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According to Razavi, \$R_{out}\$ is approximately \$r_{o2}||r_{o4}\$:

Excerpt from Razavi's "Design of Analog CMOS Integrated Circuits"

I get the same answer (assuming \$2r_{02} >> \frac{1}{g_{m3}}||r_{o3}\$):

Ralph's hand calculations of Rout

But how did he get equation 5.27? In particular, how is \$I_{xy} = I_{d4}\$. I.e. how did he get the \$2I_{xy}\$ term (where \$I_{xy}\$ is the current through \$R_{xy}\$)? I understand that M4 will copy M3's current, but won't the copying be unequal when considering channel length modulation and if \$V_{ds3} \neq V_{ds4}\$?

Ralph
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2 Answers2

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\$I_{d4} = I_{d3} + \frac{V_x}{R_{o4}}\$ because M3:M4 is a current mirror which would enforce \$I_{d3} = I_{d4}\$, except for the \$R_{o4}\$, which is the effect of channel length modulation.

None of the small-signal current flows through Iss, so \$I_{xy} = I_{d3}\$. Then, \$I_x = I_{xy} + I_{d4}\$ so \$I_x = I_{xy} + I_{xy} + V_x/R_{o4}\$.

τεκ
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  • If you neglect channel length modulation (and assuming matched devices), then yes, \$I_{d3}\$ = \$I_{d4}\$ because they have the same \$|V_{gs}|\$. But with channel length modulation, M3 and M4 must also have the same \$V_{ds}\$ in order for \$I_{d3}\$ to be equal to \$I_{d4}\$. – Ralph Mar 08 '18 at 01:05
  • @Ralph This is why there is an \$R_{o4}\$ term. That is the channel length modulation term. – τεκ Mar 08 '18 at 01:40
  • the small signal linearization of channel length modulation, that is. – τεκ Mar 08 '18 at 01:45
  • Yes, I know what \$r_{o4}\$ represents (I included it in my diagram and hand calculations; are they not legible?). When I solve for \$I_{d3}\$ and \$I_{d4}\$ using the small signal model I find no solution where they are equal. – Ralph Mar 08 '18 at 03:19
  • (unless you assume \$\frac{1}{g_{m3}} << r_{o3}\$ and \$\frac{1}{g_{m3}} << 2r_{o1}\$, which Razavi has not yet made in his derivation of equation 5.27) – Ralph Mar 08 '18 at 03:26
  • 1/gm << ro is a pretty safe assumption, for any reasonable transistor parameters. – τεκ Mar 08 '18 at 04:24
  • @Ralph Also assumed is gm4 = gm3, so gm4vgs4=gm4 ixy(1/gm3||ro3) \approx ixy. – τεκ Mar 08 '18 at 04:38
  • Yes, I put \$g_{m3}=g_{m4}\$ in my answer. Ah, I didn't think of it like that. Not sure why he doesn't mention that he ignored \$r_{o3}\$ for determining copying voltage \$v_{gs4}\$, even though he explicitly includes it in the current being copied. Update your answer and I will accept. – Ralph Mar 08 '18 at 05:44
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    @Ralph: it is a bit inconsistent with his assumption. He ignored ro3 for copying and actually he should ignore it it calculating current Ixy. However, whether ignoring it or not it doesn't affect much as ro3 is much larger than 1/gm3. – emnha Mar 08 '18 at 07:25
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Razavi forgot to mention that he's ignoring \$r_{o3}\$ while calculating \$i_{d4}\$. So in order to get equation 5.27:

$$I_x = I_{xy} + i_{d4} + \frac{V_x}{r_{o4}}$$

$$i_{d4} = g_{m4}v_{gs4}$$

As pointed out by τεκ, Razavi assumes \$r_{o3} >> \frac{1}{g_{m3}}\$:

$$v_{gs4} = I_{xy}(\frac{1}{g_{m3}}||r_{o3}) = I_{xy}\frac{1}{g_{m3}} $$

Ignoring \$r_{o3}\$ is a totally valid approximation, but Razavi never mentions it, despite including it in equation 5.27. Thus, assuming matched devices (\$g_{m3} = g_{m4}\$):

$$i_{d4} = g_{m4}I_{xy}\frac{1}{g_{m3}} = I_{xy} $$

$$ \therefore I_x = 2I_{xy} + \frac{V_x}{r_{o4}}$$

I actually emailed Professor Razavi about this and he mentioned that it's been fixed in the 2nd edition.

Ralph
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