Single Supply Gain-Offset Circuit

Discussion in 'General Electronics Chat' started by DangerousBill, Aug 25, 2010.

  1. DangerousBill

    Thread Starter Active Member

    Jul 21, 2010
    30
    1
    This circuit (attached) is the output stage for a sensor control circuit. So far, it exists on paper, designed from a guide to single-supply circuits provided by TI. The purpose of P4 is to null out the baseline signal from the sensor. P5 allows a gain adjustment over an order of magnitude. The signal is pure DC.

    My questions:
    1. Will it work as I expect? I've done the math, but I may have missed something.
    2. The parts count seems high.
    3. Will the offset and gain adjustments interact? It would be nice if they didn't but it's not fatal.

    Thanks in advance.
    Dangerous Bill
     
  2. sage.radachowsky

    Member

    May 11, 2010
    241
    38
    Could you post the schematic with the left side included, and labeled?
    What is the sensor's output voltage range, and impedance?
    What type of sensor is it?
    These things might help people help you.
    I personally don't get that resistive network, and I don't see how the pot P4 will null a DC baseline signal. But it's hard to really critique this because it's not complete.
     
  3. beenthere

    Retired Moderator

    Apr 20, 2004
    15,815
    282
    Yes, things like the connections to the inverting input would be interesting to see.

    As far as a baseline adjust goes, it is easy to apply to the non-inverting input, but with the signal of interest handled by the inverting input (very easy to set gain there, too).
     
  4. DangerousBill

    Thread Starter Active Member

    Jul 21, 2010
    30
    1
    OOPS. I posted the wrong diagram, the bit I fished out from the main diagram. Here is the labeled version.

    The input range will generally be low mV to 6 V.

    The sensor is an experimental type used to measure chlorinated hydrocarbons in air. The sensor looks like a variable resistor with negligible parallel capacitance.

    The design is based on the y=mx-b version (offset subtracted from amplified signal) in the TI document, page 12:
    http://focus.ti.com/lit/an/sloa030a/sloa030a.pdf

    DB
     
  5. Audioguru

    New Member

    Dec 20, 2007
    9,411
    896
    You drew your schematic wrong. Your inputs are backwards so the opamp has positive feedback instead of negative feedback.
     
  6. timrobbins

    Active Member

    Aug 29, 2009
    318
    16
    You may not be able to trim to 0.0mV if your ref is just taken to 0V, and you use a simple pot for the P4 divider. I assume you are using dual supply.

    Your output will rail at some stage, so you would need to keep output current relatively low to maximise span. I assume your gain will be quite low.

    You don't discuss noise - have you considered how that will be managed?

    Ciao, Tim
     
  7. sage.radachowsky

    Member

    May 11, 2010
    241
    38
    If I understand correctly, you would like some small positive voltage (a few mV) to become 0V on the output side, right?

    I think the circuit you have copied will do the opposite -- the "b" in the "mx+b" is a positive value.

    I think you want to use a diff amp configuration, with the negative input (low side input) of the diff amp set to the offset voltage. That will subtract the "b".

    Look here and note that V1 is the low side input of that configuration. So then you simply use your pot to adjust V1 and the other side (V2) takes the input voltage from the sensor. However, you will probably need to buffer the sensor voltage output, because the feedback network (R2 and Rg) will affect that voltage if it is not buffered.

    Last question -- you say your chemical sensor is a variable resistance based on the level of the test compound. How are you driving that? Are you using a fixed resistor above it, or a constant current source?
     
  8. DangerousBill

    Thread Starter Active Member

    Jul 21, 2010
    30
    1
    The sensor is heated to about 300 C and biased at a fixed voltage between 3 and 9V. The current is measured by a INA118 across a 1K resistor. There is always some standing current when the sensor is at operating temp, and I want to null that out so the resulting output is linear with sample concentration.

    (I guess you noticed that I got the op amp inputs switched in the diagram. I noticed after I hit 'send'.)

    Thanks for the help and the tip on the differential circuit. As far as I can see, it'll be cheaper and more intuitive.

    DB
     
  9. DangerousBill

    Thread Starter Active Member

    Jul 21, 2010
    30
    1
    No, it's got to be done with a single supply voltage, usually +9V.

    The sensor produces microamps of current, some standing current, plus a current proportional to chlorinated hydrocarbon. Current is measured across a resistor. Then I want to null out the voltage from the standing current.

    Most noise will be generated by the sensor. One of the purposes of the circuit is to estimate noise.

    Thanks for the advice.

    DB
     
  10. timrobbins

    Active Member

    Aug 29, 2009
    318
    16
    DB, you had said "The sensor looks like a variable resistor with negligible parallel capacitance. ". Then you indicate "The sensor produces microamps of current, some standing current, plus a current proportional to chlorinated hydrocarbon. Current is measured across a resistor."

    I guess that means you have the sensor as the upper or lower 'resistor' in a resistive divider circuit powered from 9V. I guess the sensor resistance must change by a few decades in value to span from mV to 6V signal levels. I guess the standing current you refer to relates to the mV or 6V end of the span (aka dark current).

    The TL08x is not a good choice for single supply opamp application if you want to work at mV level above 0V.

    Ciao, Tim
     
  11. DangerousBill

    Thread Starter Active Member

    Jul 21, 2010
    30
    1
    Kicking self now. Thanks.

    DB
     
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