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RF Spectrum Analysis with Analog Discovery 2


This might be dumb question, but I'd rather ask it than risk blowing up my Analog Discovery 2.

I just built up a QrpLabs QCX CW 40 Transceiver that produces approximately 5 watts of RF energy in the 40 meter band (7 megahertz-ish) and I'd love to learn more about the signal output with the Spectrum Analyzer feature of the Analog Discovery 2. I do have the device connected to a 20 watt dummy load so I won't be transmitting over the air while keying down the transceiver, but how should I properly connect it to the AD2? I don't assume the device would be terribly happy with a coax connection with 5 watts from the radio directly into the the AD2. Do I need to build / buy an attenuator? I would assume so. Can anyone point out what range I should shoot for input signals and how I might calculate the correct values to build such a circuit? Alternatively, if you have a better reference, i.e. book or video, where I could learn more about this I would appreciate those as well.

Steven Behnke

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that is V = sqrt(P*R). For 5 Watts and (assuming!) 50 ohms the voltage at your dummy load will be 16 V RMS or, multiply with sqrt(2), +/- 22.3 V peak.

Consider using a 1 : 10 probe, check against the 22.3 V with some margin. This calculation assumes a largely sine-shaped waveform  which seems a safe bet if you're FCC compliant with regard to harmonics.

If you want to use a 50 ohms input you may consider a series resistor. For example, 1 kOhm into 50 ohms is (approximately) 1:20 voltage division. A 1/4 W rating would be at the limit (16^2 / 1000 is around 1/4). Probing your load with this will cause a minimal variation in load impedance (mismatch) - this is where a power splitter would come in for "serious" RF engineering - but  most likely this can be swept under the rug. Or use 10k.

it's funny how time flies - nowadays, buy a premium mobile phone and chances are good there's a modulated DC-DC converter inside with a bandwidth of many times said "RF" frequency...


Edited by xc6lx45
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