I have a Dyna 70 clone, (it is a point to point wired version of Ned Carlson's circuit with the EF86 and the El34's triode wired) and I would like to convert it to a PP6B4g Amp.
I have read Lynn Olson's Amity article and would like to drive the output tubes with an ECC99, but rather than an input tranny I was wondering about 1/2 of another ECC99 wired as a phase splitter.
In his tubecad notes, J Broskie indicates that the Split Load Phase Splitter sets the standard for balance and suggests overcoming its limited voltage swing by placing it in the very first stage of the power amp.
The topography of the Dynacclone would work well since you would be bringing the input directly to the center dual triode and then on to the driver stage.
I would appreciate any thoughts on this idea since I have never seen it used before.
Pete
I have read Lynn Olson's Amity article and would like to drive the output tubes with an ECC99, but rather than an input tranny I was wondering about 1/2 of another ECC99 wired as a phase splitter.
In his tubecad notes, J Broskie indicates that the Split Load Phase Splitter sets the standard for balance and suggests overcoming its limited voltage swing by placing it in the very first stage of the power amp.
The topography of the Dynacclone would work well since you would be bringing the input directly to the center dual triode and then on to the driver stage.
I would appreciate any thoughts on this idea since I have never seen it used before.
Pete
If you're using a split-load phase splitter as the first stage, and since this has less than unity gain, you'll basically need another stage to have some sort of reasonable input sensitivity. Since there are now 2 phases, that means an additional two valves (or sections thereof), bringing the total to three. That means you can't use a single ECC99 for the entire input/driver section.
I think this was what rcavictim was eluding to, but I'm not sure so I'll say it anyway. You could get away with an amp with a very low input sensitivity if you had a preamp capable of large-ish voltage swings at its outputs (i.e. larger than the normal 2V or so).... but then that just becomes a three-stage arrangement anyway.
I think this was what rcavictim was eluding to, but I'm not sure so I'll say it anyway. You could get away with an amp with a very low input sensitivity if you had a preamp capable of large-ish voltage swings at its outputs (i.e. larger than the normal 2V or so).... but then that just becomes a three-stage arrangement anyway.
A cathodyne splitter does just that, it splits the phase. It does nothing else in terms of gain. (It also has limitations due to the high plate Zo vs. low cathode Zo, particularly against capacitance and thus frequency response, but I digress.) Thus putting it before or after the required gain stages makes little difference, except in the amplitude you expect it to produce.
So. Let's put it just before the output tubes. Gain comes first, as what amounts to a preamp stage. The output tubes get little drive, which may or may not be an issue - triodes have large miller capacitance, while pentodes escape that, with higher gain to boot. (Certainly, voltage output is not an issue, if you bias and supply it properly. It requires a lot of voltage, since it produces the peak to peak drive voltages simultaneously!)
Now let's put it at the front instead. It produces NO distortion ("none" is impossible, but seriously, I mean, almost total NFB here...). It splits phase and now a balanced amplifier stage is required, taking an extra tube (or two). Balance can be improved using a common cathode resistor, particularly a large one (or a CCS, i.e. LTP). This is important if HF balance is required, especially if a weak (low current) cathodyne is used (assuming triodes (high miller C) for the gain stage, which is pretty much certain). The output tubes are driven well, depending on the driver tubes. But you use two tubes instead of one.
For high gain situations, a cross between the two is used - a preamp front end, usually direct coupled to the phase splitter, which then drives a balanced amp which drives the output. The basic Williamson topology. Unfortunately... it *really* sucks. You get, at best, three phase shift elements in your loop (two pairs of coupling caps and the OPT inductance), making for motorboat city if you aren't careful.
So where does that leave us? Well, a pentode has a lot of gain, so let's use one of those. LTPs are great because they have gain and produce an unconditionally balanced output, even when driven with an unbalanced signal. And we can stand the LTP on top of a tube CCS, allowing us to direct couple the pentode to the LTP and eliminate a coupling cap from the loop. Wow, it suddenly works! And with only *TWO* tubes, at that.
(Well okay, the attached doesn't show *quite* that. You could use the 6U8 triode for a CCS instead, if you do it right. It might cost some extra voltage though.)
Tim
So. Let's put it just before the output tubes. Gain comes first, as what amounts to a preamp stage. The output tubes get little drive, which may or may not be an issue - triodes have large miller capacitance, while pentodes escape that, with higher gain to boot. (Certainly, voltage output is not an issue, if you bias and supply it properly. It requires a lot of voltage, since it produces the peak to peak drive voltages simultaneously!)
