I just want to know how to determine Q point on dc load line. I have read that intersection of AC and DC load line is the location of Q point.

How to draw AC load line ??

Also how Q point is useful ?

 

Actually, I think the way to think of that is that the AC load line _will_ intersect the DC load line at the Q point. The Q point is normally a design parameter to give you equal excursion above and below the Q point. For a grounded emitter class A amplifier this would simply be the center of the DC load line. Things get more complicated if you have an emitter stabilization resistor which is quite common. In this case you would consider the collector voltage at saturation as the collector minimum and the supply voltage as maximum and set the Q point at half the difference. This is normally done by assuming a reasonable VBE and setting the base voltage at 1/2 the emitter voltage at saturation + your assumed VBE. This is normally done with a voltage divider to set the base voltage. Good practice suggests that the current for this voltage divider be at least 10 times the base current (Ic/Beta). Assume B = 100. So, you pick the Q point and arrange your bias voltages to obtain it. You can work backwards in the grounded emitter circuit with an emitter resistor by calculating the base voltage set by the voltage divider and again assuming your VBE calculating your emitter current, (VB - VBE)/RE Since IC = IE - IB you can find the collector current and IC x RC gives you your collector voltage which is your Q point expressed either as IC or VC. These are approximations because we assumed VBE and Beta, but they will be very close with today's signal transistors.

The AC load line will be calculated with the impedances of reactive components and loads considered. This doesn't change your DC Q point, but the currents will be different under signal as some power is transferred to the load. For those current minima and maxima you consider the collector resistor and load impedance effectively in parallel, so the AC load line peak current will be larger and the line will extend through the Q point. Those two points give you the AC load line.

I hope this helps, an exact description of your amplifier stage could get you an exact answer.

Regards,
cww, Who taught electronics, once upon a time :^)

 

> Actually, I think the way to think of that is that the AC load line _will_ intersect the DC load line at the Q point. The Q point is

---- snip ----

> current will be larger and the line will extend through the Q point. Those two points give you the AC load line. <

Excellent response. I've been searching all over for this! So thanks very much for that. Can you elaborate on the following?

> This is normally done by assuming a reasonable VBE and setting the base voltage at 1/2 the emitter voltage at saturation + your assumed VBE. This is normally done with a voltage divider to set the base voltage. <

Is that ((1/2) * (the emitter voltage at saturation)) + your assumed V_(BE)? Either way, why that?