understanding Iss/2 in biasing LNA

Thread Starter

yef smith

Joined Aug 2, 2020
752
Hello,when biasing LNA i am suppose to choose a point from the plot bellow for Vsg Vds.
I read that we need to take a point where Idds/2 IE class A operation.
What does that mean regarding the plots bellow?
Thanks.

1658851189448.png
 

BobTPH

Joined Jun 5, 2013
8,954
You cut off the region on the left, where it is acting like a resistor, then choose a point somewhere near the middle. The point labeled m1 seems to be just that.
 

BobTPH

Joined Jun 5, 2013
8,954
It means half the max current. If you have a 10V supply and a 100Ω drain resistor, the max current would be 100mA and half that would be 50.
 

Thread Starter

yef smith

Joined Aug 2, 2020
752
Hello i have LNA with 50 ohm, ports on both sides.
what is my RL? 50ohms?
how do i know what Q operating point to take from the multiple plots?
Thanks.

1658854810334.png
 

BobTPH

Joined Jun 5, 2013
8,954
I don’t understand your confusion. You have shown two different graphs with the Q point marked. You then simply read the bias current (for BJT) or voltage (for FET) off the graoh.

You read the voltage correctly from the first graph. The second graph looks like Ib should be about half way between Ib1 and Ib2.
 

Papabravo

Joined Feb 24, 2006
21,225
Hello i have LNA with 50 ohm, ports on both sides.
what is my RL? 50ohms?
how do i know what Q operating point to take from the multiple plots?
Thanks.

View attachment 272261
The same concept applies to the FET, except instead of a varying base current you will have a varying gate to source voltage. Is the graph you put in your original post a depletion mode JFET? That would account for the negative Vgs.
 

Papabravo

Joined Feb 24, 2006
21,225
Hello PapaBravo, how do i choose the point location on the load line?
Thanks.
You choose the point on the load line that allows for an equal swing of the input waveform in such a way that the device does not go into "pinchoff" or the "ohmic region". Doing either of those things will result in distortion, which you do not want in a linear amplifier. If you choose the Q-point at M1 in the original diagram your input can swing ±350 mV without going into either region.
 
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