Now let's put it at the front instead. It produces NO distortion ("none" is impossible, but seriously, I mean, almost total NFB here...). It splits phase and now a balanced amplifier stage is required, taking an extra tube (or two). Balance can be improved using a common cathode resistor, particularly a large one (or a CCS, i.e. LTP). This is important if HF balance is required, especially if a weak (low current) cathodyne is used (assuming triodes (high miller C) for the gain stage, which is pretty much certain). The output tubes are driven well, depending on the driver tubes. But you use two tubes instead of one.
For high gain situations, a cross between the two is used - a preamp front end, usually direct coupled to the phase splitter, which then drives a balanced amp which drives the output. The basic Williamson topology. Unfortunately... it *really* sucks. You get, at best, three phase shift elements in your loop (two pairs of coupling caps and the OPT inductance), making for motorboat city if you aren't careful.
So where does that leave us? Well, a pentode has a lot of gain, so let's use one of those. LTPs are great because they have gain and produce an unconditionally balanced output, even when driven with an unbalanced signal. And we can stand the LTP on top of a tube CCS, allowing us to direct couple the pentode to the LTP and eliminate a coupling cap from the loop. Wow, it suddenly works! And with only *TWO* tubes, at that.
(Well okay, the attached doesn't show *quite* that. You could use the 6U8 triode for a CCS instead, if you do it right. It might cost some extra voltage though.)
Tim
Attachments
I thought that this myth had been exposed enough in this and other forums but maybe not.
The output impedances at anode and cathode of a split load phase inverter are equal given that cathode and anode are equally loaded. This has been mathematically described by several earlier and it is easy to convince one self about this by either measuring output impedance on a real circuit or by doing spice simulations.
Regards Hans
Hi,
Sure but while that would give a (more or less) equal output from the cathodyne I fail to see how that would make for an equal Zout at both ends however.
No electronics engineering book I have seems to mention this. I recall we went through this some time ago but I still fail to see to logic....
Surely resistively loading plate and cathode equally doesn't make for equal Zout at both ends as they never are. Hence "assumption", perhaps ...?
Cheers,
Sure but while that would give a (more or less) equal output from the cathodyne I fail to see how that would make for an equal Zout at both ends however.
No electronics engineering book I have seems to mention this. I recall we went through this some time ago but I still fail to see to logic....
Surely resistively loading plate and cathode equally doesn't make for equal Zout at both ends as they never are. Hence "assumption", perhaps ...?
Cheers,
It is important to realise that output impedance is the ability of a circuit to drive a certain current trough a load at a certain output voltage drop, it is not really an impedance as such that exist without load or without signal. For a split load phase inverter we have to assume that both outputs are loaded, if not it doesn´t work as a phase inverter, if we now calculate the output impedance from each output with symmetrical loading we find that the output impedances are equal as well as the gain.
However it is of course so that a cathode grounded amplifier using equal cathode and anode resistors will have unequal output impedances when each output is loaded separately but then it is not working as a phase splitter.
Morgan Jones have in later editions of his book corrected the calculations for output impedance of the split load phase inverter and now write that they are equal, otherwise the most famous text about this issue is by Preisman http://www.aikenamps.com/cathodyne.pdf
The math in Preisman´s paper can be a bit difficult to grasp but an easy way to convince one self that he is indeed correct is to either measure output impedance on a real circuit or to do simulations in Spice, then you will find out that Preisman is correct.
Regards Hans
However it is of course so that a cathode grounded amplifier using equal cathode and anode resistors will have unequal output impedances when each output is loaded separately but then it is not working as a phase splitter.
Morgan Jones have in later editions of his book corrected the calculations for output impedance of the split load phase inverter and now write that they are equal, otherwise the most famous text about this issue is by Preisman http://www.aikenamps.com/cathodyne.pdf
The math in Preisman´s paper can be a bit difficult to grasp but an easy way to convince one self that he is indeed correct is to either measure output impedance on a real circuit or to do simulations in Spice, then you will find out that Preisman is correct.
Regards Hans
Hi,
O.K., thank you.
It was precisely because I was thinking it through with each side loaded separately that I ended up confusing the issue....
Reading through, you guessed it, the Preisman article.
