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The Band-Pass Double-Tuned Transformer
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THE TRANSMISSION OF INFORMATION THROUGH SPACE, that is, by wireless, is accomplished by impressing the information upon a carrier wave whose frequency is much higher than the highest frequency present in the information to be transmitted. The process of transferring information onto a high-frequency carrier wave is called modulation, and the carrier wave is said to be modulated by the information. An unmodulated carrier wave consists of a SINGLE-FREQUENCY sinusoidal wave occupying just ONE POINT in the frequency spectrum. However, when a carrier wave is modulated new frequencies, above and below the carrier frequency, are created. These new frequencies are called side-band frequencies, and appear as a cluster of frequencies with the carrier frequency in the center.* It is for this reason that a circuit designed to handle a modulated carrier wave must be a BAND-PASS type of network. One network that is useful in this regard is the double-tuned transformer, which let us now investigate with the aid of Fig. 232.
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Fig. 232
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To begin, the component labeled FET is a solid-state device called a eld-e ect transistor. The INPUT signal voltage is denoted by Vi , which we ll take as the reference " vector. The OUTPUT voltage is denoted by Vt ; thus the VOLTAGE GAIN of the stage is " " G Vt =Vi 393
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A eld-e ect transistor has very high internal gain but very high internal resistance, and is therefore a CONSTANT-CURRENT type of generator (section 4.7). Thus the output current of a FET, for given Vi , remains very nearly constant as the value of the load
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* See note 24 in Appendix, also note 25.
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CHAPTER 10 Magnetic Coupling. Transformers
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impedance changes; this is true for all values of load impedances normally encountered in practical work. In Fig. 232 the constant-current output of the FET has the value gm Vi , where gm is a constant transistor parameter, called the transconductance, whose value depends upon the particular transistor being used. One way to begin the analysis of Fig. 232 is to convert the constant-current generator into an equivalent constant-voltage generator; one convenient way to do this is to start with Fig. 233.
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Fig. 233
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Fig. 234
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In Fig. 233, note that we ve detached the constant-current generator and the capacitor of C farads from the primary side of Fig. 232. We now wish to convert Fig. 233 into an equivalent constant-voltage generator; this can be done by making use of Thevenin s theorem (section 4.6) as follows. First, in Fig. 233, note that the open-circuit voltage between terminals (a, b) is equal to the current gm Vi times the reactance of the capacitor C; thus the voltage of the equivalent generator is equal to jgm Vi XC , as shown in Fig. 234. Next, the internal impedance of the equivalent generator is equal to the impedance seen looking into terminals (a, b) in Fig. 233 with the FET replaced by its internal impedance. Since a FET has an extremely high internal resistance or impedance, it follows that the impedance, looking into terminals (a, b) in Fig. 233 is merely equal to the reactance of capacitor C, that is, jXC . Thus, by Thevenin s theorem, Fig. 234 is the constant-voltage equivalent of Fig. 233. Next let s consider the case of parallel R and L, as shown in Fig. 235.
Fig. 235
Our object now is to convert the parallel circuit of Fig. 235 into an approximately equivalent series circuit. To do this, we begin by noting that the input impedance looking into terminals (a, b) in Fig. 235 is equal to (product of the two, over the sum)
2 jRXL jRXL R jXL RXL jR2 XL " Zp 2 2 R jXL R2 XL R2 XL
394
CHAPTER 10 Magnetic Coupling. Transformers
In the PARTICULAR APPLICATION HERE, however, the above equation can, for practical purposes, be considerably simpli ed. To do this, we must look back to Fig. 232 and note that L and C here constitute a PARALLEL resonant circuit with XL XC at the carrier frequency (the center frequency of the passband). As we found in section 8.7, the input impedance to a parallel circuit is high at and near the resonant frequency, even though the individual reactance values, XL and XC , will have quite low values at the same frequencies. At the same time, the value of the shunt resistance R must be much higher than either XL or XC , in order to prevent R from swamping out the e ect of the high-impedance parallel LC circuit at resonance, which would, among other things, cause the gain of the stage to be excessively low at and near the resonant frequency. 2 2 Thus, in practice, the value of R2 will be much greater than the value of either XL or XC , 2 and hence, for practical purposes, the denominator of eq. 394 can be written as R instead 2 of R2 XL ; thus eq. 394 becomes,
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