Calculating SSE for KT88 and KT120

[cbutterworth]

OK, my SSE is alive and well with EH KT88's. So, I'm doing some math. Please could I get some feedback on my learning.

What I have measured:

1. Loaded DC from top of optional cap (B+) = 416 vdc from 115 vac mains.
2. Cathode voltage = 40 v.

What I have calculated:

1. Cathode current (40 v with 560 ohm resistor) = 72 mA rounded up.

This is where I get more unsure of my math and reasoning:

1. Voltage seen at the anode (416 v, 85 ohm DC resistance at the primaries of the opt, 3k anode resistance taken from tube data sheet) = 405 v.
2. Anode current (anode voltage and 3k anode resistance) = 135 mA.

Does all this math/reasoning seem OK?

If so, is it only the cathode current of the KT88's and 12AT7 that counts towards my power trafo' max DC current?


Finally, if that is the case, then my KT120's should drop right in because:

1. Anode resistance is same as KT88.
2. The cathode resistor will remain at 560 ohm.

I am itching to try the KT120's but do not want to exceed the 200mA DC current of my Edcor transformer. I have checked heater current limits and can drive the heaters of the KT120's without overloading the power trafo.

Most importantly, I do not want to damage my nice new SSE!

Thanks,
Charlie

 

[w5jag]

I use 20 ma as the rule of thumb for the 12AT7 draw. So, 20 + 72 + 72 = 164 ma. If your power transformer is rated for 200 ma, you don't have a lot of room left.

I have no idea how far past the limits, if any, you can run an Edcor power transformer. Anytime you run something in excess of its ratings, you have to be prepared to accept that there can be adverse consequences.

Win W5JAG