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# INA138: INA138

Part Number: INA138
Other Parts Discussed in Thread: INA210, INA333

I am trying to replicate the D/S graph below using TINA, specifically the trace for 5K+10nF.  I am attaching the TINA simulation that shows a different frequency response surely due to my setup.  Could you guide me on how to set this up on TINA? Many thanksINA138 Sim Freq-Res.TSC

The frequency response plot I get shows a different gain for the 5K+10nF trace.

Many thanks

• The D/S graph I am trying to replicate is:

• Hello Jorge,

The trick is that you need to use a voltage input in order to see the correct gain measurement. In your original circuit, TINA is calculating the AC transfer function based on the voltage at Vout vs. the current at IG1. Because your shunt resistor is 5 mΩ, in terms of voltage gain the circuit looks like an attenuation of 0.005 V/V, or -46dB.

Using a voltage input like what I've done in the modified circuit below, TINA only calculates the transfer function of the INA. In this case you get the expected result of 0dB at low frequency.

INA138 Sim Freq-Res_iw.TSC

Best regards,

Ian Williams
Applications Manager
Current Sensing

• Hi Ian! Many thanks for your help.  Worked like a charm.

Jorge

• My pleasure! Happy to help.

Best regards,

Ian Williams

• Hi Ian, reaching out to you again, this time with a CMRR tina sim.  I am trying to simulate the effect of 1kΩ resistors on the input of INA138, we have a legacy design that has these resistors on the input to the detriment of CMRR.  The reason these were placed in the first place was for a slight decrease in gain (internal plus external R now becomes 6kΩ, decreasing the overall gain but degrading CMRR)

I am trying to quantify this but had some issues with the CMRR setup.  I know CMRR = change in Vout divided by the change in Vos due to changes in Common Mode.  To this effect I took out the signal generator and placed him outside as shown on the image and shorted the inputs while placing a voltmeter at the inputs to measure Vos.  I used then used the post processing tool to build the expression. Anyhow.  Wondering if you can steer me in the right direction.  What I am truing to do is a comparison of with and without the 1KΩ resistors.  For a G=1 (5K+10nF) I should get the D/S curve.

The TSC that I am using is attached.  Many thanks for your help.INA138 Sim CMRR.TSC

• Ian, I left out one piece of information, the 1kΩ resistors are 0.1%.

• Hello Jorge,

I've attached my test circuit for CMRR below. It uses two copies of the current sense amplifier, each configured slightly differently. The top circuit results in measuring the common-mode gain (ACM), while the bottom circuit measures differential-mode gain (ADM). The ratio of these two is the textbook definition of CMRR and gives you the expected datasheet curve. You can use the post-processor function in TINA to define CMRR as equal to ADM / ACM in order to measure the curve as a rejection.

However, the result for the INA138 model is inaccurate, as I measured 146.7 dB at low frequency. This is not unexpected, as it's a very old and very simple model. To get better results would require re-development of the model from scratch, which is not something we plan to do in the near future. More recent models are much more accurate and feature-rich.

Best regards,

Ian Williams

INA138 Sim CMRR_iw.TSC

• Hi Ian, thank you for your explanation.  Given that I cant really simulate the impact of additional of 1KΩ-0.1% at the inputs of INA138 bc the model does not allow, I have a few questions.

1. How would you recommend I make a pencil and paper quantification of the impact? this would help me quantify impact on CMRR

2. Would you recommend adding resistance at INM and INP altogether to diminish the gain? this questions steers me on the impact on CMRR qualitatively.

3. Is INA138 an instrumentation amplifier? from the block diagram I would say it doesnt look like one  but behaves like it, for example it doesnt look like an INA210 that has an OPA with matched resistors wrapped around it. Really curious on how you answer this one.

4. Any comment on end of life?

• Hi Jorge,

Thank you for your patience. I will have answers for you later in the week after the Christmas holidays.

Best regards,

Ian Williams

• Hello Jorge,

1. To determine the CMRR when using extra input resistance, use the equation below:

CMRR_ext = -20*log((2*Rtol)/100), where Rtol is the tolerance of the resistors in percent.

Quick spreadsheet calculator attached: CMRR Calculator.xlsx

For example, with ±1% tolerance, you can expect only 34dB of CMRR. With ±0.01%, this improves to 74dB.

2. Since the input resistors are trimmed to very close to 5kΩ, you can use extra resistance at INM and INP to diminish the gain if desired. However, the matching of your resistors will directly reduce CMRR as calculated in answer 1.

3. The INA138 is not an instrumentation amplifier, but a current sense amplifier. An instrumentation amplifier usually has resistor-controllable gain and consists of a difference amplifier with its inputs buffered, such as the INA333. A current sense amplifier usually has fixed gain and consists of a special input bias stage that allows Vcm > Vs, followed by a difference amplifier. The INA138 internal design is a bit different as it's a current-output rather than voltage-output device, but it behaves in a similar way as most current sense amplifiers.

4. TI has no plans to end-of-life the INA138.

Best regards,

Ian Williams