Re: Necessity of ir isolation

From: sleeper75se (sleeper75se_at_yahoo.se)
Date: 2001-12-18 05:01:13


--- In buildcheapeeg_at_yahoogroups.com, "Joerg Hansmann" <info_at_jhansmann.de> wrote:
Hi Joerg,

> You were more or less right ;-)

I got lucky this time ... I fully expect you to find any mistakes I
make in the future. ;o)

> Assuming 1% tolerances and a differential gain of 20000
> CMRR in the passband would be 120dB, what is very good.

Isn't the actual gain for our application about 5000 (10bits, 1uV
bitstep)? Of course that's still 108dB, quite respectable.
(Personally I'm going for 12 bits: gain = 1200, CMRR = 96dB)

A trimpot in the right place (Z1 or R2 in artikel7, eq 1) might not
be a bad idea either..., then we could trim the CMRR to ... er, high
values.

> The more I think about this circuit the more I am beginning to like
it...

It's pretty isn't it? :-)

> I do not quite agree. My prior posting about the necessity of
> calculating the difference between between two channels to get
> one differential channel was complete nonsense - perhaps this has
> irritated you...

Don't worry, I realized you needed more info... based on what you
got, it was completely accurate. :-)

> In EEGfig3.gif the inputs attached to V0 and V1 would form one
complete differential input.
>
> So for V0 1x LT1012 would be needed
> and for V1 1x LT1012 and 2/2 TLC1078,
> making a total of 4 opamps in 3 packages per differential channel.
>
> In order to get 2 differential channels as in the current
> INA114 based design, the above is needed twice, giving
> 8 Opamps or 6 packages.

I'm wondering if we can't remove AMP3. If we let the design be AC-
coupled, the output offset (which AMP3 corrects) is not very
interesting. We would save two amplifiers.

(Or is it the output offset of AMP2 that AMP3 corrects? To sleepy
right now to think straight...)

Also, if we go monopolar, both channels would share AMP0.

To sum it up: just two more amps per channel. Total: 7 amps in 4
packages, filtering not counted of course.

> Less noise would be better. However the LT1012 has only 1/20
> noise current of the INA114, so the current noise in the above
> mentioned protection resistors can be neglected.

Yes, I did a little early morning math-excercising. The result was
that the resistors only add about 30% more noise.

I dont have a formula for this written down but I think this is how
you calculate rms-noise over a given frequency band:

| / /\ hi \ 1/2
| | | |
| Noise voltage or current (rms) = | | N(f)^2 df |
| | | |
| \ \/ lo /

where N(f) = noise density in Volts / sqrt(Hz) and (lo, hi) are the
band limits in Hz.

Looking at the datasheet for LT1012, the current noise is a straight
line on a logarithmic scale,

with endpoints at 60 fA/sqrt(Hz), 1 Hz and 8 fA/sqrt(Hz), f = 120Hz.

Solving the line equation I get N(f) = 60 - 25*log(f), for 1 < f < 120

Plugging that into the integral above, with lo = 1 Hz and hi = 75Hz,
I got noise current = 216 fA rms
Current referred voltage noise, using a 62K resistor at the input:

A = (62K * 216) = 13.4 nV (rms)

For the voltage noise, I just set N(f) = 15 =>

B = 129 nV (rms)

The Johnson noise in a resistor is sqrt(4kTR*BW)
R = 62K, T = 300 kelvins and BW = 74 Hz gives 0.275 uV noise peak to
peak, or

C = 0.045 uV rms

Summing all noise sources:

(Current noise + voltage noise + Johnson noise) * rms-to-peak-factor =
(A + B + C) * 6 = 1.135 uV.

And since we'll be using two inputs: 1.135 uV * 2 = 2.27uV peak to
peak, where about 0.71uV can be referred to the protection resistors.

Ok, this probably wrong. ;-) Biosemi aimed for 1uV noise to get some
dithering for their ADC, and my calculation produced more than twice
that, with the noise from the resistors (that Biosemi doesn't use)
only contributing 30%. In short: I got 0.6uV too much noise. Hmm...

Maybe i should sqrt(A^2 + B^2 + C^2) * 6 * 2 instead. Returns 1.65uV
(p-p) noise. Much better. Extra 0.6uV accounted for.

> Is it better than LT1012 ?

No way, it was just the first femto-ampere-noise amp I could find. I
didnīt look to closely at the LT1012 until after what you just
wrote. :-) LT1012 appears to be perfect for the job, so we need not
look further.

Phew, long post. Hope you got time to read it carefully, I'd love to
hear your opinion, despite my sleepy ramblings.

Be well,

Andreas



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