That's possible. But remember the ccs must be able to withstand large ac voltages produced by de VT4C's . The 519 is capable of that but I should have added that the smoothing capacitors of the voltage multiplying circuit feeding the 519 should have a high voltage rating (450V or more).
Besides, the ccs as proposed by me seems to work for only a few watts; In my room it played quite loud but I was apparently a bit too enthousiastic. As several posts show I have been corrected in this respect.
The better option is the extra winding or 0-4 ohms part of the secundary to be used for the ccs which you can find at the earlier posts. It seems that for max. output the ccs must dissipate as much as the power tubes. So If you need less power you can use a lighter version but I can't give you an equation in this respect.
What is wrong about the schematic is the presumed 910V across the last electrolytic of the 519 powersupply. which should have read 90 volts. I haven't drawn it myself and forgot to look after it so it has been published prematurely. I apologize for all the misunderstandings it may have caused.
Besides, the ccs as proposed by me seems to work for only a few watts; In my room it played quite loud but I was apparently a bit too enthousiastic. As several posts show I have been corrected in this respect.
The better option is the extra winding or 0-4 ohms part of the secundary to be used for the ccs which you can find at the earlier posts. It seems that for max. output the ccs must dissipate as much as the power tubes. So If you need less power you can use a lighter version but I can't give you an equation in this respect.
What is wrong about the schematic is the presumed 910V across the last electrolytic of the 519 powersupply. which should have read 90 volts. I haven't drawn it myself and forgot to look after it so it has been published prematurely. I apologize for all the misunderstandings it may have caused.
Dave, the circuit I propose is just an ordinary SE fed from a 1000V powersupply; the floating circuit has only one function, to undo the OT of the magnitizing current of the output tubes resulting in zero voltage drop across the OT (both are drawing 0.1 A).
AC is still inductively transferred to the 8 ohms secundary. So it's not parafeed whatever way you turn it.
AC is still inductively transferred to the 8 ohms secundary. So it's not parafeed whatever way you turn it.
It doesn't eliminate the capacitor. The capacitor is the last cap of the power supply.
dave
You need to apply Kirchoff here and draw the current loops for DC and AC and you will find that this circuit is indeed parafeed. All of the DC current that traverses the 211's also traverses the high impedance supply. There is no "cancellation of current" the "floating supply" simply provides a path for the DC current and "sets" the voltage at the anodes to the same as the B+ rail. Since there is no voltage differential across the transformer, no DC current flows through it.
dave
I don't know where the misunderstanding comes from. The person drawing the schematic made one mistake writing 950V across the last supply electrolytic of the 519 powersupply.This should have read 90V. The situation is as follows. We have two electrically independant current loops one the VT4C's,OT,and 1000v pwr supply and is earthed, then the 519 ,OT,and it's 90V pwr supply which is floating The two circuits only have the primary of the OT in common. As the current loops are equal (0,1A) but running in opposite directions they cancel one another. That's all. Still ac coming from the VT4C's is transferred trough the output transformer to the speaker. This is an ordinary SE output and not parafeed.
You cannot have equal current flows in opposite directions in the same wire. The net result of your circuit is to add a floating CCS to place the anode side of the transformer at the same voltage as the B+ side. If your B+ supplies 100ma and the 211's each draw 50ma all of that 100ma of current traverses the PL519 circuit. This means that the DC is fed from a path that is in parallel with the output transformer which makes it parallel feed.
dave
dave
I disagree. There can be equal current flows through the same wire but only (as in my case) the two current loops are INDEPENDANT. The one current loop consists of the 1000V pwr supply, OT primary, through the VT4's and ground, the other which is floating hanging in the air so NOT earthed the anode current of the 519 is flowing in the opposite direction and the net result (when correctly adjusted) is zero volts across the output transformer.
This is how I built it this is how I measured it and you say it's impossible. I don't see what's so difficult about this.
Exactly the same happens in a circlotron OTL, the only difference being the load which is in the cathode circuit ( the two power halves forming a balanced circuit while both are floating and the cathode currents of each half flow through the speaker but in opposite directions thereby canceling each other). In my circuit as someone in one of the previous posts remarked you might call it a asymmetric otl except that there is an output transformer in stead of a loudspeaker. The principle however is the same.
