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

enter image description here

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?

  • \$\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

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\$).


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.


simulate this circuit

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.