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The following is from the book Design of Analog CMOS Integrated Circuit, page 207.

Find the transfer function of the circuit in Fig. 6.47(a):

enter image description here

Suppose the voltage at node A is \$V_x\$.

with the use of KCL at node B,

$$V_x g_{m1} + \frac{V_{out}} {r_o} = (V_x - V_{out}) s C_F - \frac{V_{out}} {R_D}$$

For \$V_x\$, with voltage divider (\$R_S\$ and \$C_F\$).

$$V_x = \frac{V_{in} - V_{out}} {R_S + 1/(s C_F)} \frac{1} {s C_F} + V_{out} $$

$$\Rightarrow \frac{V_{out}} {V_{in}} = \frac{(s C_F - g_m)} {s C_F R_S (g_m + 1/r_o + 1/R_D) + (s C_F + 1/r_o + 1/R_D)}$$

However, the arthor gave this.

$$G(S) = -g_m (R_D \parallel r_o) \frac{1- \frac{1} {g_m} C_F s} {1 + \biggr[(1 + g_m R_D) R_S + R_D \biggr] C_F s}$$

The zero is the same but the pole is different.

kile
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    If you check the book's derivation, you can see that the author ignores ro in their derivation of the driving point impedance. The exact Zin,0 should be [(1 + gm * Rs) * (RD || ro) + Rs]. – internet Jan 16 '24 at 19:17
  • @internet Which is more accurate? Should we ignore $r_o$ ? – kile Jan 16 '24 at 19:18
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    I don't understand your question. I provided the exact expression. How can something be more accurate than the exact one? Also, whether you can ignore ro or not depends on the relative values of RD and ro. As you can see, there is the term RD || ro. If RD is much smaller than ro, you can ignore ro. – internet Jan 16 '24 at 19:23
  • Hopefully, from this example, you can see the usefulness of EET, which can provide more insight than the brute force method you used. – internet Jan 16 '24 at 19:36
  • @internet I think when you use the exact $Z_{in,0}$, our fomula is basically the same thing, Do you agree with me? BTW, how do you know when we should ignore $r_o$? – kile Jan 16 '24 at 19:51
  • See my comment above. – internet Jan 16 '24 at 19:52
  • @internet I know that when $R_D$ is much smaller than $r_o$ when we can use $(R_D \parallel r_o) \approx R_D$. However, the question normally won't tell us whether $R_D$ is much smaller than $r_o$. (For instance, the author never mentioned in this question.) – kile Jan 16 '24 at 19:56

1 Answers1

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If you check the book's derivation, you can see that the author ignores \$r_o\$ in their derivation of the driving point impedance. The exact \$Z_{in,0}\$ should be \$(1 + g_m \cdot R_s) \cdot (R_D \,||\, r_o) + R_s\$.

Also, whether you can ignore \$r_o\$ or not depends on the relative values of \$R_D\$ and \$r_o\$. As you can see, there is the term \$R_D \,||\, r_o\$. If \$R_D\$ is much smaller than \$r_o\$, you can ignore \$r_o\$. In some cases, for example, if the load \$R_D\$ is a current source which is comparable in magnitude to \$r_o\$, then you can't ignore \$r_o\$.
Hopefully, from this example, you can see the usefulness of the Extra Element Theorem (EET), which can provide more insight than the brute force method you used.

internet
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