Changing from Split to Shared cathode what do I need to do?

Janglin_Jack · Post by **Janglin_Jack** » Sat Dec 09, 2006 4:17 am

I don't want to highjack the thread, but it sounds like you guys are interested what I was proposing. In some models, the cathode is shared 820 with some value bypass cap. In other models, there are split cathode 820 with 330uf bypass and 2.7K with .68uf bypass, (or whatever). My thought is instead of deciding between split or shared. Couldn't you leave the cathode split and just change the value of the 820 cathode to 1.6K to give you the shared value "tone". Or is there more to the shared/split cathode decision??

Jack

mightymike · Post by **mightymike** » Sat Dec 09, 2006 5:57 am

A 1.6 would put your total (if the channels were jumped) to 1004 Ohms

Because 2.7k and 1.6k in parallel would be 1004 Ohms.
So if you jumped the channels, and changed the v1a cathode from 820 to
1.6k, and left everything else the same you would end up with 1K/330.68uf.

The resistance you need on v1a would be 1177.66 ohm to equal 820, and the channels would have to be jumped.
If you replaced the 820 on v1a in a 69 circuit with a 1.2k, then 1.2k in parallel with the 2.7k would equal 830.77 Ohms, so that if you had the channel jumped with a foot switch, you would end up with 830/330.68uf split/jumped using both sides,
or 2.7k/330 split using only v1b when the foot switch was in the other position

The test would be how close 830/330.68uf split/jumped using both sides sounds the same as 820/330uf shared..

Here's some other values and what you would end up with:

If you changed v1a to

1.6k in parallel with 2.7k = 1004.65 ohms
1.5K in parallel with 2.7k = 964.28 ohms
1.4K in parallel with 2.7k = 921.95 ohms
1.2k in parallel with 2.7k = 830.77 ohms
1177.66 in parallel with 2.7k = 820 Ohms
1.1k in parallel with 2.7k = 781.58 OHM
1k in parallel with 2.7k = 729ohm ohms

Janglin_Jack · Post by **Janglin_Jack** » Sat Dec 09, 2006 4:47 pm

But if you ran straight into one input with the 1.6K cathode, (split) would it have the same tone as plugging into the shared cathode at 820?

Assuming that is true, that would give you the "shared" cathode sound in one set of inputs and the 2.7K/.68 in the other set of inputs.

Then I guess you would ge a variety of sound if you bridge the inputs.

Jack

mightymike · Post by **mightymike** » Mon Dec 11, 2006 3:56 am

Janglin_Jack wrote:But if you ran straight into one input with the 1.6K cathode, (split) would it have the same tone as plugging into the shared cathode at 820?

No

Shared is 820.

1.6k in a split setup won't become 820 if ran staright in. It would remain 1.6k, unless you jumpered the channel, then in that case it would be in parralell with what ever was on v1a, in which case 1.6k in parralell with the 820 on v1a would be less than 820ohm

SDM · Post by **SDM** » Mon Dec 11, 2006 5:54 am

Are you guys talking about jumpering the inputs? If so, jumpering the inputs of a split cathode amp does NOT put the cathodes in parallel. If you've got a 2.7K on one channel and an 820 on the other, then you'll get the 2.7K tone out of one and also seperately the 820 ohm tone out from the other, which you mix can then just mix as desired with the volumes if the inputs are jumpered. You do not get a (in this case 2.7K parallel with 820) 629 ohm shared cathode cathode result by jumpering inputs. If you meant jumpering the cathodes direct then nevermind

.

Janglin_Jack wrote:But if you ran straight into one input with the 1.6K cathode, (split) would it have the same tone as plugging into the shared cathode at 820?

Yes

, If I'm understanding you right here.
This does set the tube up to behave much as it would in a shared setup, though not neccessarily 100% exactly the same. The cathode resistor determines the bias point of the tube. When it's shared, the current for two tubes runs through it. If you want the same operating conditions to carry over to a split cathode for just one tube now, you know right of the bat that the current that will run through the now split cathode resistor will be half what of what it was before. You want to keep all else the same, so you need to double the resistance. So a shared 820 ohm tube stage will be biased the same as a split 1.64K bias stage (assuming nothing else changes). Ohm's law V=I*R.

For an example, say you measured the cathodes when a shared 820 ohm resistor was in there, and measured 1.4 Volts on the cathodes. From that, using Ohm's law you can figure out what the tubes are drawing current wise (how they are biased and thus are set up to operate). So you've got 1.4V dropped through a shared 820 ohm cathode resistor. In the formula that is 1.4=I*820, and that rearranged to find I (current) I=1.4/820 = .0017 =1.7ma. The cathode is shared, so that 1.7ma is both tubes current together. Each tube is contributing half of that (ideally, since the stages are identically set up), meaning each tube is biased at 1.7ma/2= .85ma. Okay so you want your split cathode tube to be biased at .85ma to keep it operating the same as it did in a shared cathode arrangement. The bias voltage doesn't change from before, so you still want 1.4volts on the the split cathode, and .85ma of current. Plug that into the formula you get 1.4=.00085*R, which rearranged to find R=1.4/.00085 =1647K = 1.6K .
Hope that made any sense.