# How does a Hall-based wattmeter work?

I read about the Hall effect and understand its usage in sensors like speedometers, but I also read it can be used as a wattmeter. How can it happen?

In order to calculate power (P) we need to multiply voltage and current.

How does a Hall sensor do that?

You are correct that for power measurement both voltage and current must be measured.

Figure 1. Hall effect measurement setup for electrons. Initially, the electrons follow the curved arrow, due to the magnetic force. At some distance from the current-introducing contacts, electrons pile up on the left side and deplete from the right side, which creates an electric field ξy in the direction of the assigned VH. VH is negative for some semiconductors where "holes" appear to flow. In steady-state, ξy will be strong enough to exactly cancel out the magnetic force, thus the electrons follow the straight arrow (dashed). Source: Wikipedia Hall effect.

The Hall sensor provides a means of non-contact current measurement.

Figure 2. A typical current sensing arrangement of a Hall sensor in a torroidal core around the conductor. Source: Digikey.

The energy meter will require a voltage measurement as well.

For DC the instantaneous power can be calculated as the product of the voltage and current. For AC it is necessary to continuously sample the voltage and current, get the product and then calculate the average of the power over time. Mathematically, $$\ \int_0^T VIdt \$$.

• Hmmm, i still don't understand how the circuit is going to be for using hall effect as a wattmeter ? where to do we measure to get amount of power? Jan 20, 2019 at 13:33
• You measure the voltage, measure the current and multiply them either in analog electronics or with a digital circuit. The Open Energy Monitor project may be of interest to your. Jan 20, 2019 at 16:23
• A Hall Effect sensor can be used IN a wattmeter, but not AS a wattmeter. The Hall sensor would be used to measure current, and another sensor used to measure voltage, then a microcontroller would multiply the current and voltage measurements to get power. Jan 20, 2019 at 16:44