Getting started with: 30A Ratiometric inline current sensor

The 30A ratiometric inline current sensor is a simple breakout board built around the ACS711 linear hall effect sensor. What does that mean for you? Let’s break it down:

Note: If you want to get straight to business, jump right to the next header to get started. If you’d like a little more background on this sensor, skip this for now and take a moment to read below.

Just show me how to wire it up.

Here you go:

This sensor is ratiometric, meaning that the scaling of the analog output voltage is proportional to what you provide to the VCC pin. Additionally, this analog output is offset by VCC/2 in order to output a positive voltage for both directions of current flow. The sensor has a sensitivity of 45mV/A, meaning that the final equation for the analog voltage out is:  Vout = (VCC/2) + (0.045*i)

Let’s see a quick example of this below:

Assume you’re powering the sensor off a 3.3V bus. With no current flowing, the sensor should read 1.65V nominally. Now assume you flow positive 10A through the current line. You should expect to read an output voltage of 2.1 volts. Similarly, a current of -10A should produce an output voltage of 1.2V.

A couple other important notes:

  • The usefulness of your output voltage is directly tied to the stability of your input voltage. Noise will propagate to your measurements, and should either be minimized or accounted for by scaling your ADC or analog electronics gain according to the value of VCC
  • A small 0.1nF filtering capacitor has been added to the breakout board for your convenience. If you require higher slew rates or have different filtering requirements, please desolder C2
  • The ACS711 features a fault pin, which has been broken out on this board
  • This board is not intended to provide isolation for safety purposes.

What does it do?

Want to measure the current flowing though a wire in your circuit? The 30A ratiometric inline current sensor has your back (as long as that current doesn’t exceed -30 to 30A)! With its linear analog output and minimal hookup requirements, you can be measuring current in mere minutes. It’s a simple as powering the device with 3-5.5V and reading the voltage of the analog output.

But can’t I just measure current with a shunt?

You certainly can. Shunts utilize the voltage drop of a carefully calibrated resistor to determine the current flowing through it. However, this simplicity comes with drawbacks. Power loss through a resistor = current^2 * resistance. Let’s say you want to measure a 0-5A current source. If you utilized a 0.1 Ohm shunt, you’d see 2.5W of heat loss at 5A and 0.5 volts dropped across it. This is a little wasteful, but might work in your application. Now lets assume you need to measure up to 30A. If you tried to use that same shunt, you’d burn a staggering 90W of heat and probably fry your shunt! So to compensate, you decrease the resistance by two orders of magnitude, to 0.001 Ohms (1mOhm). Now you only generate .9 Watts of heat, which is probably within your shunt’s temperature range. But you also drop 0-30mV, which might be too small for you to easily measure without amplification.

Okay, okay, so can’t I just use a low-resistance shunt with an amplification circuit?

Of course! Before you break out the dusty LM-741 op-amps though, let’s consider a few other drawbacks. Because you need to know the delta-V across your shunt to calculate current, you need to make sure that you can properly make the differential measurement between those two points. If you’re using some of the more common hobby microcontrollers out there, you may notice that the analog-to-digital converters reference the controller’s ground as the analog ground. For most practical setups, this means that one terminal of your shunt needs to be tied directly to ground. This is kind of a bummer if you planned on measuring current at any other point in your circuit. It also means that measuring bi-directional current is likely off the table.

That’s fine, I can connect one end of the shunt to ground, or I have a fully differential ADC.

Sounds like a shunt might work nicely in your application. As a last consideration though, can you ensure that the current flowing through your wire will never exceed the shunt’s capacity? If your shunt ends up fusing, you’re going to have the full destructive power of your voltage source present at the input pin of your poor ADC/other analog circuitry. If this kind of thing keeps you up at night, you might consider including an inline fuse to make sure this can never happen. Of course, now you have the expense and additional voltage drop of a fuse to consider.

The above summarizes the advantages of a linear hall effect sensor in a nutshell. It provides a isolated (to chip limits, please see datasheet) current measurement that you can put at any point in your circuit and measure with a ground referenced ADC or analog circuitry. It also automatically amplifies the analog signal to a level between 0V and your 3-5.5V bus.

Basic theory of operation

The ACS711 uses a device known as a hall effect sensor to monitor current flowing through the chip’s current path. Hall effect sensors  make use the magnetic field generated by the flow of electrons to estimate the current through your wire. In the ACS711’s case, it’s designed to modulate the output signal in direct, linear response to a changing magnetic flux density. Put simply: more current in = more voltage out.

But wait, there’s more. Because the hall effect sensor measures current based on proximity and flux density, a direct electrical connection between your current source and monitoring electronics are not required.