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Current measurement using LMP91000EVM board

Hemant Katakkar
Prodigy 10 points
Other Parts Discussed in Thread: LMP91000EVM, LMP91000

I am looking for technical support for TI
product. Following details will explain my technical difficulty. Request you to help resolving this problem.


PROBLEM 

I have LMP91000EVM board with me. I have configured it for 2 lead galvanic cell in potentiostat configuration.

In place of sensor I have connected simple resistance of 15K between WE and CE/RE terminals. I have configured transimpedence amplifier with load res of 10E and gain res of 2.75 K . 

Applied bias voltage is 0.4 V and internal zero voltage is set to 50% of supply (+5V).

I can see the current of 24 micro amp. on the galvanometer connected between resistance and WE terminal. (Galvanometer range 100 micro amp).

My problem is output voltage I see at C2 or Vout pin does not match with theoretical calculations. 

I am using equation (Vref - Vout)/RF = I
Practically I observe 0.975 V as output voltage.

Further If I vary resistance connected between WE and CE/RE terminal then current vaties , but output voltage at transimpedance amplifier does not change.

Please tell me what is going wrong.

  • 0
    TI__Genius 15760 points

    Hi,

    I am not familiar with this device. I will try to find out the right contact to support you.

    Regards

  • 0 in reply to Mayrim Verdejo
    TI__Genius 14825 points
    I would try as he has answered questions recently on this product. I'll move this thread to the precision amps forum as well.
  • 0 in reply to Jason Bridgmon
    TI__Genius 16890 points
    Hi Mayrim,

    I have forwarded this on to the engineer that supports the LMP91000.

    Mike
  • 0
    TI__Expert 5145 points

    Hi Hemant,

    Sorry for the delay in the answer.

    I tried to replicate your experiment, and it works well in my case.

    I used the LMP91000EVM connected to SPIO4 Board, in order to use the Sensor AFE software for LMP91000:

    With the use of SPIO4 Board, the supply voltage of LMP91000 is 3.3V, so I needed to change a little the settings you reported in order to reach the same condition of your experiment. In my set-up I have set:

    - I have closed the 2-wire jumper in order to short circuit CE and RE;

    - I have connected a 15kOhm resistor between WE and CE/RE;

    - Selected the Operation Mode in "3-lead amperometric cell";

    - Selected the external reference Vref = 2.5V;

    - Selected the VREF divider to "Bypassed" in order to have an internal reference at 2.5V as in your case;

    - Selected the Variable Bias to 16%VREF in order to have 0.4V (2.5*0.16 = 0.4);

    - Selected the TIA Gain = 2.75kOhm;

    - Selected the RLOAD = 10Ohm;

    - Selected the Shorting FET = "disabled"

    In this condition, the current that flows in the 15kOhm resistor is (VREF-VBias)/15kOhm=(2.5-0.4 )/15kOhm=140uA;

    This current flows from WE to RE/CE, so the Vout is VREF+I*RF=2.5+140uA*2.75kOhm=2.885V.

    Attached the screenshot of the results on my LMP91000EVM:

    Please let me know should you have any questions or needs further support.

    Best regards,

    Carmine