loadlines for newbie.

[bloozestringer]

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 for example. 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 anything till I can go through and explain/understand why everything is what it is.

Thanks

[Miles Prower]

Originally posted by bloozestringer
Looking to draw a loadline for 6EM7 section 2 for example. Steve states that the load (Z)=(max V - quiescent V)/quescient I. I'm looking at Ic=50mA and Vq=210V.

Here's your first mistake. According to the spec sheet, Pd= 10W. (210)(50E-3)= 10.5W. You're running too hot there. This will cut down the expected life of the VT. Run it cooler, with no more than seven or eight watts of Pd. The reduced power won't add up to very much, and you won't notice the reduction.

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?

The whole idea of doing a loadline analysis is to find the best load. That particular load is what the output xfmr should step up the 8.0R (or what you have) nominal speaker impedance to. If the loadline you drew shows that 5.0K works best, then you want a 5.0K : 8.0R OPT.

[bloozestringer]

Originally posted by Miles Prower


Here's your first mistake. According to the spec sheet, Pd= 10W. (210)(50E-3)= 10.5W. You're running too hot there. This will cut down the expected life of the VT. Run it cooler, with no more than seven or eight watts of Pd. The reduced power won't add up to very much, and you won't notice the reduction.



The whole idea of doing a loadline analysis is to find the best load. That particular load is what the output xfmr should step up the 8.0R (or what you have) nominal speaker impedance to. If the loadline you drew shows that 5.0K works best, then you want a 5.0K : 8.0R OPT.

Oops, my mistake. Using 35-40Ma would be better I was looking at Fred Nachbaur's design and Gary Kaufman's at the same time and got the mA switched. Fred ran his at 35mA and Gary at 50mA. 5K seems to be what most have used, saying that they can squeeze out more power at a 2.5K, but at the expense of more distortion. I'm really just starting to try and understand everything.

As far as plotting the points for the line, have I got that right, at least with the first set of numbers I had? So with 40ma at 210V the max V point = 410V ?

[EC8010]

First off, I'd just like to say that I'm delighted to see your approach - you are trying to find out how to do things and are asking for specific assistance, rather than asking someone else to do all the work. As Miles says, you're running your valve at a point that will shorten its life. Further, as he says, the whole point of loadlines is that they are a quick way of finding out the optimum load. Do bear in mind though, that nothing beats a real-world measurement. If you have the equipment to do it, setting up an output stage and testing it at 1kHz with different dummy load resistors whilst watching the distortion and maximum power is extremely useful.

I don't suppose you can scan your sheet with the loadlines?

I wish you all the best as you learn.

[Miles Prower]

Originally posted by bloozestringer
As far as plotting the points for the line, have I got that right, at least with the first set of numbers I had? So with 40ma at 210V the max V point = 410V ?

If you're loading with either an xfmr or choke, the voltage at the Ip= 0 point should be twice the actual DC voltage, in this case, 420V. The inductance acts like a phantom negative DC rail since reactive voltage drops don't represent any actual energy expended, and therefore can exceed the applied voltage. Just make sure that your minimum plate current doesn't dip way down into the nonlinear part of the characteristic at low currents.

[bloozestringer]

Thanks so far, guys. I really appreciate the help I can get. Sometimes it's easier to have someone explain in simple terms to really start to grasp a concept. I'll have plenty more questions I'm sure, but will try not to ask unless I've worked through an example and need my calc's checked or need my understanding of something clarified.

I'll try and take a few pics of my loadlines and put them up tomorrow for ya'll to set me straight with.

[bloozestringer]

Here's what I came up with on a couple of lines. Sorry, won't have access to scan till later.

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