Electronic – BJT: What exactly is an AC Load line

circuit analysistransistors

The DC Load Line is simple to explain, for a given circuit the relation between VCE and IC must comply to the KVL relationship for that circuit.
enter image description here

In the graph we can choose a Q point, and then see if we superimpose an AC signal on top of the DC voltage then the VC will follow accordingly, and we can determine the maximum output swing.

The AC Load Line confuses me however. Because when I determine this line using the AC equivalent circuit model and KVL across the vce loop, the load line now goes through zero on a graph.

ic = – vce / (RC // RL)

Then I understand to get the true load line of the circuit, I must superimpose the AC Load line on the Q point, and preferably use the middle of the AC LL as the Q point.

I then get this:

enter image description here

  • Does this mean the DC LL no longer stands?
  • How did we get the Imax and Vmax of the AC LL?
  • And how do I determine the Q point to be in the middle of the AC LL?

Thanks

Best Answer

Load lines shows the behavior linear components in the circuit. DC load line gives the I-V relationship in the DC equivalent circuit. The inverse of slope of DC line gives the DC load resistance, \$R_E+R_C\$.

AC load lines give the I-V relationship when AC equivalent model is considered. But the AC load seen by the amplifier, \$r_c=R_C||R_L\$, is different from the DC load and hence the slope of AC load line is different. But these two lines intersect at the point where AC signal part becomes \$0\$. ie, at Q point.

  • DC load line is applicable only when DC equivalent circuit is considered.

  • After calculating Q point (\$I_{CQ},V_{CEQ}\$) from DC load line, \$I_{max}\$ and \$V_{max}\$ can be calculated as, $$I_{max} = I_{CQ} + \frac{V_{CEQ}}{r_c}$$ $$V_{max} = V_{CEQ} + I_{CQ}r_c$$

  • Putting Q-point in middle means: \$I_{max} = 2I_{CQ}\$ and \$V_{max} = 2V_{CEQ}\$. Comparing this equations with equations for \$I_{max}\$ and \$V_{max}\$, it is clear that the value of \$r_c\$ should be \$V_{CEQ}/I_{CQ}\$. So after drawing DC LL, select Q-point such that $$\frac{V_{CEQ}}{I_{CQ}} = r_c$$ This will fix the Q point in the middle of the AC LL.

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