Low-side current sensing, raising the ground?

Question

One of the main reasons why we don't use low-side current sensing for PSUs is that we want the ground to be, 'ground'. adding a shunt resistor will rise the ground slightly higher.

In real world wires and even traces of PCB have resistance, I saw it in several designs and datasheets that it's suggested to use a short trace of PCB as a shunt resistor.

Does it mean in real world there's no ideal ground reference? so we can safely use a 10m ohm resistor at the low side of the circuit without being worried about rising the ground reference?

Answer 1

Answer is both yes and no.

Yes you will be "raising" the local ground reference voltage, consider the following simple circuit:

• I2: 0 - 1A switching load
• R2: Current sense resistor
• R3: Parasitic resistance between your local ground (as seen by the load control circuit) and the actual power source ground

Simulation waveform:

As you can see, the local GND bounces from 0 to 1mV. Therefore it is no more consider an "ideal" ground. However, if any load driving and current sense is connected to this local ground, the fact that it is "bouncing" does not affect the performance as they are all seeing the same "reference" voltage.

The answer is also no because the Power Source ground (at your power connector, power IC, etc.) does not see the bounce as it is "isolated" through the parasitic resistance of the ground return path from the load switching circuit.

So depending where you probe the "ground" or "circuit voltage reference" you will or will not see it vary.

Concerning the use of a 10mOhm resistor, make sure the power dissipation of the resistor does not exceed the power rating of the component you pick (P = R*I^2 with R = 0.01). Ex: if load current is 1A then pick a sense resistor with a power rating greater than 10mW