Planning my BA-6A

[dmp]

I measured the DC before the regulator and it was about 18VDC.  That is a lot of volts to dissipate.  Is this too much, do you think?

The higher the voltage, the more heat to dissipate in a linear regulator...  if you are regulating down to 6.3V and are drawing 0.6A, then the heat the regulator has to get rid of is V*I = (18-6.3)*0.6 = 7 Watts

 

[DaveP]

Matt, for some reason I missed your earlier posts.

Have you got the manual?  Its too big to attach so I've emailed it to you.

You can check the wattage of all the components in the parts list.

The caps should be closely matched within 5~10%, use some new 630V polyprop jobs from RS.  To minimise the thumping the bass balance must be near perfect, one of those old caps will have double the bass reactance of the other, bad news!

I prefer to use 2 diodes in a CT arrangement which saves 2 diode drops.  Check your regulator for the minimum voltage it needs upwind to give 6.3V.  If it only needs 3V you could use a CT and get 9V which would solve your excess heat problem which is not healthy.

best
Dave

 

[letterbeacon]

Thanks guys!

Have you got the manual?  Its too big to attach so I've emailed it to you.

You can check the wattage of all the components in the parts list.

I do have the manual, but thanks for sending it anyway -what a great support network this site is!  The reason I query what wattage pots I should use for the balance pots, was that I asked a similar question on your thread and I got this reply from lassoharp:

The balance circuit I built on the breadboard worked great with standard 100R and 10K pots. Lots of range to work with.

I couldn't find any reasonably priced 4W pots (as specified in the BOM) but I did find some 2W.  I would like to make sure that that 2W will be enough.  I know the answer is Ohm's Law, but how do I apply it for the balance pots?  I can't work it out I'm afraid.

The caps should be closely matched within 5~10%, use some new 630V polyprop jobs from RS.  To minimise the thumping the bass balance must be near perfect, one of those old caps will have double the bass reactance of the other, bad news!

Great, thank you.  I've got a matched pair of caps now, so I should be ok!

I prefer to use 2 diodes in a CT arrangement which saves 2 diode drops.  Check your regulator for the minimum voltage it needs upwind to give 6.3V.

The LM317 needs a minimum difference between the input and output voltage of about 2.5V.  I used 4 diodes instead of 2 because I knew I was going to need to drop some voltage before it got to the LM317. 

If it only needs 3V you could use a CT and get 9V which would solve your excess heat problem which is not healthy.

I have dual secondaries of 6V each -what's the best way of implementing what you describe above?

Thanks a lot!

 

[DaveP]

Matt,

Right lets get you sorted.

From my schematic, the voltage drop across the 10k pot is 123-92=31 v    /10k=3.1mA.

say 2mA goes to the screen, we call it 5ma, so 5mA x 31V = 0.155W.  So a 1W ww pot from RS will be fine.  If you want a spindle lock check the thread dia is 9.5mm.

The other pot will have ~10mA across each half so .01 x 50 ohms = 0.5V  0.5 x .01 = only 5mW!

They used over-rated components because they were on 24/7 in a hot rack.

Re the 6V transformer.

Connect the windings for 12V making sure the phase is observed e.g.

Connect the spot marked wire on one winding to the non spot marked wire on the other winding, this should then be earthed as it's the CT.  The two remaining wires go to the negative ends of the diodes and the diode positives are connected together.  This will probably give you  13V x 0.707 (inverse of root2) = 9 something volts.

best
DaveP

 

[letterbeacon]

From my schematic, the voltage drop across the 10k pot is 123-92=31 v    /10k=3.1mA.

say 2mA goes to the screen, we call it 5ma, so 5mA x 31V = 0.155W.  So a 1W ww pot from RS will be fine.  If you want a spindle lock check the thread dia is 9.5mm.

The other pot will have ~10mA across each half so .01 x 50 ohms = 0.5V  0.5 x .01 = only 5mW!

Thanks a lot!  Just so I understand it though, why do you estimate 2mA goes to the screen?

Re the 6V transformer.

Connect the windings for 12V making sure the phase is observed e.g.

Connect the spot marked wire on one winding to the non spot marked wire on the other winding, this should then be earthed as it's the CT.  The two remaining wires go to the negative ends of the diodes and the diode positives are connected together.  This will probably give you  13V x 0.707 (inverse of root2) = 9 something volts.

