BMP085 - ADC converter type ?

[syncout]

Hi,

I have read BMP085's datasheet and it says BMP085 has a 16-bit adc converter, and it takes 8MSB and convert it, and then it takes 8LSB and convert it too. I'm not 100% sure if it is correct, but if I am wrong, so is it a SAR adc? Because I don't know another adc that uses MSB and LSB to covert signals.

And, if it is as I said, why does it take first 8MSB and then 8LSB?

Thanks!

datasheet: https://www.sparkfun.com/datasheets/Components/General/BST-BMP085-DS000-05.pdf

 

[Harald Kapp]

I have read BMP085's datasheet and it says BMP085 has a 16-bit adc converter, and it takes 8MSB and convert it, and then it takes 8LSB and convert it too

Where did you read this? I can not find any reference to this in the datasheet.

 

[syncout]

Where did you read this? I can not find any reference to this in the datasheet.

Thanks for you quickly reply!

Well, I am not pretty sure about what I'm talking, but I found "this information"(probably wrong) on page 18.

 

[Harald Kapp]

What's stated there is that you read the 16 bit A/D conversion value by reading 2*8 bits.
This is no reference to the actual mode of conversion.

 

[syncout]

What's stated there is that you read the 16 bit A/D conversion value by reading 2*8 bits.
This is no reference to the actual mode of conversion.

Do you guess why does it happen? If it is a 16-bit A/D converter, why does it takes 2*8bits? It is confusing to me.

Thanks for your reply!!

 

[Harald Kapp]

Because I²C is a byte-oriented protocol.

 

[BobK]

And because they would need 8 more pins to output all 16 bits at once.

Bob

 

[syncout]

Because I²C is a byte-oriented protocol.

Thanks you!! I got it! In your link, it explains how the data is transmitted, thanks!

And because they would need 8 more pins to output all 16 bits at once.

Bob

Thanks for you reply Bob! But I didn't get these "more 8 pins", are you talking about I2C protocol? If so, I got it! Thanks!!

 

[Harald Kapp]

And because they would need 8 more pins to output all 16 bits at once.

That would be true for a parallel interface, not a serial one as I²C

 

[BobK]

Duh, I shouldn’t have posted without looking at the datasheet.