Go Back   Home > Forums > >
Home Forums Rules Articles diyAudio Store Blogs Gallery Wiki Register Donations FAQ Calendar Search Today's Posts Mark Forums Read

Tubes / Valves All about our sweet vacuum tubes :) Threads about Musical Instrument Amps of all kinds should be in the Instruments & Amps forum

Pentode Loadlines PP
Pentode Loadlines PP
Please consider donating to help us continue to serve you.

Ads on/off / Custom Title / More PMs / More album space / Advanced printing & mass image saving
Reply
 
Thread Tools Search this Thread
Old 13th July 2018, 03:21 PM   #1
mashaffer is offline mashaffer  United States
diyAudio Member
 
Join Date: Jun 2004
Location: Indiana
Default Pentode Loadlines PP

Lack of spice models is forcing me to design the old fashion way via. loadlines (this is a really good thing kind of like eating my veggies ). It seems that I ought to be able to use the published curves without the composite curves. If I am going astray at any point please slap me upside the head. Using the following curves...

6HZ8_PLoadlines.png

I like the 5K 20mA (green line) as it gets close to the knee. In Class A this would be a 10K P-P if I understand correctly. However since 20mA is a Class AB bias point that complicates things. As long as both tubes are conducting this loadline should be correct however when one shuts off the 2x relationship becomes 4x does it not? Does this mean that the 10K xformer now behaves as a 20k? How would I represent this on the curves? Would I draw a line with the 10k slope (20k P-P) from the point on the 5k loadline where the opposing tube cuts off over to the Y axis? Do I want this Y intercept to approach the knee (meaning my 10k P-P is the wrong choice)?

Apologies if this ought to be obvious.
__________________
The only reason I drive the 91 Colony Park is that I can't afford a 65. Me old school? Couldn't be.
  Reply With Quote
Old 13th July 2018, 03:55 PM   #2
jazbo8 is offline jazbo8
diyAudio Moderator
 
jazbo8's Avatar
 
Join Date: Jan 2011
Location: In Transient
Pentode Loadlines PP
You need a refresher on drawing load lines: The Valve Wizard -Push-Pull
  Reply With Quote
Old 13th July 2018, 04:03 PM   #3
smoking-amp is offline smoking-amp  United States
diyAudio Member
 
smoking-amp's Avatar
 
Join Date: Dec 2001
Location: Hickory, NC
When one tube shuts off the load becomes 1/2 the Ohms for the remaining tube.

With both tubes conducting, the load of X Ohms can be seen as two 2X Ohm loads in parallel, so each 2X Ohms seen by each tube.

So class AB takes a higher Zpri OT than class A. But on balance, for class AB, when the class B load-line sections are entered, the tube has maximized gm to handle that heavier load (from higher current). At twice the tube current, a typical square law power tube has 1.4 times the gm1. So a more reasonable OT adjustment from class A would be 2/1.4 X or 1.4 X Ohms of the class A OT case.

Note:
Tube grid2 is typically near 3/2 power law. Power tube grid1 is typically near square, 2.0, power law.
(Ip = k2 Vg2^3/2 and Ip = k1 Vg1^2 ). Then gm varies as the 1st derivative, so gm2 = k3 Vg2^1/2 and gm1 = k4 Vg1^1)

Just for laughs, I recently found that the 19 Watt 6JC5 tube has a cubic, 3.0, power law for grid 1, so its gm would be the square law of Vg1. (grid 1 is placed too near to the cathode on these) (superficially looks like the 6V6 tube for specs, but very different grid 1 characteristic)
The 6JC5 tubes are on sale on Ebay currently for $0.75, a "bargain" I guess! (use -plenty- of N Fdbk)
Have no idea why these were designed for Vertical Sweep tubes, except maybe for some weird off center CRT electron gun (side scan).

