This note on FM reactance modulator explains how severe output distortion—manifested as clipping, spurious amplitude modulation, and low-frequency breakthrough—occurs in an FM generator when the transistor in the reactance modulator stage is driven into hard cutoff.
The following shows circuit diagram of Frequency Modulator using Reactance Modulator for FM signal generation.
The reactance modulator is formed by the transistor Q1 with its biasing resistors(R3,R2,R6), RF choke(L2) and the feedback coupling capacitor (C1). The transistor Q2, along with its biasing resistors(R4,RC,R5,RE), bypass capacitor CB, LC tank circuit(L1,C5,C6) and coupling capacitors(CC1,CC2) forms the Colpitts oscillator.
Sometimes one can get distorted FM signal as shown below.
Instead of producing clean Frequency Modulation, the FM modulator circuit is now suffering from severe clipping, spurious Amplitude Modulation (AM), and low-frequency breakthrough.
The picture illustrates severe clipping by transistor with water pipe size mismatch example.
\(V_B = V_{CC} \frac{R_2}{R_2+R_3} = 5V \frac{10k\Omega}{10k\Omega+27k \Omega} \approx 1.35V\)
The ac input impedance into the base of Q1 is,
\(R_{in,ac}=R_2 || R_3 \approx 7.3k \Omega\)
and so, the ac voltage peak into the base is,
\(V_{in,ac} = V_p \frac{R_{bin}}{R_{bin}+R_1} = 2V \frac{7.3k \Omega}{{7.3k \Omega}+10k \Omega} \approx 0.84V\)
The ac signal is riding on the dc bias voltage so, the voltage peak of the input signal during the +ve half cycle is
\(V_+ = V_B + V_{in,ac} = 1.35V + 0.84V /approx 2.19V \)
and during the -ve half cycle it is,
\(V_- = V_B + V_{in,ac} = 1.35V - 0.84V /approx 0.51V \)
So, the ac signal input into the base swings between 0.51V and 2.19V. However, when a signal is fed into the base of the transistor, to turn it on, the signal amplitude must be higher than the Vbe(0.65V). If it is less than that, the signal will be cutoff. Similarly, when the base is at 2.19V, the Q1 is driven into high conduction generating high emitter current.
The Solution
The solution to this problem is hinted by the water pipe size mismatch example. That is like resizing the water pipe, we need to increase the size of the resistor R1 to attenuate the input signal before it reaches the base, and/or, decrease the amplitude of the input signal.
So here, the input signal amplitude was set to 200mV from 2V and the R1 value changed to 10k to 100k.
The resulting output fm signal waveform after the changes is shown below.
Summary
Here it is explained how severe output distortion—manifested as clipping, spurious amplitude modulation, and low-frequency breakthrough—occurs in an FM generator when the transistor $Q_1$ in the reactance modulator stage is driven into hard cutoff. Calculated via DC biasing ($V_B \approx 1.35\text{ V}$) and AC input impedance voltage division ($V_{in,ac} \approx 0.84\text{ V}$), the input signal at $Q_1$'s base swings between $0.51\text{ V}$ and $2.19\text{ V}$. Because the negative peak drops below the transistor's base-emitter threshold ($V_{BE} = 0.65\text{ V}$), the device periodically cuts off and heavily distorts the modulating drive to the $Q_2$ Colpitts oscillator stage. Analogous to fixing an oversized water hose that overwhelms a flow controller, this waveform distortion is resolved by attenuating the modulating signal—specifically by reducing the input voltage amplitude from $2\text{ V}$ down to $200\text{ mV}$ and increasing the input series resistor $R_1$ from $10\text{ k}\Omega$ to $100\text{ k}\Omega$ to yield a clean FM output.
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