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INA139: Using INA139

user5015634
Intellectual 260 points
Part Number: INA139
Other Parts Discussed in Thread: INA170

Hi there

I'm following up on a test related to the INA139 and simulations in TI-TINA.

The current that I want to  take to sample is from -2.5 A to +2.5 A, and the frequency is range from 100K Hz to 1M Hz. The current waveform is a triangular wave.

The intention is to use two INA139's in a bipolar configuration to measure the bidirectional current ,  the sense resistor is 10 milliohm. So the VSENSE=VIN+ - VIN-=-25mV-+25mV. The V+ is 5V,and Rload is 1K, so

the Gain is 1.

The device(INA139)  named A is to be used detecting positive voltage of the VSENSE,  I found that :the frequency of the current  is 100KHz, when the VSENSE is negative ,the VOUT of A is zero, but when the

VSENSE become positive ,the VOUT remained zero until 1.2 us later, then it began to grow. As the frequency increases, the phenomenon becomes more serious.  What is the reason? 

In the datasheet, FREQUENCY RESPONSE

Settling Time (0.1%) 5-V Step, ROUT = 10 kΩ 2.5 2.5 µs
5-V Step, ROUT = 20 kΩ 5 5 µs

 What does this parameter mean?

 Now I want to use the INA 170. Will this device have the above phenomenon and whether the bandwidth meets my conditions of use?

Thanks 

LIN

  • 0
    TI__Mastermind 27410 points

    Hi Lin,

    To quote Wikipedia, here is definition of settling time – “Settling time is the time required for an output to reach and remain within a given error band following some input stimulus.”

    I suspect the delay you observed is due to output saturation. If you want, please attach your simulation file so we can take a further look.

    INA170 should be better in this regard because it can be configured in bidirectional mode and the output remains in linear region.

    Regard, Guang

  • 0 in reply to Guang Zhou
    Intellectual 260 points
    Hi!


    The above phenomenon was not found in the simulation and was found in the actual test. Does the output have this "settling time" when every


    time the input differential signal(VSENSE) changes from negative to positive? In the simulation, it was found that when the input is negative, the


    output is approximately -50 mv instead of zero.
  • 0 in reply to user5015634
    TI__Mastermind 27410 points

    Hi Lin,

    Delay will happen every time the output is over driven in the negative direction. Under this condition the output NPN is cut off, base emitter junction is no longer forward biased. Besides there is an internal clamp (diode connected NPN) that is probably weakly conducting, hence the -50mV output. When the input voltage reverses to positive, both junctions need to recover accordingly. The delay is probably a combination of these two.

    The delay however should stay relatively constant regardless of switching frequency, it just looks worse proportionally at higher frequencies. Is this what you mean by “As the frequency increases, the phenomenon becomes more serious”? Or, do you observe the delay itself increases at higher frequency?

    Regard, Guang

  • 0 in reply to Guang Zhou
    Intellectual 260 points
    Hi!

    The delay is constant no matter how the frequency changes. Thanks for your answer ! So if the INA139 is set to be a bipolar configuration to measure the bidirectional current, the frequency of the current is not allowed too high?
  • 0 in reply to Guang Zhou
    Intellectual 260 points
    Hi!

    I have another question. If I use INA 170 instead of INA 136, will there be the above phenomenon? For example: If the detected current passes through the shunt resister and the corresponding voltage is -50 mv to +50 mv, the waveform is linear, then the signal is processed through the INA170, the output of INA170 is 0 to 100 mv. When the output of the INA170 rises from 0, Will there be a delay?
  • 0 in reply to user5015634
    TI__Mastermind 27410 points

    Hi Lin,

    You’re right that at higher frequencies, the delay will mess up the response.

    When INA170 is used, the above phenomenon can be avoided. The key is to create an offset current that is bigger in value than the most negative current. Specifically with the condition you specified, the -50mV input corresponds to -50uA output current. We want to have an offset current that is at least 50uA. In the attached TINA example, the offset current is set to 200uA. As a result, the input differential voltage (-50mV, 50mV) is translated to output of (150mV, 250mV).

    Regard, Guang

    INA170.TSC

  • 0 in reply to Guang Zhou
    Intellectual 260 points
    Hi !

    Tkanks for your help! I will to buy INA170 to test!

    Regard,Lin