Making sense of the Aikido calculations?
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Burnedfingers
diyAudio Member

Join Date: Nov 2002
Making sense of the Aikido calculations?

Given the following information on this link I seem to still find myself scratching my head and wondering what the hell it all means. From the emails I have sent and received I surely am not alone here. In other words how do we borrow from this wonderfully designed circuit so we can input different tubes into the mix? Can someone give an example of another tube that could be used along with the calculations and how to do them? This I am sure will put an end to the mystery.

I will attempt to attach the Aikido circuit board schematic complete with resistor numbers ( the group buy board from our friend Bas)
Attached Images
 aikido 6n19 5687 resistor numbered.gif (9.3 KB, 1176 views)

 6th January 2006, 11:15 AM #2 Burnedfingers   diyAudio Member   Join Date: Nov 2002 Anyone want to take a shot at this?
 6th January 2006, 11:21 AM #3 SY   On Hiatus     Join Date: Oct 2002 Location: Chicagoland Sure, pick a tube. The only things that have to be determined are the cathode resistors and the ratio for the divider that does the noise injection. The latter is given as 1/mu + 1/2. The former we'll get from the tube characteristic curves. __________________ "You tell me whar a man gits his corn pone, en I'll tell you what his 'pinions is."
 6th January 2006, 11:27 AM #4 Burnedfingers   diyAudio Member   Join Date: Nov 2002 Sure, pick a tube. Ok, how about the wonderful 6SL7 or 5691 just for grins for the input tube and a 6sn7 for the second tube? Is there a way to get rid of excess gain? I picked the 6SL7 tube because I like its sonic value and would like to hear what it would sound like in this circuit. SY, would you please break this down so some of us can spoon feed? Thanks, Joe
 6th January 2006, 11:42 AM #5 SY   On Hiatus     Join Date: Oct 2002 Location: Chicagoland You've hit the first design problem- the gain of the input tube shouldn't be terribly high if you don't want to take steps which will increase distortion. And a high gain input tube will cause more design problems because of Millering. You might want to use a high gain input tube and end up with low gain, but hey, I want a hot night with Scarlett Johanssen and that's not going to happen, either. So let's assume you'll rethink that 6SL7 and let's now look at the second tube. A 6SN7 has a mu of 20. From the ratio formula, the proportion of the lower resistor is 1/mu + 1/2, which comes out to 0.55. So you want 55% of the resistance in the bottom resistor, 45% in the top resistor of the divider string. If we arbitrarily decide that the total string resistance will be 200K, then the bottom resistor (R11) should be about 110K, the top (R10) should be 90K. In our next step, we'll pick a current and set the cathode resistors of the 6SN7 accordingly. __________________ "You tell me whar a man gits his corn pone, en I'll tell you what his 'pinions is."
 6th January 2006, 12:24 PM #6 SY   On Hiatus     Join Date: Oct 2002 Location: Chicagoland OK, with a 6SN7 in the output stage, you're going to need some B+. 6SN7s like volts, lots of volts. You want at least 150V plate to cathode. So, we'll set the B+ at 300V. Let's pick a current. 6SN7s need 8mA or more to run in the linear part of their transfer characteristic; we shall choose 10, since it is the smallest two digit number I can think of. 10mA, 150V, what do the 6SN7 curves tell us? They tell us that we need about 3.5V between cathode and grid. That will be provided by the voltage drop across R8 and R9. 3.5V, 10mA... Ohm tells us that we need a 350 ohm resistor in each of those two spots. As an alternative, you could connect two red LEDs in series. Well, have you chosen a new candidate for the input tube? Another 6SN7 perhaps? __________________ "You tell me whar a man gits his corn pone, en I'll tell you what his 'pinions is."
 6th January 2006, 12:39 PM #7 Burnedfingers   diyAudio Member   Join Date: Nov 2002 Ok, how did we figure we needed 3.5 volts between cathode and grid? Do I find this on the Average Transfer Characteristics graph that shows plate current and grid voltage?
 6th January 2006, 01:18 PM #8 SY   On Hiatus     Join Date: Oct 2002 Location: Chicagoland You can get it from those curves. I used the plate current versus plate voltage curves. And I cheated a bit since I happen to know that operating point off the top of my head. __________________ "You tell me whar a man gits his corn pone, en I'll tell you what his 'pinions is."
 6th January 2006, 01:29 PM #9 Burnedfingers   diyAudio Member   Join Date: Nov 2002 Quote: You can get it from those curves. I used the plate current versus plate voltage curves. And I cheated a bit since I happen to know that operating point off the top of my head. I'm just trying to follow along. I can see that experience will make the difference between a so so circuit and one that performs nicely.
 6th January 2006, 01:41 PM #10 SY   On Hiatus     Join Date: Oct 2002 Location: Chicagoland Experience doesn't hurt, but a look at the tube data sheet would give you all this info. For reference, let's use the GE datasheet for the 6SN7GTB. The "Average Transfer Characteristics" curves tell you that low plate voltages are a no-no. The 50V curve has no linear region whatever. The 100V curve just starts to straighten out when we hit the grid current region. The 150V cuve shows a nice linear region past about 6 or 7 mA. Ten is a safe number which also keeps the plate dissipation down. Anyway, as you surmised, you can immediately see the required grid voltage to achieve the desired operating point. It's closer to 3 than 3.5, my bad. __________________ "You tell me whar a man gits his corn pone, en I'll tell you what his 'pinions is."

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