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#3831 | |
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diyAudio Member
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Quote:
__________________
Een volk dat voor tirannen zwicht, zal meer dan lijf en goed verliezen dan dooft het licht…(H.M. van Randwijk) |
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#3832 |
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diyAudio Member
Join Date: Apr 2002
Location: Prague, Kitakyushu, Fukuoka
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Maybe it is only misunderstanding, and the reaction was to another post.
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#3833 |
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diyAudio Member
Join Date: Jul 2003
Location: berkeley ca
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PMA and Edmond, please lighten up. I am only trying to stay on target. Please read what I have to say, and try to understand my meaning, first, before asking redundant questions. It only slows and confuses the process.
Also, everyone, please forgive my incompleteness on occasion. I am typing on the fly, with only a cursory check on my spelling, etc. I am not teaching a course. |
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#3834 | |
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diyAudio Member
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Quote:
).Cheers, Edmond.
__________________
Een volk dat voor tirannen zwicht, zal meer dan lijf en goed verliezen dan dooft het licht…(H.M. van Randwijk) |
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#3835 |
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diyAudio Member
Join Date: Jul 2003
Location: berkeley ca
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Edmond, this is the Blowtorch thread. If I want to share the subtle qualities of fet circuit design with others, that is my choice. I already know fet circuit design. I also have a former associate looking into his class notes and he should get back to be shortly.
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#3836 |
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diyAudio Member
Join Date: Apr 2002
Location: Prague, Kitakyushu, Fukuoka
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Okay, let's wait what new equations we get from class notes of your associate.
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#3837 |
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diyAudio Member
Join Date: Apr 2002
Location: Prague, Kitakyushu, Fukuoka
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Let's go on a bit. See effect of degeneration at single JFET, lambda = 10m, source resistor = 50R
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#3838 |
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diyAudio Member
Join Date: Aug 2002
Location: Germany
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Hi Pavel,
would you mind to try lower values, say between 2.2r - 22r? Rüdiger
__________________
"I can feel what's going on inside a piece of electronic equipment. I have a sense that I know what's going on inside the transistors." Robert Moog |
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#3839 | |
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diyAudio Member
Join Date: Apr 2002
Location: Prague, Kitakyushu, Fukuoka
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Quote:
P.S.: for true diffpair (balanced in, balanced out) distortion spectrum does not change with degeneration. |
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#3840 | |
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Account disabled at member's request
Join Date: Aug 2005
Location: Toronto
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Quote:
You asked for equations, now you are asking for a phenomenologic view of the distortions in a differential stage. Until your associate comes back with something, probably in a language that you are more comfortable with, here's an gedanken experiment on why a perfectly symmetric differential stage cannot have second harmonic distortions. Imagine an perfectly symmetrical differential stage with no degeneration. Do not make any assumption about the building devices, they can be bipolars, FETs, MOSFETs, tubes, whatever you like. The only thing you need to know is the dependency of the device current i to the device input voltage u, and let's assume this dependency is an arbitrary function i=f(u). You may think of “f” as a generic, non-linear, device transconductance. Let's define the current throungh the left device as i1=f(u1) and through the right device as i2=f(u2). The differential input voltage equals the devices input voltage difference, so ui=u1-u2. The output differential signal is proportional to the difference of the currents through the devices, so let's define the differential stage transconductance as G(ui)=(i1-i2)/ui which, following the above equations, will be G(ui)=(f(u1)-f(u2))/(u1-u2) How much is G(-ui)? Following the same rationale as above, and considering the stage symmetry, it turns that: ui=u2-u1 and hence G(-ui)=(f(u1)-f(u2))/(u2-u1)=-G(ui) Which makes G an odd function. As discussed before, an odd function may generate only odd harmonics (for similar reasons, a 50% symmetric square wave has only odd harmonics). If you don't feel comfortable with the above math formalism on why G(-ui)=-G(ui) think a differential stage output in the 1st and 3rd quadrants. It is perfectly symmetrical, so if ui generates a response in quadrant 1 then -ui will generate the same response (in module) in quadrant 3. As you see, we did not make any assumptions about f(u). It doesn’t matter what type of harmonics (even, odd or both) the device (as a single ended stage) has, the differential/balanced stage may have only odd harmonics. If the single ended stage has even harmonics only, then you can think of “converting” the even harmonics to odd harmonics. If the single ended stage has both even and odd harmonics, you may think in terms of “cancelling” the even harmonics. But ultimately, this is not really important. Of course, any asymmetry (like e.g. device matching) will break the above rule, and both even and odd harmonics will appear at the output. Now, a symmetrical degeneration (or any other symmetrical model enhancement, like deviations from the ideal current JFET square law) doesn't change anything; as long as the symmetry is conserved, you may include the degeneration + lambda + ... = new f'(u). Replace above f(u) with f'(u) and nothing will change from a spectral perspective. Even if the single ended device has both 2nd and 3rd, the differential stage will have only 3rd, 5th, etc... Why do odd functions have only odd harmonic components? This is not as easy to answer. If you are interested in this detail let me know - but be prepared for a tough(er) math ride. |
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