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So in Sedra and Smith, it is stated that the common mode input resistance of a differential amplifier is

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from the diagram(s)

enter image description here

So it is stated in the book they have used the diagram on the right (equivalent common-mode half circuit) to prove 2Ricm. I've tried to prove this, and I've gotten about half of the equation, with the other half being a function of re (which is not in their answer). Would anybody be able to help me prove this and show the steps involved?

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  • \$\begingroup\$ This is on page 636, Figure 9.27, 7th Edition for anyone who wants to look up any background context. \$\endgroup\$ – user103380 Oct 9 '18 at 3:45
  • \$\begingroup\$ Please share your steps so far. \$\endgroup\$ – user94729 Oct 9 '18 at 4:42
  • \$\begingroup\$ Is this a homework assignment? (It sound like it.) Either way, you should always post what you have already done. \$\endgroup\$ – not2qubit Jun 2 '19 at 10:42
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Generally speaking, you would draw the small-signal schematic and solve for the input impedance using regular KCL/KVL equations.

But looking at the shape of the equation, I'd say they used Middlebrook's Extra-Element Theorem on \$r_0\$. In other words, you first determine the input impedance of the following circuit without \$r_0\$ (ie. \$r_0 \to \infty\$).

schematic

simulate this circuit – Schematic created using CircuitLab

$$Z_{in}^\infty = 2R_{icm}^\infty = \frac{V}{i_B}$$

$$\begin{align} i_B &= \frac{V}{2R_{EE}} - \beta i_B \\ &\Downarrow \\ i_B &= \frac{V}{(1+\beta) 2R_{EE}}\approx\frac{V}{2\beta R_{EE}} \\ &\Downarrow \\ R_{icm}^\infty &\approx \beta R_{EE} \end{align}$$

We can find the driving point impedances in a relatively simple way.

schematic

simulate this circuit

$$\begin{align} Z_n &= \frac{R_C}{1+\beta} \approx \frac{R_C}{\beta} \\ Z_d &= R_C + 2R_EE \\ R_{icm} &= R_{icm}^\infty\frac{1 + \frac{Z_n}{r_0}}{1 + \frac{Z_d}{r_0}} &\Downarrow \\ R_{icm} &\approx \beta R_{EE}\frac{1 + \frac{R_C}{\beta r_0}}{1 + \frac{R_C + 2R_{EE}}{r_0}} \end{align}$$

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