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Re: Differential amplifier for measuring Hall voltage





Robert Baer wrote:

John Popelish wrote:

Robert Baer wrote:

DG wrote:

I'm designing a differential amplifier to amplifier a small ~0.01-0.1mV
difference, where the common-mode signal is around 5-20V.  The problem
is, I need to "zero" the output of the differential amplifier before I
make the measurement.  The value I am measuring is actually the Hall
voltage.  I need to zero my output with the magnetic field turned off,
and then I need to turn the magnetic field on, and measure the voltage
at the output of the differential amplifier.  I measure the Hall voltage
between V3 and V4.  But initially, V3-V4=~0.1V, but it could technically
be up to ~1V due to assymetry in the sample contacts.  So I need to
"remove" this ~1V either from the input of the output of the
differential amplifier, before I turn on the magnetic field.  Here's my
circuit:

Initially (there is some unwanted difference here, of 0.2V.  This is a
problem):
---------o V3 = 5.1V
|      |
|sample|
|      |
---------o V4 = 4.9V

After applying the magnetic field (a Hall voltage of 0.01mV can now be
seen):
---------o V3 = 5.10001V
|      |
|sample|
|      |
---------o V4 = 4.89999V

Here's the circuit:

---------o V3 --> High Z Buffer ---     |\ differential
|      |                          |---> | \amplifier gain = 1
|sample|                                |  |--> Vout = 0.00001V
|      |                          |---> | /
---------o V4 --> High Z Buffer ---     |/

I just need some circuitry between the buffers and the diff. amp. or
after the diff. amp which can accurately and reliably tune the output to
zero (within precision <1 mV hopefully) before doing a measurement.  Can
anyone tell me of a slick way to do this?  If anyone can also suggest
some op-amps to use, that would be cool as well, although I already have
some in mind.  The resistance of my sample is about 10^13 ohms.  Thanks
for any help.

David

Well, using gain of one as shown only increases the noise and adds in thermoelectric problems also. Try a gain of 100 in the hiZ buffers and make them differential so that you can subtract out 5.0000V. This gives a "worst case" 10V output. Now if those initial values are repeatable and reliable, then subtract out the 5.1V and the 4.9V respectively, making for a difference of zero. Then make the final differencing amplifier a gain of 100, which results in a theoretical 100mV. Be damn careful about shielding, use of driven shields, same metals, isothermal layouts, etc.

What you are describing is pretty close to an instrumentation amplifier, except that they provide differential gain as well as offset adjustment. They also often include provision for signal follower outputs that are used to drive the shields of the input lines. I am not sure you will find one that handles 10^13 resistance sources, though. This is a very high resistance for a Hall effect source.

http://www.elektr.polsl.gliwice.pl/~jelon/INA_110.pdf
--
John Popelish


  So, what is wrong with a DIY similar to what i mentioned, but the two
gain buffers are FET or BICMOS or other low input current devices?
  Connect them as times +100 followers, low side of gain divider goes to
the offset voltage.
  One still has a guard voltage available on the divider.

If I use the circuit shown here: http://www.linear.com/pdf/an6.pdf with two highZ buffers in front, for guarding and perhaps some gain, then would I get the same performance? Using the offset removal method you suggested, it may require that I zero the output by adjusting the offset, frequently, and I also need to have really good precision with the zeroing. If that signal drifts at all that could be a problem as well. I think maybe the circuit in the app note avoids this problem.


Thanks,
Dave




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