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CS5490 full-scale range

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abaldur

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Hi everyone

I want to use the CS5490 energy measurement IC to measure current using the shunt resistor approach.

Can anyone help me understand the following:

In the datasheet, they write: "The voltage reference between the VREF+/- pins is 2.4V." Hence I was thinking that the full-scale range would be 2.4V.

But at the same time they write: "The full-scale signal level for the current channel is ±50mV and ±250mV for 50x and 10x gain settings, respectively."?

What I don't understand is why they write ±50mV and ±250mV and not ±40mV and ±240mV?

BR abaldur
 

Hi,

Often the decodable input voltage range is 0V ... VRef. But this is no "must".
Every IC designer is free to design their own internal circuit. .. with their own specifications.

VRef is one thing
Input voltage range is another thing.

Follow the datasheet and you are safe.

Klaus
 

Hi Klaus

Thx for the reply!

The register "Instantaneous Current (I), Page 16, Address 2" provides a normalized reading in the range -1.0 < value < 1.0.

Assuming I use the 10x gain setting, is it then correct understood that +/- 250 mV corresponds to register values of +/- 1.0?

/ abaldur
 

Hi,

It seems you use a old/different datasheet.

Please always use the latest datasheet directly from the manufacturer.
This currently is: CS5490, --> DS982F3 from MAR´13
https://www.cirrus.com/products/cs5490/

Also read the additionally provided documents.

Klaus
 

Hi Klaus

I am using the latest datasheet.

The "Page 16" is not the datasheet page, but memory page.

/ abaldur
 

Hi,

OK, now I found it.

Yes. You are correct.
In other words: it is a signed 24 bit value. Where normalized_value = binary_value / 2^23

Klaus
 

As far as I see, there's no specification of typical uncalibrated CS5490 scaling. However, the internal reference is too inaccurate to use the chip without calibration. In so far the question of uncalibrated scaling factor is interesting but doesn't need an exact answer, I think.
 

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