I love the sweetness of my 45 SE amplifier, but you know what? A great push-pull (PP) amp has a fantastic presence, bass and dynamic response. Whenever I listen to a good PP amp, I get to the conclusion that I need to have different amps ready to be played depending to the music I want to listen to! My last two years have been devoted to what Morgan Jones calls in his book “single-ended madness”. And yes, my 4-65a SE in class A2 is slowly coming to life and when ready so then I will be properly mad.
6C4C amps I listened so far made a great impression both in SE and in PP. Owning all components required, I embarked on refining a full DHT push-pull design and again, cap-less (excluding the power supplies of course).
Two pivotal components in the design are the OT and the IT. I have a pair of LL1682PP and an LL1660/20mA which can be used as phase splitter in Alt-V mode. The driver is the beloved 4P1L in filament bias which is powered from a CCS and VR array which provides isolation to the raw power supply, being this a great thing to keep the signal path around the VR:
The 4P1L is biased at 20mA and 210V which provide sufficient headroom to drive the output stage. A 6e5P option may be attractive as well in this position.
The output stage has a hybrid bias arrangement. It’s mainly fixed bias using a Rod Coleman regulator which generates a stable and quiet reference voltage across a resistor (R6) and the offset balance adjustment between valves is provided by small feedback provided by two 5 ohm filament resistors. This will help us compensating different valves.
I looked at various bias points for 6C4C and I found about 300V/50mA being quite good in terms of reduction of H3 component. Higher bias points will increase the H3 level significantly.
The LL1682 is an 8K8:8 transformer so each valve will see Zaa/2 or 4K4:
Now, how this circuit is expected to perform:
The distortion pattern is as expected in an PP amplifier. The odd harmonics ratio is high nearly to 50%. THD is about 0.22% for 4W of output power. The driving signal at full tilt is about 130Vpp and you will need to use this amp with a preamp to provide at least 16-18Vpp input:
The supply current is 136mA average due to the 100mA bias current of the two 6C4C plus the driver CCS (36mA). You can see that the peak current goes up to 156mA in each cycle as a normal operating mode in a PP amp. Each valve will suck additional current whilst the other will use less in counter-phase. The bias point is responsible for the 20mA difference in the peak that the supply needs to provide.
A promising design which will need testing as usual…
Hi Ale,
nice idea! 2A3 types in push-pull perform admirably.
However 16-18Vpp input looks like a booster rather than a power amp. How would it look if you made it as true integrated amp? I mean one like the following:
1)output stage identical
2)driver in push-pull with 2x4p1l + LL1635/5mA (the little gap will be a bonus for little unbalance + class A2 or AB2 drive and still L=130H)
3)input with 6J9P-E + LL1660/10 mA run at 150V/10mA (about -2V bias which you could get with a led)
If Idid the right maths the input signal should be around 60 mV for 1W output and about 0.22 V for a bit more than 13W ( just above12W considering about 0.35-0.4 dB insertion loss of the LL1682) with 140V pp drive.
Still not sure about how Lundahl specify primary loads. Maybe their turn ratios are approximated. Anyway, considering 32, then the reflected impedance from 8R wuold be 8192 R + 210 R (primary DC) + 256 R (secondary DC of 0.25Rx(32)^2) = 8.66K.
Cheers
Hi 45, as always great feedback and ideas!
I designed it with the components and system I have at hand. I do have a preamp but I may need a tad of more gain though.
The 4P1L diff pair option is attractive. I do have a LL1660/pp with a 5mA gap to allow for some current unbalance. May be I can try this option.
Input stage with the 6J9P-E is also good!
Regarding the OT model I just simply estimated the Lp to be around 100H per primary winding as there is no detail provided by Lundahl. The LL1682 is 5K5:5 or N^2=1100. I calculated the Ls based on this turn ratio. Since N=33.17 approx, then you are right to point out that the reflected load is 8R*1100+210R+0.25R*1100=9,285 ohms.
