# Doubts and questions about miller effect

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#### leohart

##### Full Member level 4 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?

#### paley

##### Member level 4 miller cap effect

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

### leohart

points: 2

#### A.Anand Srinivasan 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....

### leohart

points: 2

#### leohart

##### Full Member level 4 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
Your comments confirm mine with the part "at enough low freq that we can use dc gain A to calculate effective input capcitance (1+A)C",but what about high freq that we cannot just remove the C cross over input and output(leave or remove it will bring substantially change in gain at that freq)?

And what is the output capcitance exactly?

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....
Hi~Do you mean that we should always use gain at that freq instead dc gain?In order to calculate this gain,can I remove the feedback C over input/output?
I think if we use gain at particular freq then A varies with freq,so the effective miller cap varies with freq too!

I was confused when reading sedra&smith for the part of circuitry inside op741,they do remove compensation C and use dc gain of the second stage for this calculation,and the effective miller cap at input is constant and used in all freq for PZ analyze...

#### A.Anand Srinivasan 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....

### leohart

points: 2

#### psmon

##### Full Member level 3 At DC, miller effect doesn't come into play. It's only for AC.

### leohart

points: 2

#### leohart

##### Full Member level 4 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....
But miller effect comes from both side of the cap change simultanouesly,so the A used should always be the A that the cap is experiencing...At high freq.,remove miller capacitor then the "A" is not the really A which miller cap is experiencing anymore...

That's what confuses me...

#### A.Anand Srinivasan 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

##### Full Member level 4 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...
Undertand that,but this doesnt explain why can we remove miller cap(the feedback one that crossover)when we calculating A, -_-#

#### A.Anand Srinivasan try reading network analysis by van walkenberg.... actually i remember that there they have talked about this pretty clearly... it is actually a derivation and proof... i cant key in it here...

#### caosl 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.

### leohart

points: 2

#### leohart

##### Full Member level 4 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.
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"...
Maybe my lacking of insight understanding of PZ analysis...

#### paley

##### Member level 4 ### leohart

points: 2

#### force of nature

##### Newbie level 6 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]

#### BAT_MAN

##### Member level 5 We donot use miller effect when we have resistance at the emitter as it changes the conditions that are necessary for the miller theorm. tc

#### spring1860

##### Member level 5 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]
i remebered allen's book says: when there is only one signal path between input and output , miller effect isn't applied.

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