### Q

I'm asking about the AD8349 Modulator:1.Can you tell me what the impedance input of the baseband Input is? And what

it means that the Baseband must be driven from a low impedance source?

2. How filter between the DAC 9767 and AD8349 influence on the impedance

matching? And why you recommended connecting a resistor between I and Q

differential inputs?

3 . How this resistor influence on the input impedance?

### A

1. The input impedance of the baseband inputs is so high as to be negligible ifthey are driven from a low impedance source like the differential amplifiers

shown in figure 44 or even the resistor networks shown in figure 51.

2. There is no impedance matching as such being done between the DAC outputs

and the AD8349 inputs. It is assumed that the distance from the DAC to the

AD8349 is short and that the lines are not required to have characterized

impedance and it is also assumed that the input impedance of the AD8349

baseband input is much higher than the resistors R1, R2 and R3.

Looking at figure 51, R1 and R2 convert the differential current output of the

DAC into two voltages. R1 = R2 and the size of R1/R2 determines the common mode

level that the signal will be sitting at. If a filter is present between R1/R2

and R3 then the filter design must take account of the non-zero source

impedance represented by R1/R2 and the finite load impedance represented by R3.

None of this has anything to do with the impedance of the Baseband input, which

is assumed to be relatively high, and therefore negligible.

3. The resistor, R3, between the true and complementary baseband inputs

interacts with the R1 and R2 to attenuate the amplitude of the differential

signal appearing at the AD8349 input terminal. When the DAC output is zero

(midscale of a sinewave) then the DAC outputs are both outputting the same

current, typically 10mA. Therefore there is 400mV across R1 and 400mV across

R2. Therefore the differential voltage is zero so no current flows through R3.

As the DAC output moves away from midscale, one voltage will increase and the

other will decrease symmetrically. The differential load resistance R3 will

shunt away some of the current being dropped through R1 and R2, leading to a

reduction in the differential amplitude of the signal.