about transfer function?

[qutang]

how to calculate the switch-capacitor's transfer function in a continuous system, it is controled by the output square-wave as a frequency to voltage.

 

[timo67]

Hi qutang
would you refomulate your question , I dont understand it well.

 

[qutang]

to timo67:
this pic. is my system. F2V is used to trans frequency to voltage , k is the input frequency, k1 & k2 are two small pulse ,which are produced by k using the and-gate delay cell .
now my system is stable after ringing a long time, my opamp's PM is about 70 degree.
I want to get the transfer function to find a way of stabilize the system, but I cant get the function of this F2V for it is a discontinous time signal. that's my puzzle.

 

[jamesko]

I think you have to redefine the gain of transfer function of the following blocks:
(1)VCO
(2)F2V
(3)1/N
They should have a linear transfer function in continuous time domain.
Then you can use then with the OPAMP to run the phase margin simulation,

 

[timo67]

Hi qutang
The voltage signal V results on the integration of the current Ic in c2 when both k and k1 are on.
If Ic is a constant positive current V is always increasing
The dérivative of V is proportionnal to the frequency f

 

[qutang]

timo67:
sorry,
k when "0" is close, "1" is open.
k1 & k2 when "1" is close, "0" is open.
so V will be the value of Ic*c1 (v) at last.
as jamesko said :what is this F2V's linear transfer function ?

 

[jamesko]

Enquiry: Re: about transfer function?
how can I calculate the transfer function of 1/n,it just 4-frequency-divider using D flip-flop.
>> the transfer function of 1/n=Output/Input=1/n
I can't get the function of the F2V's function
>> Maybe you should specify the relation between k, k1, k2 and f first,
and then you should be able to derive the equation of Output/Input.
Because it's F2V, I think if the freq is higher, the voltage is higherm right?
Then there should be a linear relation between output freq & input voltage.
For ex: if in=1KHz, out=0.5V & in =2KHz , out=1V
then Output/Input=0.5V/1KHz=0.5(V/KHz)

 

[qutang]

jamesko:
for the frequency is higher, the voltage will be lower.
k when "0" is close, "1" is open.
k1 & k2 when "1" is close, "0" is open.
so V =Ic/(2*f*c1)
but my system can't be stable at ff corner,my opamp PM is about 75.
How can I adjust the system?
increase charge current or reduce 1/n?

 

[jamesko]

(1) V =Ic/(2*f*c1)
(2) but my system can't be stable at ff corner,my opamp PM is about 75.
How can I adjust the system?
>> I think you can descrease the open loop gain.
For ex: descrease gain of F2V or 1/n

(3) increase charge current or reduce 1/n?
Reduce 1/n should work.
Increase charge current -> F2V=Vout/Freqin=Ic/(2Ca) => It will increase gain of F2V?
So if equation (1) stands, maybe you should reduce Ic, instead of increasing Ic.

 

[qutang]

I will try it.