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Can you do a 3db Q measurement on the DUT. Set resonance a 1 unit then raise and lower the frequency until the meter reads 0.707 and record the frequencies. High frequency minus low frequency = xxx

Thanks to Everett for HP-400E Q measurement. Added to: https://www.kn5l.net/Q-Meter/ John KN5L

John, what are the low and high frequency rolloff points of your 50 to 1 #43 material transformer? Is some of that missing 50 points of Q, caused by the loss in the transformer primary? What is the Q

Hey Jeff, There is a more complete service manual on the Bama site. I had to download a djvu reader to open it, but once I did that, it's pretty nice! https://bama.edebris.com/manuals/hp/4342a/ This

I think you need to consider what happens at DUT resonance and either side of it.? Assume an ideal 50:1 turns transformer.? At DUT resonance, the impedance presented to the 1 turn will be

Hi Mikek, An example of a high Q device measurement is shown in: https://www.kn5l.net/Q-Meter/ look for "DUT = R40C1 36T 330/46 with 330pF C0G" Unfortunately my RF voltmeter prototype reads high. A

Thanks Steve for setting that up and testing. This discussion for me, is to try to understand how HP built their transformer, I think knowing what impedance drives it is part of that understanding.

If one performs a more detailed analysis, a good approximation for the output resistance should be: (rb2/beta2) + re2 + (1/gm2) + ESR I don't know the parasitic resistances of the transistor,

Despite my attempted analysis of the Impedance converter, doesn't it's output impedance have to be about 2.5Ω? Isn't the 0.001Ω output impedance because it has transformed the Impedance Converter by

And the polarity of the input cap, too, of course... -- Prof. Thomas H. Lee Allen Ctr., Rm. 205 350 Jane Stanford Way Stanford University Stanford, CA 94305-4070 http://www-smirc.stanford.edu

I believe that the 1854-0332 is a 2N3866A; your 2N5109 should indeed be a very reasonable choice here. That still leaves us with an order of magnitude discrepancy in output resistance. By any chance,

Q1 (input) is HP 1854-0091--no "2Nxxxx" equivalent transistor listed in HP equivalent list. Q2 (output) is HP 1854-0332, again no equivalent transistor listed. Both are TO-5/TO-39 case. I think a

Thanks for checking that, Steve.? Unless the 2N5109 has a surprisingly high parasitic emitter resistance, there's still about a 10x gap to be closed. --Tom -- Prof. Thomas H. Lee Allen Ctr., Rm. 205

Your analysis is spot on (well done for an FPGA designer!); it's the same as that in my first post, although I used 100 for beta of both transistors. In any case, the intrinsic output impedance is

And the output capacitor I used was 100 uF/25 volts, with measured 100 kHz ESR of 0.26 ohms. Steve

Googling Output resistance of an emitter follower, I find it's input resistance divided by (Beta + 1) plus some other factors, internal emitter resistance of Q2, etc, that I as a FPGA designer

As noted, first test was at -40 dBm. I raised the level just now to max, -17 dBm--very little change from before--the output impedance decreased about 0.03 ohms for each frequency. Steve

I was using -40 dBm. Steve

Those numbers have a believable trend, but the magnitudes seem too high. I don't know what the parasitic emitter resistance is of a 2N5109, but I would be surprised if it were bigger than an ohm. The

I breadboarded the circuit just now using 2N5109's and swept the output Z with the DG8SAQ VNWA, 10 kHz-30 MHz. The circuit draws 117 mA; the output transistor draws 84 mA. The output impedance is