Bevois Valley Design Question

cgrums · 24th February 2008, 05:01 PM

I've been working my way through Morgan Jones' book for quite some time and have a fairly clear grasp of everything up to the point of the Bevois Valley amp design.

I follow the design of the Concertina phase splitter right up to the section entitled 'Calculation of the Cathode Bias Resistor and Feedback resistor'

At what point is the 47k Rl for the first anode chosen? I don't see any mention of it before this point so I'm assuming it's something simple that I've missed.

In the same section: where is the 190V derived from to find the anode current?

Any help would be much appreciated.

TIA

cgrums

EC8010 · 24th February 2008, 06:11 PM

To answer your second question first, the 190V comes from observing that the ECC88 is very linear at around 90V. 285V - 90V = 195V.

As to the first question, 47k gives a nice loadline, but almost anything would do.

cgrums · 25th February 2008, 02:08 PM

That helped quite a bit, thanks for the info.

In the application of the feedback is the 'unmodified sensitivity' (~ 238 mVRMS - don't have the book right in front of me) referring to the sensitivity of the Concertina stage (without feedback)? If so would that 239 mVRMS be the input voltage required for an anode voltage swing of 8.6xx VRMS (required for full power output)? This seems low...am I way off?

SY · 25th February 2008, 03:03 PM

It's a combination of the gain of the input tube with the concertina. The latter is slightly less than unity.

The sensitivity numbers are as you suspect. As a check, 8.6V/0.239V = 35, about the mu of the first tube.

cgrums · 25th February 2008, 03:35 PM

Thanks SY...I'm getting there slowly.

Please forgive this next as it's probably very evident...the Concertina is composed only of the second of the two tubes shown? I misread and thought that as they were drawn as one device that both tubes comprised the Concertina. So the device shown is actually an input stage and a phase-splitter in one envelope? Is the input stage analyzed as a common cathode stage?

SY · 25th February 2008, 04:09 PM

Yes on all points. See, you're quicker than you thought!

cgrums · 25th February 2008, 04:19 PM

Ha! Fantastic! Coming from a mechanical engineering background this is a whole different way of thinking but I think I'm slowly making sense of it all.

cgrums · 25th February 2008, 06:11 PM

My current plan is to use this circuit for the basis of a birthday present for my brother. He currently has a pair of JBL 2600 bookshelfs (89 db) and we'll be looking in to making him a more efficient pair in the future.

As of right now I'm not planning a pre-amp and was going to use a rotary source selector switch as well as a pot for volume attenuation. What considerations need to be taken into account in the selection of the resistance of the pot? I know it's related to the input impedance of the input stage but am not quite sure how to put that relationship in numbers.

EC8010 · 25th February 2008, 06:17 PM

100k will do you nicely.

cgrums · 25th February 2008, 06:23 PM

Thanks! Could you explain how you arrived at that conclusion please?

24th February 2008, 05:01 PM	#1
cgrums diyAudio Member Join Date: Dec 2006	Bevois Valley Design Question I've been working my way through Morgan Jones' book for quite some time and have a fairly clear grasp of everything up to the point of the Bevois Valley amp design. I follow the design of the Concertina phase splitter right up to the section entitled 'Calculation of the Cathode Bias Resistor and Feedback resistor' At what point is the 47k Rl for the first anode chosen? I don't see any mention of it before this point so I'm assuming it's something simple that I've missed. In the same section: where is the 190V derived from to find the anode current? Any help would be much appreciated. TIA cgrums

24th February 2008, 06:11 PM	#2
EC8010 diyAudio Moderator Join Date: Jan 2003 Location: Near London. UK	To answer your second question first, the 190V comes from observing that the ECC88 is very linear at around 90V. 285V - 90V = 195V. As to the first question, 47k gives a nice loadline, but almost anything would do. __________________ The loudspeaker: The only commercial Hi-Fi item where a disproportionate part of the budget isn't spent on the box. And the one where it would make a difference...

25th February 2008, 03:03 PM	#4
SY diyAudio Moderator Join Date: Oct 2002 Location: Austin, TX Blog Entries: 1	It's a combination of the gain of the input tube with the concertina. The latter is slightly less than unity. The sensitivity numbers are as you suspect. As a check, 8.6V/0.239V = 35, about the mu of the first tube. __________________ “Listening to records is like ****ing a picture of Brigitte Bardot.” - Sergiu Celibidache

25th February 2008, 04:09 PM	#6
SY diyAudio Moderator Join Date: Oct 2002 Location: Austin, TX Blog Entries: 1	Yes on all points. See, you're quicker than you thought! __________________ “Listening to records is like ****ing a picture of Brigitte Bardot.” - Sergiu Celibidache

25th February 2008, 06:17 PM	#9
EC8010 diyAudio Moderator Join Date: Jan 2003 Location: Near London. UK	100k will do you nicely. __________________ The loudspeaker: The only commercial Hi-Fi item where a disproportionate part of the budget isn't spent on the box. And the one where it would make a difference...

25th February 2008, 02:08 PM	#3
cgrums diyAudio Member Join Date: Dec 2006	That helped quite a bit, thanks for the info. In the application of the feedback is the 'unmodified sensitivity' (~ 238 mVRMS - don't have the book right in front of me) referring to the sensitivity of the Concertina stage (without feedback)? If so would that 239 mVRMS be the input voltage required for an anode voltage swing of 8.6xx VRMS (required for full power output)? This seems low...am I way off?

25th February 2008, 03:35 PM	#5
cgrums diyAudio Member Join Date: Dec 2006	Thanks SY...I'm getting there slowly. Please forgive this next as it's probably very evident...the Concertina is composed only of the second of the two tubes shown? I misread and thought that as they were drawn as one device that both tubes comprised the Concertina. So the device shown is actually an input stage and a phase-splitter in one envelope? Is the input stage analyzed as a common cathode stage?

25th February 2008, 04:19 PM	#7
cgrums diyAudio Member Join Date: Dec 2006	Ha! Fantastic! Coming from a mechanical engineering background this is a whole different way of thinking but I think I'm slowly making sense of it all.

25th February 2008, 06:11 PM	#8
cgrums diyAudio Member Join Date: Dec 2006	My current plan is to use this circuit for the basis of a birthday present for my brother. He currently has a pair of JBL 2600 bookshelfs (89 db) and we'll be looking in to making him a more efficient pair in the future. As of right now I'm not planning a pre-amp and was going to use a rotary source selector switch as well as a pot for volume attenuation. What considerations need to be taken into account in the selection of the resistance of the pot? I know it's related to the input impedance of the input stage but am not quite sure how to put that relationship in numbers.

25th February 2008, 06:23 PM	#10
cgrums diyAudio Member Join Date: Dec 2006	Thanks! Could you explain how you arrived at that conclusion please?

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