# Widlar current source 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.

• 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. ;) – 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.

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?

Guessing:

$V_{GS1}$ 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?

• 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. – CaliSax Dec 31 '14 at 6:12