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Radio Frequency Integrated Circuit Design - Webs

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370 <strong>Radio</strong> <strong>Frequency</strong> <strong>Integrated</strong> <strong>Circuit</strong> <strong>Design</strong><br />

added capacitor C A . Thus, it is possible to obtain close to 100% efficiency<br />

even in the presence of parasitics.<br />

10.7.1 Analysis of Class E Amplifier<br />

Several simplifying assumptions are typically made in the analysis [4]:<br />

1. The radio frequency choke (RFC) is large, with the result that only dc<br />

current I dc flows through it.<br />

2. The Q of the output circuit consisting of L o and C o is high enough<br />

so that the output current io and output voltage vo consist of only the<br />

fundamental component. That is, all harmonics are removed by this<br />

filter.<br />

3. The transistor Q 1 behaves as a perfect switch. When it is on, the<br />

collector voltage is zero, and when it is off the collector current is zero.<br />

4. The transistor output capacitance co , and hence C, is independent of<br />

voltage.<br />

With the above approximations, the circuit can now be analyzed. Waveforms<br />

are shown in Figure 10.24. When the switch is on, the collector voltage<br />

is zero, and therefore the current iC through the capacitor C is zero. In this<br />

case, the switch current is = I dc − i o . When the switch is off, i s = 0. In this<br />

case, iC = I dc − io . This produces an increase of collector voltage vC due to<br />

the charging of C. Due to resonance, this voltage will rise and then decrease<br />

again. To complete the cycle, as the switch turns on again, C is discharged and<br />

collector voltage goes back to zero again. If the component values are selected<br />

correctly, then the collector voltage will reach zero just at the instant the switch<br />

is closed, and as a result, there is no power dissipated in the transistor.<br />

We cannot explicitly solve for voltage and current waveforms over the<br />

entire cycle. We can, however, determine the collector voltage waveform when<br />

the switch is off:<br />

vC (� ) =� I dc �<br />

B � y − 2� + Vom BR sin (� − y) + I dc<br />

B � + Vom BR<br />

cos (� + �)�<br />

(10.21)<br />

where I dc is the dc input current, I om is the magnitude of the output current<br />

io , Vom is the magnitude of the output voltage and is given by the product of<br />

Iom and R, � is the phase of v o measured from the time the switch opens, 2y<br />

is the switch-off time in radians (e.g., y = �/2 for 50% duty cycle), and B is<br />

the admittance of the capacitance C.

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