Just to show that I'm still struggling with the concept; am I correct in think that ultimately Zout of the cathodyne splitter is limited by the inherent capability of the anode of the tube chosen for this duty?
Again thanks for the clarification,
O.K., thank you.
It was precisely because I was thinking it through with each side loaded separately that I ended up confusing the issue....
Reading through, you guessed it, the Preisman article.
Just to show that I'm still struggling with the concept; am I correct in think that ultimately Zout of the cathodyne splitter is limited by the inherent capability of the anode of the tube chosen for this duty?
Again thanks for the clarification,
Math blows. I'm lighting up a 12AU7.
+400V supply
330k to 100k grid bias voltage divider (approx. 1/4 = 95Vg out of 404V supply)
22k plate and cathode resistors
Single triode (first, of a 6189 to be exact)
Cathode voltage: 104V = 4.7mA and 9V grid bias, seems a lot but it might just be the high supply voltage. Fits the GE 12AU7 graph within 10%.
25Vrms 1kHz applied to grid via 0.1uF coupling capacitor.
Plate output: 21.2V, cathode: 21.6V. (Ripple negligible at <10%, 150mV to be exact.)
This proves middle frequency balance is good. IOW... the damn thing works.
25Vrms applied to cathode via 0.1uF capacitor and 100k series resistor (to make generator appear nearly constant current). 25V at signal generator, 0.42V at cathode. Voltage across 100k = 24.6 = 0.246mA AC, Vk / Igen = Zk = 1.7kohms (a bit on the high side for a cathode follower, but remember it's only half follower).
FWIW, Vp = 4.8V -- amplified as a cathode input amplifier with gain = 21.2dB.
25Vgen applied to plate with same hookup and conditions: 4.2V plate (0.24V at cathode, as a result of plate resistance changing Ip and thus Ik). 25 - 4.2 = 20.8V across 100k R = 0.21mA = 20kohms. Since the plate resistor (22k) is comparable to this, I will dig up the parallel resistor formula to find effective Rp of the tube:
R = R1*R2 / (R1+R2) --> R*R1 + R*R2 = R1*R2 --> R*R1 = R1*R2 - R*R2 --> R*R1 = R2*(R1 - R) and finally,
R2 = R*R1 / (R1 - R)
R1 = 22k, R = 20k (measured), R2 = (20k)*(22k) / (22k - 20k) = 220kohms plate resistance. This is far greater from the approx. 15kohm plate resistance of the tube with a bypassed cathode at this operating point, and indicates 23dB current NFB at the plate (which happens to correspond to the approximate gain reduction of the system).
Now for the on topic experiment: signal attenuation at high frequencies due to capacitance. I will test equal and unequal capacitors.
First: generator at 10kHz, 25Vp-p (since I can't read this frequency accurately with my voltmeter, I will read off the scope). Given the Zo noted above, I will choose capacitors approx. 10kohms reactance; 1nF = 0.001uF = 1000pF will suffice. This is equivalent to 100pF at 100kHz, so is quite reasonable.
Voltage at cathode: 23Vp-p. This is reasonable given the Zo of the cathode compared to the reactance (20% smaller). Voltage at plate: here's what you've all been waiting for: 22Vp-p. Slightly less, but within tolerances of this experiment.
Second: no capacitor at plate, 1nF at cathode. Same conditions. Vk = 23Vp-p, Vp = 33Vp-p. Plate gain increases because the cathode isn't as "soft", reducing NFB.
Third: no capacitor at cathode, 1nF at plate: Vp = 14, Vk = 22V. Cathode voltage remains constant because the capacitor never gave it a second thought in the first place, but Vp is lower due to its high Zo.
Fourth: absurd values, deep into the cutoff region in other words. Generator to 50kHz, 25Vp-p, and 2.7nF for plate and cathode. Cathode: 18Vp-p, with significant slope (discharge) limiting. A rounded sawtooth. Plate: 17Vp-p, inverse appearance.
Conclusions: due to the NFB and condition of it seen by each, plate and cathode Thevenin impedances are widely different. Cathode voltage is nearly immune, however plate voltage can be easily affected, independent of the cathode's actual voltage (due to cathode input amplifier behavior). Because the tube is a mostly one-way device (mostly because remember there was a little feedthrough from plate resistance), plate will not affect cathode or grid voltages.
In summary, cathode current unequal to plate's will cause a difference.
This can cause very interesting behavior, for instance when a cathodyne's cathode suddenly finds itself providing grid current, the plate does a wild bit of movement and your amplifier's output waveform grows a tit on top.
