EMRFD Message Archive 15378

15378 2019-08-24 09:21:20 peter_dl8ov Free (RF) Energy ?? Message Date From Subject No, I don't believe in this free energy stuff either, which is why I need your help solving a puzzle.I have a 10 MHz GPSDO that is connected to a distribution amplifier built from op amps, output impedance is about 5 ohms per channel and I see a 5.25V P-P signal when I plug it directly into the scope, the scope is assumed to be 1M ohm and 20pF but I haven't measured it.This weekend I built a 10 MHz bandpass filter to clean up the signal. A parallel tuned circuit, a series tuned circuit and a second parallel circuit on the output. The capacitors are fixed but the inductors are variable. When I connect the filter between the distribution amplifier and the scope and tune for maximum smoke I see 6.5V P-P on the scope.Best guess is that the filter is providing a nominal match between the op amp output and the 1M/20pF of the scope input, 20pF is about 796 ohms at 10 MHz. However, the series tuned circuit in the middle uses a 3pF capacitor which is about 5300 ohms at the working frequency so I expected the filter to be lossyWhere is the voltage gain coming from?Peter DL8OV Impedance transformation gives voltage gain. In this case there is a stackup of impedance mismatches, one from the GPSDO to the filter, then since the filter is unterminated, to the scope input impedance. There is a reason we transform to 50ohm before measuring, using terminated instruments. EMRFD ch 7 goes into great details and are worth a read. 73 de Thomas LA3PNA. Sendt fra min iPhone24. aug. 2019 kl. 18:21 skrev peter_dl8ov@yahoo.de [emrfd] : No, I don't believe in this free energy stuff either, which is why I need your help solving a puzzle.I have a 10 MHz GPSDO that is connected to a distribution amplifier built from op amps, output impedance is about 5 ohms per channel and I see a 5.25V P-P signal when I plug it directly into the scope, the scope is assumed to be 1M ohm and 20pF but I haven't measured it.This weekend I built a 10 MHz bandpass filter to clean up the signal. A parallel tuned circuit, a series tuned circuit and a second parallel circuit on the output. The capacitors are fixed but the inductors are variable. When I connect the filter between the distribution amplifier and the scope and tune for maximum smoke I see 6.5V P-P on the scope.Best guess is that the filter is providing a nominal match between the op amp output and the 1M/20pF of the scope input, 20pF is about 796 ohms at 10 MHz. However, the series tuned circuit in the middle uses a 3pF capacitor which is about 5300 ohms at the working frequency so I expected the filter to be lossyWhere is the voltage gain coming from?Peter DL8OV See embedded comments. re: "So 5.25Vpp * 4/pi = 6.68Vpp, somewhere close to the measurement."Correct!  A filtered square wave reduced to its fundamental always results in a sine wave with a larger voltage peak amplitude.Andy