Current mirroring in diff amplifier

Question

Hi , I tried to design simple current mirror (load) based differential amplifier. I was trying to make it for optimum swing limit. In the first image as you see I used an ideal current source . I was able to make all transistors in saturation with ideal current source having an input common mode of 400mv. When I tried to use current source via current mirror instead of ideal current source at tail, I found out that I am unable to sink the same amount of current when I used an ideal current source . Can anyone tell me why is the reason for it. What I tried was to increase the current in fig 2. by raising input common mode to 625mV. Is any intuitive reason for the same ? When I used ideal current source AC gnd point i.e. point X in fig 2 was at 90mV. VGS of MO was 370mV and VT around 325m. Even though I mirrored with proper ratio with respect to current source , I was unable to match my requirement. When I tried with increased common mode I was able to achieve my goal but I couldn’t understand the intuition behind it.

One more question is while designing opamp do we need to maintain specific voltage overdrive for maximum swing limit or how is it chosen? If it’s specific should we use formula for the design ?

PS.:I used min length for current source

Answer 1

It would be easier to see a specific simulation, but you gave enough information here to at least figure out the issue.

All devices need to have a minimum amount of \$V_{ds}\$ (drain source) voltage to be in saturation, otherwise your mirror will not mirror properly. The fact that you noted you originally had a very small drain source voltage at the mirror drain (90mv) and after lifting the common mode, it worked, tells me you likely did not have enough voltage to have \$M_{0}\$ saturate. You need at least 1 \$V_{GS}\$ for \$M_{1,2}\$, and 1 \$deltaV\$ across \$M_0\$ to saturate.

Some ways to deal with this.

1. Increase common mode voltage until you have enough to at least provide 1 \$V_{GS}\$ to \$M_{1,2}\$ (that's 1 \$V_{T}\$ + 1 \$deltaV\$) and one \$deltaV\$ across \$M_{0}\$. This is what you did, but didn't understand why.
2. You can increase the device widths to lower \$deltaV\$ requirement for the same current on each side of mirror.

Note. \$deltaV\$ is the overdrive or \$V_{GS}\$-\$V_{T}\$ of Fet. Where \$deltaV >= V_{GS}\$-\$V_{T}\$ is a requirement for saturation

Other questions. Swing limit is maximized by lowering \$deltaV\$ in device design. And at least 2-3 times minimum length is usually used for mirrors for matching and impedance reasons.