Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

help needed about miller effect and feedback theory

Status
Not open for further replies.

qslazio

Full Member level 3
Joined
May 23, 2004
Messages
175
Helped
18
Reputation
36
Reaction score
7
Trophy points
1,298
Activity points
1,420
This is from gray's book<analysis and design of analog integrated circuit>page 567 figure 8.15b.

Now as I calculate the output impedance of the circuit with feedback following the steps in gray's book.
the output impedance will be decreased to (Zo//Rf//RL)/(1+T),and the T is loop gain of the circuit.

But on the other hand,when i follows miller equivalence,the resistance looking into Rf from output is Rf/(1-1/A),the A is the circuit voltage gain without Rf.Then the output impedance of the circuit with Rf should be (Zo//(Rf/(1-1/A))//RL),which is very close to Zo//Rf//RL when A is large.

Now these two method has inconsistency with the output impedance of circuit with feedback.

So I'd like to know why miller equivalence's result does not match the feedback one.

This is the schematic of the circuit
**broken link removed**
thanks!
 

The feedback resistor looks like Rf/Av to a source *feeding* the amp. Looking from the output the feedback resistor is just Rf.
 

Beetroots said:
The feedback resistor looks like Rf/Av to a source *feeding* the amp. Looking from the output the feedback resistor is just Rf.

yes,if the A >>1 so Rf/(1-A)~Rf/(-A) and Rf/(1-1/A)~Rf
this the miller effect impedance
but as the feedback point, negative feedback will reduce the output impedance by (1+T),which is not consistent the miller effect above ,so that is what i can not figure out yet
 

you mixed up the standing points. Who is your instructor?
 

qslazio's caculation is right.

It is an approximate result in the text book. As normal, the a is larger than ten, so the result in the text book is no worse than the result by qslazio.

What the text do is for simple the result.
 

Status
Not open for further replies.

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top