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In Reply to: RE: what 2 tubes will replace a 6BM8? posted by DAK on December 12, 2015 at 13:52:58
A 5751 is about as close as you can get to the triode, but its gm is low. AFAIK, nothing comes close to the power pentode. You can try the more capabable 6BQ5/EL84.
Eli D.
Follow Ups:
Hi Eli, what about the 6v6gt to replace the pentode unit? Although it is a beam pentode, it has some similar attributes. I have literally hundreds of 6v6. This is the schematic of the amp I am considering.
http://www.single-ended.com/6550se.htm
If you abandon the triode option, you can run the amp without the second stage and just half a 12AX7 driving each 6550. If you want to do both, use half a 12AT7.
The choke driver would be great for pushing into A2, but I don't believe you'll have enough room on the high side of the load line to justify the trouble.
If you are insistent about the directly coupled follower drive, use an 811, remove the choke, remove the regulator at the output, and make the 22K power supply resistor ~7K.
...and I still don't know what numbers you used to determine your findings. I used the UL graph and referring to the schematic of the "cathode drive amp" the only number given is the plate voltage of 360vdc. How do you know what the plate current will be? cheers, Dak
Hi CB. A couple questions. You state;
The choke driver would be great for pushing into A2, but I don't believe you'll have enough room on the high side of the load line to justify the trouble. ot
Installing a choke is really not much trouble at all, as long as you have the room. But, I don't know what you mean by ....enough room on the high side of the load line.... Could you clarify that?
Also, using the 811 tube. Don't this transmitter tube require high voltages to operate successfully? The B+ in the circuit and that I am considering is around 400volts maybe up to 450v. cheers, Dak
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.
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
The 6BM8's pentode section is set up as a choke loaded, triode wired, cathode follower. A 6V6 should do the job too.
BTW, that 80 H. loading choke will cost a fair amount of money. Perhaps CCS loading of the CF is an affordable alternative.
Eli D.
Thanx for the help Eli. I thought of that too when I first ran across the schematic. I have been using more SS devices in my builds but i am still learning. I have some LM317 and LM337 devices. Do you know of a design using those or other devices that is a good candidate for this? cheers, Dak
What does choke loaded mean? I have thought it meant could only accept certain constant voltage signals, but I am just guessing and am prolly totally wrong.
The second term is "A cathode follower" I have no idea what this means? I thought the signal would be directed a specific outlet pin only in order to dissipate the left over signal that isn't being used for anything, not to something important, a refuse bin in common vernacular. Again I'm prolly hopelessly lost and confused!
It is better to have blurbed and blown it than to have never blurbed at all.
Run a Google search for NEETS, which is U.S. Navy training material. Learn the basics of electronics.
Eli D.
I think he WAS trying to learn by asking valid questions, but didn't get his questions answered.
I was hoping for an easy way but I guess that just aint gonna happen. I will look at the NEETS Navy material. I once knew college level physics, chemistry and did pretty well in calculus etc., but my brain hasn't worked out in many years and is out of shape -so to speak.
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