Hi Qui, I know I have asked about this topic before, but after recently reviewing it I'm not sure what I implemented was correct...

Referencing the document you wrote titled voltage gain vs power gain for demodulators July 10 2012 i have a few queries...

On P.2 under point C: you calculate the power delivered to the 450R load based on an I+-I- differential voltage measurement, but you use 500R in the calc. I realise you've done this because you've included the 2x25R internal resistors, but is this correct? Does the voltage gain of the device include the 25R resistors? I'm confused because you say the 450R comes from the 50R spectrum analyser load translated across the 9:1 transformer, but the spectrum analyser is not measuring the power dissipated in the 25R legs, only the power dissipated in the translated 450R load.

I'm thinking its possibly better for me to work in voltage as this is obviously what my ADC is measuring, which in that case means working out the voltage gain at my frequency and for a given load. We can then convert back from output voltage to demod input voltage and work out the power there as we know the input is 50R. Does this work? What is the calculation? I know from the datasheet that at 900MHz the voltage gain is 3dB for a 200R load, what if my frequency range is 1600-2200MHz? with a TBD load? e.g. could be 300, 400, 500 etc. we could increase the load if we need a higher voltage swing...

Sorry for all the questions

Thanks

Tom

Hello Tom

I can’t seem to log into Engineering Zone, so if you don’t mind can you please paste my answer into the system and close the thread if you feel that your question has been answered?

Yes, most definitely work in voltage because it will be so much easier to get everything straight. I would highly recommend you use a differential probe and oscilloscope to make these measurements to convince yourself of how the theory directly applies to the measurement. All of this can get very confusing but once you convince yourself that the measurements match the theory you’ll feel a huge relief.

Here is the theory explanation. The internal 50 ohms of the ADL5382 forms a voltage divider with the external load. Let’s start off with the numbers already on the datasheet. Let’s choose a frequency that is almost mid-band to your desired operating range; 1900 MHz. On the datasheet, at 1900 MHz the voltage conversion gain is 3.9 dB and the load is 450 ohms. Using the voltage divider rule, the ratio of the impedance is 450/(50+450) = 450/500 = 0.9

Now if you change the impedance to 300 ohms, the ratio changes to 300/(50+300) = 300/350 = 0.857.

Let’s compare these two ratios. We will be using the numbers on the datasheet at the reference. 20* log (0.857/0.9) = -1.3 dB. This is saying that with a 300 ohm load the voltage gain will decrease to 3.9 dB – 1.3 dB = 2.6 dB

Let’s do the same for the 400 ohm and 500 ohm load

400/450=0.89

20 * Log (0.89/0.9) = -0.1 dB

Voltage gain for 400 ohm load = 3.9 – 0.1 dB = 3.8 dB

500/550 = 0.91

20 * log (0.91/0.9) = +0.087 dB

Voltage gain for 500 ohm load = 3.9 + 0.087dB = 3.987 dB

Now convince yourself that this theory is true by making the measurements in the lab. Use a diff probe and measure the input voltage and the output voltage of the demod for the different loads. Measure from the RF input to either I or Q. Make this measurement differentially across I+ and I- or Q+ and Q-. The results should be the same if you measure I or Q.

I hope I answered your question

Qui