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24 February 2009
Author: Giorgos Lazaridis
BJT Transistor theory

Output characteristic, Load Lines and Quiescence point

To get a first clue about the Load Line, I will use an example with a Common Emitter amplifier with voltage divider biasing for the base. More details are yet to come for each connection type. Here is the circuit:

For an instance, let's forget about the input portion of the circuit (the voltage divider) and let's work only with the output portion. According to Kirchhoffs' law we have:

VCC = IC x RC + VCE + VE

Let's now calculate the first point of the load line. This point (like any other point in a x-y Cartesian system) has 2 terms: an x and a y term. We need to find a VCE and a IC pair of values which corresponds to the x and y terms respectively. We can make this calculation much easier with a simple and common trick: We will calculate a VCE value for IC=0. This way, the first pair will be on the VCE axis. Let's solve the previous equation for VCE:

VCE = VCC - IC x RC - VE

First thing that we notice is that IC x RC is zero, since IC is zero. VE is zero as well, because IE is -almost- equal to IC, and VE is equal to IE x RE. The equation can be re-written as follows:

VCE = VCC - 0 - 0 => VCE = VCC = 10V

Now for the second pair. Similarly, we will calculate a IC value for VCE = 0. So, the second point will be on the IC axis. Let's solve for IC:

VCC = IC x RC + VCE + VE = IC x RC + VCE + IC x RE = IC (RC + RE) + VCE =>
IC = (VCC - VCE) / (RC + RE)

And since we defined that VCE = 0:

IC = VCC / (RC + RE) = 10 / 4600 = 2.17 mA

So, now we have the 2 points that we need to draw the load line. The points are:

Point A (10, 0)
Point B (0, 2.17)

The green dots indicate the 2 points of the load line, and the red line is the DC load line itself.

The Operation Point, AKA Quiescence point - Q

We will continue the previous example and we will calculate the Q point. There is something that you need to make clear: The Q point is a point on the Load Line. The Load Line is calculated (as we saw before) by finding two points in the cut-off and saturation area. The Q point is usually placed within the linear area. The Q point is determined by the DC biasing of the transistor. The fact that the circuit uses VDB (Voltage divider Bias), allows us to neglect the base current in our calculations. So, the base voltage is:

VB = VCC x R2 / (R1 + R2) = 10 x 2200 / 12200 = 1.8 V

We can now calculate the emitter voltage as follows:

VE = VB - VBE = 1.8 - 0.7 = 1.1 V

And the emitter current is:

IE = VE / RE = 1.1 / 1000 = 1.1 mA

And since the collector current is almost equal to the emitter current, we can calculate the voltage drop across RC as follows:

VRC = IC x RC = IE x RC = 1.1 x 3600 = 3.96 V

Finally, the Collector-Emitter voltage is calculated as follows:

VCE = VCC - VC - VE = 4.94 V

Now we have everything we need to set the Q point:

IC = 1.1 mA
VCE = 4.94 V

Let's analyze for one moment what we've done so far. First, we found two points to draw the load line. The first point was found on the VCE axis by zeroing the IC current, and the other point was found on the IC axis by zeroing the VCE voltage. Both points were calculated with the same equation:

VCC = IC x RC + VCE + VE

For the first point we solved this equation for VCE, and for the second we solved it for IC. Then, we calculated the Q point. For the Q point, we need another pair of IC-VCE values. These values are calculated from the DC transistor bias. If the DC bias is not in the cut-off or in the saturation area, then it must be somewhere on the DC load line that we draw before. That is true since both the load line points and the quiescence point are calculated with the exact same equation. The only difference is that for the load line, we choose (for our own ease and only) to find points on the two axis, whilst for the quiescence point we solve the equation for the DC values defined by the biasing resistors.

Since the quiescence point is only one, and since the IC and VCE values of the quiescence point are critical, we usually use the pointer "q". So, for the quiescence VCE voltage we use the symbol VCEq, and for the quiescence IC current we use the symbol ICq.

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