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In Reply to: RE: what 2 tubes will replace a 6BM8? posted by DAK on December 13, 2015 at 12:06:42
Draw the load line for the 6550 in that amplifier, then look at how much positive swing you have vs. negative swing available. If you move along the load line and hit the bunched up curves and 0mA plate current much more quickly than you'd hit the 0V grid voltage line, then you don't gain anything by being able to drive the amp into grid current, as you would already be driving the 6550 to the point where it cuts off anyway.
The 811 can work OK on lower voltages if you accept some global nfb.
Follow Ups:
I have never drawn "load lines" before, but I have gotten loaded in my younger days. All kidding aside, it sounds like this exercise can only be done on a completed amp.
No, a loadline would be drawn long before the amplifier would be constructed.
..and they have a bunch of graphs showing the tube characteristics under various conditions. Is this the load lines you are referring to?
nt
Hi CB, thank you very much for the link. cheers, Dak
The schematic says 75mA current, 38V bias, 318V on the plate (you have to take the bias voltage away from the voltage shown), so you can put a dot on the 6550 triode curves here.
The output transformer used is 2.6K, which will allow you to draw the loadline.
I've done a quick sketch of your loadline, if you swing along that load line, things get pretty ugly as you pass the -50V line, and beyond salvation at -70. This is 32V above your quiescent operating point, which isn't more than the bias voltage, so trying to swing below 0V on the grid doesn't look to be all that useful.
A few questions again.
The 75ma indicated on the schematic is the the tube quiescent current at the cathode. This would be the value for the anode as well?
How does the 2.6k output transformer value establish the loadline that you drew.
The OPT I want to use for my amp has primary impedance 3.5K
Thank you for guiding me thru this process. I have a hard time relating to these procedures on an intuitive basis. regards, Dak
Edits: 12/14/15
For a triode idling in class A1 (most designs), cathode current is equal to plate current. With pentode and ultra-linear designs, there is some screen current that adds with the plate current, and this total is the cathode current.
Ohm's Law says that V=I*R, we know that R=2600 Ohms. If we arbitrarily say that we want to move up 100V on the X scale, then 100V=I*2600, so we must move down 38mA. Similarly, you can move down 100V on the X scale, then move up 38mA.
..the circuit, we can make the load line more horizontal? In my case, using a 3500 ohm primary impedance OPT, the current calculated is 28ma. Using the same graph the load line would be flatter and should improve linearity with voltage swing, correct?
Yes, the load line slope will decrease a bit. You will be able to swing further before you hit the nasties, however, this will require more drive voltage, and you'll be stepping this voltage down more to your speakers, so you'll generally end up with a little less power and a little less distortion (to a point).
If you have a 3.5K transformer, have a look at the 2A3 curves and draw the loadline at 325V/45mA.
I really don't want to go into other tubes right now. I would have to get new sockets as well, and have to deal with the AC or DC filament question on the power tube. So I very much want to stick with the power tube and circuit layout and find ways to improve it. Your suggestions and guidance has been extremely helpful.
Any other ways to improve the circuit is most appreciated. thanks a million, Dak
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