A simple question that I am not able to find any solution online or in the literature I have. What does it mean when it something is written as dbFS / LSB, for example The lock level is stored in −dBFS in a resolution of 1 dB/LSB or resolution is 0.5 dB/LSB

Any explanation would be helpful.



1 Answer 1


This sounds like something from a datasheet. It means, that a dBFS-Value is stored (in some register). So, not a linear value, but a dB-Value.

The resolution in dB/LSB can be thought of as a multiplicator. With 1dB/LSB (Least Significant Bit), the dBFS-Value can be directly obtainend by reading the register. So, an 8Bit-Register holding the value 0x20 with resolution 1dB/LSB yields the dB-Value -32dBFS. With 0.5dB/LSB, it would be -16dBFS.


The HEX value 0x20 is 32 in decimal. The resolution is given in dB/LSB This means, adding one LSB (0x01) to the register represents a change of just this much in dB. Put differently, one needs to multiply the register value by the resolution value to obtain the desired value.

$32\cdot1=32$, so the register represents a value of -32dBFS (as the unit of the register is stated as -dBFS). The resolution provides a mapping, to go from the digital register value to an actual value, in this case in dBFS. You may be thinking about this to complicated. Here is an analogy: Some figure of money is to be stored in an 8Bit register. You could just map directly (1USD/LSB), then 0xFF in the register would mean 255USD. Or you could say use 5USD/LSB, then 0xFF would mean 1275USD.

  • $\begingroup$ would you explain as to how 0x20 gave the value of -32 when we 1db/LSB? $\endgroup$
    – beenvan
    Commented Jan 7, 2022 at 7:51
  • $\begingroup$ I want the second mapping in your last sentence-- unless it is a bill that owe! $\endgroup$ Commented Jan 7, 2022 at 13:50
  • $\begingroup$ I can agree to that. 😁 $\endgroup$
    – Max
    Commented Jan 7, 2022 at 14:08

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