Millers Impedance ?

[Kristine Hyvang]

When there is an impedance connecting the input side to the output side
across an amplifier, there is an expression saying :


Zmiller,in = ZF/(1-Av)

where ZF is the impedance connecting input to output, and Av is the voltage
gain from input to output. Now the Millers impedance is from input to ground
(without any impedance connecting the input to the output), and the model
looks different. That's allright.

but.......

when looking on the output side,

Zmiller,out=(Zf*Av)/(Av-1) (according to the book)

......what I don't grasp is the denominator .... Av-1.

When I analyze this I get:

ZFmiller,out=(Zf*Av)/ (1-Av)


models :

Impedance connecting input and output:

IF------->
-------------------ZF---------------------
---
|
|
|
|
|
|
Vin-------------
Av -------------------Vout




gnd-----------------------------------------gnd



Millers impedance:

Vin -----------
v -------------------Vout
|
|
|
|
Zmiller,in
ZMiller,out
|
|
|
|


gnd-----------------------------------------gnd


If I add a minus to the ZMiller,out-expression it adds up,

but that means that the current enters the amplifier from both ends, and
that doesn't make sense !

Or is this a definition of input- and output- impedances..........that the
current allways enters the amplifier from both ends ?

 

[Alan Rutlidge]

"Kristine Hyvang" <kristine.hyvang@chello.no> wrote in message
news:8DxWb.3902$O41.96162@amstwist00...

When there is an impedance connecting the input side to the output side
across an amplifier, there is an expression saying :


Zmiller,in = ZF/(1-Av)

where ZF is the impedance connecting input to output, and Av is the
voltage
gain from input to output. Now the Millers impedance is from input to
ground
(without any impedance connecting the input to the output), and the model
looks different. That's allright.

but.......

when looking on the output side,

Zmiller,out=(Zf*Av)/(Av-1) (according to the book)

.....what I don't grasp is the denominator .... Av-1.

When I analyze this I get:

ZFmiller,out=(Zf*Av)/ (1-Av)


models :

Impedance connecting input and output:

IF-------
-------------------ZF-------------------
--
---
|
|
|
|
|
|
Vin-------------
Av -------------------Vout




gnd-----------------------------------------gnd



Millers impedance:

Vin -----------
v -------------------Vout
|
|
|
|
Zmiller,in
ZMiller,out
|
|
|
|


gnd-----------------------------------------gnd


If I add a minus to the ZMiller,out-expression it adds up,

but that means that the current enters the amplifier from both ends, and
that doesn't make sense !

Or is this a definition of input- and output- impedances..........that
the
current allways enters the amplifier from both ends ?

Consider visiting this site -
http://www-inst.eecs.berkeley.edu/~ee105/fall03/handouts/lectures/Lecture19.pdf
Slides 16, 17 and 18 may help explain your observations.

Cheers,
Alan