Geoff,

 Blair here in Aust,

I do lots of (work) stuff with instrumentation amps, matching is certainly
easier using some adjustable components...0.01% resistors are about $30 each
here!

However, the increase in noise from using a 3 opamp IA config compared to a
traditional single input amp has to be weighed against the better common
mode rejection with a true instrumentation amp.
If you don't have to deal with Common Mode noise, then don't go down the IA
path..
There is much literature on the pluses and minuses of single opamp input
verse 3 opamp IA input designs from a seismic point of view.

With instrumentation amps (infact with all opamps), there has to be some
bias current path from the inputs back to ground, usually provided by the
excitation circuitry (strain gauge stuff) but if you are just 'going in'
with a floating signal from a coil, you will need to provide some bias
resistance / reference to ground, probably a pair of 1meg resistors would
suffice , one from each input to ground (0 volts) depending on the IA input
resistance, ofcourse, these could be your dampening resistors, the total
value being split and the centre going to ground...

I prefer the 1 megs as any mismatch in low resistance values here will
affect your CMR a lot.

So the dampening resistor is across the inputs with a couple of 1megs to
ground.

Need also to be careful to balance the capacitance of your input leads to
ground as well, but if you are using good quality 2 core screened microphone
cable (with the screen tied to grounded only at the amplifier end) that
should be fine for the frequencies of the signals you are looking at.
The other end of the screen (seismo end) would normally be tied to the
sensor's ground and should not be 'hard grounded'(eg. earth staked) unless
you are expecting lots of trouble from lightning.

I'd not bother with having a guard amp driving the screen for seismic
signals.

With careful design and correct choice of components, CMRRs of 150db at low
frequencies with respect to output are achievable without trimming but if
you have that much common mode noise you should do something about it!

The 1.25 hz periodic noise... what is your sampling rate and do you have
some sort of low pass filter that lobs everything off before you get to the
nyquist frequency (approx 1/2 your sampling rate)?

Also note that induced magnetic fields at the coil end are not common mode..

Blair

-----Original Message-----
From: psnlist-request@.............. [mailto:psnlist-request@...............
On Behalf Of Geoff
Sent: Friday, 1 July 2011 3:59 AM
To: psnlist@..............
Subject: Re: Damping CDR for HS10-1

I think I just wasted a lot of time
trying to get rid of a signal which
is real differential and not
common mode,
there seems to be a machine
owned by a neighbor which is not
always used.

I am totally unable to rid the artifact of
about 1.25 Hz which is periodical.

Creating the multiple resistor pairs
was a waste of time.
A pot, most likely, is the beat way
to balance the two against the ground.
Like 100 Ohm or 10 Ohm 15 turn
between two resistors matched already.
1 ohm is the best my DMM can do
without help.


I have trouble keeping my website
to be real time. So its not
100% reliable.

I think my website shows the
instrumentation amp with the geophone
on the negative leads of the instrumentation
amp, will have to change that
with a new schematic.


Regards,
geoff



-----Original Message----- 
From: Geoffrey 
Sent: Saturday, June 25, 2011 8:39 PM 
To: psnlist@.............. 
Subject: Re: Damping CDR for HS10-1 

Interesting Bob,

But I'm using an instrumentation amplifier.
In such an arrangement of three op amps
you are using two positive inputs which means
the input impedance is mega ohms to giga ohms.
The only input is the the resistors which are
split against ground. So in my case the you
have verified my numbers to be basically correct.

I have learned something new to myself in the past
few days about this input.

There seems to be common mode signals
of an electrical nature coming in on the
geophone input. The only way to balance out
this unwanted signal has been to
make several pairs of identical split resistors
and see which pair will after installed eliminate the problem.
It seems my test equipment can not resolve the measurements
fine enough to properly match these two resistors.
Therefore it is a matter of chance that the right
combination can be achieved.

I have never been able to do this balancing
act with any configuration other than an instrumentation
amplifier.

It is my ignorance in combination with
people who simply refuse to talk about this
which has caused me years of headaches.

In my case the Ge seems to reduce to
(2.99 * 1302)/1742 or 2.234 v/(in/sec)
But this is not how I handle this figure.
I treat it as an overall loss of 20log(2.234/2.99) or -2.53dbv
when calculating the final amplifier gain.

Thanks for your feedback.

Regards,
geoff



-----Original Message----- 
From: Bob McClure 
Sent: Saturday, June 25, 2011 6:11 PM 
To: psnlist@.............. 
Subject: Re: Damping CDR for HS10-1 

For whatever it is worth, here is my computation of the shunt resistance to
be applied to the HS-10 geophone to obtain a 
damping coefficient of 0.707. It confirms Geoff's latest results, but also
allows for the loading provide by the amplifier itself.

HS-10 properties

Sensitivity, E = 2.99 V/ips = 117.7 volts per meter per second
Natural Frequency = 1 Hz = 2*PI radians per second
Natural damping = 0.031
Inertial Mass = 33 oz = 0.936 kilogram

Erhard Wielandt, in his chapter "Seismic Sensors and their Calibration"  in
the Manual of Observatory Practice 
presents a formula for electromagnetic damping.

The formula is h = (E^2 / 2* M * wo * Rd) , where
   E is the output in volt-seconds/meter,
   h is the damping coefficient (0.5/Q),
   M is the effective pendulum mass in kilograms,
   wo is the natural frequency of the pendulum in radians/sec, and
   Rd is the total shunt resistance.

The recommended total damping is 0.707. Since the HS-10 has an open circuit
damping of 0.031, we want the electromagnetic
contribution to be 0.707 - 0.031 = 0.676.

so,

Rd = E^2 / (2*h*M*wo) = (117.7)^2 / (2 * 0.676 * 0.936 * 2 * PI) = 1742 ohms

Let us say the coil resistance is 440 ohms. The input resistance of the
amplifier and its applied shunt resistor must then 
equal 1742 - 440 = 1302 ohms. The 1302 value is that of the external shunt
resistor in parallel with the amplifier input 
resistance. 
Say the amplifier input resistance is 10K ohms.
1/Rext = 1/Rt - 1/Ramp
1/Rext = 1/1302 - 1/10000 = 0.000768 - 0.000100 =  0.000668 

Rext = 1497 ohms

The applied load will reduce the sensitivity of the geophone. The output
will be Rshunt/(Rcoil + Rshunt) times the open 
circuit value.






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