# Circuit to convert a voltage output of a Hall device to a current

Posted on Feb 6, 2014

The approach you have shown would dissipate (12V)(3A) = 36 watts. This is quite a lot of dissipation so the complete circuit will get HOT. There are two possible alternatives: 1) Run the op amp from a lower supply voltage, if available. 2) Use a class-D type amplifier. The Hall Device voltage represents a current in a separate conductor of 5 Amps.

That 5 amp current produces a magnetic field in a gapped toroid core of about 30 G and the Hall device sensitivity is 5 mV/G. The requirement is to measure the current over a narrow range (4 to 6 A) very accurately (<0. 1%). The Hall device is an Allegro A1321 and the output is push-pull. I don`t know if it is possible to avoid the ground reference for the output resistor. The current measuring instrument is a Yokogawa WT 1600 power meter which has a current transformer input stage with 25 milli-Ohms of resistance. The 50 milli-Ohm includes some added resistance to achieve the correct output current. It sounds as if the 0. 05 ohm load is a winding of a current transformer, correct It would seem that this load could then be floating and would not need to be ground referenced. This could simply the circuit used to drive the 3A current. Your power calculation is the power dissipated by the 0. 05 ohm load which is small by comparison. The power that must be dissipated by the amplifier drive circuitry is 36 watts at 12V operation. (Actually about 35. 5 watts, accounting for the power delivered to the 0. 05 ohm resistor. ) This will require a large finned heat sink. I just want you to consider whether this is practical before we design a circuit. Another comment. the spec on the Allegro Hall device shows an operating voltage of 5V with an absolute max of 8V. It should not be powered from 12V as you show it. Furthermore it is...

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