What is the voltage to RX IQ sample conversion factor?

Greetings,

I’m trying to determine a conversion between IQ sample received and the actual voltage that the IQ sample represents.

From the datasheet, I see there is a parameter saying that the input voltage of the ADC is 0.8 Vpp.

My question is if this means that the ADC reads from -0.4 to 0.4 volts without saturating, i.e., -2048 corresponds to -0.4V and 2047 corresponds to 0.4 V.

However that would not make much sense since it is possible to change the RX gain.

So, how do I convert RX IQ sample to Voltage?

Thanks in advance

You are correct - if you are trying to relate the I/Q numeric values to an actual voltage at the antenna port, you have to take ALL the RX gains into account. So it’s not like a 2048 always means 0.8VRMS at the antenna port; you have to multiply by gain. Moreover, if you have an AGC in the loop that becomes almost impossible.

Thanks for the reply. That was clarifying. But the input at the ADC is always converted the same way, is it not? I mean, when you receive, for instance, 2048, it does mean that the voltage input at the ADC was indeed 0.8Vpp, however it tells me nothing about the voltage at the antenna port, like you mentioned. Am I correct to assume this?

Exactly. The voltage at the A/D is the same, but that’s just a point within the chip - you couldn’t even put a wire there. You need the settings for the gain stages.

1 Like