Measuring Current in Combination Circuit with a Shunt

I've done a fair bit of looking/asking around for help with this but I'm struggling to come up with an answer ....

I'd like to measure and record current in a circuit. There is a 12 VDC / 2 A power supply which provides power to two instruments. They draw 75 mA, and 640 mA at 12 V respectively. They're in parallel with the supply, like this:

Instrument 2 is a water pump which operates at a nominal rate of 3 L/min. However, if it gets clogged for some reason, it draws more current in order to try maintain that 3 L/min output.

SO, I would like to be able to measure and log the current supply to these devices. From my research, it appears as though a shunt is a possible option. I would place it just before the negative lead on the power supply, and record the voltage drop across it using an external DAQ device (which measures from -10 to 10 V), much like this:

I guess my questions are: 1) Is this the best way to solve this problem? If not, what would be a better alternative? 2) If it is a good option (or even if it isn't, maybe humour me a bit ....) what "type" of shunt should I get? Is this the right way to wire the shunt. Unfortunately, the instruments themselves cannot be re-configured.

I would be happy to provide more detail as required.

• Thanks for the responses, guys! I have some comp support here at work, so we'll try roll with these suggestions. – RIz Jul 22 '16 at 22:38

There are very convenient modern IC solutions for measuring high-side current. For example consider the TI INA169 60-V, High-Side, High-Speed, Current Output Current Shunt Monitor. You can even get a breakout-board ready to use from vendors like SparkFun, et.al.

It is generally not desirable to put a shunt in the ground-return side because of possible interactions with external ground connections. Unless the entire circuit is within your control (no external connections).

Your drawing is plenty good enough, and you do not need a shunt for such low currents. A precision resistor of .1 ohm .1% 5 watts maybe cheaper than a shunt. Precision shunts can be expensive, while .1% tolerance resistors are becoming common. 1 amp flowing through a .1 ohm resistor = 100mV.

Also, you do not need a bipolar DAC if your not reading a negative current flow. A low cost 12 bit DAC will give you 3-1/2 digits of resolution. You may consider a digital panel meter by Modutec for a direct reading and use the DAC with a Raspberry Pi board as a data logger, though I do believe some versions of Raspberry Pi have analog input options.

You are on the right track but generally we try not to disturb or "load" the circuit too much when taking measurements. That usually means that when measuring voltages we use a high input resistance meter and when measuring currents (as in this case) we use low resistance shunts. As you've indicated on your drawing, the shunt could cause a drop of 10 V leaving only 2 V for the motor! In practice the shunt would cause so much voltage drop that the motor current would drop too and the 2 V situation would not be reached.

To meet our criteria of not disturbing the circuit too much we might decide that we could tolerate a 0.2 V drop at 1 A. From that we could calculate the shunt resistance as $R = \frac {V}{I} = \frac {0.2}{1} = 0.2~\Omega$. That's the shunt sorted out.

Now we have the problem that 0.2 V isn't going to give us good resolution on our 0 to 10 V DAC so an amplifier is required. We can calculate the gain, $G = \frac {V_{OUT}}{V_{IN}} = \frac {10}{0.2} = 50$. This would usually be done using an opamp in non-inverting amplifier mode.

You could put a shunt into the leg of each load to measure both independently.

As an alternative you could investigate Hall effect current sensors.

Figure 1. The Hall effect generates a voltage across a current carrying conductor exposed to a magnetic field.

These sensors run a small, known current through a portion of the chip which is placed in the magnetic field around the current carrying conductor. The small voltage induced across the conductor is amplified to give a signal suitable for direct interface with a micro ADC circuit.

Figure 2. To enhance the magnetic field a toroidal core is usually employed and the Hall effect sensor inserted in a small gap. One or more turns of the current-carrying conductor are passed through the core to obtain the required ampere-turns ratio to optimise the resolution of the complete circuit.