mosfet widlar circuit

I need help solving this problem. I believe the current in the left side of the circuit equals the current in the right side. Therefore, $$K(V_{gs1} - V_t)^2 = 4K(V_{gs2} - V_t)^2$$

I also know \$V_{gs1} = V_{d1}\$, the voltage from the drain of M1 to the bottom rail $$V_{gs2} = V_{d1} - V_r$$ where \$V_r = I_{d}R\$

$$g_m = 2\sqrt{I_{d}K}$$

but there seems to be no way of solving the equation. Any help would be greatly appreciated.

  • \$\begingroup\$ Pro tip: an exclamation mark before the placeholder for the image will display it instead of the link. You can see it in the revision history. ;) \$\endgroup\$ – clabacchio Nov 5 '14 at 14:41

I am not sure if this actually answers the question the right way, but it might help to think about \$M_2\$ as a source follower:

The current through \$M_2\$ is \$V_{s2}/R\$. This current is mirrored by the cascode PMOS current mirror, so:

$$ I_{ds1} = \frac{V_{s2}}{R} $$

Now \$V_{gs,2}\$ is pretty constant. (If \$M_2\$ is acting like a source follower.) Therefore:

$$ g_{m1} \equiv \frac{\partial I_{ds1}}{\partial V_{gs1}} = \frac{\partial I_{ds1}}{\partial V_{g2}} \approx \frac{\partial I_{ds1}}{\partial V_{s2}} = \frac{1}{R} $$

I suppose there are a few conditions necessary for that to work, but when they are met it seems like \$g_{m1}\$ is (almost) independent of W/L matching in the bottom NMOSes.

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Assume the current flow in M1 and M2 are equal to \$I_{D}\$, and M1 and M2 are matched.

$$ V_{GS1} = V_{GS2} + I_{D}R\\ \sqrt{\frac{I_{D}}{K}}+V_{T}=\frac{1}{2}\sqrt{\frac{I_{D}}{K}}+V_{T}+I_{D}R $$ Solve for K $$ K=\frac{1}{4I_{D}R^2}\\ V_{GS1}=\sqrt{\frac{I_{D}}{K}}+V_{T}=2I_{D}R+V_{T}\\ V_{GS2}=\frac{1}{2}\sqrt{\frac{I_{D}}{K}}+V_{T}=I_{D}R+V_{T} $$

So, my result is

$$ g_{m1}=\frac{1}{2R}\\ g_{m2}=\frac{1}{R} $$

What's wrong with me, or what's wrong with the book?


\$V_{GS1}\$ is equal \$V_{GS2}\$ plus the resistor voltage drop, then if \$V_{GS1}\$ changed, so the current, so \$V_{GS2}\$ must change too. If the book is right, that is, \$g_{m1}=\frac{1}{R}\$, then \$V_{GS2}\$ should not change!!!! Is this possible?

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  • \$\begingroup\$ That is the entire point of the circuit. A negative feedback loop is created to ensure that the current does not change with PVT variations, thereby making it a good current source. \$\endgroup\$ – CaliSax Dec 31 '14 at 6:12

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