Last edited by smoking-amp; 13th July 2018 at 04:27 PM.
  Reply With Quote
Old 13th July 2018, 07:16 PM   #4
6A3sUMMER is offline 6A3sUMMER  United States
diyAudio Member
 
Join Date: Jun 2016
1. Consider a 10k Ohm plate to plate transformer. Using the whole plate to plate winding, it is 10k Ohms. As mentioned earlier in this thread, using 1/2 of the total plate windings (plate to the center tap) it is 2.5k Ohms (Z is proportional to the square of 1/2 of the turns). In class A push pull, the plates work in conjunction, so each plate effectively sees 5k Ohms.

2. Consider a single ended amp, with an air gapped transformer that is 5k Ohms. I will use voltage numbers that may not be real, but the concept is real, and with this method you will be able to approximate the output power you would get with push pull on a class A amp. Do not forget to also calculate the reduction of the output power caused by the loss of the output transformer (i.e. might be 0.5 dB or 1 dB). This method only works if no tube goes into cutoff (it would now be class AB).

Taking a hypothetical set of tube curves, and applying a 5k load line.
Suppose the quiescent plate voltage is 300V, and the grid bias is -10V.
Suppose that with 0 Volts on the grid, the the plate voltage is 100V.
Suppose that with -20V on the grid, the plate voltage is 400V.

The plate moved from 300V to 100V, a 200V change.
(200V squared)/(5k Ohm * 2) = 4 Watts rms
The plate moved from 300Vk to 400V, a 100V change.
(100V squared)/5k Ohm * 2) = 1 Watt rms

This would be extremely distorted as single ended. But for 10k plate to plate, and class A push pull we get approximately: 4 Watts + 1 Watt = 5 Watts rms, with very little 2nd harmonic distortion; and the 3rd harmonic distortion would be dominant.

Last edited by 6A3sUMMER; 13th July 2018 at 07:23 PM.
  Reply With Quote
Old 13th July 2018, 07:22 PM   #5
kward is offline kward  United States
diyAudio Member
 
Join Date: Sep 2013
Location: Western USA
Or in other words, when both tubes are conducting, each tube sees 1/2 of the end to end impedance. But when one tube shuts off, the other tube sees 1/4 of the end to end primary impedance.

So there are two load lines to draw in an AB amp: the load line when both tubes are conducting, and the load line when only one tube is conducting. The slope of the load line changes based on whether one or both tubes are conducting.

Last edited by kward; 13th July 2018 at 07:34 PM.
  Reply With Quote
Old 13th July 2018, 07:38 PM   #6
6A3sUMMER is offline 6A3sUMMER  United States
diyAudio Member
 
Join Date: Jun 2016
kward,

Yes, good point! You teach me a better way to look at that. I like the idea of the second load line. But . . . There is not a single exact point where one tube is cut off. It is gradual. Is it cutoff at 10mA, 1mA, 100uA, etc.? For push pull, you might say the one tube is cut off when it is perhaps 1/4, or 1/10 of its quiescent current. That is because at the same time, the other tube's plate current may be 1.4 times or 2 times its quiescent current.

For Pentode operation, it is when the dictated plate current is no longer effectively controlled by the grid voltage (you see that on the family of curves).

For Triode operation, you might consider when the slope of the curves (plate resistance) is much lower (higher plate resistance), than the quiescent plate resistance. At the same time the other tube's plate resistance is now lower than it's quiescent plate resistance.

Where does class A quit, and where does class AB begin? This is an analog world, not a digital world. "Some things nearly so, others nearly not" (King of Siam?).

Last edited by 6A3sUMMER; 13th July 2018 at 07:44 PM.
  Reply With Quote
Old 13th July 2018, 08:50 PM   #7
mashaffer is offline mashaffer  United States
diyAudio Member
 
Join Date: Jun 2004
Location: Indiana
Thanks guys. Jazbo, I should have known that Merlin could make it clear.
__________________
The only reason I drive the 91 Colony Park is that I can't afford a 65. Me old school? Couldn't be.
  Reply With Quote
Old 14th July 2018, 01:08 PM   #8
mashaffer is offline mashaffer  United States
diyAudio Member
 
Join Date: Jun 2004
Location: Indiana
Naturally my talent for getting things exactly backwards showed itself again. Using Merlin's method (assuming I didn't get that turned about) I get the dark blue line for 10 P-P 300V 20mA bias. If so it looks reasonable.

