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Q&A

# Find Q point in common base BJT configuration

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Suppose we have this graph:

How can we find the operating point (Q)?In common emitter collector we just draw a line from the vertical axis (Ic) at the point (0,VCC/RC) to the horizontal axis at the point(VCC,0) and where that line intersects with the IC/VCE curve for some specific IB , there is our operating point but now 1)we have VCB instead of VCE and 2)VCB may take negative values.

Here is the circuit:

Help appreciated!

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## Formal "explanation"

In common emitter collector we just draw a line from the vertical axis (Ic) at the point (0,VCC/RC) to the horizontal axis at the point(VCC,0) and where that line intersects with the IC/VCE curve for some specific IB, there is our operating point…

This explanation is generally true but the problem is that it does not explain anything. It only states WHAT and HOW it was done but does not say WHY it was done exactly that way. This is because it is formal. It works for a mathematician who wants to find the graphical solution of this electrical circuit without understanding exactly what it does… but for a technician this is not enough. They need a "physical" (circuit) explanation to reveal the basic idea of ​​the solution.

## Intuitive explanation

The so-called "load line" is not simply a line inclined to the left and drawn as you described. It is an IV curve of a network of two elements (the ideal voltage source V+ and the resistor Rc) in series. You can think of this network as a real voltage source with voltage V+ and internal resistance Rc.

# Graphical solution

The graphical representation by means of the load line is based on the belief that any (no matter how complex) electrical circuit can be reduced by (equivalent transformations) to two 2-terminal parts connected to each other (in parallel and in series). At least one of them contains a source.

In this configuration, the voltage Vout across and the current Ic through the two parts are the same. So, to find them (the graphical solution or the so-called operating point), we draw (superimpose) them on the same coordinate system; the intersection point gives the solution.

So, in the CB graphical representation, the intersection point of the IV curves of a real voltage source and a transistor current source gives the solution (Q point).

## CB graphical solution

In your CB stage, the one (top part) of the pair is the real voltage source (V+, Rc) and the other (bottom) part is the transistor acting as a current source. The intersection point of their IV curves gives the solution (operating point).

… but now 1)we have VCB instead of VCE and 2)VCB may take negative values.

Using the collector-base voltage VCB as Vx here is a misconception. Instead, the output voltage Vout (referenced to ground) should be placed on the X axis. Really, the base is DC grounded (0 V) here but, in other applications, it can be at some voltage. Someone can say that the base is AC grounded… but more simply is just to use the real ground as a reference.

## CB vs CE representation

So there is nothing supernatural in the CB graphical representation; it is the same as in the case of CE. The only difference is that the transistor output characteristic moves in the horizontal direction when the input (emitter) voltage changes. Of course, at the same time, it also moves vertically (like in the case of CE).

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