Hi

AD9161/9162 data sheet on Page 80 mentions "If the AD9161/AD9162 are programmed for an IOUTFS = 40 mA, its ideal peak ac current is 20 mA and its maximum power, delivered to the equivalent load, is 10 × (RINT/(RINT + RLOAD) = 8 mW (that is, P = I2R)." We don't understand why the '10' is suddenly appeared and why this equation is true.

Is this the discussion about a 1:1 balun with 50 Ω differential source termination e.g. Fig 191 in AD9161/AD9162 data sheet Page 81 as follows?

Could you help me about this issue? Our customer waits for the reply for one month.

Regards,

Hiroyuki

There are two flaws in this section of datasheet which we are planning to correct with the new version of Datasheet:

If we consider the following simple circuit, the 200 Ω impedance shows the DAC output differential internal pull-up resistors.

The source load (left side of the balun, including 2xR

_{INT}) seen from the DAC ac current source is equal to [( 100Ω+100Ω) || (50Ω+50Ω) || 50Ω]=29Ωnot 50Ω.However, let’s assume the total source load seen from the ac current source is 50Ω (which we already know that it is 29Ω not 50Ω). Then:

where is the peak current delivered to the load, is the peak voltage measured across the total load and is the total RMS power delivered to the total load. However, power share of actual load (excluding R

_{INT}) is a dependent on the portion of the current that flows into the load (including the source termination).Because the source and load resistance seen by the 1:1 balun are equal, this power is shared equally. Therefore, the output load receives 4 mW, or 6 dBm maximum power.

Again, we are considering a new revision to this datasheet. It that, simpler calculations and corrected values will be included.

Sorry for making this confusion!