That particular transistor current source has a poor temperature stability. Suggestion withdrawn.
Tony Williams.
Interesting sensor design problem
Feb 11, 2005
44 Replies
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
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.
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
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Edmondson Engineering
Unique Solutions to Unusual Problems
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.Join the Discussion
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