INA138: About the INA138 sch confirm

Hi Bruce, 

Thanks for using E2E!

The capacitance on the output for this device does not have a dictated maximum for stability or an optimal value as you ask. In the simplified diagrams shown in the datasheet, this is meant more to mimic potential load capacitance that may be present in your load. Best practice here is to keep this value as low as possible for best performance as stated in section 7.3.2 of the datasheet, as large amounts of output capacitance will cause potentially undesired shifts in the bandwidth of the output:

Hi AE,

       Since our RL is expected to be around 330 kΩ, do you have any recommendation for the CL value we should use?

       Or could you provide a guideline or calculation method for determining the appropriate CL value?

Best Regards

Bruce, 

Do you have an idea of how much bandwidth you need here? For a 330kΩ resistor, the gain as per this plot would be 36.4 at DC. 

With no load capacitance, I would estimate BW to be roughly 12-15kHz. Addition of 100pF of load cap would reduce this to around 5kHz, and so forth as per the chart. The roll off points are calculated as per standard first order cutoff for an RC LPF (1/(2*pi*R*C). 330kΩ and 100pF yields a 4.822kHz cutoff. Increasing this to 1nF truncates that BW to  482Hz. You can see this correlates to the above curve, and you can use this to set your 3dB down point for the chosen resistor / load cap combo. 

Hi AE,

       Thank you for your reply.

        But I would like to clearly understand how to choose the proper CL value when RL is 330 kΩ or 499 kΩ.

        If we are only reading the current and do not require a fast response, is it acceptable to set the bandwidth in the range of about 1.45 k to 2.12 kHz ?

        If I use a 330 kΩ or 499 kΩ resistor, I can select a CL value between 150 pF and 220 pF to set the bandwidth in the range of about 1.45k to 2.12 kHz, correct?

Best regards

Bruce, 

As the plot above shows, the resistor will set the DC gain level. As 5kΩ is normalized at 0dB, we can establish the correction on this to be 20*log(5k) = 73.98. Then, if 20*log(330000) =  110.37, normalizing this into the plot yields 110.37-73.98 = 36.4, which is the blue line I drew across the plot. 499kΩ is close enough to the 500kΩ plot that I would just use that. 

Then, dependent on resistor choice, you use the formula I gave you above to set your bandwidth. From the above, dependent on resistor choice, you would have 4 distinct bandwidths based on the 2 capacitors and 2 resistors you have listed:

330k and 150pF: 3.215kHz

330k and 220pF: 2.192kHz

499k and 150pF: 2.126kHz

499k and 220pF: 1.45kHz