  24 February 2009
Author: Giorgos Lazaridis
BJT Transistor theory

Let's take a look at the DC and AC equivalents of a common emitter amplifier: Let's remember the equation that we used before to draw the load line:

VCC = IC x RC + VCE + VE

It is obvious that if the resistor RC is changed, the slope of the load line will also change. As you see from the equivalent circuits above, the RC resistor of the DC equivalent is different from that of the AC equivalent. That is because the output of the amplifier has a load coupled through a coupling capacitor. This load (RL) takes no part on the DC equivalent since the capacitor acts as an open circuit in DC, but the same load is connected in parallel with the collector resistor RC on the AC equivalent. If the resistance of the load is many times higher than the collector's resistor, then the parallel total resistance (RC//RL) is almost equal with the collector's resistance RC. Otherwise, the resulting total parallel resistance is significantly smaller than RC. This means that the saturation current is increased and the VCE voltage is decreased. Let's see what changes in the scene. In the AC equivalent, we can add the voltages according to Kirchhoff's law:

uce + ic x rc = 0 => ic = - uce / rc (1)

The minus sign means that the current is reversed, but for now we can simply omit it. The AC collector current is given by the following equation:

ic = ÃIC = IC - ICQ (2)

And the AC collector voltage:

uce = ÃVCE = VCE - VCEQ (3)

We can replace the equations 2 and 3 to the equation 1 and extract the following equation:

IC = ICQ + (VCEQ - VCE) / rc

This is the new equation from which we get the 2 points for the AC load line. To find them, we do the same trick as we did for the DC load line: First we zero the IC current to extract the VCE voltage, and then we zero the VCE voltage to extract the IC current:

VCE(cut) = VCEQ + ICQ x rc (for IC=0)

And now let's reset the VCE to get the IC:

IC(sat) = ICQ + (VCEQ / rc)

Drawing the DC and AC load lines - An example

Here is a typical common emitter amplifier: Rg is the AC generator's internal resistance and RL is the load resistance. Let's first draw the DC load line. For this, we will be needing the DC equivalent circuit. Very quickly we have: VB = 12 x 2200 / 12200 = 2.16 V
VE = 2.16 - 0.6 = 1.56 V
IE = 1.56 / 800 = 1.95 mA
IC = 0.99 x 1.95 = 1.93 mA
VC = 1.93 x 2200 = 4.24 V
VCE = 12 - 4.24 - 1.56 = 6.2 V

The Q point is at VCEQ = 6.2V and ICQ = 1.93mA. For the Load line, we have:

For IC=0: VCE = 12V
For VCE=0: IC = 12 /(2200 + 800) = 4 mA

Now we can draw the DC load line and set the Q point: Now we will work on the AC equivalent to draw the AC load line: The 1k8 resistor is the result of the 10k (R1) parallel to the 2k2 (R2), and the 1k4 is the result of the 2k2 collector resistor parallel to the 4k load resistance. First the VCE:

VCE(cut) = VCEQ + ICQ x rc = 6.2 + 1.93 x 10-3 x 1400 = 8.9 V

And for IC:

IC(sat) = ICQ + (VCEQ / rc) = 1.93 + (6.2 / 1400) x 103 = 6.35 mA

We will now draw the AC load line with green color on the same characteristic with the DC load line (red color): Relative pages
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