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How to replace Standard.JFT in Mojave?
Hi, I have installed LTSpice on my main computer (Mojave) and this week too ?on my laptop (High Sierra)?- both latest version.?The component lists differ. I need a 2SK170 (or 2SK389). Example: I select JFT; right-click (pick a new one) - the Mojave version has a list of n-fets with just a few entries ending in U309; the HighSierra laptop ends with three more - added is the 2SK486a/b/c . No 2SK170. On the desktop I have found the file-location with the?Standard.JFT?in the package contents,?replaced?this file with the one from this LTwiki?(which includes the 2SK170)?, restarted, but nothing changes after restarting. So still no ".model J2sk170¡± that I can select.?No 2SK486a/b/c either on the desktop. The drop-down list remains the same. Under documents I found a LTSpice entry with "LTspiceXVII" that LTSpice makes, also replaced the standard.jft file ?there, still no effect. What should I do? Yes, I reinstalled LTSpice twice; I deleted the old directories under Documents // - nothing helps. There is no entry under Application Support for LTSpice. albert |
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Re: Looking for models of tubes
Alan, I'm sure Bordodynov has an excellent model collection.? He always does.? If interested in the models in the group's files, try searching the group's Files archives (starting with the Table of Contents file, all_files.htm) for these keywords: "Koren" (that's not "Korean"), "Duncan", "valve", "triode", and "pentode". You can also find the NormanKoren and DuncanAmps SPICE models on the Internet. Don't bother searching from the Yahoo site.? Download all_files.htm and search through that. Regards, Andy |
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Re: Question on loop gain and phase simulation
Andy, Thanks for the reply.? The thing that threw me off is that in IV version the method of placing the DC operating point value is different than XVII.? In XVII it is much more obvious and in IV you have to dig down into the right click menu to find it. Dan |
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Re: Looking for models of tubes
Hi. You will find these models on my page. Sometimes there are variants of the model (up to three pieces). ? Bordodynov. 05.12.2018, 09:25, "alan.revera@... [LTspice]" :
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Re: Question on loop gain and phase simulation
Alan, I believe you are trying to use the Middlebrook method of measuring open loop gain while the loop remains closed.? I believe that method depends on the fact that there is a zero current (high impedance) feedback node that accepts all the feedback.? But there is no such node in this circuit so I don't think that simple method will work for this circuit. You can probably break up the two loops using another op-amp and then find out the open loop Bode plot for each feedback loop independently.? But I don't think the Middlebrook method can give you the combined open loop response. In fact, the combined open-loop response doesn't really make a lot of sense in this case. Also, note that there is no overall DC feedback (no integrator term) so there is a lot of DC offset in the output. Good luck, Dan |
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Re: Question on loop gain and phase simulation
The local feedback of the opamp inside the forward path set the gain of the opamp as a unit with closed loop gain. I am trying to get the global loop gain of the opamp and the two tubes. Is there a way to measure that. This is like an amplifier with 3 gain stages, then the global feedback determine the gain after the global negative feedback and phase margin. That's the one I want to plot to look at the system stability. Thanks |
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Re: Question on loop gain and phase simulation
Alan wrote, "I want to plot the LOOPGAIN, that is the Openloop gain minus the Closed Loop gain. or LG=OLG-CLG ( in log term)." I think you have a problem because there is more than one feedback loop.? Someone might argue that there is no clear concept of what is "openloop gain".? Is it the gain when both loops are open?? When only one of them is open?? Technically, from the point of view of the overall loop's outer feedback path, the inner op-amp has only an open-loop gain because its "closed loop gain" is effective only from its non-inverting input pin. Regards, Andy |
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Re: Question on loop gain and phase simulation
Alan, There is a notable difference from moving the voltage source (Helmut's V3) to the other side of the resistor R15.? It moves note B to the other side of that resistor.? Therefore, when you plot V(Out1)/V(B), you are plotting something very different.? In your case, node B was the inverting input of the high-gain op-amp.? In Helmut's case, it was the opposite end of the feedback resistor.? So of course the plotted "voltage gain" will be very different. Regards, Andy |
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Re: Question on loop gain and phase simulation
Alan wrote, "But you can see the opamp local loop is closed." But that is irrelevant. The thing you've been plotting, is V(Out1)/V(B).? The feedback resistors around the op-amp do not affect the gain V(Out1)/V(B).? V(B) is the -IN input to the op-amp itself, where the feedback resistors don't affect the voltage gain. If you want to include the effects of the closed-loop op-amp's gain, perhaps what you should be looking at is V(Out1)/V(N007) ... or instead of N007, whatever is the nodename of the?+IN pin of the op-amp. Regards, Andy |
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Re: Question on loop gain and phase simulation
Yes, I got rid of that and still works. What is the reason. All you did is move the AC source from one side of R15 and C1 to the other side. I have been putting the AC source on the summing junction and it always worked. Why the difference? Thanks Helmut. |
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Re: Question on loop gain and phase simulation
Thanks Helmut You plot V(out1)/V(b) also? Yes, this looks right. The gain of U6 in the system is -1 because R22/R15=1. The opamp just making up the closed loop gain of 21. The loop gain is just the gain of the two tubes. I see you move the AC source and added one more voltage source. Can you explain why you do this and why it works? Thanks Alan |
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