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LMP90080: LMP90080 GAIN configuration

Rajesh PL
Prodigy 70 points
Part Number: LMP90080


Support Path: /Reference designs/Technical support help for a specific technical issue with a reference design/

Hi , We have a product designed with LMP90080, we have used internal gain 8 using the register CHx_CONFIG. We have used 0x76  for setting gain 8,it is working fine.We would like to increase the gain to 32 , for this we used 0x7B  for gain 32, but it is not reflecting anywhere. We can set upto gain 8. But from 16 onwards we are unable to set the gain using this register. Can anyone guide me ,whether we are missing any steps? or we need to set any other registers as well for setting the gain above 8.

Thanks in advance.

  • 0
    TI__Genius 16890 points
    Hello Rajesh,

    Writing 0x7B should change the gain to 32. Can you explain what you mean by it not reflecting anywhere so that I can better understand?

    Some other data that would help is: what pins are INP and INN, what is the voltage on INP and INN, what is the reference voltage, which reference are you using?

    Mike
  • 0 in reply to Michael Stout
    Prodigy 70 points
    Hello Mike,

    Thanks for your immediate response and time.

    I printed the ADC data out value with different gain. So values are getting multiplied when I increased the gain. Gain 1,2,4,and 8 are giving values with proper multiplication . But when I tried with 16 and above , i am getting the value lower than the value received with 8 gain.

    INP and INN are connected to pin 2 and 3 of the chip
    Voltage on pin is 60.66mV (minimum) 138.12mV ( Maximum)
    VRef=5V

    As per our calculation gain can be given upto 32 ( which will give value 4.42 Volt, which is under Vref)

    Hope this inputs will help to understand the issue.
  • 0 in reply to Rajesh PL
    TI__Genius 16890 points
    Hello Rajesh,

    Just to check, the voltages you listed above (60.66mV and 138.12mV) are the differential voltage between pins 2 and 3? Can you tell me the actual voltages on pin2 and pin3 with respect to ground? I am wondering what your common mode voltage is. Also, do you have VREFN1 connected to ground?

    Mike
  • 0 in reply to Michael Stout
    Prodigy 70 points

    Hi,


    Thanks for your support.

    For your information ,we are connecting 3 wire RTD  as input . Please refer the attached circuit for your reference.

    60.66mV and 138.12mV are the voltage between pins 2 and 3.  Please note that while connecting  3 wire RTD , pin 3 of the IC  will be connected to GND since we are using RTD with 2 white wires are shorted internally inside the RTD sensor element. So while connecting 3 wires to the 3 connector terminals,  pin 3 of the IC will  be getting connected to ground. So voltage levels at pin 2 and 3 while connecting RTD are given below.

    Pin2 :60.66mV ( Min )/ 138.12mV (Max)

    Pin3 : 0V  ( always)

    We have connected VREFP1 to 5V and VREFN1 to GND

    Hope this will help.

    RDT connection.pdf

  • 0 in reply to Rajesh PL
    TI__Genius 16890 points

    Hi Rajesh,

    I see the problem now.  When you switch to a gain above 8x, it turns the buffer on.  When the buffer is on the input range must be at least 0.4V above ground.  When the buffer is off the input range can be down to ground.  See the Analog Input specs on page 8 of the datasheet.  That is why the output is decreasing when you go to a higher gain, because part of the input is getting cut off. 

    You will need to get the common mode voltage of the input higher.    Try connecting the RTD as shown in Figure 75.  The top of the RTD goes to VIN0, one of the two wires that are shorted together goes to VIN1 and the other wire goes to VREFP.  Connect a resistor between VREFP and VREFN.  You can start with a value of 1k and see how that works.  This should be a precision resistor, 0.1%, as it sets the VREF voltage.  Doing this will lift your input common mode voltage above the 0.4V limit.

    Mike

  • 0 in reply to Michael Stout
    Prodigy 70 points

    Hi Mike,


    Many thanks for a valid information, we will try this out and update you.

    We have one more issue. When we give some voltage as input with internal gain set at 8,we are getting the ADC values as the half of the value that we are expecting


    For example when we give 0.5V as physical input voltage ( across RTD connected across chip pins 2 and 3 )with internal gain 8, ADC should give the ADC value equivalent to 4V ( 0.5 X 8). We are expecting a value 52428 as adc value . But we are getting half (26214)of this value.

    We are using VREFP1 as 5V and VREFN1 as GND.

    Can you please help for this issue too

  • 0 in reply to Rajesh PL
    TI__Genius 16890 points
    HI Rajesh,

    The ADC output code formula is shown in section 10.1.3.

    For your values: ADC out = ((0.5V x 8)/(5V-0V)) x 2^15 = 0.8/32768 = 26214.

    Mike