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Steve.
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January 7, 2023 at 12:36 am #27262
SteveModeratorI built the SWAH v5 and it is not performing like it’s supposed to. I am in the process of verifying all the components and came across a small discrepancy of sorts. The build doc and schematic states that R17 is a 47k but in the “Populated Board Image and Resistor Chart for Troubleshooting” image, it is showing that R17 is a 22k resistor. I built mine using the 47k but just wanted to clarify that R17 is in fact supposed to be a 47k.
January 7, 2023 at 6:42 am #27264
PlaysforfunModeratorHey Steve, here is some info from an older build doc. Just substitute R17 where it says R18 as you read. In short, the 47K allows better use of the sensitivity pot rotation. You are good to go.
“In Regards to adjusting R18 as well as much more here is the explanation from Tonmann. short answer – increasing R18 will allow you to set the Sensitivity more towards the 12 o’clock position. I’d go with a 47kΩ resistor here. The only advantage is being able to turn the signal down a bit further (50% to 0% is better than 20% to 0%) at the expense of turning the signal up (20% to 100% is better than 50% to 100%). In regards to R18 a quick look at the “green” stage might be in order. Nothing more than a gain stage. The gain is determined by whatever the Sensitivity pot is set at divided by the value of R18. If the pot is set to 50% rotation, the resistance value of the pot (for a 100kΩ linear pot) is 50kΩ. Divide 50kΩ by 20kΩ – the value of R18 – and you have a gain of 2.5. If the input signal is 100mV, the output signal is 250mV. So how did I arrive at a value of 47kΩ in the short answer?. It would appear from the above posts that setting the Sensitivity pot at the 9 o’clock position gives good results. Since “9 o’clock” is about 20% rotation, the value of the Sensitivity pot will be about 20kΩ.The gain is therefore pot value divided by R18 which is 20kΩ / 20kΩ giving a gain of 1.If I set the pot to 50% rotation, the pot value is now 50kΩ and if I change R18 to 50kΩ (47kΩ nearest standard value), the gain is 50kΩ / 50kΩ which is also a gain of 1.And just to show you what a clever monkey I am, it is possible to leave the value of R18 as it is (20kΩ and change the pot to a 47kΩ pot and still achieve the same results.”
January 7, 2023 at 12:08 pm #27267
BarryKeymasterThat is an image error I have to correct. The BOM and schematic are correct.
January 7, 2023 at 12:50 pm #27268
SteveModeratorThank you Playsforfun (great explanation BTW) and Barry for clearing that up! I will continue troubleshooting.
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