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 1st September 2007, 09:36 PM #1 cbutterworth   diyAudio Member   Join Date: Oct 2006 B+ with or without tubes? I am often asked at what voltage I run my Aikido. I usually answer around 290V. However, I was messing around with the PSU today and measured B+ without the tubes installed. I know that I have done this in the past, but I didn't really take much notice. Anyway, B+ read 330V. Which is the voltage to report when asked about B+? I expect that it would be 330V, as that is the voltage presented to my tubes, which when current is drawn gets reduced to the 290V. Also, in order to estimate current draw, I measured the circuit resistance from B+ in on the PCB to GND. I read 16.9K. When I feed this into Ohm's Law for 330V, I get 19.5mA. Does this calculation represent the actual current drawn by the tubes? Or do I need to figure other things? At least this figure seems close to what Broskie reports for 6SN7s. Charlie
 1st September 2007, 10:03 PM #2 Brett diyAudio Member     Join Date: Jan 2002 290V loaded, 330V unloaded.
 1st September 2007, 10:15 PM #3 HollowState   diyAudio Member     Join Date: Aug 2006 Location: Taxland, New Jersey If asked, I would say 290 volts since that is the actual level. Others should realize that the source needs to be higher to accommodate the loading effect. Measuring static DC circuit resistance will not reflect tube current which is determined much more by voltages applied to grid(s), plate etc. as well as the load it sees. B+ to ground only measures all stray paths including filter capacitor leakage, if any, which could increase as voltage across them increases. It was by coincidence that you came close. Victor __________________ 1944: 18 year olds storm the beach of Normandy into almost certain death. 2017: 18 year olds need a safe place because words hurt their feelings.
 1st September 2007, 11:38 PM #4 cbutterworth   diyAudio Member   Join Date: Oct 2006 So, 290V it is! How do I calculate tube current? I took a look in the Morgan Jones book, but he doesn't really go into detail on the operation of tubes other than discuss the pros and cons of different design topologies. I expect that the 290V equates to 145V to each side of the tube. Then I expect that I need to use the values for cathode resistor, etc alongside charts for the tubes. Would someone be willing to write-up a step-by-step guide to doing this that could be left as a "sticky" topic for reference use? I have another question, so a new topic..... Charlie
 1st September 2007, 11:51 PM #5 kevinkr   diyAudio Moderator     Join Date: Sep 2004 Location: Boston, Massachusetts You can calculate the cathode current for one tube in each stage by measuring the voltage drop across a convenient cathode resistor in that stage. You only need to do it on one channel and multiply by 2 for stereo or you can measure both channels and get a small sense of possible parametric variations depending on the tubes used. __________________ "To argue with a person who has renounced the use of reason is like administering medicine to the dead." - Thomas Paine
 1st September 2007, 11:55 PM #6 cbutterworth   diyAudio Member   Join Date: Oct 2006 Kevin, So, I simply put the negative probe to PSU GND and the positive to one side of the cathode resistor, take note, then put +ve probe to the other side and take note? My GND from the PCB is lifted by 20R, so I should use the PSU GND (on final cap) rather than IEC/chassis GND? Charlie
 2nd September 2007, 12:23 AM #7 kevinkr   diyAudio Moderator     Join Date: Sep 2004 Location: Boston, Massachusetts Put the meter directly across the cathode resistor to measure the voltage across it, simpler as long as you are careful about it. __________________ "To argue with a person who has renounced the use of reason is like administering medicine to the dead." - Thomas Paine
 2nd September 2007, 12:43 AM #8 cbutterworth   diyAudio Member   Join Date: Oct 2006 Success! I simply reference the GND to PSU GND and probed both sides of the cathode resistor. Broskie recommends running the output tubes hotter than the input for the Aikido. Input tube = 4V drop over a 1K resistor = 4mA Output tube = 3.3V drop over a 470R resistor = 7mA So...a total current draw (assumming both L and R sections are equal) of 22mA. That was EASY! Charlie

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