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  • #46
    Did I miss something? 1V in with 1K ohm source impedance and 850mV out? Doesn't that imply a input resistance of 0.85/(1.0-0.85) * 1k= 5.66K ohms?
    Experience is something you get, just after you really needed it.

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    • #47
      Originally posted by nickb View Post
      Did I miss something? 1V in with 1K ohm source impedance and 850mV out? Doesn't that imply a input resistance of 0.85/(1.0-0.85) * 1k= 5.66K ohms?
      Ah Ha! I was just testing you to see if you were still payin' attention.... you passed. No, you're totally right. I just did the math and completely dropped out a decimal point or something.
      That's what I get for using a figure like "1.002V" and "961.82Ω". I'm sure we can all go through the rest of our day just fine without me having to provide an accuracy down to two hundredths of an RMS volt!
      But since I've already done it, technically, I got an input impedance of 5.27kΩ. But who's counting

      Thanks for catching that quickly. It's a good thing you guys are here to pick up the pieces.
      If I have a 50% chance of guessing the right answer, I guess wrong 80% of the time.

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      • #48
        I am trying to figure out why especially the cathode impedance of the left tube seems to be much higher than with a normal CF. I think one of the reasons is that the left 470k grid resistor couples most of the cathode signal to the left (input) grid.
        So I would like to ask you to repeat the cathode input impedance measurement with the left grid AC grounded via a 0.1µ cap to eliminate (positive!) cathode to grid feedback.
        If - what I am expecting - cathode impedance drops, this will show that cathode input impedance depends on the source impedance at the grid.

        BTW, what is a 6C67? I thought you would use the 6111?
        Last edited by Helmholtz; 08-26-2019, 05:01 PM.
        - Own Opinions Only -

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        • #49
          @HH

          Some observations using the 6111 model above, I see the input Z as
          2.5K for the circuit shown
          If you AC ground the grid of the second stage the Zin goes down to 600 ohms. Not a surprise since it's now got a much bigger signal going into it.
          If you remove the second stage altogether, the Zin is 2.6K
          If you also remove the AC ground (i.e the 20uf to ground) from the plate of the first stage the Zin goes up to 6.1k
          If you now replace the second stage the Zin is 4.9K i.e roughly double where we started (the two tubes are in effect in parallel).

          @SF if the objective is to get a high input Z then drive the grid instead of the cathode. Then you'll get more like 5meg. If you drive a common cathode stage from the plate of the first the gain go from a very sad gain of 1dB to a happy 24dB. It seems to be that the circuit as it stands doesn't do anything very useful at all and you'd get only slightly less gain from a bit of wire.
          Experience is something you get, just after you really needed it.

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          • #50
            If you remove the second stage altogether, the Zin is 2.6K
            Thanks, but what is the cathode input impedance of the first stage alone with its grid AC grounded? I would expect something like 250 Ohm IIRC.
            - Own Opinions Only -

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            • #51
              Originally posted by Helmholtz View Post
              Thanks, but what is the cathode input impedance of the first stage alone with its grid AC grounded? I would expect something like 250 Ohm IIRC.
              About 820 ohms. Remember the model might be poor....

              PS: Ignore that. I should have AC bypassed the 10k resistor so the DC conditions were changed to be closer the the answer you needed. Zin by this method is 116 ohms. The plate current is 4.8mA, the gm at 4.8mA is 2.9mA/V
              Last edited by nickb; 08-26-2019, 06:11 PM.
              Experience is something you get, just after you really needed it.

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              • #52
                Originally posted by nickb View Post
                About 820 ohms. Remember the model might be poor.
                Thanks a lot!

                Even though the sim results seem high (indicating a too low model gm) , they help me to get a better understanding of the interactions.
                Last edited by Helmholtz; 08-26-2019, 06:22 PM.
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                • #53
                  Originally posted by Helmholtz View Post
                  Tanks a lot!

                  Even though the sim results seem high (indicating a too low model gm) , they help me to get a better understanding of the interactions.
                  Please note my revision above. I made a boo boo...
                  Experience is something you get, just after you really needed it.

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                  • #54
                    I should have AC bypassed the 10k resistor so the DC conditions were changed to be closer the the answer you needed. Zin by this method is 116 ohms. The plate current is 4.8mA, the gm at 4.8mA is 2.9mA/V
                    No, I actually meant AC shorting the grid only, second triode disconnected. The 6111 datasheet gives me gm = 3.15mA/V @ 4.8mA/Vp = 150V. As actual plate to cathode voltage would be more like 200V, gm = 2.9mA/V seems reasonable (the 6111 is a rather non-linear tube meaning that tube parameters strongly depend on plate voltage and current). From this I calculate an internal cathode impedance of 345 Ohm. this would be shunted by 10.47K and 470k, resulting in 334 Ohm.
                    Last edited by Helmholtz; 08-26-2019, 07:49 PM.
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                    • #55
                      Originally posted by Helmholtz View Post
                      BTW, what is a 6C67? I thought you would use the 6111?
                      I'm glad you asked. The 6C67 is the tooth fairy's favorite tube, in that it doesn't exist.
                      I'm sorry, that should be 6CG7. Of course, I was going to tell you that a 6CG7 has very similar characteristics to the 6111 and 6SN7 tubes; that it was designed to be a 9-pin equivalent to the 6SN7. But I suspect that actually know all this, now that it's clear that I really need to be wearing glasses more often.

