AM modulator that claims 8 W from an 8 V peak carrier
A student builds a simple AM modulator for a 1 MHz carrier. With no audio, the carrier envelope is 8.0 V peak. With a 1 kHz sine input, the envelope measures 14.4 V maximum and 1.6 V minimum. The RF load is 50 Ω. The antenna is a 1.5 m whip. A spectrum display shows components at 999 kHz, 1000 kHz and 1001 kHz. The student claims the modulation index is 0.8, the total transmitted power is 8 W, and the antenna is a quarter-wave radiator. The audio gain is then increased until the negative envelope peak just touches zero; the received audio becomes distorted and the spectrum shows extra components. The modulator transistor is rated 100 mA collector current.
1. Calculate the modulation index and check the student's 0.8 claim. Calculate the carrier power and total power.
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For an AM envelope, m = (Emax − Emin) / (Emax + Emin) = (14.4 − 1.6) / (14.4 + 1.6) = 12.8 / 16 = 0.8. The claim is correct. The carrier power is found from the unmodulated peak voltage: Pc = Vp² / (2R) = 8.0² / (2 × 50) = 64 / 100 = 0.64 W. Total power at m = 0.8 is Pt = Pc(1 + m²/2) = 0.64(1 + 0.32) = 0.845 W. The student's 8 W claim confuses peak volts with power; even the instantaneous peak power, 8² / 50 = 1.28 W, is far below 8 W. The correct total is less than 1 W.
2. Calculate the total sideband power and the bandwidth. Explain the three spectrum components.
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The total sideband power is Pc m² / 2 = 0.64 × 0.32 = 0.205 W, so each sideband carries about 0.102 W. The bandwidth is 2fm = 2 × 1 kHz = 2 kHz. A 1 MHz carrier modulated by a 1 kHz sine therefore produces a carrier at 1000 kHz and sidebands at 1000 ± 1 kHz, i.e. 999 kHz and 1001 kHz. That matches the spectrum display. If the audio level is increased past m = 1, the negative envelope peak is clipped, extra side frequencies appear, and the receiver produces distorted audio and adjacent-channel splatter. At exactly m = 1 the total power would be 0.64 × 1.5 = 0.96 W.
3. Evaluate the quarter-wave antenna claim. Check whether the modulator transistor current rating is adequate at m = 0.8 and m = 1.
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At 1 MHz, λ = 300 / f(MHz) = 300 m. A quarter-wave antenna is 75 m, so a 1.5 m whip is only λ/200. It is far too short to be a quarter-wave radiator and will have very low radiation resistance; it needs a loading coil or a proper matching network. The transistor current is also a concern. At 0.845 W into 50 Ω, Irms = √(P/R) = √(0.845/50) ≈ 0.130 A = 130 mA, which already exceeds the 100 mA rating. At m = 1, P = 0.96 W and Irms ≈ 139 mA. The modulator transistor is therefore overstressed even before overmodulation; a higher-current device or a different output stage is needed.