really dumb question about SE 6bq5 operating point (triode)

[zobsky]

as i take a break from thinking about push pull , i decided to revisit my SE amp a bit.

i was trying to walk through one of norman crowhurst's tube power stage calculations using a 6bq5 (data sheet @ http://www.retrovox.com.au/STC6BQ5.pdf) as opposed to his example using a KT66 triode wired.

question:
1. working from the suggesting operating points for the 6bq5, ......

given that the power dissipation of my tube in triode mode is 2W, my operating voltage is 250V and considering the load impedance to be 3500 ohm (=> Ip = 70mA), why does my preliminary operating line cut the plate dissipation curve so severly, .. seemingly suggesting less than optimum operation

2. looking back at the original schematic from which I built the SE amp in (pentode mode), .. is it really possible (or optimal) to use a 300 ohm OPT in series with a 3K9 resistor to make up for the rest of the plate load in an output stage.

3. also, what's with the 14W dissipation curve for triode mode?


thanks

 

[zobsky]

section of amp schematic

 

[EC8010]

You have drawn your loadline as if it was a resistor and imply Ia = 0 at Va = 250V. However, if you have set Pa = 2W (seems far too low) then the valve passes 8mA. Shift your loadline vertically by 8mA. You will notice that your loadline now implies that the anode can swing above 250V. This is correct - it really can. Energy is stored in the output transformer.

Adding a 3k9 resistor would waste most of your audio power as it would form a potential divider in conjunction with the reflected load from the transformer. You need the correct transformer.

In triode mode, you see the addition of Pa and Pg2 for pentode mode. I would suggest that you allow a much higher anode dissipation, at least 10W, and possibly 14W, then start choosing loadlines.

 

[G]

Here is a item at Ebay that has the transformers you need. I would run the EL84s at about 36-40 mA with transformers with a 5K primary. Hope this helps.

P.S. A 6AQ5 and a 6BQ5 both use a output transformer with a 5K primary so the transformers in this link should work well for you. And they are cheap.

http://cgi.ebay.com/ws/eBayISAPI.dll?ViewItem&item=3089002392&category=39783

 

[zobsky]

You have drawn your loadline as if it was a resistor and imply Ia = 0 at Va = 250V. However, if you have set Pa = 2W (seems far too low) then the valve passes 8mA. Shift your loadline vertically by 8mA. You will notice that your loadline now implies that the anode can swing above 250V. This is correct - it really can. Energy is stored in the output transformer.

Adding a 3k9 resistor would waste most of your audio power as it would form a potential divider in conjunction with the reflected load from the transformer. You need the correct transformer.

In triode mode, you see the addition of Pa and Pg2 for pentode mode. I would suggest that you allow a much higher anode dissipation, at least 10W, and possibly 14W, then start choosing loadlines.

thanks for all the replies, ...

continuing from my previous account of things, ..

suppose i do shift the load line vertically by 8mA (ie. operating point at 250,8), ...
1. will i not still be operating above max. plate current,
2. my operating point seems like a fairly sane choice, .. yet looking at the way the load line cuts the grid curves, i see compression at the bottom around the -16 mark. in fact, on the version of the chart i use in this post, .. the -18 and -20 grid curves are not even included.
what would the correct way to determine the correct operating point.

P.S. my speakers are > 96db fostex rear-horns so efficiency is not a prime concern, .hence the exercise in low power amp. design.

thanks, .. happy weekend.

 

[EC8010]

The maximum power you can safely dissipate in the anode of a triode-strapped 6BQ5 is 14W. Thus, the maximum quiescent current it can safely pass with a 250V HT is 14/250 = 56mA.

As you have already discovered, if you don't position your operating point correctly, you can only move a very limited distance in one direction before the valve cuts off (Ia = 0) or you hit grid current (near Vg = 0). The ideal operating point (for SE Class A) allows equal swings between these two restrictions.

Have another go. You will probably find this post helpful:

http://www.diyaudio.com/forums/showthread.php?s=&threadid=31858