Tim
+400V supply
330k to 100k grid bias voltage divider (approx. 1/4 = 95Vg out of 404V supply)
22k plate and cathode resistors
Single triode (first, of a 6189 to be exact)
Cathode voltage: 104V = 4.7mA and 9V grid bias, seems a lot but it might just be the high supply voltage. Fits the GE 12AU7 graph within 10%.
25Vrms 1kHz applied to grid via 0.1uF coupling capacitor.
Plate output: 21.2V, cathode: 21.6V. (Ripple negligible at <10%, 150mV to be exact.)
This proves middle frequency balance is good. IOW... the damn thing works.
25Vrms applied to cathode via 0.1uF capacitor and 100k series resistor (to make generator appear nearly constant current). 25V at signal generator, 0.42V at cathode. Voltage across 100k = 24.6 = 0.246mA AC, Vk / Igen = Zk = 1.7kohms (a bit on the high side for a cathode follower, but remember it's only half follower).
FWIW, Vp = 4.8V -- amplified as a cathode input amplifier with gain = 21.2dB.
25Vgen applied to plate with same hookup and conditions: 4.2V plate (0.24V at cathode, as a result of plate resistance changing Ip and thus Ik). 25 - 4.2 = 20.8V across 100k R = 0.21mA = 20kohms. Since the plate resistor (22k) is comparable to this, I will dig up the parallel resistor formula to find effective Rp of the tube:
R = R1*R2 / (R1+R2) --> R*R1 + R*R2 = R1*R2 --> R*R1 = R1*R2 - R*R2 --> R*R1 = R2*(R1 - R) and finally,
R2 = R*R1 / (R1 - R)
R1 = 22k, R = 20k (measured), R2 = (20k)*(22k) / (22k - 20k) = 220kohms plate resistance. This is far greater from the approx. 15kohm plate resistance of the tube with a bypassed cathode at this operating point, and indicates 23dB current NFB at the plate (which happens to correspond to the approximate gain reduction of the system).
Now for the on topic experiment: signal attenuation at high frequencies due to capacitance. I will test equal and unequal capacitors.
First: generator at 10kHz, 25Vp-p (since I can't read this frequency accurately with my voltmeter, I will read off the scope). Given the Zo noted above, I will choose capacitors approx. 10kohms reactance; 1nF = 0.001uF = 1000pF will suffice. This is equivalent to 100pF at 100kHz, so is quite reasonable.
Voltage at cathode: 23Vp-p. This is reasonable given the Zo of the cathode compared to the reactance (20% smaller). Voltage at plate: here's what you've all been waiting for: 22Vp-p. Slightly less, but within tolerances of this experiment.
Second: no capacitor at plate, 1nF at cathode. Same conditions. Vk = 23Vp-p, Vp = 33Vp-p. Plate gain increases because the cathode isn't as "soft", reducing NFB.
Third: no capacitor at cathode, 1nF at plate: Vp = 14, Vk = 22V. Cathode voltage remains constant because the capacitor never gave it a second thought in the first place, but Vp is lower due to its high Zo.
Fourth: absurd values, deep into the cutoff region in other words. Generator to 50kHz, 25Vp-p, and 2.7nF for plate and cathode. Cathode: 18Vp-p, with significant slope (discharge) limiting. A rounded sawtooth. Plate: 17Vp-p, inverse appearance.
Conclusions: due to the NFB and condition of it seen by each, plate and cathode Thevenin impedances are widely different. Cathode voltage is nearly immune, however plate voltage can be easily affected, independent of the cathode's actual voltage (due to cathode input amplifier behavior). Because the tube is a mostly one-way device (mostly because remember there was a little feedthrough from plate resistance), plate will not affect cathode or grid voltages.
In summary, cathode current unequal to plate's will cause a difference.
This can cause very interesting behavior, for instance when a cathodyne's cathode suddenly finds itself providing grid current, the plate does a wild bit of movement and your amplifier's output waveform grows a tit on top.
Tim
Hi Tim,
You have done the same "mistake" as many others when you tried to measure the output impedances of the split load inverter, you measured the anode and cathode separately from each other i.e they where unequally loaded. The result you get is exactly the expected one that anode and cathode impedances are unequal.
