Interesting sensor design problem

[Tony Williams]

BTW, is that a diode in the emitter bias circuit?
Or something else....

They are diodes, compensating for the dVbe/dT of
their respective transistors. Diode-connected
transistors work slightly better, as below.

| |
| +----+
\| | |/c
npn1|--+--|npn2
e/| b |\e
| |
| |
[8k2] [6k8]
| |
+--------+----

Connect npn2's base to collector.

Good thermal coupling is required to avoid drifts
due to stray breezes. Mount the two transistors
alongside each other and glue them together. A
dollop of hot-melt glue over the pair is useful.

 

[Fred Bloggs]

You know, the fun thing is, one of the tasks I decided to do was
simulate this thing, and see what helped. I got the latest INA122 spice
model from TI/BB, and tried this. Finally figured out that none of the
reference voltage changed did any good, got just about the same results
every time, or results that didn't make any sense. Ralized that they
modeled it as a single opamp, not as an instrumentation amp.

Right- forget their dumb SPICE model. Throw the equivalent circuit into
SPICE using ideal OA's, then do a DC sweep on Vdiff from say -15mV to
+15mV, stepping VCM as a parameter from 0.5V to 2.5V, and plot the two
OA outputs. This will tell you how the INA122 will do.

Now I realized I don't have any matching PNP/NPN pairs in the junk box,
just a pile of 2N2222's, so will have to run by RatShack later and pick
some up.

BTW, is that a diode in the emitter bias circuit? Or something else....

Thanks for all the help! I figured an actual DESIGN question wouldn't
be TOO off topic...

On second thought, the transistor current source offset is not so good
in this application for two reasons: 1) it is not ratiometric to the
battery voltage, and 2) it takes a lot overhead ~ 1.3V to each rail
which does not leave a lot of room for VCM variation- especially when
you consider battery drain down. It is better to go with the INA122
followed by an OA with Vo=2(Vin-Vbatt/2) as in my original circuit- this
is ratiometric and has no CM limitations. You can conserve current by
going with a dual RRIO OA and using an inverted mode transistor like so:

View in a fixed-width font such as Courier.





gnd Vbatt
| |
INA122--------------+
| | RRIO
+ | |\ | +-R-+
V -------|--------|+\ | | | |\
in | | >---+--|-+-R-+------------|+\
| ,---|-/ | | | | >--+-> 0<V <3V
| | |/ | | R +----|-/ | out
| | | | | | |/ |
| | | | | | |
| +---100K----- | +---R---+-----R----+
| | | | |
| | | | 3V R R=10K
,----|---+---25K-----+ | | |+ |
| | | | +--||---+--Vbatt
(957)RG | | | | | |
| | | | | |
'----|---+---25K ----+ | | 443K
| | | | | |
| | | | | |
| | | | | |
| | | | +--10K--+
| | | | | |
| | | | gnd |
| | |\ | | |
| +-----|-\ | | |
- | | | >-- | |
V ------|---|-----|+/ | |
in | | |/ | |
| 100K | |
| | | |
| | | |
+------------------+ |
V | |
ref +----10K----+-------. |
| |100n | |
(66.2mV) e === /| | |
\| | /-|--' |
|--6.8K---+-< | |
/| \+|-----------+
| 2N3904 \|
gnd

 

[Tony Williams]

On second thought, the transistor current source offset is not so
good in this application for two reasons: 1) it is not
ratiometric to the battery voltage, and 2) it takes a lot
overhead ~ 1.3V to each rail which does not leave a lot of room
for VCM variation- especially when you consider battery drain
down.

That particular transistor current source has a poor
temperature stability. Suggestion withdrawn.

 

[Fred Bloggs]

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

 

[Fred Bloggs]

That particular transistor current source has a poor
temperature stability. Suggestion withdrawn.

You can still do it with two Howland type current sources, but by the
time you add in all those components, it may be simpler to just leave
that alone and use the Vref method.

 

[Charles Edmondson]

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...

 

[Fred Bloggs]

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.