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TMCS1108-Q1: Not able to get voltage on IC Leg

Fenil Ghoghari
Prodigy 10 points
Part Number: TMCS1108-Q1

Hello,

I am using TMCS1108A1U-Q1 IC to measure voltage as well as current in channel. According to my Design I have connected Current IC output to my 16-Bit Resolution ADC Channel. So, I have main issue with my current ic is when my load is connected and consuming some of voltage nearly 2.8V those i have measure with my DMM but when i place check this voltage in IC output i just got only 0.28V this is not a actual voltage after getting this voltage i used some of equation mention in a Datasheet and calculate current i got -1 A.

Calculation:

Vout = 0.28 V

S = 50 mV/A

Vout,oa = Vin * 0.1

Vout,oa = 3.3 * 0.1 = 0.33

So,

Vout = S x I(in) + Vout,oa

0.28 V = (50mV/A) * I(in) + 0.33 V

I(in) = (0.28 - 0.33) / (50mV/A)

I(in) = - ( 0.05)*(1000) / 50

I(in) = -1A

I have used as this equation for current calculation.

CASE 2:

Once i disconnect my all load and measure voltage on current IC i can get 0.33 V this is correct voltage i get as per datasheet because there are have one equation to calculate voltage in 0A 

Here, Input Voltage is 3.3 V

so,

Vout,oa = 0.1 * Vs

Vout,oa = 0.1 * 3.3 = 0.33

Also i have mad one calculation using datasheet value and checked maximum voltage what i am getting i have used for that,

I(in) = 54.5 A

S = 50 mV/A

Vout,oa = 0.33 V

So,

Vout = S * I(in) + Vout,oa

Vout = (50 mV/A)*(54.5A) + (0.33 V)

Vout = 3.1 V

Here also i have checked i can't get maximum Vout as 3.3 V but i have given 3.3 to so i am assuming i need maximum voltage get as 3.3 V in Vout also.

Ideally i am assuming that when i connect my load i need to get some current in positive as well as match voltage with the DMM and calculation formula in so i am requesting you to please provide any formula for linear current calculation, also I have attached schematic for your reference.

Thanks & Regards,

Fenil

NOTE: I have application on Linear Reading based so i am taking input and all parameter accordingly.

  • 0
    TI__Mastermind 18790 points

    Fenil,

    where in the diagram is the load? Are you referring to R4? If this is the case, in the diagram, the way I interpret this, I would think that current would flow up through the FET Q2 when it is on in the bridge, which would be negative current flow through the TMCS the way you have the device configured. While our "U" designation parts are optimized for unidirectional operation, they can swing slightly in the other direction, which it looks like is happening here, as 0.28V is still well above the worst swing limits of the device. Was the current you were expecting in your test approximately 1A in the forward direction,? Have you attempted swapping the terminals of the device and remeasuring?

  • 0 in reply to Carolus Andrews
    Prodigy 10 points

    Hello

    Thanks for your reply, I want to clear some of your points.

    -> We have connected load in between MOSFET so out load connection pins is LOAD O/P N and LOAD I/P IN.

    also i am trying to measure current at Vout pin in IC, In your point about input swap but this is not possible as i am using PCB so.

    Kindly suggest a solution or provide any schematic design and current conversion formula for unidirectional IC.

    Thanks,

    Fenil Ghoghari

  • 0 in reply to Fenil Ghoghari
    TI__Mastermind 18790 points

    Fenil,

    Is it possible to place an ammeter in the load line to get a read on the actual current system here? Knowing the actual current against what you are measuring may help here. The current conversion formula you use above is correct. From the diagram, it seems like you are simply using the optocoupler to turn on both FETs and expect current to flow through the load, so this should be relatively straightforward.

  • 0 in reply to Carolus Andrews
    Prodigy 10 points

    Hello 

    Thanks for the reply.

    I have tested with ammeter so below i mention my ammeter reading,

    Voltage Current
    0.33

    0mA
    0.340 412mA

    This above reading i have taken multiple time now i calculate current according to input current,

    so,

    Vout = S * I(in) - Vout,oa

    CASE 1

    Vout = 0.33 & Vout,oa = 3.3 * 0.1 = 0.33

    so,

    I(in) = (0.33 - 0.33)/50

    I(in) = 0 mA this is corrent calculation i got and justify with datasheet

    CASE 2

    Vout = 0.340 & Vout,oa = 0.33

    So,

    I(in) = (0.340 - 0.33)/(50mA/V)

    I(in) = 0.2 A = 200 mA

    Here, this result is not justify with datasheet as well as ammeter so in this case ammeter showing current consumption of 412mA

    Summary,

    Voltage (DMM) Current (Ammeter)(Expected) Current Calculation using formula Acceptance
    0.33 0 mA 0 mA Yes
    0.340 412 mA 200 mA No
  • 0 in reply to Fenil Ghoghari
    TI__Mastermind 18790 points

    Fenil,

    This device sounds like it is working correctly from a functional point of view. The issue is the error between expectation and realized measurement. There's a few things to point out here:

    First, are you taking the offset of the device into account? Each device is going to come with a current offset similar to the voltage offset of an amplifier. This could be adding error to the measurement: 

    Next, you are testing the part outside of the sensitivity specification given in the datasheet. While I don't expect the accuracy to be as bad as what is shown here, you are testing outside of this range. Have you tried larger values to see if these are in line with expectation?

    Finally, you also need to take the noise density of the part into consideration here. Hall effect devices are inherently noisy, and you may need to reduce the bandwidth of the device dependent on how small of a signal you are hoping to measure here. This measurement as it stands is buried in the noise floor if you are operating at 80kHz.