current mirror Cascode Structure

[sys_eng]

1)From this picture, why they label M2 VGS 0.9-1.5V? Shouldn't that be VDS instead of VGS?

2)From the equation why VGS1=VT1+VDSAT1?





3)From above, why they said voltage swing of cascode is reduced?? I seen the output voltage got increased. Vo , increased substantially.

 

[kemiyun]

1) In this configuration (assuming simple model for transistor) Vgs of M2 is equal to Vds of M2. Follow the biasing path, the source of M3 sees Vgs of M1, and drain of M3 sees Vgs of M1 + Vgs of M3. If M3 and M4 are matched meaning they have the same overdrive, the source of M4 will be Vgs of M1, which is equal to Vds of M2. But you are right it's not a nice way of annotating it.

2) Because that's the definition of Vgs, Vgs is the threshold + Vdssat of a transistor. Of course this only applies to simpler models.

3) I think that refers to double cascoding. The output resistance goes high naturally, but when they talk about swing they talk about the minimum voltage that can be applied to the drain of M4 without making any of the transistors go out of saturation. This voltage is significantly lower than it was in a single cascode.

This exercise is probably given to help you understand biasing of cascodes. You need to bias them properly so that you can let the output node take a wide range of voltages, because that's what is expected from a current source. Output current should be independent of the output voltage (in this case both nodes are drain of the cascode). If your devices leave saturation they can't maintain same current for different voltages.

 

[sys_eng]

1) In this configuration (assuming simple model for transistor) Vgs of M2 is equal to Vds of M2. Follow the biasing path, the source of M3 sees Vgs of M1, and drain of M3 sees Vgs of M1 + Vgs of M3. If M3 and M4 are matched meaning they have the same overdrive, the source of M4 will be Vgs of M1, which is equal to Vds of M2. But you are right it's not a nice way of annotating it.
.

well Vds never equal to Vgs. In saturation Vds=Vgs-Vt.
I don't see how Vfs of M2 =Vds of M2.

 

[kemiyun]

Vds never equal to Vgs? Vds of M2 is forced by the Vgs of M4. Vds = Vgs-Vt is the limit for saturation, the gain of the transfer functions starting from drain are really low so drain voltage does not enforce Vgs. Rather you make sure that Vds never reaches its limits.

Just solve the DC bias problem and you'll see that the source of M4 (drain of M2) will be at the voltage equal to Vgs of M1. And Vds of M2 is therefore equal to Vgs of M1 which is the way it's annotated in the picture which is a bad annotation. This means Vds of M2 will be equal to Vgs of M1 which is a waste because it doesn't need that much of a headroom, it just needs Vgs-Vt to make sure that it stays in saturation. If only there had been a way to force the drain of M2 at a lower voltage (there is).