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I would look into the schematic and see voltages and currents to compare FF vs TT to find the reason. The bias current that biases badgap (opamp that is used in the bandgap) varies with corners and it affects the final bandgap voltage.
If I remember correctly the .dc sweep must be done from min to max value (voltage range on X-axis) with a given step and then back. .The op point calculated for a given step should be used to calculate .op for the next step.
In my opinion, they are equal. Make distortion on Y-axis and X-axis and see that both matchings compensate the distortion in the same way. For example distortion on X-axis (rises in the right direction by 1):
1 2 3 4
1 2 3 4
and the distortion on Y-axis (rises down by 1):
0 0 0 0
1 1 1 1...
For opamp design, yes but you need PGA so an opamp with programmable gain as:
- Vin = +/-1 V -> Vout = +/- 3.3 V -> gain = 3.3V/V
- Vin = +/-3.3 V -> Vout = +/- 3.3 V -> gain = 1V/V
For comparator your picture is true, except the fact that output should be a rectangle signal...
If you want to design CMOS IC opamp (because it is not clear to me what are you want to do) an example list of simulations may be as follows:
Source: https://payhip.com/b/5Srt ("Preview" button in the top right)
It is rather full list of possible simulations. It depends on the design, to...
It is a hard question due to the complexity of your architecture. Start with L = Lmin or L = 2 x Lmin so for high voltage domain 3.3V that would be 0.35 um or 0.7 um, respectively.
Why you are using such architecture? Cannot you start with the simpler variation of your architecture and then...
You would rather want to size them in a following manner:
P7 = P8
N5 = N6
By selecting W/L of N6 and P8, you set the gate voltages of output transistors N9 and P11. Hence you are rather looking for such situation, where:
- ID_N6 = ID_P8 = 1/2 ID_P6
- output quiescence currents ID_N9 = ID_P11...
All IPs are used. Otherwise, companies would not sell them. Try looking for companies that offer IPs to see what is the market. USB, MIPI, HDMI ... so on, so on.
It's like with your current mirror in another topic. Just depends and different persons use different numbers. Just go with 10, then try 20 and see whether everything is ok. In case of layout if you break W = 100 um into 10 x 10 um, everything should be ok. However, if you break W = 100 um into...
Hard to say for me, as the last time I used BJT except bandgap was years ago during studies. I would need some time to verify your calculations. Just try to run simulation and verify them.
If there is anything else I could help you, please write. :)
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