LM25056A: Current sensing using LM25056A

Hello Hari,

Welcome back. The LM25056A should really not be used for measuring microamps (uA) of current because its input pins for the shunt resistor (VS+ and VS-) have input bias currents (or "IB+ and IB-") of 20uA typically. So 20uA is being sourced into each pin. In addition to this, the input full-scale range for the shunt resistor voltage sensed by VS inputs is only 60.88mV, which is much smaller than the 100,000:1 (100mA to 1uA) range you need.

You could look into the INA190. It has an IB of 0.5nA typical and 10nA maximum. It has a linear 100:1 maximum full-scale output ratio (or range) when powered with 5-Volt supply and measuring one shunt resistor. The linear output for the device is optimally 50mV to 4.9V, but the device can swing much closer to rails (VS-45mV) and GND+5mV, which gives you a 910:1 output range, but any linearity error seen from this range could be potentially calibrated out. Overall, this means you need multiple current sensing devices or some network of switches to dynamically reduce shunt resistance for higher currents. The INA190AxRSW is a low-power current shunt monitor with 42V to -0.3V common mode voltage range. Similar to LM25056 it also has an ENABLE pin.

I would recommend looking into the INA233 (163mV/10uV = 16384:1 ratio) even though they have 8uA typical input bias currents into input pins. Programmable-gain amplifiers (PGA) or other analog front end parts with internal ADCs with greater resolution or programmable gain would provide you maximal input dynamic range. Examples are PGA300 or ADS122C04.

Please reply back with any questions.

Best,
Peter Iliya
Current Sensing Applications

Hi Peter,

Thank you very much.For INA190 you told it has a linear 100:1 maximum full scale output ratio.May I know how you obtained it.Could you please explain me with the help of maths so that I can understand easily.

If we are using multiple sense resistors how to design a switch network that switches between switch resistors.

Regards
Hari

Hello Hari,

The maximum full-scale linear output ratio I calculated was a quick calculation of Vout_max/Vout_min = 5V/0.05V = 100. This is actually not techincally accurate, but just for comparison. Any voltage output amplifier will have a linear region of operation for its output stage. This range can be determined by inspecting what testing conditions the datasheet specifies for linearity or gain error specifications. For the INA190, gain error is guaranteed for GND+0.1V to VS - 0.1V. So if power = 5V = VS, then the most linear output range of device is 100mV to 4.9V. So really the maximum linear output is 4.9V/0.1V = 49:1, but this is conservative and as I mentioned earlier you can extend this range to 4.95V to 10mV which would be 4.95/0.01 = 495:1. You can read about amplifier output swing in this post: e2e.ti.com/.../703088

As for more circuit idea please help yourself to these reference designs which show ways to increase dynamic range. There could be more on the website.
www.ti.com/.../TIDA-00328
http://www.ti.com/tool/TIPD135
http://www.ti.com/tool/TIPD156
http://www.ti.com/tool/TIPD129
http://www.ti.com/tool/TIPD104

We also have current sensor training videos here: training.ti.com/getting-started-current-sense-amplifiers

Best,
Peter Iliya
Current Sensing Applications

Hi Peter,
Good day...

I will study these reference designs and will try to reach a solution.
In my initial inspection I understood that all these reference designs use single shunt resistor.
will contact you if any further clarifications is needed.

Thanks&Regards
Hari

Hi Peter,

I watched some videos of current sense amplifiers by TI.
I have a question,do we need to use instrumentation amplifiers when current to be measured starts from 1uA.
The video says current sense amplifiers is used when the current to be measured starts from 10s of uA

Regards
Hari

Hello Hari,

This was true until we released the INA190 current shunt monitor, which uses a different internal topology compared with our other current shunt monitors. This allows it to have an input bias current that is 3 orders of magnitude less than most analog current shunt monitors. These designs were made before the INA190 and thus they could not have known about this device. Before the INA190, instrumentation amplifiers and transimpedance amplifiers were what we suggested for measuring very small (nA to uA) currents.

Hope this clarifies your question.

Best,

Peter Iliya

Hi Peter,
Is it possible to amplify 1uV DC signal using INA190.It's offset voltage is showing 15uV.If the input voltage is less than offset voltage is it possible to amplify 1uV signal.
Regards
Hari

Hari,

Resolving/measuring a 1uV DC signal could be possible with the INA190A5 (500V/V gain; 20kHz BW), but would be tough. Aside from calibrating out the offset and needing a stable power supply, reference voltage, and input signal, you would need to filter the output as well. The input referred noise is 70nV/rtHz, so the output noise will be 500V/V*70nV/rtHz*sqrt(1.57*20kHz) = 6.2 mVrms output noise. So a 1uV input gained up to a 0.5mV output shift would be buried in the noise.

If you add a 2.5kΩ and 1uF capacitor to the output, then you can reduce this 6.2 mVrms output noise to around 360 uVrms. Note that the input filter with 500Ω and 1uF do not contribute to the output filtering. This filter would only be necessary if you input voltage signal is noisey and you need to filter out this input noise so it does not get gained up by the device.

Best,

Peter Iliya

Hi Peter,

Good day...
This IC is not available with Digi key or other suppliers.
May I know you have any alternate parn numbers.
Any amp is avialable with pV offset voltage

Regards
Hari

Hi Peter,

We have almost finalized our specs

Minimum current to be measured=.2mA
Maximum Current to be measured=100mA
Rsense=.1Ohm
Minimum Current Step=0.01mA
I think now I can manage with single resistor

Regards
Hari

Hi Peter ,

I need one more clarification
My required step is 10uA so ADC and sense resistor is .1Ohm,so ADC should detect 10uA*.1OhmVolts..correct me if I am wrong.
May I know is this possible?
Can any amplifier will be able to amplify this voltage (10uA*.1OhmVolts)

Regards
Hari