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MrCurwen
diyAudio Member

Join Date: Feb 2010
Location: Helsinki

I was just wondering, how could I calculate the plate load impedance of the attached stage?
Attached Images
 6SN7 driver.png (15.5 KB, 231 views)

 24th September 2010, 09:24 AM #2 SY   On Hiatus     Join Date: Oct 2002 Location: Chicagoland It will depend on the resistance at the other end of the line going to the right. The plate load contribution will be rp + (mu + 1)Rk, with rp and mu being the plate resistance and amplification factor for the top tube. This will be effectively in parallel with whatever load is off to the right. __________________ "You tell me whar a man gits his corn pone, en I'll tell you what his 'pinions is."
 24th September 2010, 10:31 AM #3 MrCurwen   diyAudio Member   Join Date: Feb 2010 Location: Helsinki It is direct coupled to a 1626 grid. How do I figure out the the load impedance of that?
 24th September 2010, 10:56 AM #4 SY   On Hiatus     Join Date: Oct 2002 Location: Chicagoland Is there a resistor to ground? If not, it's just the input capacitance of that stage. __________________ "You tell me whar a man gits his corn pone, en I'll tell you what his 'pinions is."
 24th September 2010, 01:11 PM #5 MrCurwen   diyAudio Member   Join Date: Feb 2010 Location: Helsinki Both triodes are 6SN7. Data quotes rp as 6.7k at 90V / 10mA and 7.7k at 250V / 9mA. Both triodes have about 150V across them with 9.6mA, so I'll guestimate that the rp is about 7.5 k. mu = 20 rp = 7500 rk = 312 (the bottom triode has a diode in there, but the voltage drop is equal to this resistance) rp + (mu + 1)Rk = 7500 + (20 + 1) x 312 = 14052 Now I must have misunderstood something, this can't be - with 14k plate load the DC conditions don't add up; at 9.6mA the voltage drop measured is about 150V, but calculated it's only 134.9V - and I've understood that this kind of arrangement has a higher AC load than DC load, at least it should be equal, right? Where did I go wrong?
SY
On Hiatus

Join Date: Oct 2002
Location: Chicagoland
Quote:
 Originally Posted by MrCurwen with 14k plate load the DC conditions don't add up; at 9.6mA the voltage drop measured is about 150V, but calculated it's only 134.9V - and I've understood that this kind of arrangement has a higher AC load than DC load, at least it should be equal, right? Where did I go wrong?
It's an AC load, not a DC load. But you've calculated it correctly; a triode run that way is an amazingly poor CCS. That's why people have added extra bits to run mu or beta followers. Or used MOSFETs...
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"You tell me whar a man gits his corn pone, en I'll tell you what his 'pinions is."

 24th September 2010, 03:34 PM #7 Bandersnatch   diyAudio Member   Join Date: Jul 2003 Location: Ann Arbor, MI hey-Hey!!!, One solution is to deliver a voltage reference to allow a significantly higher cathode R OTO 5-6k Ohms, which when multiplied by the mu of ~20 delivers some significant plate resistance. Or get a higher mu tube, or both, and/or stack a pentode on top of the triode to make a tube cascode...but that will take two biasing voltages... cheers, Douglas __________________ the Tnuctipun will return
MrCurwen
diyAudio Member

Join Date: Feb 2010
Location: Helsinki
Ok, so I put some extra bits in there; now it's a mu follower. How do I calculate the plate load for this? (Still direct coupled to 1626 grid.)

Sure does sound a lot better!
Attached Images
 mu follower.png (18.4 KB, 136 views)

 24th September 2010, 07:00 PM #9 DF96   diyAudio Member   Join Date: May 2007 Same formula as before. All you have done is separate the AC and DC arrangements for the upper triode. Rk is now 10.1K.
SY
On Hiatus

Join Date: Oct 2002
Location: Chicagoland
Quote:
 Originally Posted by DF96 Same formula as before. All you have done is separate the AC and DC arrangements for the upper triode. Rk is now 10.1K.
Yes, but that increases the source resistance. It's still mediocre, but 200k is better than 14k, and is close enough to a true CCS (in relation to the rp of the lower tube) that the gain will be very close to mu and the distortion very close to minimum.
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"You tell me whar a man gits his corn pone, en I'll tell you what his 'pinions is."

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