DAC FULL SCALE POWER A

Hello,

Referring to the equation above along with default setting of 0x28 (Decimal 40), the default IOUTFS setting (assuming no gain error) is 15.625 mA + (40/256)*25 mA or 19.53 mA as described in text above equation. 

To achieve a different setting up to 25.977 mA,    one would re-arrange the equation above so as to solve for the decimal value of FSC_CTRL as shown below.

FSC_CTRL =  (IOUTFS-15.625) * (256/25).

For instance, to achieve a max value of 25.977 mA the decimal value of FSC_CTRL is 106 with the hex value being 0x6A loaded into Reg 0x05A.

what should be consider here I_OUTFS ???  

IOUTFS is any value you choose to be between 15.625 and 26 mA.

Please do clear my doubt . from data sheet in dc specifications gives the analog output  FULL SCALE output current is min(23.6 mA)  and max(28.8mA)  its value why cant use here ??

please clear my doubt . 

And one more doubt is given below formula . what is here take Value of

" N" .

What is here DACCODE & what is its value take ??????  please i need your support .

Hello,

IOUTFS determines the maximum full-scale current that can be set between nominal 15.625 and 26 mA.  It represents the summed current outputs of IOUTN and IOUTP when the two outputs are combined.  Obviously, one would not tie the outputs together since one is trying to generate a signal by changing the DACCODE value so that the IOUTP and IOUTN current output varies depending on the code setting as described in your screenshot equations.  Note that best AC performance is achieved when configuring the DAC output for differential operation via a transformer/balun

Please i have one more question  sir/mam .

• I have take DAC 16 bit , And IOUTFS = 16mA 

Then i got the P_LOAD(dBm) = 4.539 .

•  If i take the 16 bit DAC , And I_OUTFS= 26mA 

Then i got the PLOAD (dBm) = 8.588

please  verify my answer and reply soon .

Hello,

The "peak" output power delivered to the load (typically 50 ohms) when using balun depends on the IOUTFS setting of DAC as well as the matching characteristics between the DAC output (differential 100 ohms) and the load.  Note that "peak" refers to when DAC code is at +FS or -FS (i.e. code of  2^16 -1  or 0) while the RMS value is delivered to load is dependent on the peak-to-rms (PAR) ratio of the waveform being constructed (i.e. square wave has PAR=0 dB while sine wave has PAR=3 dB).   

Looking at circuit below where a 2:1 impedance ratio balun is used to perfectly match the 50 ohm load reflected back to the 100 ohm DAC differential source impedance hence the DAC output AC current source will effectively see a combined 50 ohm load resistance (i.e. 50 ohms = 100 ohms // 2x50 ohms). If the IOUTFS is set to 16 mA than the peak IOUTFS differential current will be 8 mA when DAC code is at -FS or +FS setting.   With the combined load of 50 ohms, the DAC will deliver a total of 3.2 mW  (5 dBm) of power of which 1/2 is dissipated in the 100 ohm DAC termination and the other 1/2 in the 50 ohm load or 2 dBm since the load is matched to the source.  Note that for a sine wave having a PAR of 3 dB, this means that one would suggest that one would read -1 dBm power (i.e. 2 dBm Peak Power - 3 dB PAR) on a spectrum analyzer or power meter (assuming ideal lossless 2:1 balun).  

If the current is increased to 26 mA (i.e. factor of 1.625 relative to 16 mA), the peak power will increase by factor of 2.64  (since P=I^2 *R) or 4.2 dB such that the "peak" power is now 6 dBm............again with a sine wave rms power being 3 dB less or  3 dBm.