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AD911x driving an IQ modulator

Question asked by andpas on Mar 21, 2012
Latest reply on Mar 22, 2012 by andpas

Hello.

I need to use an AD911x to drive an IQ modulator that requires a VCM of 1.3V and a level around 1.0Vpp differential.

I plan to put a resistive network to change the VCM from the 1.2 max allowed by the DAC and the 1.3V required by the modulator,

then a low-pass filter as anti-aliasing.

The input impedance of the modulator is high, so basically the true differential impedance will be fixed by a resistor.

The DAC will be powered at 3.3V, allowing higher common-mode voltages.

I saw some posts about problems and particular requirements using the CMLx pins to have higher common-mode voltages,

so I'm asking if my solution is good for the purpose.

As fas as I understand from other posts, to obtain +/- 0.5V swing (1Vpp signal) I need to use 8mA DAC full-scale,

as with 20mA this swing is impossible to obtain... Is that right?

Supposing not to use internal resistors, I would use a passive network using three resistors for each differential signal, R1 from DAC output to ground, R2 from DAC output to filter input, R3 from filter input to +3.3V; the LPF will propagate DC, so VCM will be reported to the modulator input.

I did some calculation to find the values for this network, allowing right impedance on the DAC port for the DC and for the signal.

I found: R1=150ohm, R2=33ohm, R3=560ohm, that should give 1.15V VCM on the DAC, 1.27V VCM on the modulator, and an impedance at the

filter input (for the single rail) of 137ohm, so a filter could be designed for this impedance, and a shunt resistor of 270ohm could be used at the modulator input to let the filter see the right impedance at the output.

Could you confirm that these values are good for the AD911x?

Considering the use of the internal resistors, R1 could be obtained from the internal 62.5ohm resistor and an external 43ohm on the CML.

In the case I would have the ability to tune the DAC full-scale current of a small amount by software, but without changing the VCM,

I would need to use also the internal Rcm, but in this case the monimum value is too high, and also you wrote in precedent posts

that I would need to put the CML pin to 0.5V.

This part is a little difficult to understand to me... Are you telling that I need to put an external voltage reference to generate the 0.5V, or

I can simply put an external resistor (maybe bypassed with a capacitor) that will operate at the same time in parallel with the Rcm,

so the parallel value will give around 0.5V?

In the case I need the external regulator, should this work as source-sink or source only? How much current?

Please clarify this...

Thanks

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