If you look at your top circuit you will notice that there are shorts where the input, emitter and output capacitors were - this is the first step to doing an AC analysis. The caps are assumed to pass AC without hinderance so thay are shorted. Resistors are presumed to to attenuate so these are left in.
Whether a resistor is connected to Vsupply or ground is irrelevant - The power supply is assumed to be one big capacitor and so this is also shorted to ground (note the ground symbols on R1, Rc and Rout1).
You then use the Hfe of the transistor (current gain of the transistor) to calculate the signal-gain as it goes from base to collector.
So, on the base is a fraction of Vin and this fraction is initially determined by the attenuation produced by Rs, R1 and R2. BUT importantly it is mainly dictated by the B-E of the transistor - it can be assumed to be forward-conducting and sometimes it is easier to assume it is a short circuit hence the current into the base is purely Vin / Rs.
This current is amplified by Hfe (current gain) and produces a voltage on the collector that is dictated by the parallel combination of Rc, Rout1 and Rout2.
Thus you can determine the approximate signal gain for AC.
Me, I use a simulator - it's quicker especially if you are trying different values out AND takes all the transistor details into account and can give a fairly accurate frequency response too.
There are a few mistakes in the question:
- The original reference design called for 4K7 (4.7K) ohm resistor for R1. This should fix your problem. Note that the specs for hfe (DC current gain) are typical and not guaranteed, so you need to design in some margin (as was done in the reference design).
- There was a mistake in your equation: vc=(12.38−25m∗47k) and later vc=(12.38−25m∗10k). You need to use the value of R2 (1K in your schematic). Also note that the result of the calculation is negative but the actual value of Vc will not go below Vce-sat (all other things being correct, a negative value would indicate saturation and the actual Ic would be lower.
The specified 4.7K resistor for R1 should correct the problem.
Edit:
One way of working out an appropriate value for R1 would be to work backwards from the collector. Assuming you want R2 to be 1K (for adequate edge rates for the IGBT in this case):
- Calculate the saturation current (we'll ignore Vce and call it 0) ... Ic = 12V / 1000 ohm = 0.012A
- Calculate the necessary base current with some margin (we will use hfe=10 to be safe) ... Ib = Ic / hfe = 0.0012A
- Calculate the value for R1 needed to achieve the above base current (0.7V assumed for Vbe) ... R1(max) = (12V -0.7V) / 0.0012A = 9417 ohm. With some more margin (allows for temperature, capacitor leakage or ?), 4.7K is a good choice for R1.
Best Answer
In linear circuits with both AC and DC components, one can split the circuit, using the Superposition Theorem, into a DC equivalent circuit to model the effect of the DC components and an AC equivalent circuit to model the effect of AC components. By "components" I mean voltage sources and current sources.
For AC analysis, you are only concerned with time-varying voltages and currents. Therefore, any DC voltage sources are considered shorted (0 V of voltage drop across them), while DC current sources are considered open circuits (0 A of current through them). Of course, the DC sources are included in DC analysis while all the AC sources are made 0. It is exactly the same as applying the Superposition Theorem to circuits with multiple DC sources - you consider the effect of one DC source at a time, and negate all the others.
This is the best link I found for explaining AC analysis so far.