Hi all! I'm currently working on a class A Audio amplifier. And I am messing around with my final design. I am wondering why my quiescent emitter current has to be greater than or equal to my output current. I understand that if my biasing current is not big enough, then my signal will be distorted but why? Thanks for your help!

 

What is your definition of a "Class-A" amplifier?

 

What is your definition of a "Class-A" amplifier?

I designed this amplifer with a linearity of 99% close to 100%.

 

I designed this amplifer with a linearity of 99% close to 100%.

That still doesn't tell me your definition of a "Class-A" amplifier.

 

That still doesn't tell me your definition of a "Class-A" amplifier.

I'm not sure what you mean. I'm kind of confused what the true definition of a "Class-A" amplifier is besides that it uses 100% of the input source if I am correct.

 

We did one like his just a couple of days ago. The answers are in this Thread.
http://forum.allaboutcircuits.com/threads/clarification-with-common-collector.130176/

 

Take a look at this:

http://www.electronics-tutorials.ws/amplifier/amplifier-classes.html

 

Take a look at this:

http://www.electronics-tutorials.ws/amplifier/amplifier-classes.html

Ok

 

Take a look at this:

http://www.electronics-tutorials.ws/amplifier/amplifier-classes.html

I'm not sure what you want me to take away from that. I have done countless google searches, and have stumbled on that before. SO I guess is what I am saying is I have already looked at that, numerous times.

 

So what does it tell you about a "Class-A" amplifier?

 

So what does it tell you about a "Class-A" amplifier?

"Generally class A amplifiers use the same single transistor (Bipolar, FET, IGBT, etc) connected in a common emitter configuration for both halves of the waveform with the transistor always having current flowing through it, even if it has no base signal. This means that the output stage whether using a Bipolar, MOSFET or IGBT device, is never driven fully into its cut-off or saturation regions but instead has a base biasing Q-point in the middle of its load line. Then the transistor never turns “OFF” which is one of its main disadvantages."

So what I am taking away from this is that, my signal will be distorted because if my biasing current is not large enough then it can't enter cut off or saturation?

 

So what I am taking away from this is that, my signal will be distorted because if my biasing current is not large enough then it can't enter cut off or saturation?

You've got that backwards. You do not want to enter saturation or cut-off.

... is never driven fully into its cut-off or saturation regions but instead has a base biasing Q-point in the middle of its load line. Then the transistor never turns “OFF” ...

 

You've got that backwards. You do not want to enter saturation or cut-off.

OK! Thank you, you have been great great help!

 

Read about quiescent current, Q-point and transistor load line:

http://hyperphysics.phy-astr.gsu.edu/hbase/Electronic/loadline.html#c1