Going on reading "The Art of Electronics", I am still stuck on another input impedance calculation for a transistor amplifier. This is really something I should overcome, maybe your explications will make me understand based on the following example:
The author writes "The input signal sees R1's resistance effectively reduced by the voltage gain of the stage [300Ohm to the ground in this case]". Why? How do we get this result, analitically and/or visually?
The current through a resistor is proportional to the voltage across it:
$$I = \frac{V}{R}$$
This can also be written in terms of the change of current and the change of voltage:
$$\Delta I = \frac{\Delta V}{R}$$
If the collecter end of R1 was connected directly to a fixed voltage source VC, the voltage across it would be Vin – VC, and any changes to this voltage would be attributed only to Vin. Therefore, the current changes through it would be:
$$\Delta I = \frac{\Delta V_{in} - \Delta V_C}{R} = \frac{\Delta V_{in}}{R}$$
since ΔVC is zero. We could calculate the effective resistance as:
$$R_{eff} = \frac{\Delta V_{in}}{\Delta I} = \Delta V_{in}\cdot \frac{R}{\Delta V_{in}} = {R}$$
All of this is pretty obvious, but what if VC varies, and does so in proportion to Vin:
$$V_C = A_V \cdot V_{in}$$
Now we have to write:
$$\Delta V = \Delta V_{in} - \Delta V_C = \Delta V_{in} - A_V \cdot \Delta V_{in} = \Delta V_{in}(1 - A_V)$$
Therefore:
$$\Delta I = \frac{\Delta V_{in}(1 - A_V)}{R}$$
and:
$$R_{eff} = \frac{\Delta V_{in}}{\Delta I} = \Delta V_{in}\cdot \frac{R}{\Delta V_{in}(1 - A_V)} = \frac{R}{1 - A_V}$$
Keeping in mind that AV is a negative number (a common-emitter amplifier inverts the signal), this tells us that the effective resistance is the real resistance divided by the gain of the amplifier.
In other words, if Vin varies by a little bit, the far end of the resistor swings in the opposite direction by a much larger amount, causing the current to be much larger than it would otherwise be, which makes the resistor seem much smaller than it actually is.
