ICL7107 zero reading Isuue

Discussion in 'The Projects Forum' started by akisnas, Mar 3, 2014.

  1. akisnas

    Thread Starter New Member

    Mar 3, 2014
    5
    0
    Hello there, i'm a new user & i'd like to ask you about the icl7107 possible zero reading trigger-offset.
    I made an amp meter 2 years ago & it's still working perfect but from the power on status i have a minimum reading current 14ma this measurement is the result of a minimum device consumption, i want to remove it from the panel meter i mean there is no need to see this measurement the better is to have this consumption but the meter it must read zero. I tried to offset the icl beteewn the pins 30-32 but without success. Is it possible, can i read zero even i have 14ma current through the panel meter?
    Thank you for your possible help.
    Cheers
     
  2. #12

    Expert

    Nov 30, 2010
    16,343
    6,828
    Just re-wire your sensing wires so the operating current is not measured. Completely external to the 7107 circuit. Can you post a drawing?
     
  3. akisnas

    Thread Starter New Member

    Mar 3, 2014
    5
    0
    Thank you so much for your reply.
    Here is the schematic, i can't deviate the 1ohm shunt resistor or to place it to an another point, in that case it will cause a non accurate current reading or the measurement could be affected from possible EMI-RFI signals.
    This is a shunt resistor build-in to a solid state circuit which measures & Works like an overcurrent sense resistor.
    Maybe there is a way to make an offset between the +5vdc ICL (pin 1) with a resistor 390k in series with a small trimmer 100k the other end point of the trimmer it will be the input (pins 32,35 must be connected at this point) for the current meter & the variable point it will be placed to the In-Lo (pin 30) (removing the pin 30 from 32 & 35 pins Connection as i wrote before).
    Sorry for my poor english.
    Kind Regards

    http://s129.photobucket.com/user/akisnas/media/amperometer_zps062c256e.jpg.html?sort=3&o=0
     
    Last edited: Mar 5, 2014
  4. #12

    Expert

    Nov 30, 2010
    16,343
    6,828
    The ICL7107 is not causing the 14 ma of current. Something external to the 1 ohm resistor (R3) is causing the current. There must be more to this than I can see. What is connected to the 1 ohm resistor? It isn't getting 14 ma from inside the ICL7107.

    This, "solid state circuit" is what I need to see. There are usually ways to do this, but not without seeing what I have to work with.

    ps, your English seems to be working very well.
     
    Last edited: Mar 5, 2014
  5. akisnas

    Thread Starter New Member

    Mar 3, 2014
    5
    0
    Thank you so much for your interest, really it will help me to clear out the issue, yes you' re right, i mention it to my first message but it wasn't clear sorry, there is a permanent current from the main circuit 10-14ma not from the ICL7107, the circuit is a "black box" sealed with black epoxy resin, which means there is no access & i can't change the topology, only 2 output pins for current measurement supplied, of course i can measure directly the output but if i do it (as i wrote before) it will be a hudge mistake, EMI & RFI signal can cause more problems than the expected & the measurement it will be wrong. The measurement is about a psu & located to the negative rail which using a shunt resistor 1ohm for variable current limitation & not affected when the icl7107 circuit is present, this resistor is carefully attached behind a very good EMI -RFI output filter & the measurement is not affected from these signals.
    The solution is an offset to the measurement unti but i'm not sure if it's really work, i wonder if i supply more possitive voltage to the In-Lo it will eliminate this current from the measurement, am'i right?
    Cheers
     
  6. #12

    Expert

    Nov 30, 2010
    16,343
    6,828
    Apparently I am working with this:
     
  7. #12

    Expert

    Nov 30, 2010
    16,343
    6,828
    Look on page 14, figure 18 to see a "zero adjust" circuit.
    page 11 top right paragraph, the last 2 sentences.
    "This offset reading can be conveniently generated by connecting the voltage transducer (R3) between IN HI and COMMON and the offset voltage between COMMON and IN LO."

    This seems to require disconnecting ICL7107 Pin 32 (COMMON) from pin 30 (IN LO) as shown in Fig. 18

    I am struggling with this. Any help is requested.
     
    Last edited: Mar 6, 2014
  8. R!f@@

    AAC Fanatic!

