help me with SC feedback and Settling!

[qslazio]

how to calculate feedback factor

I'm reading the Ph.d. thesis of Arnold Feldman.
"High-Speed, Low-Power Sigma-Delta Modulators for RF baseband Channel Applications"

On Page 45 and 46(you can see it in the below figure)

He assumed a simple single transistor model for class A SC integrator.
And in equation 3-22. He said the feedback factor of the circuit is
f=Cf/(Cf+Cs+Cgs).
It really confused me.
First, I think it is a shunt-shunt feedback configuration. So the feedback is shunt and the feedback sensing is shunt, too. For the feedback network, the input is voltage(Vout), the output is current(If). The unit of the feedback factor should be transconductance(G). But the thesis said it is unitless.

Secondly on page 46, equation 3-24, tau=Cl,tot/(f*gm)
How comes it?
In my opinion, in a closed loop system, if the gain is reduced by (1+T), the -3db pole should move to (1+T) further. Right? So the tau should be Cl,tot/((1+T)*gm)
(T is the loop gain)

Can anyone explain these two question to me?
That will help a lot.
Thanks you guys!

 

[rambus_ddr]

First, when in the input is no other loading except these three capacitor, you can use the feedback current multiple the impedance to get the feedback voltage at input of mos. Which is caluated the ouput impedance at drain of mos for convenience.

second, use the above results, the output impedance at the drain of mos is 1/(gm*f), and there capacitor is Cl, so t=r*c=Cl/(gm*f).

 

[ezt]

Hi.
I think that's simple. don't confuse yourself with feedback equations. As you can see in that figure, you can calculate feedback factor by a simple voltage division. feedback factor (f) equals the voltage across Cs to ground so we have

f=Zs/(Zs+Zf)
where Zs is the total impedance between the gate of ransistor and ground and Zf is the total feedback impedance. So
Zs = 1/(Cs+Cgs)ω because Cs and Cgs are in parallel.
Zf = 1/Cfω
Hence f=Cf/Cf+Cs+Cgs

And by the way for the second equation, as you know
time constant = 1/ω-3dB = 1/f*ωu
because ωu=f*ω-3dB
and we know
ωu=gm/CL,tot
Hence
time constant = CL,tot/f*gm

Regards,
EZT

 

[jordan76]

Hi ezt,

You might mean ωu=ω-3dB/f, right? :wink:

regards,
jordan76

 

[mists]

I think it is true for feedback factor equal 1.

Hi ezt,

You might mean ωu=ω-3dB/f, right? :wink:

regards,
jordan76

 

[qslazio]

First, when in the input is no other loading except these three capacitor, you can use the feedback current multiple the impedance to get the feedback voltage at input of mos. Which is caluated the ouput impedance at drain of mos for convenience.

second, use the above results, the output impedance at the drain of mos is 1/(gm*f), and there capacitor is Cl, so t=r*c=Cl/(gm*f).

thank you guys!

But I'm still confused about the series feedback. It should have voltage input ,and the feedback is voltage, right? So you will find the error voltage which is vin minus vf.But I cannot find it.
So in this configuration, I'm still wondering how we can regard it as a voltage feedback.
Thanks a lot!