6N6P cascode calculation
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 11th September 2007, 07:09 AM #1 croccodillo   diyAudio Member     Join Date: Mar 2005 Location: Bergamo 6N6P cascode calculation Good Morning to all, I'm trying the following circuit: 6N6P Cascode The upper valve grid is set to about 170V, while the lower one works at about 160V. The circuit works, but has a too low gain, for my application. I know, from Morgan Jones book explanation, that the lower valve sees a very low load, since the upper tube has a very low internal, dynamic impedance. My question is: How can I calculate the load impedance for the lower valve, in manner to better understand how to change the circuit (maybe a different lower valve, or a parallel of two 6N6P triodes) to suit my needs? Thanks in advance, Giovanni Albergoni __________________ In Nomine Libertatis Vincula Edificamus, In Nomine Veritatis Mendacia Efferimus.
 11th September 2007, 03:45 PM #2 jimmyd53   diyAudio Member     Join Date: Jan 2006 Location: EUGENE, OREGON Good Morning Giovanni, Here is a copy of an email that Lukasz from the lampizator web site sent me about using the 6h6p triode in a parallel SE configuration. Hope it helps. James Hi, I never actually built it but I think Ewgennyi has a prototype. The 6H6P can give out even tens of miliamps. So just operate it in the highest possible operating point (say 160 V DC power supply, via primary of transformer - on anode. Cathode to ground. Separate circuit for bias - try minus 6,3 V adjustable with a pot. You can steal -6,3 V from the heater circuit and make minus 4 from it. Just make heaters with DC and ground the plus not minus. Makes no difference. then paralell as many tubes as you desire power. Of course - paralell the halves too. Set 15 mA per half tube. 30 mA per one paralelled triode. the only trouble is the speaker transformer. It has to be bought somewhere. Just remove it from old broken tube radios from 1950's or 60's. Yoy also need gigantic power supply for the heaters - 5 watts per tube. If you decide to use 4 tubes per channel - the heaters must be powered by 40 watts (maybe a 100VAC transformer) Regards and good luck. Lukasz
 11th September 2007, 04:56 PM #3 EC8010   diyAudio Moderator Emeritus     Join Date: Jan 2003 Location: Near London. UK The lower valve sees as its anode load: (RL + ra)/(u + 1) Where: RL = the load resistance in the anode circuit of the upper valve ra = anode resistance of the upper valve at the chosen operating point u = amplification factor of the upper valve at the chosen operating point 1 = one __________________ The loudspeaker: The only commercial Hi-Fi item where a disproportionate part of the budget isn't spent on the box. And the one where it would make a difference...

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