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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 Art of Electronics, 2nd edition, p86

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?

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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.

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  • \$\begingroup\$ Really clear explanation, exactly what I was hoping for, thanks. Since I'm trying to understand the concept underneath as well, is it general practice to inject small \$\Delta V\$ across resistors which are not obviously wired (to ground or Vcc) and then work out the effective resistance? Also, effective resistance wired to what, Gnd (why?)? So the input impedance is a high pass [C value, 300||6.8k] ? \$\endgroup\$ – user42875 Apr 1 '14 at 17:00
  • \$\begingroup\$ Sorry can't vote up, but I'll accept the answer as soon as I get answers to those obviously. \$\endgroup\$ – user42875 Apr 1 '14 at 17:01
  • \$\begingroup\$ Yes, the concept of small deltas in voltage and current is a key concept in what is known as "small signal analysis", in which the operation of a circuit is "linearized" around an operating point. This greatly simplifies the math, since any fixed sources such as power supplies effectively drop out of the equations. \$\endgroup\$ – Dave Tweed Apr 1 '14 at 17:16
  • \$\begingroup\$ And yes, the effective resistance is with respect to whatever the reference for V_in itself is; in this case, ground. The overall input impedance of the amplifier is the effective resistance of R1 in parallel with R2 in parallel with the input resistance of the transistor itself. Since the first of these is by far the smallest value, it dominates. All of this is in series with the coupling capacitor. \$\endgroup\$ – Dave Tweed Apr 1 '14 at 17:19

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