06-11-2016, 01:03 PM
So.....How would you calculate ra for a pentode from published valve curves?
Lawrence.
Lawrence.
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Pentode ra....How to...
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06-11-2016, 01:13 PM
Look at the slope of Ia vs. Va.
But, ra isn't a well-controlled characteristic of a pentode, varying over a 2:1 range. Fortunately, it's so high anyway, that it doesn't matter in practice.
06-11-2016, 02:13 PM
Sorry folks I owe you an apology, I meant to say amplification factor (mu) not ra
 I'm looking at RDH4 chapter 2.2, 2nd graph, I can understand why the Va scale has to be extended and I can understand the top tangent but the bottom tangent is confusing me, why is there no equal gap between the tangent line and the 6 volt -ve curve at say 120 volts either side of the 180 volt reference point?P.S I'm not up to speed with geometry... http://www.paleoelectronics.com/RDH4/CHAPTR02.PDF Lawrence.
06-11-2016, 05:40 PM
Hmmm... I'm not quite sure what you're asking Lawrence!
The mutual conductance is change in anode current, divided by change in grid voltage, for constant Va. So, go vertically from Q, each side, you get just over 4mA change for 4V grid change, and that's just over 1mA/V. The anode resistance is change in anode voltage divided by change in anode current, for constant grid voltage. That's the same as the slope of the tangent at Q. The fact that the tangents at A, at Q, and at B are not quite parallel, says that ra isn't quite constant. The mu is just ra x gm, which you can work out by multiplication. Alternatively, it's just change in anode voltage divided by change in grid voltage, for constant anode current. It is given by the intersections of the tangents with the horizontal, constant-current line at the operating point Q, divided by the grid voltage change needed to shift from one tangent to another. But it's a rather complicated way when just simple multiplication would be easier!
06-11-2016, 05:50 PM
Yes I understand that but....getting back to that graph the lower tangent doesn't look right to me, also the voltage of 447 volts quoted between points C & D doesn't seem to tally up with the scale presented, try measuring it with a ruler and you will see what I'm getting at unless I've got something wrong.
Lawrence.
06-11-2016, 06:55 PM
I don't have a ruler, but... Point D looks about +340V (so swing of 160V above the operating point, 180V). Posted net C looks about -50V, so about 230V below the operating point.
The fact that these aren't equal says that here would be some distortion (there would be anyway, because the curves deviate massively from the tangents for a grid swing as much as this). But crudely, the change in Va is 160V + 230V = 390V, yes I see what you mean, not quite the 447V in the text. If the manufacturers had given curves about 0.25V apart, it would have probably been more meaningful... No way would this valve cope with 2V p-p swing on its grid! But they didn't, so we have to use them, and extend by straight lines.
07-11-2016, 08:19 AM
Must have been a drawing error of some kind, glad you spotted it too, I thought I might be going bog eyed.
Lawrence.
07-11-2016, 11:31 AM
Discrepancy solved, managed to find an online copy of RDH3, the edition previous to RDH4, book page number 271 shows the same set of curves but the lower tangent is drawn a lot better, the 447 volts quoted (point C to D) now scales to a ruler measurement correctly, also note that Ib and Eb in RDH4 is shown as Ip and Ep in RDH3.
Link to RDH3: http://www.tubebooks.org/Books/RDH3.pdf Scan of the relevant page from RDH3: Lawrence.
07-11-2016, 01:02 PM
(06-11-2016, 06:55 PM)Kalee20 Wrote: I don't have a ruler, but... Point D looks about +340V (so swing of 160V above the operating point, 180V). Posted net C looks about -50V, so about 230V below the operating point. Sorry, predictive text kicked in while typing on an iPad in a layby! (How did "Point C" get changed to "Posted net C"?) |
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