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.

I mostly understand what you are saying, but I'm confused about how 
where the guard voltage can be taken from.  I thought the guard voltage 
had to be taken from a x1 follower?

BTW, when you say +100 follower I assume you mean an opamp in 
non-inverting configuration with a gain of 100.  I just always thought 
the word "follower" was only used to describe something with gain=1, but 
maybe you use "follower" to describe something non-inverting.

When you say "gain divider" you mean the voltage divider which connects 
the output to the negative input for the feedback right?  I'm sorry I 
don't know all the terminology, unfortunately they don't teach us that 
in school...  So we connect the offset voltage (5-10V let's say) to the 
bottom of that voltage divider, and we can take the guard from the 
negative input of the opamp?  I am right?

I know your method works, but I can't get the equations right.  The 
output should be:

Vout = (Vpos - Voffset)*(1+R2/R1)

where Vpos is the voltage at the positive terminal of opamp and Vneg is 
at the negativ terminal of opamp but I'm not getting that.  I get that 
the current from Vneg to Voffset is I1=(Vneg-Voffset)/R1.  So Vout = I1 
+ Vneg.  Therefore vout = Vpos(1+R2/R1) - Voffset(R2/R1), where I 
replaced Vneg with Vpos since they should be the same.

Thanks for your help.