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I'm working on a brushed dc motor driver and I have to monitor the current. so I use ACHS-7121 current sensor. This sensor convert +-10A into 0V to +5V and have +2.5V in zero current.

Now I want to shift this to 0v to 10v without including the current direction. It means that the sensor shows that 0V to +2.5V means the current is negative and +2.5V to +5V the current is positive. But I want the op-amp map the values that include negative current (0V to +2.5V) on positive values and ignore the current direction. \$ V_O = |V_I -2.5| \times 4 \$

I found difference amplifiers such as AD8276 but I'm not sure about that if it works for me or not if I use single positive supply. Can anyone suggest another solution or confirm that AD8276 works fine?

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  • \$\begingroup\$ The transfer function you require is \$ V_O = |V_I -2.5| \times 4 \$. Is that correct? If so then add it into your question using MathJAX: \$ V_O = |V_I -2.5| \times 4 \$. \$\endgroup\$ – Transistor Feb 12 '20 at 7:46
  • \$\begingroup\$ Are you saying that you want the following mapping from in->out, 0v->10v, 2.5v->0v, 5v->10v? This is what I think you mean by ignore direction and include negative values. If so, that involves rectification, taking the absolute value of the input after shifting by 2.5v, which needs a more complicated circuit than simply a difference amplifier. /edit/ which is words for exactly what Transistor expressed in an equation //edit/. Google for 'absolute value circuit', or here on stackexcahnge \$\endgroup\$ – Neil_UK Feb 12 '20 at 7:47
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It sounds like you need a precision full wave rectifier like this: -

enter image description here

Where the ground connection in the above circuit is at 2.5 volts. This converts input voltages above 2.5 volts and input voltages below 2.5 volts to a positive voltage above 2.5 volts DC.

Then you would add a gain stage that incorporates subtraction of the 2.5 volts DC to give you an output that went between 0 volts and 10 volts. A circuit very similar to this one will subtract an offset whilst applying gain.

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  • \$\begingroup\$ Actually I'm using "DC" voltage, not a signal. are you sure this diagram will work for me? \$\endgroup\$ – parisa-far Feb 12 '20 at 11:03
  • \$\begingroup\$ Your DC voltage varies over time because you are measuring it in order to determine what it is hence, it is a signal. As with any diagrams, you are urged to try and understand them or check them out using a simulator. Using a sim enables you to add the gain stage and get full end to end performance limits. \$\endgroup\$ – Andy aka Feb 12 '20 at 11:09

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