# AD8302 Application using Eval board

Hello,

I am trying to use the AD8302-EVALZ board to measure the gain of a sensor. Right now, I have a VCO outputting around 930 MHz, and I'm dividing the magnitude of that signal using a voltage divider, AC coupling it, and then passing it across a sensor and a reference. I've removed the two shunt 52.3 ohm shunt resistors to ground from the board, and so the connections look like this: I just want to check that this setup even makes sense. From the datasheet, with the SW1 and SW2 switches in their default positions I expect the VMAG output to be 30mV/dB + 900 mV, so if I match the impedances as shown in the schematic I expect an output of 30mV/dB * 20 * log(2) + 900 mV = 1.08 V. This is not what I see when I try this, but my test setup is not ideal so I want to make sure that the circuit itself is okay.

I appreciate your input and recommendations.

## Top Replies

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• Hi there,

We see R5= 22k resistor in series with the input. Lets just estimate the stray capacitance to ground at the jcn. of R5, R6, and C1 to be 1pF. It might be more, but its probably not less. If so, the capacitive reactance at 930 MHz is -j171 Ohms or lower. This is a lot lower than R6=1k, so be sure to take the capacitive loading into account.

Moving on, it looks like Rs and Rr represent the sensor. So yes, you are using the AD8302 roughly correct. By lifting the on-board termination resistors, the voltage level at each IC input channel becomes more dependent on frequency, and RF cable lengths. It's probably best to put those resistors back on the board, especially if there are longer cable lengths connecting to the EVALZ PCB. For example, if the cables are Zo= 50 Ohms and 1/4 wave electrical length, the AD8302 3k Ohm input resistance transforms down to 0.83 Ohms, not a very favorable input impedance. And that's neglecting the input capacitance of 2pF, which will just make things worse! A far better alternative is to leave the 52 Ohm resistors on the board, which keeps input impedance low but matched to the cable transmission line impedance, and relatively constant with respect to frequency.

To help keep voltage level at the two inputs to the AD8302 more balanced, consider a fixed resistive divider for the INPA channel also. This way, the AD8302 VMAG output voltage is nominally 0.9V, but goes up or down depending on the sensor.

Reply
• Hi there,

We see R5= 22k resistor in series with the input. Lets just estimate the stray capacitance to ground at the jcn. of R5, R6, and C1 to be 1pF. It might be more, but its probably not less. If so, the capacitive reactance at 930 MHz is -j171 Ohms or lower. This is a lot lower than R6=1k, so be sure to take the capacitive loading into account.

Moving on, it looks like Rs and Rr represent the sensor. So yes, you are using the AD8302 roughly correct. By lifting the on-board termination resistors, the voltage level at each IC input channel becomes more dependent on frequency, and RF cable lengths. It's probably best to put those resistors back on the board, especially if there are longer cable lengths connecting to the EVALZ PCB. For example, if the cables are Zo= 50 Ohms and 1/4 wave electrical length, the AD8302 3k Ohm input resistance transforms down to 0.83 Ohms, not a very favorable input impedance. And that's neglecting the input capacitance of 2pF, which will just make things worse! A far better alternative is to leave the 52 Ohm resistors on the board, which keeps input impedance low but matched to the cable transmission line impedance, and relatively constant with respect to frequency.

To help keep voltage level at the two inputs to the AD8302 more balanced, consider a fixed resistive divider for the INPA channel also. This way, the AD8302 VMAG output voltage is nominally 0.9V, but goes up or down depending on the sensor.

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