## PSN-L Email List Message

Subject: Re: two-stage Roberts circuit amplifier
From: ChrisAtUpw@.......
Date: Sun, 27 Dec 2009 23:48:21 EST

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In a message dated 28/12/2009, barry_lotz@............. writes:

Chris

"   Chris
I found the  abstract and filter diagram on John Lahr's web site. 1) I am
not sure  how the  max gain is chosen. Is it based on the difference between
the actual sensor low frequency corner and the desired low frequency
corner given a 40 db /decade dropoff?

Hi Barry,

If you  wished to get 20 seconds from a 1.5 second pendulum, you would
need a  total gain at 20 seconds of 178. Each stage will need a gain of  13.3

2) I assume the two  single pole circuits identical

That is  correct

3)  Do the other values in  the example circuit remain the same except as
you have commented  on?

Correct.

4) If I want a lower  corner frequency of 0.03 hz is the "lower turnover
frequency (1/2 Pi C1  R2)" = 0.015 hz? It is probably obvious I don't have
the full  report."

See above.   correct

I realize that R3,R7 &  R11 are offset resistors.

Correct
R5 = R6 = R8 = R9

R10 may be chosen to give any gain that you need. I use a gain of  at least
x2 per stage

How does one determine the value of R5 and R8 or does it  matter?

amplifier, but you want to use values which have  low noise. 150 K is too much in
my opinion.

This will give a high maximum  gain and noise below 0.03 Hz, so you
couple the circuits with a 2 pole high pass  filter set at about 0.015 Hz.

Regards,

Chris  Chapman

In a message dated 28/12/2009, barry_lotz@............. writes:

Chris" =20
Chris     I found=
the=20
abstract and filter diagram on John Lahr's web site. 1) I am not=
sure=20
how the  max gain is chosen. Is it based on the difference be=
tween=20
the actual sensor low frequency corner and the desired low frequen=
cy=20
corner given a 40 db /decade dropoff?

Hi Barry,

=
;If you=20
wished to get 20 seconds from a 1.5 second pendulum, you would nee=
d a=20
total gain at 20 seconds of 178. Each stage will need a gain of=20
13.3

2) I assume the=
two=20
single pole circuits identical

=
;That is=20
correct

3)  Do the other=
values in=20
the example circuit remain the same except as you have commented=
=20
on?

Correct.

4) If I want a lo=
wer=20
corner frequency of 0.03 hz is the "lower turnover frequency (1/2=
Pi C1=20
R2)" =3D 0.015 hz? It is probably obvious I don't have the full=20
report."

See above. =20
correct
I realize that R3,R7=
&=20
R11 are offset resistors.

Correct
R5 =3D R6 =3D R8 =3D R9

R10 may be chosen to give any gain that you need. I use a gai=
n of=20
at least x2 per stage

How does one determine the value of R5 and R8 or does it=20
matter?

I choose values of about 20 K.=
=20
You don't want to overload your amplifier, but you want to use values whic=
h have=20
low noise. 150 K is too much in my opinion.

This will give a high max=
imum=20
gain and noise below 0.03 Hz, so you couple the circuits with a 2 pole hig=
h pass=20
filter set at about 0.015 Hz.

Regards,

Chris=20
Chapman
```