When performing a large-signal analisys in a CE amplifier, the load (RL) is being neglected, since it is said that the BJT will be biased so the DC collector current will be equal to the DC supplied current from VCC (V+ in the picture), and therefore RL would be seen to have an inifinite resistance.

I dont understand this since if you connect the load to the amplifier, then according to KVL: V_RL = VOUT.
If V_OUT = 2.5V DC for example, then according to Ohm's law, I_RL = 2.5V / RL.

So how come it is said that no current is flowing through RL?

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Hi,


I dont see any RL in that diagram, so perhaps they are talking about setting
the bias point. The DC bias point for an AC coupled output would not include
the load resistance.

 

here Vout = Vce = V_RL. If the BJT is biased so that Vce is zero, then there is no drop across the load.

 

Thank you very much guys.
What does it mean AC coupled output?

 

hi,
Its where the output signal is connected to following circuit via a capacitor.

The 'dc' voltage level is effectively blocked by the capacitor, which only allows the 'ac' signals to pass thru.

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Eric " Good enough is Perfect "
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I see thank you.
I always thought that what you said refers to DC coupled output.
What is DC coupled output then?

 

hi,
A 'dc' coupled circuit can be as simple as piece of wire from the output of one stage to the next or a resistor coupling.

__________________
Eric " Good enough is Perfect "
I will NOT answer PM's requesting technical help, please use the Forum
PIC tutorials: Nigel's www.winpicprog.co.uk/ Bill's: www.blueroomelectronics.com/

 

Thank you