This is how I built it this is how I measured it and you say it's impossible. I don't see what's so difficult about this.
Exactly the same happens in a circlotron OTL, the only difference being the load which is in the cathode circuit ( the two power halves forming a balanced circuit while both are floating and the cathode currents of each half flow through the speaker but in opposite directions thereby canceling each other). In my circuit as someone in one of the previous posts remarked you might call it a asymmetric otl except that there is an output transformer in stead of a loudspeaker. The principle however is the same.
Actually what has been done is adding a load parallel to the primary of the transformer lowering its resistance. Then you have a better match to the triodes. You could measure the frequency response to see if this is circuit improved that. But in the end all that matters is how it sounds, right?
You're right. I used a switch to turn the added circuit on or off. Without it the sound was thin and somewhat distorted, switched on there was bass and less distortion. So I didn't measure anything simply listened.
On the other hand the low dissipation of the ccs I used limits the output power as Chris H. has clearly pointed out.
On the other hand the low dissipation of the ccs I used limits the output power as Chris H. has clearly pointed out.
I am not debating whether the circuit works or not, I believe that the circuit works. My point is that this is a simple parafeed circuit with an added twist that the output transformer is referenced to B+ rather than DC ground.
There cannot be as you state 100ma in one direction offset by 100ma in the other direction in the same piece of wire. Current flow requires a voltage offset and the second you place both ends of the primary at the same voltage no current flows. All of the DC current that flows through the 211's also flows through the PL509.
Since this is a parafeed circuit the issue becomes the compliance of the CCS circuit becomes the limiting factor for AC swing which explains your limited power output. This means that the parallel 211's could be replaced by a single 300B and the low distortion power output would not change appreciably.
here is a spice grab of your basic circuit showing the currents through various points.
dave
Kees I get wat you are doing. The principle is OK, no DC current through the OPT. Philosopically, maybe the current goes around in a circle and only the difference goes to the OPT. I think you made a sort of unbalanced/single ended Circlotron. But indeed, the PL 519 would need the same (independent) voltage or it would become reverse-biased and cause clipping.
You can not replace the tubes with one 300b and omit the CCS as the transformer wil be about 6k. Just a bit too high for the troides (for almost any triode with low voltage on the anode). When adding the CCS the power is limited and the 300b will not suffice here. The amp is not very efficient, however: if it sounds right who cares? There are many OTL monsters in the field that are less efficient!
At the moment 100ma of the VT4's flow in one direction and 100ma of the 519 flow in the opposite direction the voltage across the primary of the OT (normally in the order of 30V when just the VT4's are used) becomes zero .
In a circlotron as built by Ralf Karsten the two opposing cathode currents flow through the loudspeaker resulting in a zero voltage drop. Depending on the number of output tubes it can be 2A but as they are flowing in opposite directions the resulting voltage across the speaker is zero by varying the negative grid voltages of the upper and the lower tubes (6AS7G's) . And wether this is a loudspeaker or the primary of a output transformer doesn't matter.
In a parafeed circuit the output tube and a choke are in series ; from the connection between the two a capacitor is connected to the output transformer. Instead of the choke a tube can be used as Philips used to do in its 800 ohms amps. A capacitor was used to connect the tubes with the speakers. They also sold 800 ohms OT's so 8 ohm speakers could be used.
Opposing but equal currents through a choke result in zero voltage across the choke. It can very simply be tested. Take two low voltage powersupplies, Take a choke with a dc resistance of say 150 ohms. Connect the choke across supply no.1 ; now connect supply no.2 the opposite way. The supplies are independant of course. Each of the leads of the used pwr supplies must contain a resistor of equal value as a protection. Swith on one supply and measure the voltage across the choke. Now the other supply reversely connected with the choke is swithed on . Now measure the voltage across the choke. When the currents are the same the voltage across the choke will be zero. Both supplies will now deliver their maximum current.
I repeat my standpoint that my circuit is a simple SE one with only one difference - a floating circuit which is parallel on the OT creating zero volts across it.
