How does Hall based wattmeter work?
$begingroup$
I read about Hall effect and understand it's usage in sensors like speedometers but i also read it can be used as a wattmeter. how it can happen? in order to calculate power (p) we need to multiply voltage and current.
how hall sensor does that ?
sensor watts hall-effect power-meter power-measurement
asked 37 mins ago
Pooya EstakhriPooya Estakhri
1455
$endgroup$
add a comment |
$begingroup$
I read about Hall effect and understand it's usage in sensors like speedometers but i also read it can be used as a wattmeter. how it can happen? in order to calculate power (p) we need to multiply voltage and current.
how hall sensor does that ?
sensor watts hall-effect power-meter power-measurement
asked 37 mins ago
Pooya EstakhriPooya Estakhri
1455
$endgroup$
add a comment |
$begingroup$
I read about Hall effect and understand it's usage in sensors like speedometers but i also read it can be used as a wattmeter. how it can happen? in order to calculate power (p) we need to multiply voltage and current.
how hall sensor does that ?
sensor watts hall-effect power-meter power-measurement
asked 37 mins ago
Pooya EstakhriPooya Estakhri
1455
$endgroup$
I read about Hall effect and understand it's usage in sensors like speedometers but i also read it can be used as a wattmeter. how it can happen? in order to calculate power (p) we need to multiply voltage and current.
how hall sensor does that ?
sensor watts hall-effect power-meter power-measurement
sensor watts hall-effect power-meter power-measurement
asked 37 mins ago
Pooya EstakhriPooya Estakhri
1455
asked 37 mins ago
Pooya EstakhriPooya Estakhri
1455
asked 37 mins ago
Pooya EstakhriPooya Estakhri
1455
asked 37 mins ago
Pooya EstakhriPooya Estakhri
1455
asked 37 mins ago
Pooya EstakhriPooya Estakhri
1455
1455
add a comment |
add a comment |
1 Answer
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$begingroup$
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 $.
answered 25 mins ago
TransistorTransistor
82k778176
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add a comment |
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
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 $.
answered 25 mins ago
TransistorTransistor
82k778176
$endgroup$
add a comment |
$begingroup$
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 $.
answered 25 mins ago
TransistorTransistor
82k778176
$endgroup$
add a comment |
$begingroup$
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 $.
answered 25 mins ago
TransistorTransistor
82k778176
$endgroup$
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 $.
answered 25 mins ago
TransistorTransistor
82k778176
answered 25 mins ago
TransistorTransistor
82k778176
answered 25 mins ago
TransistorTransistor
82k778176
answered 25 mins ago
TransistorTransistor
82k778176
82k778176
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add a comment |
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