TRF370417: Testing TRF370417 with DC IQ signals for phase noise measurement

Kang so to drive I=1, Q=0, I should drive IP = 1.2V, IN = GND, QP = 1.2V, QN = 1.2 or should QP = QN = Gnd

Wouldn't this be Q = -1, creating the point 10 in the diagram. To have point 11, wouldn't we want Q to be fully off? Qp=Qn?

I think there is a misunderstanding. My understanding is, QPSK or pi/4QPSK follows the phase circle as the diagram provides. To represent the '11' point on the circle, phase must be 0. In the IQ domain that means I should be 1 and Q = 0. So to have the Q be off I'm assuming the difference between P & N must be "0" since we need a common mode dc of 1.7V, I'm assuming we want both inputs to be 1.7 for Q allowing for only the I signal to be present.

When I = 1, Q = 0, we get a phase of 0

When I = 1, Q = -1, we get a phase of 315

Hi Daniel,

Ok, yes, I misunderstood. Please give your suggestion a try. You will need to have the Qp/Qn to be at 1.7V for 0 offset. 

Kang,

I tried this approach and it was partially working. My output looked like a very funky square wave, any idea why? When I put a 100 MHz LPF (LO is 100 MHz) it appears as a sine wave as expected. When I change the I & Q Values the phase shifts. My LO is within the limits of the device. I'm using the evaluation kit and for DC a bunch of low noise regulators.  It is snowing today so I cant get a scope diagram til tomorrow.

Hi,

You may be driving the input level too hard. You have to back-calculate the DC level (in RMS) based on the P1dB specification. You can then translate to input RMS equivalent based on the gain specifications per band.

-Kang

Kang,

Could you help me work through this?

Since we are at 100MHz, P1 is around 7.3dBm. This gives us a Vrms of 0.52V. Does that mean we should not exceed a 0.52V on the input or should Ip - In < 0.52Vrms?

we should find Vp to be 0.73V.

So for I =1, we should still maintain the 1.7V CM so should IP = 1.7 + Vp or 1.7 + Vrms/2, IN = 1.7-Vp or 1.7 - Vrms?

I can play it safe and try for I = 1, IP = 1.8V & IN = 1.6V giving a DC voltage of 200mV.

Hi,

I did not back calculate your math. The input DC level should be differential Ip-Im. 

Daniel Lukach said:

should Ip - In < 0.52Vrms?

At DC level, Vrms = Vdc, which is a differential voltage in your case. 

So I tried with Ip=1.8, In=1.6, still have an output that does not resemble a sine wave unless I filter. This gives me 0.2Vdc. I attached the scope diagram, the purple trace is the output, the yellow is the LO

with 100MHz BW

Hello,

There are perhaps RF harmonics (i.e. 2xLO and 3xLO) in the output spectrum and cause such distortion. There are datasheet parameters on these specs that you can double check. See if you can capture RF spectrums on spectrum analyzers and double check against datasheet specs. 

I think you're right, I will check on the spectrum analyzer. Do you know where the specs are in the datasheet, can't seem to find them

After measuring with the S.A. I see the highest peaks aside from the LO to be 3xLO & 5xLO, so you're right.