    Apr 2, 2009
    8,754
    760
    Are you trying to make it. #12
     
  9. akisnas

    Thread Starter New Member

    Mar 3, 2014
    5
    0
    I'm talking about this, pls take a look i made a few changes, maybe it's a possible solution.
    Just follow the link to the first post the second image is the one i'm talking about.
    #12 i checked the figure 18 & i belive it's more complicated for this application (i mean temperature calibration), the possible solution is somewhere in the middle i believe.

    http://s129.photobucket.com/user/akisnas/media/amperometer_zps062c256e.jpg.html?sort=3&o=0

    Thanks for your kindness to reply.
    Cheers
     
    Last edited: Mar 6, 2014
  10. ifixit

    Distinguished Member

    Nov 20, 2008
    639
    108
    Hi akisnas,

    If I under stand the situation, you want to null out the minimum 14mA reading to make it zero. 14ma thru R3 (1 ohm) is 14 mV of offset to get rid of.

    To do this you will need to inject an -ve offset current into the IN HI pin 31. The current will cause a -14mv drop across R4, 10K and cancel the +14 mV voltage from R3. 1.4 uA thru R3 is needed to cause a 14 mV drop. This tiny current will also flow thru R3, however the offset caused can be neglected.

    To do this, connect a 3.6 Meg resistor from -5V to pin 31. (0.014/10,000=1.4 uA). -5V/1.4 uA= 3.6 Meg.

    The 1.4 uA should constant regardless of the voltage drop across R3 cause by additional current thru R3. If the expected max current is not too large then just a resistor should be good enough.

    What is the maximum current you expect thru the 1 ohm R3?

    I have not tested this, it just out of my head, however, it is easy and safe for you to just try it. Tweak the value to get it just right

    Good Luck,
    Ifixit
     
    #12 likes this.
  11. #12

    Expert

    Nov 30, 2010
    16,343
    6,828
    I believe I found a method in the datasheet, but it requires cutting on the circuit board and a good understanding of how the ICL7107 is built internally. I'm having a hard time understanding the symbols used in the ICL datasheet and have been ill for over a week.
    I just can't seem to get the light bulb (in my head) to go on for this one.

    The external method proposed by ifixit seems much easier and less risky about breaking something. COMMON and IN LO are locked at 2.8 volts below the +5V supply. Therefore, at idle, IN HI and IN LO will be 7.2 volts above the -5V supply.
    With R4 = 10K, I see 1.4 ua to discard as ifixit said.
    7.2V difference / 1.4 ua = 5.142 Meg to the -5V supply. Try 5.1 meg. (a disagreement with ifixit)
    Both of these discussions depend on the idea that the + and - 5V supplies are 5.000 volts, and they won't be. Expect some tiny error.

    This change will become (7.2V + the input voltage) / 5.1 meg, and that is an error term. The circuit board conveniently provides a full scale adjustment called, P1. Adjust that for correct display at full scale input voltage and you're done.

    Many thanks to Ifixit.
    While I believe I contributed, I could not get this to come together at all without a bit of help.
     
    Last edited: Mar 6, 2014
    ifixit likes this.
  12. #12

    Expert

    Nov 30, 2010
    16,343
    6,828
    I found the, "second image" and it seems to follow the instructions in Fig. 18 correctly. However, the ifixit method is easier and I think it will work.
     
  13. akisnas

    Thread Starter New Member

    Mar 3, 2014
    5
    0
    Thank you so much for your reply, i forgot to mention the expected max current it's over 450ma.
    Thanks #12, yeah the simple is the best i'll try megs as you said both of you.

    Kind Regards
     
    Last edited: Mar 7, 2014
  14. ifixit

    Distinguished Member

    Nov 20, 2008
    639
    108
    I agree with #12, I neglected to consider the +2.2 volt offset from ground, so 5.1 Meg would be a good value to start with. A 1 Meg pot in series with 4.7 Meg would give you some adjustment range if you want to get fancy.

    This simple method will cause a different offset depending on the current across R3, but I suspect you don't care about this small effect.

    E.G. 450mA * 1 ohm = .45V, so 7.2 + .45 = 7.65V and 7.65 / 5.1M = 15 mV offset instead of 14.

    Ifixit
     
  15. #12

    Expert

    Nov 30, 2010
    16,343
    6,828
    That's the error I was talking about in post #11
    (Thanks for doing the math.)
    That error can be adjusted out with the on-board high end calibration potentiometer P1 (22k).

    or not. It's only 1/4% error. 1 count in the least significant digit.
     
Loading...