A square law n-channel FET (Field Effect Transistor) will pass a drain-source current. (. )2 p .... will be just one-half of the carrier amplitude (voltage). ..... Charge carriers can't ...... This brings the two frequencies together, and removes this 'beat-.
Punčochář, Mohylová: TELO, Chapter 9, Modulation, demodulation, signal sampling
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9. Chapter: Modulation, demodulation, signal sampling Time of study: 6 hours Goals: the student should be able to •
describe basic modulator principles
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describe basic demodulator principles
• describe basic signal sampling principles Text
1. Amplitude modulation circuits (AM), demodulation Amplitude modulation of a sine or cosine carrier results in a variation of the carrier amplitude that is proportional to the amplitude of the modulating signal. A modulating signal should produce an AM wave of the form S (t ) = A0 [1 + m{t}]⋅ cos(2πf c t ) m{t} - modulating signal A cos(2πf ct ) - carrier signal (A – amplitude; fc – frequency; φ – phase)
Rd A.cos (2πfct)
Rg
Rg
Square low FET amplitude modulator – FET is nonlinear element
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Punčochář, Mohylová: TELO, Chapter 9, Modulation, demodulation, signal sampling
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As an example let us describe FET amplitude modulator. A square law n-channel FET (Field Effect Transistor) will pass a drain-source current I ds = K (V g + V p )
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where Vp is the FET's pinch-off voltage and Vg is the gate voltage. It is evident that (superposition theorem) Vg =
m{t} + A cos(2πf c t ) 2
thus K (m{t}+ A cos(2πf c t ) + V p )2 4 K 2 K K 2 = m {t } + V p2 + 2V p m{t } + 2m{t}A cos(2πf c t ) + 2V p A cos(2πf c t ) + [A cos(2πf c t )] 4 4 4
I ds = I ds
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This produces a drain voltage (output) of VD = Vb − Rd I ds We know that cos 2 α = (1 + cos 2α ) / 2 . Hence the last term in expression for Ids is a combination of a steady current and a fluctuation at the frequency – 2fc. For simplicity we can arrange that the frequencies with which m(t) fluctuate are all 1, always. If 1 + m′{t}> T we have discharging current between each peak and the next almost constant ≈ Vmax/R. Now we can determine a change of charge: ΔQ ≈ T.Vmax/R. Then the change of voltage between successive peaks (Ripple) is ΔV ≈ ΔQ/C = T.Vmax/(RC) = Vmax T/τ = Vmax/( τfc) A sudden, large reduction in the amplitude of the input AM wave means that capacitor charge isn't being ‘topped up’ by each cycle peak. The capacitor voltage therefore falls exponentially until it reaches the new, smaller, peak value. To assess this effect, consider what happens when the AM wave's amplitude suddenly reduces from Vmax to a much smaller value. The capacitor voltage then declines according to V (t ) = Vmax ⋅ exp(−t / τ ) This produces the negative peak clipping effect where any swift reductions in the AM wave's amplitude are ‘rounded off’ and the output is distorted. Here we've chosen the worst possible case of squarewave modulation. In practice the modulating signal is normally restricted to a specific frequency range. This limits the maximum rate of fall of the AM wave's amplitude. We can therefore hope to avoid negative peak clipping by arranging that the detector's time constant τ > τ >> 1/fc to minimize the signal distortions caused by these effects. This is clearly only possible if the modulation frequency fm
