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If you consider the feedback as a voltage signal (i.e output of the source follower), then the input must also be a voltage. To implement the feedback, the input voltage and the feedback voltage must be added. Two voltages cannot be added by connecting them together (as in this circuit), instead...
Loop gain is 1+gm3*R. Input impedance is (1/gm2) /(1+gm3*R). Output impedance is R/(1+gm3*R). The circuit is a current controlled voltage source. Voltage is sensed at the output and current is fedback at the input. You can also find the impedances by usual methods of KVL/KCL. But while...
The code seems to be right ( except perhaps a typo in the first line in Output stage- (vcm) instead of V(vcm) ).
With resistive feedback, there should not be any problem. However you cannot use an opamp as in your test bench, without any feedback for DC. However, if you just want to see the...
In addition to the condition,Gm*R2>>C1*CL/(C2*C2), the other condition for the expression to be valid is that both C1 and CL >> C2. Examples you have taken do not satisfy these. If you do not want the constraint on CL (as in your examples), the approx transfer function has to be modified.(...
An approximate transfer function gives an intuitive understanding of the location of poles and zeros. Thus there location can be found by just looking at the circuit , without doing much maths.
To have reasonable passband fh>>fl gives Gm*R2>>C1*CL/(C2*C2).
Also under the condition C1 and CL >>...
The given transfer function is valid only under certain condition. Some observations
1) Even though there are three capacitances C1 ,C2 and CL, it will be a second order system (as pointed earlier) as the the capacitances are in series and hence only two independent conditions can be...
If you neglect the base currents, PTAT current expression comes as follows:
Same current I flows through Q1 and Q2, So
Vbe1 - Vbe2= Vt ln(A3/A1) , where Vbe1 is the base -emitter voltage of Q1 and like wise.
or Vb1 -Ve1 -Vb2+Ve2=Vt ln(A3/A1) or Vb1-Ve1-Vb2=Vt ln (A3/A1)
Similarly for Q3 and...
You may try the following things
1) Chose the test input signal bin such that the third harmonic also lies in the signal bandwidth.
2) Now verify with ideal DAC that you get the SNDR.
3) In the real DAC, replace the two mirrors with ideal current sources. Does the SNDR degrade?
4) In the real...
Sampling instant is determined by Φ1p only. The reason being after Φ1p goes down, voltage across the sampling capacitor Cs cannot change because one of its terminal is floating. This is also the reason why there is no charge injection from the sampling switch.
You refer to Razavi book. (chapter on noise and mismatch).
Its not "more" precise. Random variables are characterized by both mean and variance. You can also search about gaussian random variables.
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