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|15th July 2011, 11:03 PM||#11|
I wired two of my bulbs (GE Crystal Clear 300) in parallel, and with 54V across them they draw 2.8A, for about 20 Ohms.
I also cleaned up my bench prototype and double-checked everything to be sure my results were repeatable. Then I replaced the 2SK82 with another, just for good measure and the results were the same.
I now feel pretty confident that I'm giving you good information with respect to the pinout as well as the THD results. Now I'm off to work up the plywood prototypes.
|16th July 2011, 05:58 AM||#13|
Join Date: Feb 2007
When using 2SK82 in the common source stage, we might need to care of possible limitations. I do not have a datasheet for K82, but looking at this japaneese site
I have realised that they struggle against the huge capacitances of K82, something (as a guess from the google-translated text) near Ciss + Crss = 4000pF, and they use four-stage schematics for driving them properly.
Apart from the THD and enjoyment from simple schematics, we should assess what can be expected soundwise, IMHO.
|16th July 2011, 07:04 PM||#14|
The one and only
|16th July 2011, 07:06 PM||#15|
The one and only
gate stoppers on them.
|16th July 2011, 08:42 PM||#16|
Join Date: Jun 2004
Working with Nelson's K82 graph, I have drawn two loadlines for illustration. (see attach.)
You need to use a 'millimeter ruler' and position it in such a way, that the spacing in mm on the ruler will be equal between the adjacent Vgs lines
from zero Vgs to the max Vgs.
First, the top line.
We cannot go below -14V because of the distortion, so the the Vgs bias is 14/2=7 V.
At the loadline -7Vgs will produce 2.4A Id and 18V Vds. 2.4*18=44W dissipation at idle.
The drain will swing between 7.5V and 27V.
so one half will have higher amplitude = even harmonics, mainly second.
Load resistance calculated from coordinates crossing.
So, the optimal load will be 4 Ohm speaker. To make load look like 4 Ohm
with the 8 Ohm speaker, we will have to parallel it with 8 Ohm resistor, which will seriously compromise the efficiency.
Another way is to use a 8 Ohm:4 Ohm (1:1.414) transformer.
Taking 10V output amplitude, output power will be 10²/2*4=12W.
Obviously, we can reposition the LL to cross the -14Vgs at 28Vds, or more. In any case, having some of the second harm. will help to balance the spectrum against the unavoidable third.
Voltage gain (deltaVd / deltaVgs) with this loadline will be ~1.2-1.5 (3dB). The input signal will have ~7V amplitude.
With source follower configuration the gain will be ~0.6, or so.
The lower LL is another try. Here we have to limit the Vgs swing to only 13V, because -14V is already distorted too much.
At this Vgs the Id=2.2A, Vd=15V. Power diss. at idle 2.2*15=33W.
From LL crossings 23V-5V/5.1A=18/5.1=3.5 Ohm.
Now, we can use 4 Ohm speaker with the calculated resistor (lightbulbs) in parallel to produce the desired load value for the V-JFET. Obviously, same limitations here for an 8 Ohm speaker.
Another issue to keep in mind is the speaker low freq. resonance impedance (much higher, than 4 Ohm).
So, the optimum drain biasing solution will be a low inductance CHOKE with Rdc=4 Ohm, and staying at that AC impedance level at all low end freq., where the speaker impedance is still too high. In other words:
X=2*pi* (~150 Hz)*L=4 Ohm, and
L=4/6.3*150 ~=4 mH
Clearly, all these V-JFETs will be different, and will require individual measurements (curve tracing).
The truth is out there. Who can handle the truth?
Last edited by wintermute; 17th July 2011 at 11:34 AM. Reason: edited at request of StephenOH
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