0
\$\begingroup\$

I have a system similar to the one given by schematic. There is a 12V battery used for supplying MCU board and couple of distributed boards(inverter/measurement network etc.). Since I want to measure their current consumption I added low side shunt resistor.

In configuration like this I would also measure MCU board power consumption which is why I didn't connected MCU board GND to "earth" ground.

I need to sense currents from 0 to 30A with ~10mA resolution(12bit ADC is used).

Will this configuration work? Is it better to switch to Hall sensor due to it's hysteresis and bigger measurement error?

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$
2
\$\begingroup\$

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. Common ground for all logic.

I haven't come across that chip before but I'd suggest a layout like this. It avoids any sneak paths between ground and common - such as when programming your MCU through a USB port on your laptop, etc., and is common practice in current measuring applications.

The gain is only 20 so if you want, for example, 3 V out at 30 A then your shunt would need to drop \$ \frac {3~V}{20} = 0.15~V\$ and its resistance would be \$ \frac {0.15~V}{30~A} = 5~m\Omega\$.

Perhaps a better option based on your comment is to use high-side monitoring.

enter image description here

Figure 36 of the datasheet shows a high-side monitor. (They seem to have the battery symbol upside-down.)

High-Side Current Sense with a High-Side Switch

This configuration minimizes the possibility of unexpected solenoid activation and excessive corrosion (see Figure 36). In Figure 36, both the switch and the shunt are on the high side. When the switch is off, the battery is removed from the load, which prevents damage from potential shorts to ground, while still allowing the recirculation current to be measured and providing for diagnostics. Removing the power supply from the load for the majority of the time minimizes the corrosive effects that can be caused by the differential voltage between the load and ground. When using a high-side switch, the battery voltage is connected to the load when the switch is closed, causing the common-mode voltage to increase to the battery voltage. When the switch is opened, the voltage reversal across the inductive load causes the common-mode voltage to be held one diode drop below ground by the clamp diode.

\$\endgroup\$
  • \$\begingroup\$ Thank you for answer. Using 5mR is 4.5W dissipation which is a lot! Where will those "sneaky currents" actually flow? Configuration you proposed is a problem because there will be communication interface between MCU and other boards and they will reference "wrong" GND \$\endgroup\$ – Bip Feb 27 '16 at 12:29
  • \$\begingroup\$ See high-side monitoring update. The 4.5 W is just the physics of the thing. If your ADC is more sensitive you can scale down. \$\endgroup\$ – Transistor Feb 27 '16 at 12:47
  • \$\begingroup\$ There is a great chance though that in the near future 12V source will be updated to 42V and using AD8207 on high side will have only 80dB CMRR giving ~4mV output error. For now I see Hall sensor as only solution. Do you know what are the drawbacks when using Hall sensors? \$\endgroup\$ – Bip Feb 27 '16 at 18:13
  • 1
    \$\begingroup\$ Sorry, I haven't a clue. I know how they work but never used them in this way. \$\endgroup\$ – Transistor Feb 27 '16 at 18:18
0
\$\begingroup\$

resolution with a 12bit adc and 30A full range:

resolution = 30 / 2^12 = 7.5mA if you can use full range of the ADC

with this kind of current hall effect based current sensors are logical solutions although if you want to get 10mA resolution noise free, you need a lot of filtering since hall effect sensors are quite noisy.

if you need bandwidth go with resistor sensing based methods,since your instumentation amplifier can handle high common mode voltages you can move the sense resistor to the high side and avoid gound shifting problems

\$\endgroup\$
  • \$\begingroup\$ 30A/4096 < 10mA. \$\endgroup\$ – Bip Feb 27 '16 at 12:27
  • 1
    \$\begingroup\$ oops my bad, fixed it \$\endgroup\$ – Ali80 Feb 27 '16 at 12:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.