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It is known that it is possible to get 2Vcc output voltage in a BJT Common Emitter amplifier by using an RF Choke (instead of a simple resistor) connected to the collector.

Precisely, we will get at the output terminals a signal with VCC mean value and a VCC peak voltage (with a total value equal to 2VCC).

enter image description here

(there is VCC connected to the collector, although it is not written in the image)

With this configuration, we will have this following graph:

enter image description here

With a precise load resistance RL it will be possible to get 2VCC, because the slope of the dynamic load line depends on it. But I was told that, regarding the output current dynamic, the maximum current peak is equal to IQ (= 0.5 Imax). Why?

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  • \$\begingroup\$ How can the maximum current peak equal half of itself? \$\endgroup\$ – Andy aka Jun 24 at 12:37
  • \$\begingroup\$ "It is known that it is possible to get 2Vcc output voltage in a BJT Common Emitter amplifier by using an RF Choke" well, first time I hear it. Your interpretaion in confused, what is VCC and what 2VCC seen by you? \$\endgroup\$ – Marko Buršič Jun 24 at 12:39
  • \$\begingroup\$ Vcc corresponds to the DC voltage applied between Collector and Emitter, since the RF choke is a short circuit at DC. At small signal analysis it will be open and so the dynamic load line will have slope equal to -1/RL. So with the right RL we get 2 Vcc at output. About the current, I was trying to understand why its peak is equal to IQ (this is said in the slides I found the images). The peak is said to be equal to IQ and so, since IQ is also the DC current, we have a total current equal to 2IQ = I max. But I do not understand why IQ, in addition to be the output current, is also the peak. \$\endgroup\$ – Kinka-Byo Jun 24 at 13:03
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The energy stored in the RFC, as the collector pulls toward ground, is dissipated into Rload once the transistor is turned off. Given by your definition this is a linear system WITH NO RESONANCES, the I_V trajectory must remain on the resistive loadline, and energy is dissipated.

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The Vbe (DC) controls the Ic (DC) ( from some Vcc)

The collector voltage DC = Supply Vcc.

The AC coupled input signal modulates this collector current such that the AC coupled load R cannot draw more average signal current than the DC Q current.

For linear operation, saturation is not permitted.
But it can swing higher than 2Vcc when Vce=Vce(sat) becomes more and more saturated. With a pulse input, it operates as a current pulsed flyback voltage much greater than Vcc.

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  • \$\begingroup\$ Why do we say this "The AC coupled input signal modulates this collector current such that the AC coupled load R cannot draw more average signal current than the DC Q current."? \$\endgroup\$ – Kinka-Byo Jun 24 at 13:55
  • \$\begingroup\$ AC to modulate about Q point. But a collector cannot swing more than 2x max DC current as that would saturate it form AC output swing. If the input was DC coupled that would change/disturb Q point from the change in average Vbe \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jun 24 at 13:58
  • \$\begingroup\$ Is this due to the fact that if the AC current peak was higher than IQ, its negative peak would be -IQ and so the total collector current would be IQ (DC) - IQ(AC) =0? \$\endgroup\$ – Kinka-Byo Jun 24 at 14:20
  • \$\begingroup\$ you got it..... \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jun 24 at 14:22
  • \$\begingroup\$ Perfect, thank you! \$\endgroup\$ – Kinka-Byo Jun 24 at 14:25

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