# 2VCC in a BJT C.E. amplifier

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).

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

With this configuration, we will have this following graph:

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?

• How can the maximum current peak equal half of itself? Commented Jun 24, 2019 at 12:37
• "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? Commented Jun 24, 2019 at 12:39
• 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. Commented Jun 24, 2019 at 13:03

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.

• 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."? Commented Jun 24, 2019 at 13:55
• 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 Commented Jun 24, 2019 at 13:58
• 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? Commented Jun 24, 2019 at 14:20
• you got it..... Commented Jun 24, 2019 at 14:22

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.