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INA214

Other Parts Discussed in Thread: INA214, INA213, TIPD135, INA210, INA212

Hi Engineers, 

Recently I found the chip INA214 and this chip support very well my current shunt applicantion, but I tried to simullate on Tina my circuit but without sucess. My simulation appeared a saturation. 

My circuit is a power supply with 3V peak sinuidal symetric power supply. My shunt resistor is 100R/0.1% and my load is a variable resistence between 0 and 300R. working with bidirectional

This current sensor will be send analog signal to ADC of Tiva129x, then I need adapt my signals to 3V Peak. This mode I need a Shunt amplifier with gain 50 to price Vpeak 1.5 + 1.5DC offeset.

Then I consider  Rload = 0R then my current will be 30mA, and My ADC will read 3V and 

I attached in design of circuit on TiNa. Someone can help me solve the problem. Thanks for you attention  and best regards. 0181.correntesense.TSC

  • Carlos,


    I tried your simulation and it didn't work, as you said.  The first thing I saw was that you are using the INA213 instead of the INA214 as you said, but you wanted the gain of 50 which is the INA213 so this is a correct device choice in the simulation.

    The next thing I see that is really odd is that you are using an enormous shunt resistor, which is certainly going to be a problem, from a performance and a practical standpoint.  If you have a 100 ohm shunt and a 300 ohm load, you are already cutting your power supply down to 75% of its original value.  If you have a 100 ohm shunt and a 1 ohm load, you're barely getting 1% of the voltage to the load resistor.

    Typically people use very small shunt resistors to measure currents with shunt monitor devices, in the range of 1-10mOhms.  This is because they want as little voltage drop across the shunt as possible, and they have large currents that will generate large voltages across even small resistors.

    Because this is the typical application, the design targets are different from a true instrumentation amplifier and the front end of the current shunt monitor has significantly larger input bias (Ib) currents.  With small shunt resistors, the effects of an imbalance of Ib is negligible. With large resistors, this becomes a significant source of error (Ib+ - Ib-) * 100ohm >> (Ib+ - Ib-) * .010ohm.


    The information I need to help you make a device selection and set up a good simulation would be the target current to measure and the level of accuracy needed.  Having such a dynamic load value (0-300ohms) makes me think that you might want to investigate a PGA based solution where you can select different ranges to make your large measurements without clipping the output and then switch down and maintain decent accuracy as you make small measurements.  I think I understand the desired output requirements so I can work with that once I have the input requirements.  TI Design TIPD135 is a good example of this.

    Thank you,

    Jason Bridgmon

  • Hi Mr. Jason Bridgmon
    Thanks for you attention and your time. 
    Then about your suggestions and considerations, I agree with a lot of topics. But I need explain my device. Im finish to build a electronic to a chemistry sensor . And the condiction to this sensor works is 3,2V ( Vpeak) with 100R  resistor to limit current. When sensor is in operation it variable resistence between 100 and 330R. 
    Im my original project I read voltage be 100R resistor. But in my original device, But now this device is too much expensive because I used OPA27G and other Amplifiers to instrumentations to read voltage and current. And mysensor do a  frequency sweep between 100Hz and 10Khz with steps of 500Hz.  
    According with datasheet page 7 the maximum frequency operation to INA214 is  30Khz. 
    Considering this situation: 
    In my worse situation I can consider a short circuit in sensor then my maximum current will be: 
    Ioutmax =  Vout(peak)/100R  = 3.2V/100R = 32mA
    In my better situation I can consider sensor with 330R then
    Ioutmin = Vout(peak)/430R = 3.2V/430R  = 7.44mA 
    With our sugestion about Shunt resistor I found a Resistor shunt on Digikey with 0.0022R (RHM.0022BRCT-ND)
     Using this resistor SMD CHIP we have:
    Vshunt max = 0.0022R*32mA= 70uV
    Vshunt min = 0.0022R+7.44mA= 16uV
    With this shunt voltage INA212 with gain 1000 will provide 70mV/32mA and in datasheet Offset Voltage: ±35 μV (Max, INA210) then offset is in same scale of Vshunt. 
    To increase a shunt voltage I consider resistor shunt with 1R( RHC2512FT1R00CT-ND) then we have:
    In my worse situation I can consider a short circuit in sensor then my maximum current will be: 
    Ioutmax =  Vout(peak)/100R  = 3.2V/101R = 31,6mA
    In my better situation I can consider sensor with 330R then
    Ioutmin = Vout(peak)/430R = 3.2V/431R  = 7.42mA 
    now voltage of Shunt resistor will be 
    Vmax = 31.6mV and Vmin: 7.42mV
    choosing INA213 ( gain 50)  voltage output will be 
    Vmaxout =  31.6mV*50 = 1.58V
    Vminout =   7.42mV*50 = 0.371V
    Now considering that my current is sinuidal and with symetric we add the offset voltage, and this mode ADC of TIVA processor can read.  The voltage offset will be 1.65V, this mode with maximum current my ADC will work between 0.07V and 3,23V.
    then in pin REF I will tie power supply with 1.65. 
    If I build this project on TINA (TI) and try simulate my output is completly wrong. Then I attached my example again. Can you help me solve this problem. I dont understand where Can I doing a mistake?
    Thanks for you attention and time 2664.correntesense.TSC
  • Carlos,

    The design you sent me has the input voltage sin wave going from -3.2V to 3.2V with reference to ground.  While the device does work in a bi-directional mode, the input pins cannot exceed the GND-0.3V limit.  You would have to offset your input voltage by at least +2.9V to have the circuit work.

    Best regards,

    Jason Bridgmon

  • Perfect, work very well. Thanks for you attentions