EMRFD Message Archive 15378

Message	Date	From	Subject
15378	2019-08-24 09:21:20	peter_dl8ov	Free (RF) Energy ??
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 lossy Where is the voltage gain coming from? Peter DL8OV
15380	2019-08-24 09:31:09	Thomas	Re: Free (RF) Energy ??
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 iPhone 24. aug. 2019 kl. 18:21 skrev peter_dl8ov@yahoo.de [emrfd] <emrfd@yahoogroups.com>:   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 lossy Where is the voltage gain coming from? Peter DL8OV
15381	2019-08-24 12:25:23	Gary Chatters	Re: Free (RF) Energy ??
See embedded comments.
15382	2019-08-24 20:54:33	Russell Shaw	Re: Free (RF) Energy ??
15383	2019-08-25 21:41:24	Andy	Re: Free (RF) Energy ??
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