12AT7 Phase splitter

[Mikaeel]

Hello!
I understand tubes/valves are not ideal. But, my experience is beyond this problem.
The following splitter I calculated has near unity gain amplification ratio.



The only problem is that the amplitudes of the signals do not match:


Lower half:



I know it's a noob question, but is there any way to minimise these differences?

 

[ruffrecords]

This is because the two output impedances are different and the loads are too heavy. Change both loads (R8 and R5) from 10K to 100K and see what happens.

Cheers

Ian

 

[Matador]

https://valvewizard.co.uk/cathodyne.pdf

I think the bottom phase will always be ~5% less due to the extra 400 ohms, yes?

Notice in Merlin's explanation, the gain equation assumes that the load impedance is >> the product of uZ (mu times the anode/cathode impedances), however you have a 10k load which means the gain will be impacted.

 

[Mikaeel]

Change both loads (R8 and R5) from 10K to 100K and see what happens.

Unfortunately not helps:



Only the DC component on outputs raises:

Notice in Merlin's explanation, the gain equation assumes that the load impedance is >> the product of uZ (mu times the anode/cathode impedances), however you have a 10k load which means the gain will be impacted.

I'f understand it right, I'll be unable to have 10k output impedance...
If my calculations are correct, with Ra and Rk = 50k , Zout-anode=8.33kΩ and Zout-cathode= 819.6 ohm....
I ommited the additional 400 ohms but the difference raises too:

 

[ruffrecords]

Some of this may be simulation artefacts. I notice the waveforms are not symmetric which they should be if capacitor coupled. Can you post the .asc file

Cheers

Ian

 

[Mikaeel]

Thanks for the help Ian,

Please see attached zip archive.

 

[ruffrecords]

OK, I realise now that your plate and cathode resistors of 50K and your load resistors are 10K. This is the sorce of the problem and increasing the loads to 50K will not help much. But increasing them to 470K does pretty much fix the problem.

I also changed your trans sim to 50mS duration and 1uS step size as this give a much more readable display. I attach the modified .asc file (with a .txt extension). If you run the trans sim and plot V(out+) + V(out-) you get a plot of the difference between the two waveforms. I get about 80uV which seems close enough to me.

Cheers

ian

 

[Mikaeel]

Thanks Ian!

This is the sorce of the problem and increasing the loads to 50K will not help much. But increasing them to 470K does pretty much fix the problem.

Only what I noticed, if we increase duration to 200ms, the difference keeps raising again to 6mv.

On my first screenshot I used additional 400 ohm resistor, to match the level of the signal that comes from cathode. It gives the same difference of 6mv, but I'm afraid such load may make tube get too hot, and will such a stage be safe to use. Could you pllease check attached file below:

 

[merlin]

Your imbalance has nothing to do with the loading. The balance of the inverter is *perfect* as long as the loads are identical; Ohm's law guarantees it. The apparent imbalance is an illusion caused by the coupling caps charging from time zero. If you run the simulation long enough, balance will settle at perfection.