Help With an Ohmmeter Reference Design

Discussion in 'The Projects Forum' started by rk073190, Feb 19, 2014.

  1. rk073190

    Thread Starter New Member

    Feb 19, 2014
    3
    0
    Hello All,

    I'm working on a project that calls for a linear response 1-3% accurate ohmmeter in an embedded form factor (very small, very low power consumption). In my research I settled on the circuit pictured below but I don't feel I completely understand how it works. It seems like the LM334 and LM336 are set up in a configuration which will provide an minimum temp coefficient voltage reference which is then supplied to an (manually) adjustable voltage divider. The output at the central node of this divider is then supplied to a voltage follower op-amp circuit. I believe that the purpose of connecting the negative feedback loop of the voltage follower to the zener diode (LM336) anode is to "feed-forward" the voltage reference such that the reponse to increasing R_sense is linear rather than negatively exponential. Is my interpreation correct? Any help understanding how this circuit works or how to break it down into seperate functional elements would be greatly appreciated. Thanks!

    [​IMG]
     
  2. alfacliff

    Well-Known Member

    Dec 13, 2013
    2,449
    428
    ther circuit works as a bridge circuit, measuring the difference between the fixed resistors switched by the rotary switch and the unknown resistor.
     
  3. crutschow

    Expert

    Mar 14, 2008
    13,056
    3,245
    I think you understand the basics.

    My understanding--

    The LM334 generates a constant current to simulate an infinite resistance. A resistor could be substituted for the LM334 but that would increase the circuit error slightly (as determined by the dynamic resistance of the LM336 in relation to the resistor value). Don't see how its change of current with temperature helps the circuit accuracy though. It would actually seem to degrade it ever so slightly.

    The LM336 generates a stable 2.5V across the sense resistors to generate a constant current through the unknown Rx. Thus, for example, selecting the 2.5kΩ resistor gives a current of 1mA through Rx giving the linear scale factor of 1V/kΩ.

    The bottom of the LM336 is bootstrapped by the op amp output voltage so that the 2.5V appears across the sense resistors independent of the output voltage due to the value of Rx, maintaining the constant current through Rx. The 10kΩ pot is to adjust its voltage to exactly 2.5V.

    That all make sense?

    Edit: Note that those resistors are not standard resistance values (for example 2.49kΩ is the closest 1% standard value to the 2.5kΩ shown) so you would just have to adjust the 2.5V voltage to 2.49V to compensate.

    Also the LM308 is an old op amp that has a maximum input offset voltage of 10mV, which will reduce the circuit accuracy. You might want to select an op amp with a lower offset voltage (many are available with offsets below a mV).
     
    Last edited: Feb 19, 2014
  4. rk073190

    Thread Starter New Member

    Feb 19, 2014
    3
    0
    Thank you for the lightning fast responses! I think (emphasis on the maybe) that I understand what you mean by bootstrapping the LM306 such that there is always 2.5V across the sense resistors (by sense resitors do you mean the set of fixed resistances on the dial i.e. 2.5 kOhm, 25 kOhm, etc?).

    If you look at it backwards (which granted may be invalid but is helping my thinking). An Rx of 1 kOhm should mean the potential diff across it is 1V and we see 1V at V+ on the opamp. The negative feedback loop wants to maintain V- at 1V as well. This will "feed-forward" the Voltage reference supplied by the LM336/LM334 to 3.5V at the input to the selection/sense resistors (2.5 KOhm, 25 kOhm, etc). That works out (for a voltage divider 1 kOhm / 3.5 kOhm * 3.5 V = 1 V, and for 2 kOhm we have 2 kOhm / 4.5 kOhm * 4.5 V = 2 V and so on presumably until the opamp saturates). Is that interpretation essentially correct?

    Turns out the LM308A is a "precision" version of the more common LM308 that is actually very hard to get and has a maximum offset of 0.5 mV (this circuit is from the LM336 datasheet's applications section). I was thinking of substituting the UA741CP (super cheap) general purpose opamp which has a typical input offset voltage of 1 mV and maximum of 5 but can be adjusted by attaching a pot to the null offset adjustment pins.

    P.S. By "minimum temperature coefficient" I meant that the paired configuration of the LM334 current source and LM336 Vref is supposed to create a Vref that is extremely insensitive to temperature changes.
     
  5. crutschow

    Expert

    Mar 14, 2008
    13,056
    3,245
    That sounds correct.

    The cost of the pot will likely be greater than getting a better op amp (and you wouldn't have to do any tweaking of the offset). Also the better op amp will likely have a lower offset voltage change with temperature.

    Another consideration is that if you get a rail-rail type precision single supply op amp you wouldn't need two supply voltages.

    Well I didn't look at the effect of temperature of the two together so can't comment on that.
     
  6. rk073190

    Thread Starter New Member

    Feb 19, 2014
    3
    0
    Point taken. I'm planning to get the LMC6081 now. It has a 0.15 mV bias voltage ($2.40 for 10+ pcs at Digikey as of writing)...
     
  7. crutschow

    Expert

    Mar 14, 2008
    13,056
    3,245
    Looks like a good choice for low offset and being able to operate from a single supply.
     
Loading...