Which is the rule that allows to calculate the load impedance (and so, the impedance of the output transformer) of two or more valves connected in parallel ?
Is it possible to connect in parallel two or more valves having a different internal resistance (i.e. 1000 and 2500 ohm) ? ...If so...is the rule the same ?
With anticipated thanks.
Is it possible to connect in parallel two or more valves having a different internal resistance (i.e. 1000 and 2500 ohm) ? ...If so...is the rule the same ?
With anticipated thanks.
Posted by Paul D. Scearce on March 22, 2002 at 10:52:01
In Reply to: Load line posted by Bas Horneman on March 22, 2002 at 03:43:34:
From Ohm's law, impedance is V/I. To add an impedance line to a tubes plot, you need to find two points on the load and draw a straight line through them.
For a resistance coupled tube, start with the B+ voltage. If the plate of the tube is at that voltage, the voltage accross the load resistor is 0. From Ohm's law, current is V/R, in this case 0 ma. So the first point is at B+ v,0 ma. You can than pick any voltage less than B+ to find your second point. Lets pick 0 volts. If 0 volts appears at the plate of the tube, the voltage accross the load is B+ volts. Using Ohm's law again, current is B+/R, and your second point is B+/Rma,0v.
For example say B+ is 200v and the load is 20kohm. Draw one point at 200v,0ma and another at 0v,200/20ma(0v,10ma in other words). Then draw a straight line through these points. That is the load line.
The process is similar for a transformer coupled tube, except now there is a fair amount of dc current flowing when the plate is at B+. Using your output stage for an example, the bias conditions are probably about 100 volts on the plate at 60ma per section. (Since grid voltage is represented as a negative number on the plots, plate voltage is read from the plate to the cathode.) So draw a point at 100v, 60ma. When the plate is at 0 volts, 100 volts would appear accross the ac load, so current will be 100v/3.2Kohm or 31.25 ma. (Two sections are used in parallel, so each section will see a load of about twice the primary impedance). Add the ac current to the bias current, and draw a second point at the resulting current along the 0volt line on your graph-about 0v,90ma. Now draw a straight line through the two points for your load line.
This line passes through the grid voltage lines of the tube, so you can tell what the plate voltage will be for any grid voltage. Bear in mind this is only an approximation, but it should give you a fair idea of how the tube will opperate.
Hope this helps
Paul
Don't think it is possible...well it could be done... but won't sound good ...methinks
In Reply to: Load line posted by Bas Horneman on March 22, 2002 at 03:43:34:
From Ohm's law, impedance is V/I. To add an impedance line to a tubes plot, you need to find two points on the load and draw a straight line through them.
For a resistance coupled tube, start with the B+ voltage. If the plate of the tube is at that voltage, the voltage accross the load resistor is 0. From Ohm's law, current is V/R, in this case 0 ma. So the first point is at B+ v,0 ma. You can than pick any voltage less than B+ to find your second point. Lets pick 0 volts. If 0 volts appears at the plate of the tube, the voltage accross the load is B+ volts. Using Ohm's law again, current is B+/R, and your second point is B+/Rma,0v.
For example say B+ is 200v and the load is 20kohm. Draw one point at 200v,0ma and another at 0v,200/20ma(0v,10ma in other words). Then draw a straight line through these points. That is the load line.
The process is similar for a transformer coupled tube, except now there is a fair amount of dc current flowing when the plate is at B+. Using your output stage for an example, the bias conditions are probably about 100 volts on the plate at 60ma per section. (Since grid voltage is represented as a negative number on the plots, plate voltage is read from the plate to the cathode.) So draw a point at 100v, 60ma. When the plate is at 0 volts, 100 volts would appear accross the ac load, so current will be 100v/3.2Kohm or 31.25 ma. (Two sections are used in parallel, so each section will see a load of about twice the primary impedance). Add the ac current to the bias current, and draw a second point at the resulting current along the 0volt line on your graph-about 0v,90ma. Now draw a straight line through the two points for your load line.
This line passes through the grid voltage lines of the tube, so you can tell what the plate voltage will be for any grid voltage. Bear in mind this is only an approximation, but it should give you a fair idea of how the tube will opperate.
Hope this helps
Paul
Don't think it is possible...well it could be done... but won't sound good ...methinks
I'm no tube specialist, so not sure on the way to calculate it, BUT I am pretty sure that whatever it is, you aend up with very unequal load sharing by the tubes. If you do this, at least use same tubes to parallel. That way, the total output power can be made the sum of the individual output powers.
Jan Didden
Jan Didden
Hehe.. nice one there 
On a more serious note, I realize the difficulties in not just differing plate r, but also the drive requirements of triodes vs pentodes. But my point was, that everybody *knows* it can't be done, so nobody does it. And so there is no progress. One of these days I'm gonna make a lab-amp with 10 spare sockets just for testing these crazy ideas when the urge comes over me...
Everybody *knows* that flea-power triodes have too high distorsion to be of any good. But the sound is still to die for.
On a more serious note, I realize the difficulties in not just differing plate r, but also the drive requirements of triodes vs pentodes. But my point was, that everybody *knows* it can't be done, so nobody does it. And so there is no progress. One of these days I'm gonna make a lab-amp with 10 spare sockets just for testing these crazy ideas when the urge comes over me...
Everybody *knows* that flea-power triodes have too high distorsion to be of any good. But the sound is still to die for.
TOO LATE...
Hi,
Well...I've done similar cocktails with various valves but it didn't yield I nice Manhattan...
Maybe I should shake a little harder...?
Seriously, I vaguely remember that there was once some experimenting done in a PP design to obtain better distortion characteristics...
I think I read this somewhere in an Audio Anthology issue, not sure though.
In my OTL amps I once mixed (per pair), 6080s, 6336As and 6528As...It worked, just not convincingly so.
Any other crazy ideas?
Cheers,
Hi,
Well...I've done similar cocktails with various valves but it didn't yield I nice Manhattan...
Maybe I should shake a little harder...?
Seriously, I vaguely remember that there was once some experimenting done in a PP design to obtain better distortion characteristics...
I think I read this somewhere in an Audio Anthology issue, not sure though.
In my OTL amps I once mixed (per pair), 6080s, 6336As and 6528As...It worked, just not convincingly so.
Any other crazy ideas?
Cheers,
I can't remember who, but someone actually did a hybrid pentode-triode output stage in a commercial design. Obviously, it was not a commercial success
It might be interesting to experiment with the tube equivalent of current dumping- have an output stage with a pair of smaller triodes biased for class A, then have a set of big-boy pentodes biased at class B (or even a bit beyond) to cut in when the triodes start running out of steam.
It might be interesting to experiment with the tube equivalent of current dumping- have an output stage with a pair of smaller triodes biased for class A, then have a set of big-boy pentodes biased at class B (or even a bit beyond) to cut in when the triodes start running out of steam.
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