Electronic – negative current

In an AC system the mean voltage and the mean current are both zero but just because the mean of a quantity is zero doesn't mean that the quantity doesn't exist.

As an analogy consider two hydralic cylinders connected together by a pipe. I push one cylinder in and pull it out again. There was no net flow in the pipe but there was clearly flow when I was pushing the cylinder in and also flow in the opposite direction when I was pulling it out. If I place a load on one cylinder and move the other up and down i'm clearly using the flow of water to transfer power even though there is no net water flow.

Getting back to electricty consider a resistor connected to a sinewave AC power supply. For simplicity lets assume that our voltage source has a peak voltage of 1 unit and our resistor has a resistance of one unit. We can write equations for the instantaneous voltage, instantaneous current and instantaneous power.

V=sin(2πft)
I=V/R=sin(2πft)
P=IV=(sin(2πft))<sup>2</sup>

If we calculate the mean values of those functions we quickly see that V and I have a mean of 0 but P doesn't, it has a mean of 0.5.

So if our mean voltage and mean current are zero how do we meaningfully measure voltage and current? We use something called the "root mean square" value. That is we take the voltage or current waveform, square it, take the mean and then take the square root. Voltages and currents in AC power system are nearly always measured and discussed in RMS values.

We can use these RMS voltage and current values in the equations P=I²R and P=V²/R equations to calculate average power delivered to a resistive load.

It's tempting to also think we can apply RMS values to the P=IV equation but it is important to remember that this only applies if the voltage and current waveforms are proportional.

Further reading: https://en.wikipedia.org/wiki/Root_mean_square#Average_electrical_power

Both the Hall sensor and the op-amp are powered from 0 - 5 V supply. It is impossible for either of their outputs to go negative or > +5 V.

Figure 1. LM324 output stage (image from my answer in the post referenced in the OP's question) is typical of an amplifier output. The output can only be pulled high towards V+ by Q6 or to ground by Q13. There is nothing to pull it negative.

How does pull down 10k resistor effect the Op Amp input conditioning? Will it change the above calculations?

It won't have any affect.

How to add a protection from reverse hookup of sensed wire (voltage from sensor would fall below +2.5v)?

It's not needed the opamp will go towards +5 V or 0 V but will recover.

I intend to use two sensors for two ranges of sensitivity (100A and 300A), therefore one (100A) is intended to be for a long time in saturation. Will saturated op amp overheat?

The opamp won't but your sensor might if the current actually goes through it. Check the datasheet.

Figure 2. Principle of operation of a Hall effect current sensor. Note that the core gap would typically be very small - just enough to slide the chip in and is open wide here for illustrative purposes. Source: Tamura Current Sensor Products.

Figure 2 shows a Hall sensor installed in the gap of a magnetic core.

• When the magnetic flux is vertical through the sensor as shown a reference or control current in the y-direction will produce a voltage at right-angles in the x-direction.
• The voltage will be proportional to the product of the reference current, Ic, and the magnetic flux.
• Since the magnetic flux in the core is proportional to the ampere-turns on the prmary conductor being measured we can amplify the Hall voltage to give a signal we can work with.
• Some on-board electronics (not shown) carry out the amplification and add an offset so that the device can sense current in both directions while being powered from a single supply.

Tamura sensor provides only two current specs: Saturation current (looks like a primary current, in this case 100A or 300A) and also "consumption current" Icc which is probably "secondary" current producing the output Icc<=15mA.

• Saturation current is the current at which the core is magnetically saturated. Any further increase in current will produce little or no increase in flux (or sensor output). Note that the saturation current is 1.25 times the rated current to give some headroom before saturation starts to take effect. (It's not a sharp limit.)
• The 15 mA will be made up of two parts.
  • Some current to power the on-board electronics.
  • The control current, Ic. This will be the main current draw and will be a fixed value chosen as a trade-off between increased sensitivity and power consumption.
• Power consumption will be given as \$ P = VI = 5 \times 15m = 75~mW \$ as you correctly calculated.