6HZ8PentodeLoadlines.png

Then the driver triode with 51K load and 560 Rk lands us at about Vg of 1.5V and Ia of 2.8 mA.

6HZ8TriodeLoadline.png

For each I look at the output voltage swing for a 1V p-p input. the gain of 50 on the triode seems a bit optimistic for a tube with mu of 70. of course there will be degeneration from the un-bypassed cathode resistor (planning for P-K feedback). The output stage swings about 40V for a 1V input. So the voltage gain from driver grid to output plate should be a bit less than 2000. If so then we want the feedback to drop the gain to somewhere around 200.

If it were an op-amp the feedback resistor would be 560*200=115k. So perhaps a 120K feedback resistor might be a reasonable starting point.
__________________
The only reason I drive the 91 Colony Park is that I can't afford a 65. Me old school? Couldn't be.
  Reply With Quote
Old 14th July 2018, 05:45 PM   #9
kward is offline kward  United States
diyAudio Member
 
Join Date: Sep 2013
Location: Western USA
Assuming the triode is driving an infinite load (which it is not), the load line you drew will have a gain of about 50x if cathode is fully bypassed, and about 30x if the cathode is not bypassed. Actual gains will be less than that depending on the load the stage is driving.

The gain reduction you wanted (from your quoted 2000x to 200x) is a factor of 10, or 20 dB. But if you bring the feedback back to the unbypassed cathode (where in that condition stage gain will be some amount less than 30x), you will need maybe 14 dB of feedback to achieve your targeted forward voltage gain, and that would require a feedback resistor of approximately 135KΩ.

But we're not done yet. If feedback is DC coupled, that condition will inject about 2 mA of additional current into the 560Ω cathode resistor, which will skew bias of the first stage by 1.2V or so. You would have to account for that in the first stage biasing. Alternatively you could look into AC coupling the feedback but that brings with it an R/C time constant that will need to be dealt with.
  Reply With Quote
Old 14th July 2018, 06:04 PM   #10
mashaffer is offline mashaffer  United States
diyAudio Member
 
Join Date: Jun 2004
Location: Indiana
Thank you Kward. My first inclination is to AC couple and adjust input caps to counter the rising low bass response. That way I can more easily fiddle with the Rfb to fine tune the reaults.
__________________
The only reason I drive the 91 Colony Park is that I can't afford a 65. Me old school? Couldn't be.
  Reply With Quote

Reply


Pentode Loadlines PPHide this!Advertise here!
Thread Tools Search this Thread
Search this Thread:

Advanced Search

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is On
HTML code is Off

Forum Jump

Similar Threads
Thread Thread Starter Forum Replies Last Post
Help with loadlines in p.s.e. gsd Tubes / Valves 3 26th July 2013 04:59 PM
Loadlines for Parallel SE outputs Richard Ellis Tubes / Valves 3 26th February 2013 03:36 AM
Steve Bench PP loadlines Klimon Tubes / Valves 0 8th November 2006 02:25 PM
loadlines for newbie. bloozestringer Tubes / Valves 6 4th September 2006 07:41 PM
I've put loadlines aside for a while... Mikael Abdellah Tubes / Valves 22 5th May 2003 07:17 PM


New To Site? Need Help?

All times are GMT. The time now is 05:42 PM.


Search Engine Optimisation provided by DragonByte SEO (Pro) - vBulletin Mods & Addons Copyright © 2018 DragonByte Technologies Ltd.
Resources saved on this page: MySQL 14.29%
vBulletin Optimisation provided by vB Optimise (Pro) - vBulletin Mods & Addons Copyright © 2018 DragonByte Technologies Ltd.
Copyright ©1999-2018 diyAudio
Wiki