I can't figure out why drain voltage is rectangular in the following circuit: Class F power amplifier circuit

I tried this:

$$ V_{ds}=V_{DD}-I_{fund}R_lsin(\omega_0t)-I^{(2)}Z_{in}^{(2)}sin(2\omega_0t)-I^{(3)}Z_{in}^{(3)}sin(3\omega_0t)... $$ $$ I^{(k)}= \int i_{d}sin(k\omega_0t)dt $$ $$ Z_{in}^{(2k)}=0\\Z_{in}^{(2k+1)}=\infty\rightarrow I^{(2k+1)}=0 $$ $$ \rightarrow V_{ds}=V_{DD}-R_lI_{fund}sin(\omega_0t) $$ So the output I get is sinusoid but the reference I studied states it is rectangular. Would anyone please help me with this?

Edit: The transistor is assumed to act as switch.

  • \$\begingroup\$ You must also state your assumptions for Q of each filter and drain current for each level and linked reference to whom said it. \$\endgroup\$ Jun 17 '19 at 4:34
  • \$\begingroup\$ I'm refering to "The design of CMOS radio frequency integrated circuits" by Thomas H.lee. The input is at w0 frequency. "Q is assumed high enought to act as short at all frequency outside bandwidth". And transistor is assumed to act as switch. \$\endgroup\$
    – Farzad
    Jun 17 '19 at 6:17

The quarter-wave acts as an impedance transformer, producing a near-short at the FET drain. With strong gate drive producing a constant current Idrain???, then the Vdrain will indeed be a rectangle.

  • \$\begingroup\$ Well actually that is the part I can't understand. I know the transformer produce short at even harmonics and open circuit at odd harmonics. But I can not figure out how it leads to recfangular voltage at drain. \$\endgroup\$
    – Farzad
    Jun 16 '19 at 23:56
  • \$\begingroup\$ To make it work the FET would need to be driven with a strong gate drive of 50% duty cycle. That plus the impedance at the drain would create the rectangular waveform there. \$\endgroup\$
    – TimWescott
    Jun 17 '19 at 0:12
  • \$\begingroup\$ @TimWescott so is this conclusion I draw correct? : Although odd harmonics of current are zero, the infinite impedance seen at this frequencies, contributes non zero voltage change at drain. \$\endgroup\$
    – Farzad
    Jun 17 '19 at 6:38
  • \$\begingroup\$ That's a valid way of looking at it, yes. \$\endgroup\$
    – TimWescott
    Jun 17 '19 at 14:45
  • \$\begingroup\$ Thanks for the help. \$\endgroup\$
    – Farzad
    Jun 17 '19 at 15:23

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