Guitarnut Posted October 30, 2010 Share Posted October 30, 2010 I'm trying to get my head around tube amp design and operation. I was making good progerss in my studies this morning until I started looking at the PS. The rectifier in particular. I understand the concept of a diode tube, and how electrons only flow from cathode to anode. That the AC supply off either end of the center tapped seconday PT is out of phase with the other end and the rectifier only conducts on the positive side of the sign wave. Filter caps fill in the gaps but for a bit of ripple that is handled by the choke. My big question is what exactly is going on in the tube. From this schematic, I see that 5 VAC is going to the heater and the HT is connected to each of the plates. But I aslo notice that the resulting DC output is connected to the 5 VAC line. How is it possible that these 2 are on the same "buss"? Is it because the output of the tube is only the positive side of an AC voltage and is only considered DC when in a circuit that connects it to ground? And, if electrons only flow from neg to pos, or cathode to anode, how is a higher voltage created on the cathode? Is the cathode pos in this case? Peace, Mark Link to comment Share on other sites More sharing options...
mtheory Posted November 1, 2010 Share Posted November 1, 2010 Great questions! Here's some good information that might help answer: http://www.vacuumtubes.net/How_Vacuum_Tubes_Work.htm http://en.wikipedia.org/wiki/Rectifier Link to comment Share on other sites More sharing options...
fuse Posted November 3, 2010 Share Posted November 3, 2010 Let me first say that I ma not the best person to explain this..... IMO,the power supply, rectifier, B+ filter, is the most complex circuit in the amp. There is probably a better way to draw the schem, and that would make things easier to understand, but it certainly is correct as drawn. The filament supply shares a connection with the cathode. It is kind of similar to a plate resistor, where the AC signal rides on the DC B+. The two do not intermingle. OK, I will do my best....... The "load" of the first filter stage forms a series connection with the cathode and anode of the rectifier tube. The cathode is common, so the series connection changes from V1 to V2 as the phase flip-flops on the PT secondary. The current path of travel is from the CT, to the capacitor/resistor stage, to cathode, to anode, and back to the PT secondary where it is split in half by the CT. So, there is a voltage potential at the node that is formed by the junction of that first cap/resistor stage, and Pin 8. This potential is relative to ground and the PT secondary. So you get the peak value of the DC ripple and then the filtering smooths it out to almost flat line DC. With tube diodes you have a pretty big voltage drop (compared to solid state diodes), so you loose a little voltage there. That is one way to shave some voltage off your B+ if you need to. You just have to make sure the PT can handle the current draw of the different rectifier tubes. Like I said, this is a total "Layman's" synopsis. I do not have any kind of electronics education, or background. I am sure my explanations is embarrassingly simple. But I think it is basically correct. Good Luck Link to comment Share on other sites More sharing options...
Guitarnut Posted November 4, 2010 Author Share Posted November 4, 2010 Thansk for the replies and info. Much appreciated. I think I have my head around this now. I'm a visual learner so I did a quick animation to map it out and see it "in action". Hope I have it correct. If not, feel free to corrct me and I'll update the animation. Link to comment Share on other sites More sharing options...
fuse Posted November 5, 2010 Share Posted November 5, 2010 Wow! I am duly impressed! I believe the meters are showing (reading) correctly. Let ME ask YOU a question...... If your animation showed the DC Ripple meter with its negative terminal connected to the same ground as C35 and C36, do you think it would be easier for some people to grasp the concept? Again, that is an AWESOME drawing and I am truly impressed. I believe it to be schematically, and electronically correct. But I wonder if some might find it easier to understand if the ripple meter were to be "Physically" connected in a different manner. Who knows, some might find my suggestion to be more confusing. For the purpose of explaining the dual diode, you were probably smart to pretend the caps do not exist. Great Job..... Link to comment Share on other sites More sharing options...
fuse Posted November 5, 2010 Share Posted November 5, 2010 Man! I cannot get over your "active" schematic. I am pretty much computer illiterate. The INTERNET is just a library for me, and a means of communication. I cannot do any of the graphic stuff that some of you guys are capable of. So now I have two more questions: 1. Is there some kind of "software" that you use to do those drawings? 2. Can you do a drawing that shows the effect of the caps on the DC ripple? I am not necessarily asking you to do that. It might be a time consuming PITA for the purposes of this forum. But if you needed to......? Thanks Link to comment Share on other sites More sharing options...
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