Input Power calculation

[DFomi.1]

Hello. I use board st25ru3993-eval and application ST25RU3993 Reader Suite. When tags are readed we can see RSSI LOG Channel I & Q and Input Power. RSSI LOG Channel I & Q we can find in RSSI Display Register in AS3993, but i do not understand how calculate Input Power. I found the document "AN4971 Application note" and there are the formula and table in it.

If i calculate by this formula my values are not match those values from Reader Suite.

Please, explain me how to calculate Input Power. Thanks!

[Cedric Dalban]

Hello DFomi,

as described in AN4971, the input power is calculated with the formula:

  PowerInput = 2.1*(RSSI(I)+RSSI(Q))/2 - G

RSSI(I) & RSSI(G) are the values reported in the GUI.

G calculation is a bit more complicated.

As mentioned in the AN4971, this value is constant during the full inventory, it is calculated at the inventory beginning.

The formula used in GUI is not only taking into account the values from table 1 but also the graphs in figures 1 & 2 as well as the content of registers 0x0A and 0x09 (please refer to DS11840).

The final formula can be summarized as below:

  gain = (reg0A >> 6) * 3

  if (reg0A[bit5] == 0)    gain = -gain

  if (reg0A[bit1-0] == 01b)  gain = gain-8

  if (reg0A[bit1-0] == 10b)  gain = gain+10

  G = gain + 60

  if (reg09 == 0x3F) || (reg09 == 0xBF) || (reg09 == 0x90) G = G + 1   // BLF 160kHz

  if (reg09 == 0x24) || (reg09 == 0x27)           G = G + 2   // BLF 320kHz

  if (reg09 == 0x34) || (reg09 == 0x37)           G = G + 3   // BLF 256kHz and BLF 213kHz

  if (reg09 == 0x04) || (reg09 == 0x07)           G = G + 4   // BLF 640kHz

best regards,

Cedric.

[DFomi.1]

Hello, Cedric.

Thank you very much. I understood it, except 2 moments:

1) Why G = gain + 60 ? According the table in the AN4971 G = gain + 71 ( reg 0A = 0x00 ). Or do I think wrong?

2) if (reg09 == 0x24) || (reg09 == 0x27)           G = G + 2

Why? I think, that G = G + 2,5. This statement from AN4971

best regards, Denis.

P.S. If Q = 10 and I = 0 i can see that Input Power = -55. reg0A = 0xE0. If I use our formula I have that Input Power = 10*2.1 - 82.5 = -61.5 (if G = gain + 71 + 2,5).

I often have such a big error if gain = 6 or 9 but one of the I, Q < 3. Can you help with this problem? I hope I described clearly

[Cedric Dalban · Accepted solution]

Hello Denis,

After a double check with our RF team, it seems there is indeed a discrepancy with the AN4971,

thanks for the finding.

The G calculus formula in this document has been established on an old board,

and as mentioned in the AN page 6 "It is recommended to check the calculated power against

the actual input power and adjust the G constant accordingly for every PCB design.".

The formula "G = gain + 60" is the one related to EVAL and HPEV boards,

unfortunately the AN4971 has not been updated accordingly and it needs to be corrected.

The Input Power reported by the GUI is anyway correct if you are using an EVAL/HPEV board.

If you are using another board, as G is a constant all along the inventory session, the absolute value for Input Power may be indeed wrong

but the reported Input Power value provides anyway a consistent information on relative power strength.

best regards,

Cedric.

[Cedric Dalban]

Hello Denis,

Concerning the PS in your second message, let's do the calculus.

Reg0A = 0xE0, so this means you probably setup the receive gain to 9dB, could you confirm ?

You did not provide the Reg09, let's assume you used the default value (256KHz), so it should be 0x34, could you confirm ?

For info, all registers can be read by clicking on "Register Map" button (the one located aside "Reader Settings").

G:

reg0A= 0xE0   => gain= bit[6-7]*3 = 9

reg0A[bit5] = 1 => gain = +9

reg09 = 0x34  => G = 60 + gain + 3

=> G = 60 + 9 +3 = 72

Input Power:

Q = 10 and I = 0

=> IP = 2.1*(10+0)/2 - 72 = -61.5 dB

It is indeed not consistent with the reported Input Power (-55 dB),

I think the values you reported for reg0A and/or reg09 are erroneous.

Could you please check again the values of registers 0x09 & 0x0A

and redo the calculus with the formula provide above ?

thanks and regards,

Cedric.