# Calculating signal input from AD8333 output

Discussion created by mattwright481516 on May 26, 2016

Hi,

We are currently using the AD8333 IQ demodulator in a product we are developing, but I am having trouble making sense of the results I am getting from the AD8333 evaluation board. My current set up is pictured below:

The SMA cables on the right are going directly to a scope from I1 and Q1 and the probes shown are connected to TP5 and TP6, the differential signal input going directly to RFIP and RFIN on the AD8333. The other SMA inputs are for the LO and RF signal. The scope output is as shown:

The blue (3) and purple (4) trace are the probes at TP5 and TP6 and the light pink trace (M1) is the differential signal (trace 3 - trace 4). Since the AD8333 has differential input, this is the signal I would like to recreate. It is characterized by:

f = 5MHz

3.82 Vpp

The yellow(1) and green (2) traces are I1 and Q1 respectively. Because the clock (shown below) and RF signal are coming from the same signal generator, there are no oscillations on these traces. From this I get

Vi = 0.797 V

Vq = 2.639 V

From the datasheet, I am assuming the AD8333 provides a transconductance gain of 2.17 mS (mA/V) and the transimpedance amplifiers converting the AD8333's output current to voltage have gains of 787 V/A based on Rfeedback.

I am trying to work back from these values to recreate the 3.82Vpp input signal (1.91V amplitude). This seems like it should be straight forward but I do not understand the results I am getting:

Vin,i = Vi / (787 V/A * 0.00217 A/V) = 0.797V / (787 V/A * 0.00217 A/V) =0.467 V

Vin,q = Vq / (787 V/A * 0.00217 A/V) = 2.639V / (787 V/A * 0.00217 A/V) =1.545 V

Taking the square root of the sum of the squares I get:

Vin = sqrt( Vin,i ^ 2  + Vin,q ^ 2 ) = 1.614 V

which I cannot relate back to any input trace, let alone the differential input.

If I first take the square root of the sum of the squares of Vi and Vq and then try using the gains to get Vin, I get the same result of 1.614 V.

For completeness, the LO clock input is also pictured below:

The blue trace (3) and purple trace (4) are on either side of R1 and show the clock signal going directly into the AD8333. As shown they have an offset of ~1.2V and and amplitudes of ~460mVpp. The light pink signal (M1) is the differential signal (trace 3 - trace 4). The frequency is 20MHz. I am confident his clock is working effectively. If I alter the input RF signal so that LO != 4*RF, then I see sinusoidal oscillations on I1 and Q1, as expected. All the results shown in this example are taken when LO = 4*RF and no oscillations are seen on I1 and Q1.

I would very much appreciate any help I could get on this issue, I have been experimenting with the evaluation board for some time and cannot figure out where I am going wrong.

Thanks

- Matthew