If you have time you can do what I did when I measured output impedances, I connected different values, (but same for cathode and anode) of high and low load impedances and calculated the output impedance by measuring the difference in output voltage, this give the same value of output impedance as in Preismans article, actually much lower then rp, see equation 12 in Preismans article for actual value.
Another way of realising that the output impedances are equal and much lower then you would expect is to check the frequency response of the amplifier and determine the poles which you will find are located at much higher frequencies then you would expect given the output impedances you measured in your setup.
As you write you have to be careful when using a split load to directly drive power tubes if there is a risk that you go into grid current as the output impedances are affected by unequal loading but this behavior is well known and described by Preisman as well of by Morgan Jones.
There is an unbalance for very high frequencies as the loads will be unequal due to different capacitances but this can be fully compensated for and is not a problem at frequencies below a few MHz in most cases.
Regards Hans
You have done the same "mistake" as many others when you tried to measure the output impedances of the split load inverter, you measured the anode and cathode separately from each other i.e they where unequally loaded. The result you get is exactly the expected one that anode and cathode impedances are unequal.
If you have time you can do what I did when I measured output impedances, I connected different values, (but same for cathode and anode) of high and low load impedances and calculated the output impedance by measuring the difference in output voltage, this give the same value of output impedance as in Preismans article, actually much lower then rp, see equation 12 in Preismans article for actual value.
Another way of realising that the output impedances are equal and much lower then you would expect is to check the frequency response of the amplifier and determine the poles which you will find are located at much higher frequencies then you would expect given the output impedances you measured in your setup.
As you write you have to be careful when using a split load to directly drive power tubes if there is a risk that you go into grid current as the output impedances are affected by unequal loading but this behavior is well known and described by Preisman as well of by Morgan Jones.
There is an unbalance for very high frequencies as the loads will be unequal due to different capacitances but this can be fully compensated for and is not a problem at frequencies below a few MHz in most cases.
Regards Hans
fdegrove said:
Thus my hint! It doesn't matter WHAT the load is, as long as it's symmetrical, the frequency response at each side will be identical (Kirchoff's Law forces this)- so one is inevitably forced to conclude that the source impedences are the same.
The source resistance is Rr/([mu+2]R+r), where R is the load resistor value and r is the plate resistance.
Edit: Hans, I was about to respond to Tim- you saved me the trouble, thanks. Tim, go get two 10x probes and a two channel scope. You'll convince yourself quite quickly.
My thanks to all who contributed to this thread.
Based upon your comments I believe that I will seek another circuit configuration to convert my Dynaclone 70 to a 6B4G PP, probably something along the lines of the Ralph Power's, Stigla's Project or Chris Lillja's "leapfrog" amp published in Valve Mag.
Although things did not turn out as expected, I certainly learned a great deal about Phase Splitters and that's a plus.
Thanks again and Happy New Year!
Pete
Based upon your comments I believe that I will seek another circuit configuration to convert my Dynaclone 70 to a 6B4G PP, probably something along the lines of the Ralph Power's, Stigla's Project or Chris Lillja's "leapfrog" amp published in Valve Mag.
Although things did not turn out as expected, I certainly learned a great deal about Phase Splitters and that's a plus.
Thanks again and Happy New Year!
Pete
tubetvr said:
If this were a newsgroup you'd be plonked by now. You can't be convinced so nuts to you.
The whole point of Zo of plate and cathode is because, if for some reason, the impedances ARE mismatched, your output goes to pot. PERIOD.
With a circuit like this, where the cathode effects the plate, you can't see an unbalanced output without unbalancing the load.
You'll note I conducted the experiment with absurd values of capacitance, namely 1447 ohms reactance, for both plate and cathode, and the result was: GUESS WHAT, the voltages are EQUAL.
Sy: I have a dual channel scope but didn't feel like connecting the second channel because I, like everyone else, already know that the signals are equal and opposite (minus the unbalanced load situations).
I will leave it as "an exercise for the student" to determine the phase of the signal measured on the opposite terminal when testing Zo. Reading what I have, the results may suprise you...
Tim
Tim,
I must assume that you are not interested in knowing the output impedance of the split load inverter when it is working as a phase splitter.
Can you then tell me how you calculate the poles in the frequency response that comes from the phase splitter output impedances and output tubes input capacitances?
Regards Hans
I must assume that you are not interested in knowing the output impedance of the split load inverter when it is working as a phase splitter.
Can you then tell me how you calculate the poles in the frequency response that comes from the phase splitter output impedances and output tubes input capacitances?
Regards Hans
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