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Loadlines [message #9801] Thu, 31 August 2006 09:47 Go to next message
Shane is currently offline  Shane
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Registered: May 2009
Illuminati (3rd Degree)
I am learning how to do loadlines from MJ's book and Steve Bench's webpages and just want to make sure I'm understanding it.

Looking to draw a loadline for 6EM7 section 2. Steve states that the load (Z)=(max V - quiescent V)/quescient I. I'm looking at Ic=50mA and Vq=210V. Is the load the primary impedance for the OPT? So if I'm using a 5K OPT the max V on the loadline at 0 plate current would be 460V. Then draw a line from the 460V,0mA point through the 210V,50mA point and figure all the other parameters from there?

I refuse to build anyhting till I can go through and explain/understand why everything is what it is.

Thanks

Re: Loadlines [message #9802 is a reply to message #9801] Thu, 31 August 2006 10:23 Go to previous messageGo to next message
Manualblock is currently offline  Manualblock
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Registered: May 2009
Illuminati (13th Degree)
Good work Shane. Damir is on vacation but he gets back this week. So you bought Morgan Jones book? Is it worth it or is all that info on the net? I am very interested.

Re: Loadlines [message #9803 is a reply to message #9802] Thu, 31 August 2006 10:29 Go to previous messageGo to next message
Shane is currently offline  Shane
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Registered: May 2009
Illuminati (3rd Degree)
I have both of Mr. Jones's books, the 3rd edition Valve Amps and the newer Building Valve Amplifiers. I find them both worth every penny. Very informative. Unfortunately my math skills are lacking and I'm having to refresh the old noodle. I found that Steve Benches explanations on loadlines was a bit easier for me to understand, but MJ goes into much more detailed explanation of the "why" something is done.

Re: Loadlines [message #9804 is a reply to message #9803] Thu, 31 August 2006 20:37 Go to previous messageGo to next message
Manualblock is currently offline  Manualblock
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Thanks; time to whip out the plastic.

Before you whip out the plastic! [message #9805 is a reply to message #9804] Fri, 01 September 2006 08:08 Go to previous messageGo to next message
Shane is currently offline  Shane
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Registered: May 2009
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I looked at them both using inter-library loan here at the local public library. Maybe see if you really would want them or not.

Re: Before you whip out the plastic! [message #9806 is a reply to message #9805] Fri, 01 September 2006 10:13 Go to previous messageGo to next message
Manualblock is currently offline  Manualblock
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Your library system has them? Thats great; let me check that out; although somehow I doubt it. Our libraries are a little behing the times here; we're lucky of they carry Grob's textbook.

Re: Before you whip out the plastic! [message #9807 is a reply to message #9806] Fri, 01 September 2006 10:47 Go to previous messageGo to next message
Shane is currently offline  Shane
Messages: 1116
Registered: May 2009
Illuminati (3rd Degree)
My library didn't have them, but was able to pull them from another state that did.

Duh! [message #9808 is a reply to message #9801] Fri, 01 September 2006 14:14 Go to previous messageGo to next message
Shane is currently offline  Shane
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Registered: May 2009
Illuminati (3rd Degree)
Um 50mA is a bit much, huh? Let's revise that to 40mA where the Max V point = 410V. Better?

Re: Loadlines [message #9809 is a reply to message #9802] Tue, 05 September 2006 08:09 Go to previous messageGo to next message
Damir is currently offline  Damir
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Registered: May 2009
Illuminati (2nd Degree)

Hi, I`m back...
Load lines - simplified:

-the OP you chose (say, 200V/50mA/-30V - with respect to max Ia and Pa values) "determines" your "optimal" load (yes, this is your primary impedance - simplified). The AC current can "swing" from quiescent value (Iaq=50mA) to the double value, max. 100mA, and from "quiescent" Ug value (-30V from Ua/Ia/Ug diagram for 6EM7-2) to the Ug=0V, max. peak in class A1.

-draw the line through our "O" point (200V/50mA/-30V) and through another point, "A" (Ia=100mA, Ug=0V, and we read Ua=63V)

-point "B" can be another "extreme", (about 292V/16mA/-60V)

-our Ra, or primary load is (from Ohm`s Law) Ra=Ua/Ia, or "voltage swing" divided with "current swing", or:
Ra = (UaB - UaA) / (IaB - IaA) = (292-63)/(0,1-0,016) = 2,7 kOhms

-of course, you can just "extend" your line to the apsis and ordinate, to simplify Ra graphical finding, where Ia=0 and Ua=0, and you have
Ra = 335/0,125 = 2k7

-there`s a more, we want a symmetrical swing "around" your "O" point (min. distortion)...more horizontal LL is closer to this goal (increasing of the "minimal" Ra=2k7) on the less power "expense"

-for more, see this two messages:

http://audioroundtable.com/GroupBuild/messages/1111.html
http://audioroundtable.com/GroupBuild/messages/1113.html

Trying these for 5K [message #9810 is a reply to message #9809] Tue, 05 September 2006 17:07 Go to previous messageGo to previous message
Shane is currently offline  Shane
Messages: 1116
Registered: May 2009
Illuminati (3rd Degree)
Here's what I came up with on a couple of lines. Sorry, won't have access to scan till later. This is using Steve Bench's sheets. Picking a 5K OPT since this is what seems to be most commonly used for this amp. Mainly just want to see if I'm getting the math right and understanding a little more of the concept.

At 210V/40mA 5K:8 ohm OPT

-34V bias voltage
Va= 50V
Vq= 210 V
Ve= 333V
Ia= 72mA
Ib= 54mA
Ic= 40mA
Id= 27mA
Ie= 15mA

Po= 2W
HD2%=6.25
HD3%=1.78
HD4%=0.89


At 210V/30mA 5K:8 ohm OPT

-37V bias
Va= 45V
Vq= 210V
Ve= 330V
Ia= 63mA
Ib= 45mA
Ic= 30mA
Id= 16mA
Ie= 6mA

Po=2W
HD2%= 7.84
HD3%= -0.58
HD4%= 1.45


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