I am looking at an accelerometer(data sheet is BMA280 ) which has capabilities of measuring temperature from -40 to 85 degrees celsius. It also specifies a Temperature Sensor Slope of 0.5K/LSB, where it fills up an 8-bit register. It's centered on 23 degress celsius when the register reads 0.

Obviously this is a signed number. one question I have is how does the slope play into this. If it really is 0.5K/LSB then it means per single bit I have increments of -272.65 celsius which doesn't make sense to me. What do they mean with the 0.5K/LSB slope?

If I take 125 (the overall temperature range) and divide that by 256 (8 bits) I get 0.488 which is about 0.5/LSB. What I don't understand is the 'K'. why is it a K and not a C.

This is in all the documentation and they are consistent except for a few locations, does K mean Celsius in some countries? This also spans into other datasheets as well. So I am beginning to suspect I don't know something I should.

  • \$\begingroup\$ The expression ".5K/LSB" is a short hand way of saying 0.5 Kevins per LSB. \$\endgroup\$ – Dwayne Reid Feb 6 '15 at 3:37

Kelvins are exactly the same size as degrees Celsius. The Kelvin temperature scale is merely offset by 273.15.

A difference of 1K is exactly the same as a difference of 1°C.

So if you take the 8-bit number (which would have a range of -128 to +127 if it's 2's complement) as 0.5K per LSB you'd have a range from -41C to +86.5C.

Kelvins are often used in place of °C when differences (such as temperature rise) are involved, probably to keep separate the temperature and the difference-in-temperature units. In most electronics contexts (not all!) it's obvious you're not talking about temperatures near absolute zero.

  • \$\begingroup\$ Thanks, I completely missed the Kelvin Offset difference when compared to Celsius. \$\endgroup\$ – Zayzoon Feb 6 '15 at 13:43
  • \$\begingroup\$ No problem. A similar potential ambiguity arises when dealing with °F/°C. A temperature of 30°C is 86°F, however a difference in temperatures of 30°C is a difference of 54°F. \$\endgroup\$ – Spehro Pefhany Feb 6 '15 at 13:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.