Charles Edmondson
Guest
|
 Posted:
Thu Feb 17, 2005 3:32 am Post subject:
Re: Interesting sensor design problem |
|
|
Fred Bloggs wrote:
| Quote: |
Fred Bloggs wrote:
in | | |/ | |
| 100K | |
| | | |
| | | |
+------------------+ |
V | |
ref +----10K----+-------. |
| |100n | |
(66.2mV) e === /| | |
\| | /-|--' |
|--6.8K---+-< | |
/| \+|-----------+
| 2N3904 \|
gnd
That should be changed to below for 120dB attenuation of common mode
noise on the sensor at line frequency with ultimate attenuation of 60dB
at high frequency, using something like an OP291:
View in a fixed-width font such as Courier.
| | | | | |
| | | | | 3V |
| | | | | |+ |
| | | | +--||---+--Vbatt
| | | | | | |
| | |\ | | | |
| +-----|-\ | | | 443K
- | | | >-- | | |
V ------|---|-----|+/ | | |
in | | |/ | | |
| 100K | +--10K--+
| | | | |
| | | gnd |
+------------------+ |
V | |
ref +------+------------10K----. |
(66.2mV) |680n | | |
=== e 2N3904 /| | |
| \| /+|--' |
100 |---6.8K----< | |
| /| \-|----'
| | \|
gnd gnd
Hi Fred, |
Decided to try simulating this, and just simplified it as a 66.2mV
voltage source tied to Vref. All this does is move the OUTPUT up the
corresponding voltage, doesn't move it down at all. If I inject a
negative voltage here, it lowers the output, but it takes a big voltage
to make much difference...
I guess I am missing something...
--
Charlie
--
Edmondson Engineering
Unique Solutions to Unusual Problems
|
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Fred Bloggs
Guest
|
| Posted:
Thu Feb 17, 2005 6:11 am Post subject:
Re: Interesting sensor design problem |
|
|
Charles Edmondson wrote:
| Quote: | Fred Bloggs wrote:
Fred Bloggs wrote:
in | | |/ | |
| 100K | |
| | | |
| | | |
+------------------+ |
V | |
ref +----10K----+-------. |
| |100n | |
(66.2mV) e === /| | |
\| | /-|--' |
|--6.8K---+-< | |
/| \+|-----------+
| 2N3904 \|
gnd
That should be changed to below for 120dB attenuation of common mode
noise on the sensor at line frequency with ultimate attenuation of
60dB at high frequency, using something like an OP291:
View in a fixed-width font such as Courier.
| | | | | |
| | | | | 3V |
| | | | | |+ |
| | | | +--||---+--Vbatt
| | | | | | |
| | |\ | | | |
| +-----|-\ | | | 443K
- | | | >-- | | |
V ------|---|-----|+/ | | |
in | | |/ | | |
| 100K | +--10K--+
| | | | |
| | | gnd |
+------------------+ |
V | |
ref +------+------------10K----. |
(66.2mV) |680n | | |
=== e 2N3904 /| | |
| \| /+|--' |
100 |---6.8K----< | |
| /| \-|----'
| | \|
gnd gnd
Hi Fred,
Decided to try simulating this, and just simplified it as a 66.2mV
voltage source tied to Vref. All this does is move the OUTPUT up the
corresponding voltage, doesn't move it down at all. If I inject a
negative voltage here, it lowers the output, but it takes a big voltage
to make much difference...
I guess I am missing something...
|
It is not how small the signal is, it is how far removed from 1.5V it
is. This is 66.mV-1.56=-1.43V. Since the IA has a gain of
(5+200k/957)=214, this -1.43V corresponds to an equivalent differential
input of -1.43V/214=-6.7mV- in other words the output due to a 6.7mV
input has been subtracted from the output so that when the input is
6.7mV,the IA will be 1.5V. Then for every volt beyond that threshold the
output is gained by 214 and added to 1.5V. By the time you reach 13.7mV
that is 13.7mV-6.7mV=7mv *beyond* 6.7mV so IA output is 214*7mV+1.5V=3V.
Your IA transfer function therefore runs linearly from 1.5V to 3V as the
input differential runs from 6.7mV to 13.7mV. You then want to drive
this into that final stage OA which does the 2*(Vin-1.5V) to translate
the output excursion to 0V to 3V for the same input range. If you want
some offset other than 6.7mV, say Voff, then this would be
Voff=(1.5V-Vref)/214, within limits, where Vref is voltage at Vref input. |
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