# Current Sensor Resolution

I have an application where the maximum DC current is 160A. However in reality the actual current flow never exceeds 40A. I want to measure the current with an ADC (like one found on the Arduino or a higher resolution ~16 bit ADC). I originally wanted to use a hall effect sensor, specifically the ACS770ECB-200U-PFF-T.

Typically, my application uses around 5A. I am aiming for a .01A resolution. I looked at the sensitivity (mv/A) and thought I could get a higher resolution ADC to get more accurate current sensing. However when I looked at the data sheet it said the following:

The noise floor is derived from the thermal and shot noise observed in Hall elements. Dividing the noise (mV) by the sensitivity (mV/A) provides the smallest current that the device is able to resolve.

This means that with the typical noise found in the datasheet of 6mv, and the sensitivity being 20mv/A, I would only get a .3A current resolution.

Am I correct? How can I achieve a >.01A current resolution? I can use a shunt resistor if I have to, or a 50A current sensor instead. But the 50U version gives 20mv of typical noise and 80mv/A sensitivity, giving a 0.25A current resolution.

Also, is this noise the ripple voltage of the power supply for the current sensor?

I'm also a little confused as to what the datasheet means by bandwidth

120 kHz typical bandwidth

As I don't see what is oscillating.

• Also something to note, there will be low current conditions ~.5A - 2A, which I've heard is a problem with hall effect sensors as they obviously rely on the current to generate a field which they measure. – Peter Jun 8 '16 at 15:51
• If you don't need high bandwidth, you can filter the noisy signal (low pass, average, etc.) to get a less noisy, slower changing signal. – corecode Jun 8 '16 at 22:47

I'd recommend the ACS722 10AU sensor. However, you will likely not get the accuracy you want from a Hall effect device. A shunt resistor is always more accurate, but that comes at the cost of BOM lines and potentially more expensive amplifiers/current shunt's (at least at the very good sensitivities/very high currents). The integrated Hall-effect current sensor market is reducing module sizes and BOM cost by making compromises on total accuracy. If it works for your application, great, otherwise use a shunt resistor.

The bandwidth figure you see is the maximum current frequency the sensor can resolve.

• How would I implement a shunt resistor on a pcb able to give me my required sensitivity? How would I connect my ADC to the resistor? – Peter Jun 8 '16 at 16:06
• You use a shunt resistor and an amplifier, the amplifier connects to the ADC. The current through the resistor, and thus the current in the circuit is given by I=V/R. Remember, P=I^2R, so your resistor's power rating will need to be correct for the amount of current flowing through it. This usually means using a lower value resistor, thus you need the amplifier to amplify the small voltage developed. – Brendan Simpson Jun 8 '16 at 16:09
• Thanks for your answer! Electronics are so cool, it must be a good feeling when you know a lot about it and can get things right from the start and understand everything in data sheets. – Peter Jun 8 '16 at 16:13
• My pleasure. And in this case it does help that I worked as a test engineer for Hall effect current sensors at Allegro. – Brendan Simpson Jun 8 '16 at 16:16
• That's so interesting, far out. I'm straying from the question but I just find the people and engineers on this website generally interesting in their jobs and knowledge. It's rather inspiring. – Peter Jun 8 '16 at 16:18

As a designer, you get to set the dynamic range. This comes at a cost of sensitivity, because the wider the range the less gain you will need. Noise and ADC resolution sets a lower limit in a system to what can be measured, so when you increase the dynamic range and lower the gain, you will get lower sensitivity per bit. Noise also needs to be considered on the ADC because usually the lower bits of the 16-bits are noisy.

When analyzing noise its best to choose a point in the circuit to make your comparisons, like the input to the ADC. So if you have no gain in your system and your ADC has a 3.3V range $3300mV/\frac{20mV}{A}= 165A$ So far so good. Now lets check what the sensitivity at the input would be.

The ADC @ 3.3V has a resolution of $\frac{3.3V}{2^{16}} = 50uV$. Now working backwards $0.05mV/\frac{20mV}{A}= 0.0025A$

If that doesn't meet your requirments then you need to change the gain of the system, you can do this with an op amp in between the hall effect sensor and the ADC, lets say it has a gain of 2V/V or 2mV/mV

$3300mV/\frac{20mV*2mV/mV}{A}= 3300mV/\frac{40mV}{A}= = 82.5A$
and $0.05mV/\frac{40mV}{A}= 0.00125A$

Gain comes with a sacrifice. There is less dynamic range but more sensitivity. This can be over come by using a system that changes the gain by switching OR using a multi sensor approach with two channels one with high gain,low range,high sensitivity and a channel with low gain,high range,low sensitivity.

Any sensor has bandwidth, the sensor acts like a low pass filter and can only 'observe' currents less than 120Hz, anything faster will be filtered out. So if you had a line with a 240Hz sine wave AC the sensor would read close to zero volts on the output.

• Sure, the ADC can get that resolution, but the actual sensor itself cannot, from what I've gathered from the data sheet. I've considered using the ADS1115 since it has a programmable gain amplifier, which I could scale with the current. Or am I wrong and data sheet is saying it cannot detect currents lower than .3A but anything more than that and it will work very accurately? – Peter Jun 9 '16 at 0:34

The bandwidth of the device, since it is DC coupled, is the frequency for which the output will be 3 dB down for a constant input current. In your case, since you are only interested in the DC component, you might be able to achieve a higher resolution by filtering the output of the sensor to reduce the wideband noise level. The datasheet gives no information on the frequency spectrum of the noise so it is difficult to determine how effective filtering would be.