paley said:because the input node is a dominate pole generally, the gain of amp in the dominate pole frequency be equal to DC gain approximately, so the input capcitance is (1+A)C.
but the output capcitance isn't equal to C+Cout
A.Anand Srinivasan said:i dont think A is the dc gain.... actually the equivalent input capacitance is C(1-A) and since most amplifiers are inverting it becomes C(1+A) due to negative gain.... this is because the change in voltage between input and output is 1-A times Vin...
The DC gain is calculated without the cap because the cap has no purpose at that time....
A.Anand Srinivasan said:A is the gain of the system without the miller capacitor.... so all that you have to do to find the value of capacitor to use at the input at that frequency is to remove that miller cap and find the value of gain at that frequency and then use the multiplication factor and place the capacitor....
Undertand that,but this doesnt explain why can we remove miller cap(the feedback one that crossover)when we calculating A, -_-#A.Anand Srinivasan said:assume that you have found the A at high frequency and now you are introducing a (1+A)C at the input... obviously the impedance of such a cap is gonna be very low at high frequency and obviously apply a voltage divider rule between the source resistance and the capacitors impedance you will find that the input is very attenuated.... this is why the 1+A component is introduced...
the 1+A factor takes care of reducing the input and makes the output what it would be if the miller cap was present...
leohart said:Hi,when we consider a miller capcitance C connect between input and output,what we do is remove the capcitance and calculate the dc gain -A then the miller capcitance at the input is effectively (1+A)C.
Here comes my question,why should we use dc gain(so can we remove miller capcitance when calculating gain) here?If the frequency is high enough,the gain will change a lot if we remove the miller capcitance,so does it mean the effective miller capcitance equals (1+A)C seen at input only valids at low frequency?What should we do if we are dealing with very high F?
Seems reasonable to me and something is there,but I cannot grab the connection between "the main pole of whole opamp is much smaller than the pole of second stage" and "so we can use dc gain to calculate the effective miller capacitance"...caosl said:leohart said:Hi,when we consider a miller capcitance C connect between input and output,what we do is remove the capcitance and calculate the dc gain -A then the miller capcitance at the input is effectively (1+A)C.
Here comes my question,why should we use dc gain(so can we remove miller capcitance when calculating gain) here?If the frequency is high enough,the gain will change a lot if we remove the miller capcitance,so does it mean the effective miller capcitance equals (1+A)C seen at input only valids at low frequency?What should we do if we are dealing with very high F?
The gain should be full frequency gain, but actually, after miller effect, the main pole of whole opamp is much smaller than the pole of second stage, as shown in the figure. So, we often use dc gain to calculate the miller effect equivalent capacitance.
force of nature said:Hello all,
I have a question regarding the applicability of the Miller effect (multiplying by 1-Av to put in the input node). I remember that there was a situation where one can not apply Miller effect but I can not remember. If anyone can help, I appreciate. Also, can we apply Miller effect to the circuit that I draw in the link below?
Thank you very much. [/img]
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