 MHz for the low band and 10600 MHz for the high band, to an IF of 950-2150 MHz which is what the STB tuner actually sees over the coax and which it selects by 13 V/18 V and 22 kHz tone signalling.
On the safety and servicing questions, the candidate says the set can be opened once the power switch is off, picks the chassis up with the mains connected, and does not mention discharging the CRT anode or using an isolation transformer.
The questions are short, so the rescue procedure looks like obvious common sense and the exam answer is left as 'switch it off', while the lab habit comes from what the candidate usually does on the bench.
What to do instead
Write a full procedure: unplug the mains, then discharge the CRT's anode after waiting, using an insulated probe through a 1-2 kΩ resistor to chassis earth, because the aquadag capacitance holds a dangerous charge long after switch-off. Service any mains-referenced chassis only through an isolation transformer (or a 1:1 isolating variable transformer), keep one hand in your pocket, and treat the high-voltage cover and flyback wiring as live until proved dead.