                      Originally posted by nickb View Post
                      @SF if the objective is to get a high input Z then drive the grid instead of the cathode. Then you'll get more like 5meg. If you drive a common cathode stage from the plate of the first the gain go from a very sad gain of 1dB to a happy 24dB. It seems to be that the circuit as it stands doesn't do anything very useful at all and you'd get only slightly less gain from a bit of wire.
                      Hang on a second mr 24dB. First, the circuit was an a hypothetical experiment to better understand terminal impedance characteristics, -- more specifically, how and if non traditional use of bootstrapping Techniques could be used to limit losses that due to impedance loading.
                      Second, I disagree that its a useless circuit. The purpose of this circuit is to serve as a mixing stage for out of phase signals without phase cancellation, so I am using the grid as an input. Secondly, I suspect that this also offers good channel isolation without having to pad the shit out of the inputs.
                      Since my application specifically does not require any voltage gain, I wasn't concerned with it. Thirdly(?), setting up this experiment and running tests is highly beneficial to me. Plus, I got this Heathkit up and running quite nicely now. – New pamona banana jacks, precision capacitors in the multiplier, new filter caps, and a sweet violet indicator LED to replace the burned up neon.
                      Click image for larger version

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                      Pretty sweet, eh? (bad photo, but you get the idea.)
                      Lastly, this is an evolving idea. I haven't ironed out all the kinks yet... Having said all that, I've been thinking that maybe driving the plate of the first triode is really a much better option. Your observations seem to indicate this too.
                      If I have a 50% chance of guessing the right answer, I guess wrong 80% of the time.

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                      • #56
                        that maybe driving the plate of the first triode
                        Plate input??

                        Why don't you reveal what kind of signals you're going to mix? Might help to suggest a solution.

                        Maybe you just need a differential amplifier (like a LTPI)? Reducing input impedance is easy.
                        Last edited by Helmholtz; 08-26-2019, 08:21 PM.
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                        • #57
                          Originally posted by Helmholtz View Post
                          No, I actually meant AC shorting the grid only, second triode disconnected. The 6111 datasheet gives me gm = 3.15mA/V @ 4.8mA/Vp = 150V. As actual plate to cathode voltage would be more like 200V, gm = 2.9mA/V seems reasonable (the 6111 is a rather non-linear tube meaning that tube parameters strongly depend on plate voltage and current). From this I calculate an internal cathode impedance of 345 Ohm. this would be shunted by 10.47K and 470k, resulting in 334 Ohm.
                          Ah! Okay doke. Done like that Zin is 220 ohms.
                          Experience is something you get, just after you really needed it.

                          Comment


                          • #58
                            Originally posted by SoulFetish View Post
                            I'm glad you asked. The 6C67 is the tooth fairy's favorite tube, in that it doesn't exist.
                            I'm sorry, that should be 6CG7. Of course, I was going to tell you that a 6CG7 has very similar characteristics to the 6111 and 6SN7 tubes; that it was designed to be a 9-pin equivalent to the 6SN7. But I suspect that actually know all this, now that it's clear that I really need to be wearing glasses more often.



                            Hang on a second mr 24dB. First, the circuit was an a hypothetical experiment to better understand terminal impedance characteristics, -- more specifically, how and if non traditional use of bootstrapping Techniques could be used to limit losses that due to impedance loading.
                            Second, I disagree that its a useless circuit. The purpose of this circuit is to serve as a mixing stage for out of phase signals without phase cancellation, so I am using the grid as an input. Secondly, I suspect that this also offers good channel isolation without having to pad the shit out of the inputs.
                            Since my application specifically does not require any voltage gain, I wasn't concerned with it. Thirdly(?), setting up this experiment and running tests is highly beneficial to me. Plus, I got this Heathkit up and running quite nicely now. – New pamona banana jacks, precision capacitors in the multiplier, new filter caps, and a sweet violet indicator LED to replace the burned up neon.
                            [ATTACH=CONFIG]54914[/ATTACH]
                            Pretty sweet, eh? (bad photo, but you get the idea.)
                            Lastly, this is an evolving idea. I haven't ironed out all the kinks yet... Having said all that, I've been thinking that maybe driving the plate of the first triode is really a much better option. Your observations seem to indicate this too.
                            Let's wait and see it in the final context. It seems that this is just a part of the picture.

                            BTW, I'm actually more envious of the drill press.
                            Experience is something you get, just after you really needed it.

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                            • #59
                              Originally posted by Helmholtz View Post
                              Plate???
                              Yeah, instead of a grounded plate input stage. For instance... if you are mixing two out of phase signals -- one feeding the inverting input (grid), and the other feeding the non-inverting input (cathode), might it be better to take the output of the plate of that stage?
                              Or maybe something like this? Just thinking aloud here, so everybody calm down if it sucks.
                              Click image for larger version

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                              If I have a 50% chance of guessing the right answer, I guess wrong 80% of the time.

                              Comment


                              • #60
                                Originally posted by nickb View Post
                                Let's wait and see it in the final context. It seems that this is just a part of the picture.

                                BTW, I'm actually more envious of the drill press.
                                Oh, dude. The drill press is really nice (Jet). Picked it up second hand in great shape. I've got a set of precision bearings lined up for the next real maintenance cleaning and teardown. I'm looking a getting a LLambrich keyless chuck. Or an Albrecht if I can find on used in really nice shape, so I'll probably get into all that when I readjust and measure the runout.
                                If I have a 50% chance of guessing the right answer, I guess wrong 80% of the time.

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