Current to voltage conversion with level shifting for the current

Thread Starter

Ajay1989

Joined Feb 1, 2016
6
Hi Guys and Gals,

I am working with a sensor which outputs current in the range of +/- 72 mA.
I want to amplify (1.1976mA to 1.2024 mA) of the sensor output to (0 to 5V).

I am comfortable with current to voltage amplification but I am not sure how to do level shifting for the current i.e. 1.1976 mA corresponds to 0V and 1.2024 mA corresponds to 5V.

What I was thinking was, to connect the output from the sensor to a 1.1976 mA current sink (so that my 1.1976 mA to 1.2024 mA becomes 0 to 4.8 uA), and then apply 10^6 amplification using op-amp transimpedence amplifier. But I can not find any more information if this can be done.
Any seniors could shed some light if this can work? Or is there any other simpler way to achieve this?

Thank you in advance and please let me know if you need any more clarification from my side

Regards
Ajay
 

PeterCoxSmith

Joined Feb 23, 2015
148
Take a look at the instrumentation amplifier AD623. Apply a reference of 2.5V and the output will swing from 0 to 5v, centred on 2.5V, with an input scaled to +/-0.5V and a gain of 5.
 

Thread Starter

Ajay1989

Joined Feb 1, 2016
6
Thank you Peter for your reply.
AD623 will very well work if the output from the sensor is voltage (please correct me if I am wrong)

But my sensor generates +/- 72 mA and I have to extract 1.976 to 1.204 mA and amplify to 0 to 5V.

Thanks
Ajay
 

PeterCoxSmith

Joined Feb 23, 2015
148
Thank you Peter for your reply.
AD623 will very well work if the output from the sensor is voltage (please correct me if I am wrong)

But my sensor generates +/- 72 mA and I have to extract 1.976 to 1.204 mA and amplify to 0 to 5V.

Thanks
Ajay
use a resistor to convert current to voltage, calculate the resistance using ohm's law to scale +/-72mA to +/-0.5V...
R=0.5/0.072...6.94R...nearest value is 6.98R 0.1% tolerance.
 

Thread Starter

Ajay1989

Joined Feb 1, 2016
6
Dear Bordodynov,

Thank you very much for your diagram. I assume I1 you drew is current sink of 1.2 mA. And I understood the 1 stage amplification circuit.
But I still have couple of doubts.

1. What is the use of R2, R3, U4 and BAT54S in this circuit
2. How do we realize 1.2mA current sink in actual circuit i.e. how to realize I1. I saw one sample design and is attached here. Will this work?
upload_2016-2-2_9-27-9.png

Thank you

Ajay
 

dannyf

Joined Sep 13, 2015
2,197
I would add it. Analog solutions are dicy, as the resolution is high.

An act with 20x gain plus Pam can do this easily, in a 8dip package.
 

Thread Starter

Ajay1989

Joined Feb 1, 2016
6
I would add it. Analog solutions are dicy, as the resolution is high.

An act with 20x gain plus Pam can do this easily, in a 8dip package.
Thank you Danny for the reply. But I didn't understand your explanation here. Would you be able to elaborate a bit more?

Thank you
Ajay
 

Bordodynov

Joined May 20, 2015
2,628
On the twin Schottky diode limiter made. This protects the input of the amplifier. When operating in the linear range of the amplifier at the inverting input voltage is substantially zero. Therefore there is no need for precision current generator! Current generator can be replaced by a resistor. Used reference diode napyazheniem ~ 1.22V. For example LM4041-1.2, ZXRE125. With stable voltage across the resistor goes stable current. For the flowing currents of the current scheme is necessary to turn the resistor. It is also necessary to add an inverting amplifier. What is the current direction you?
Current resistor I have divided into two parts. This is the usual resistor and multi-turn potentiometers.
If you wish to limit the output voltage range (0V-5V), you'll need to add a precision limiter with operational amplifiers.
 
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