In order to determine the quiescent point on the load line, I need to find the saturation point. How is this determined? Is it somewhere on the datasheet? I know that at the saturation point it is Vs/Rc, but in order for me to determine Rc I need the Ic(sat).

Typical circuit w/ 9V Vs.

 

I don't know why you need the saturation point in order to determine the quiescent point since the quiescent point isn't anywhere near saturation (in most cases).

Saying that the saturation current is Vs/Rc implies that the full supply voltage appears, at least as an approximation, across Rc. That's true in some circuits but not in this one.

The quick and dirty way to estimate the conditions in saturation is to assume that the transistor is ideal and that Vcesat = 0V. That means, in this case, that the current in saturation is Vs/(Re+Rc). If you want something closer (but still easy), assume that Vcesat is something on the order of 0.2 V to 0.3 V. To get better, look at the data sheet and see what the saturation voltage is at the current that the prior estimate resulted in and adjust the results.

To get the quiescent point, you do something completely different. Assume your signal input is zero and analyze the circuit. Normally you first assume that the base current is negligible (but you need to confirm this later). In this case, that gives you a bias circuit current of 9 V / 1220 Ω or 7.38 mA. It also makes the base voltage 1.62 V. Assuming that the Vbe is 0.7 V, that means that the emitter voltage is 0.92 V, making an emitter current of 4.20 mA.

Assuming that the beta is about 100, that means that the base current is about 40 μA, which is a tiny fraction of the bias circuit current so our assumption that it is negligible is reasonable.

The collector current is about 4.2 mA means that the collector voltage is going to be about 4.8 V, making Vce about 3.9 V, which is well out of saturation.

 

The saturation point is usually only of concern is switching circuits, not linear circuits.

 

What is hanging me up is the initial determination of Rc. From that point, I can calculate the rest. So If I know what the value of Ic(sat) is and use Vs then I can approximately calculate Rc? I can see that Vs is actually being divided between the base and collector sides of the circuit.
I often see the circuit shown as the Thevenin equivalent with separate Vcc and Vbb.

I have also seen Vce = 1/2Vs, Vc = 3/8Vs and Ve = 1/8Vs but still need Ic to determine Rc & Re.

I can get the Q point off the net for Vs = 9V, but what I want is to be able to determine it for any Vs so if I can determine the required Ic then I can find the Rc.

"assume that the base current is negligible (but you need to confirm this later). In this case, that gives you a bias circuit current of 9 V / 1220 Ω or 7.38 mA."

Where did the 1220Ω come from? Rc + Re on my diagram? Those are just fill in numbers. My problem is finding true value for Rc.

Thanks, Sam

 

See

 

My problem is finding true value for Rc.

What do you mean by saying "true Rc value"?

Why are you trying to find a formula for Rc? I know that is is very common for the "beginners" to look for the "formula" for everything.
But you don't find "formula" for every component in the circuit. Because of the fact that the infinite number of solutions exist.
This explains why we often have to choose some component values and parameters. And later use trial and error to see if we meet our design goals.
And this makes that circuit design is a really challenging task: To find the "best" solution as a trade-off between several (often conflicting) requirements.

But you do not give us any circuit requirements you are trying to fulfill.

Where did the 1220Ω come from?

In your diagram we can see R1 = 1kΩ and R2 = 220Ω hence R1 + R2 = 1220Ω

value of Ic(sat)

What is your definition of an Ic(sat)?

 

But you do not give us any circuit requirements you are trying to fulfill.

I am trying to learn what I consider to be the starting point for how to bias a transistor in this basic circuit. I think I understand most of the process, but I am confused as to how I select the Ic to calculate the Rc which seems to be the starting point for the rest of the values based on the supply voltage and the transistor specs. From what you say it seems to be a SWAG to be determined as possibly correct later?

"What is your definition of an Ic(sat)?"

The collector to emitter current at which the transistor goes into saturation.

I am totally new to amplifiers and trying to learn transistor biasing from scratch. Not just following someone else's circuit diagram.

All help in achieving this goal is greatly appreciated.

Thanks, Sam

 

The saturation point is usually only of concern is switching circuits, not linear circuits.

Isn't the saturation point the upper limit of current for the load line?

 

Well the general rule is to pick Rc << RL or Ic >> IL
Example here
https://forum.allaboutcircuits.com/...r-amp-doubts-and-questions.91084/#post-663327
And to make sure that BJT work in the linear region and to achieve a large voltage swing at the output we need to set the bias point (voltage at collector) equal to about 0.5Vcc (and this assumption automatically ensures that the BJT will work in the active region).
https://www.physicsforums.com/threads/what-exactly-is-maximum-symmetrical-swing.633117/

And to for good thermal stability we usually choose Re = (0.1 ... 0.4)*Vcc/Ic or we pick Ve voltage large then 1V or at least Ve>>Vbe.

And we select the voltage divider so that voltage divider current is 5....20 times large then the required base current (Ibq).

https://electronics.stackexchange.c...ing-in-common-emitter-amplifier/436341#436341
https://electronics.stackexchange.c...ias-the-base-voltage-in-a-commo/276559#276559

 

Isn't the saturation point the upper limit of current for the load line?

In theory, yes, but in practice, the transistor will start to distort the output signal much earlier.
This is why some design prefer to additional headroom voltage.

For example Vce_min > 1V and Vce_max = Vcc - 1V or more.

 

Thx Jony let me read and digest this.

Thanks, Sam