Breaker! Beaker!

Should an input signal voltage to common emitter amplifier should reach saturation (that is out volt/current)when the signal input voltage is amplified or increased to an ampltude that is half the D.C. bias voltage and current?Is this true?
.If a.c. is riding on D.C.the signal input must produce a result that is half of the of the Vce, VRL, IRload? Is this correct understanding ? That is, gain in voltage or current must be only half of the quiescent point of operation since a.c. is riding on D.C., realizing that a.c. added to D.C. should produce Quiescent Bias parameters. Which seems reasonable to think that a specify gain current or voltage that uses only half of the D.C. Quiescent operation and the other half is the signal voltage or current.

 

Is there a question there?

 

yes its a question been stated.Let me repeat it.If the input reaches a specify
amplitude that it produces voltage thru the gain of the transistor that equal 1/2 of Supply voltage can be thought as satuation point since the a.c. is riding on D.C. which is the bias point. ac +DC = Vs

 

Actually what you describe is cut off. Saturation would be DC - AC = ground.

 

In this situation ,useing a meter at the point ,say output across a resistor load it he would read zero for both a.c and D.C voltage or current ?

 

You should probably draw a schematic of the circuit or we will all be confused. But having said that with no signal at the input there should be no AC at the output and the DC voltage should be 1/2 the supply voltage.

 

I don't have a schematic The question emerged from the bias situation you just stated.Theoretically speaking, I wonder if the input is 10 milivolts and the bias across Vce=5volts,VRc=5 volts ,if the signal is amplified to 5 volts would this be the limit before clipping begins ? or It can be amplified to 10 volt which is the Source Voltage before clipping begins ?

 

A little less than 10 volts peak to peak before clipping, but distortion occurs before that.