Thanks
Ale
P.S.
Ideally the PP shouldn’t have that dominant H2. So you THD should be even lower if you can find the source….
Maybe this is caused by that current source to bias the 6C4C.
I would try the bias circuit in the following link which is specifically thought for grid current flow.
http://softone.a.la9.jp/english/2a3/2a3amp.htm
Cheers
I believe this is the driver H2 component which is not cancelled by the output stage? I remember simulating it with a simple voltage source in the secondary CT and getting same harmonic profile.
I will run the simulation again and look at the distortion of the driver.
Thanks for the link and will read this.
For A2 operation in push pull, what topology would you recommend?
thanks
Ale
For the LL1682 the specs are quite close at 5586 R (turn ratio of 32 which gives 5120 R with 5R load + 466 R total DC resistance). Its real turn ratio might be a bit lower, for example 31.85.
The LL1660 PP/5mA is good as well. You might be right about the H2 as I see that the amp in the link above behaves just like yours. I have never tried the SE drive+IT for PP output stage. Usually I can see dominant H2 only at lower output with PP driver. Anyway is not an issue being the total THD quite low!
The SE driver will work if well designed but I am not sure about the bias of the output stage because if you want to swing into positive grid then I can only see the grid current going through the 56K resistor and that will modulate the bias enough to reduce the efficiency and thus the output power.
The 3 stage amp I was thinking of was just to get more sensitivity to be driven from any source, including MC phono stages which usually have 200 mV or lower output. I usually consider 350 mV as average output from CD players. Especially with Classical music recording levels can be not so high to allow more dynamics…..
P.S.
The 6C4C has a lower gain in comparison to the 2A3. I remember that I got about 58V at 300V/50mA. So you might need some more swing to get those 13W I mentioned in the previous post.
Hello Dear Ale,
Just my 2c, you state that “The LL1682 is an 8K8:8 transformer so each valve will see Zaa/2 or 4K4” which is not at all the case in PP. Actually, each 6C4 sees 1/4 of the plate to plate impedance which is 8k8/4=2k2 per tube.
Thank you!
Hi Panos29,
I have to say that I disagree with you here. Check Menno Van Der Veen’s “High-end Valve Amplifiers 2”, “The triode construction of B.J. Thomson”, Page 73.
The Zaa/4 is dI * dVa1. Both triodes in class A will ‘see’ same impedance and this is Za1=Za2=dVa1/(0.5*dI) = 2dVa1/dI or Za1=Za2=Zaa/2.
The total current dI through the whole of the primary winding ‘sees’ 1/4 Zaa.
Hope this is clear?
Ale
Check Steve Bench at http://diyaudioprojects.com/mirror/members.aol.com/sbench102/composit.html
Push Pull Load Line
“The same rules we used in the single ended case apply, with one exception: The impedance you plot is the impedance AS SEEN BY A SINGLE TUBE. Thus it is 1/4 the plate to plate load impedance. Lets use a 8800 ohm plate to plate load. In this case, each tube sees 2200 ohm load, and one “point” on the loadline is the 300 volt “0” mA quiescent value. Another convenient point is at “0” volts (namely, 300 volts drop across the 2200 ohm load). By ohms law, this is i = e/r = 300/2200 = 136 mA.”
Precisely. The impedance seen by the end to end current is Zaa/4, however each individual valve sees half of the primary as the HT is a short to ground from an impedance point of view. Hence the load per valve is Zaa/2. Steve, as well as Thomson use the Zaa/4 to calculate the output power of the push pull system.
To be correct you need to draw the composite valve. Doing this you will get a device which has 1/2 Rp and sees 1/4 of the plate-to-plate load. Drawing the composite valve will automatically account for class A and AB as the curves are summed point to point starting from the quiescent conditions. So different quiescent conditions will result in different curves!
Indeed. I built the composite curves in the updated post here:
http://www.bartola.co.uk/valves/2013/04/28/push-pull-fascination-part-2/