In a circlotron as built by Ralf Karsten the two opposing cathode currents flow through the loudspeaker resulting in a zero voltage drop. Depending on the number of output tubes it can be 2A but as they are flowing in opposite directions the resulting voltage across the speaker is zero by varying the negative grid voltages of the upper and the lower tubes (6AS7G's) . And wether this is a loudspeaker or the primary of a output transformer doesn't matter.
In a parafeed circuit the output tube and a choke are in series ; from the connection between the two a capacitor is connected to the output transformer. Instead of the choke a tube can be used as Philips used to do in its 800 ohms amps. A capacitor was used to connect the tubes with the speakers. They also sold 800 ohms OT's so 8 ohm speakers could be used.
Opposing but equal currents through a choke result in zero voltage across the choke. It can very simply be tested. Take two low voltage powersupplies, Take a choke with a dc resistance of say 150 ohms. Connect the choke across supply no.1 ; now connect supply no.2 the opposite way. The supplies are independant of course. Each of the leads of the used pwr supplies must contain a resistor of equal value as a protection. Swith on one supply and measure the voltage across the choke. Now the other supply reversely connected with the choke is swithed on . Now measure the voltage across the choke. When the currents are the same the voltage across the choke will be zero. Both supplies will now deliver their maximum current.
I repeat my standpoint that my circuit is a simple SE one with only one difference - a floating circuit which is parallel on the OT creating zero volts across it.
Kees, it is not that simple. If the CCS part would be active then your idea is correct. But now it acts like a load only and will give a lower R with less current throu the primary when in rest. If a signal is applied the current will not be equal anymore as the current of the CCS will not be able to keep increacing. Then the empedance will increase sudddenly for the VT4 tubes and the tubes will draw all the current. The same as in a class B push pull amp will happen. The distortion will go up rapidly. Plus the load match is lost. As long as you keep in class A (low power) you will be fine. Maybe removing 1 VT4 and increase the ability of the CCS will increase efficiency if you care about it.
I've thought about increasing the dissipation of the ccs (same current, higher voltage) to have more usable output power. My main interest has been to have enough power to fill the room with (only a few watts), not max. possible power which would require a ccs dissipating the same amount of power as the power tubes (and at high voltage of course). I thought the funny thing about this arrangement would be that you can choose for yourself how much undistorted power you would want, calculate the appropiate voltage and dissipation of the ccs and enjoy the VT4 sound. I'm not interested in squeezing every drop of power out of these tubes and apart from this, simply using what you have at hand. I have to admit I was too optimistic the moment I published this but I'm glad I did because of the insightful information given by you and others who have reacted on this thread.
So, we have the new special circuit that can do SE with a push pull transformer. Then, we have parafeed SE with a push pull transformer. But you also have . . .
A 6.6k push pull transformer (a little bit too low of an impedance for a VT4 / 211 push pull amp. But try it anyway. Put B+ on the center tap. It would actually work, you would get some power, and you only need a phase splitter and driver stage(s) for that.
Every extra milliamp, and every lesser milliamp from the other VT4 will go to drive the OPT.
In class A push pull (not AB), each tube would effectively be in parallel with 6.6k/2. So each tube would see 3.3k load. As you start to approach AB, one tube rp would be increasing, but the other tube rp would be decreasing.
You have 3 alternatives to use the push pull transformer. Pick one. Go.
Nuff said?
A 6.6k push pull transformer (a little bit too low of an impedance for a VT4 / 211 push pull amp. But try it anyway. Put B+ on the center tap. It would actually work, you would get some power, and you only need a phase splitter and driver stage(s) for that.
Every extra milliamp, and every lesser milliamp from the other VT4 will go to drive the OPT.
In class A push pull (not AB), each tube would effectively be in parallel with 6.6k/2. So each tube would see 3.3k load. As you start to approach AB, one tube rp would be increasing, but the other tube rp would be decreasing.
You have 3 alternatives to use the push pull transformer. Pick one. Go.
Nuff said?
Yes, if power is needed and single ended not a requirement. If so then just omit the phase inverter and have 1 VT4 as current ballancer. In this case you would stil have more power output and no need for an additional CCS. That simpifies the amp and is stil single ended. Anyway i like the original idea also because it stirred things up a bit (food for thought so to say). Thanks for sharing, Kees!
- Status
- This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.