At the moment I have the 0V of the first winding connected to the 6V of the second -and isolated. Then I take the 12V between the 6V of the first winding and the 0V of the second.

Just so I'm clear, my transformer doesn't have a CT on it at the moment -we're creating a CT, right? And this is using a 2 diode arrangement, as per your 6.3VDC circuit on your BA-6A?

I'm off to bed now, but I'll give it a go tomorrow -thanks again!

 

[DaveP]

Matt,

When you look at pentode data sheets, the ratio between plate current and screen current is always about 4:1.  So if the total current is 10mA then they divide about 8:2. 

Yes, you connect the connection you isolated to earth and follow the rest of the diagram.

You can make a CT from any transformer that has two identical windings.

best
DaveP

 

[letterbeacon]

Gotcha -thanks Dave!

 

[letterbeacon]

Dave, I rebuilt the transformer/ bridge rectifier part of the PSU as per your instructions and... hey presto!  It runs nice and cool.  Thanks a lot!

Now, onto the fun stuff - the actual BA-6A build!  Now I'm in the build part of the project, perhaps I should start a build thread under Dynamic Processors?  Maybe not actually, probably best to keep everything in this one thread.

 

[DaveP]

Matt,
The Drawing board is supposed to be as you've used it, asking questions and sharing ideas, someone will correct me if I'm wrong, but builds are probably best in the right section.  There are thread mover members who will tidy you up otherwise!
best
DaveP

 

[letterbeacon]

I've started the build thread over here: http://www.groupdiy.com/index.php?topic=49637.0

 

[letterbeacon]

I started building the 25k input attenuator today and made a very very silly error.  Hopefully it's not the end of the world, but I'm hoping someone here will be able to confirm that or not.

I followed the the Goldpoint instructions here: http://www.goldpt.com/r_series.html to make a 2x 25K ladder attenuator.  However, as I was soldering the second to last resistor on (2 hours later) I just realised that my switch is a 23 position switch and NOT a 24 position like the ones Goldpoint uses.  Idiot! This means I cannot solder on the last 5.11k resistor, meaning the total resistance of the switch is 20k rather than 25k.

Is a 5k difference going to affect things too much?  Will I get away with using an attenuator adding up to 20k rather than 25k?

Thanks!

 

[lassoharp]

Is a 5k difference going to affect things too much?  Will I get away with using an attenuator adding up to 20k rather than 25k?

I can't speak much of the changes on the sec side of things, but with the lower total resistance your input impedance will now be lower than the 600r-ish impedance than you would get with full 50K load on sec.  I wouldn't call that a major issue - it will just load the output of whatever is driving it a little heavier.

It's actually a 10K total difference (20K + 20K = 40K vs 50K)

Don't feel bad, I've made this same mistake making stepped attenuators.

 

[DaveP]

If it was mine, and it would take re-soldering, I would take them all off and replace the string with values 25% higher.  Its more authentic/important to keep the correct input Z than to worry about which resistors it took to get it.

best
Davep

 

[letterbeacon]

Thanks for the replies.

I'm trying to think of a cheap solution to this as I've already put a bit of money into this project.  If I make it a linear taper, then I'm buying resistors of the same value, right?  Therefore much cheaper.

So how about this -I divide 25k by 22 = 1136.  The nearest value is 1.13k.  If I make a linear taper attenuator made up of 1.13k resistors, each ladder will have a total resistance of 24860.  Is this close enough?

 

[DaveP]

It's certainly close enough, even 1.2k's would give you 26.4K which is only 5% out, but in the right direction.  Having got the total resistance in the right ballpark, the next most important thing is that the resistors are matched as far as possible on either side, to keep the push pull in balance.
best
DaveP

 

[letterbeacon]

Thanks Dave.

If I was to use 23x 1.2k resistors or 23x 1.13k, how do I go about calculating the dB attenuating for the scale? Is there a formula?  I've had a Google but I'm getting slightly lost!

 

[DaveP]

Matt,

This is a simple if tedious application using the voltage divider calculation (R1+R2/R1) x 20LOG

You've got 23 positions right? maybe the last is grounded or a similar resistor to ground? I don't know the details.  Anyway, taking the first position from full on as an example.

Full on = 23x1.2k=27.6k, next position is minus 1.2k =26.4k

27.6/26.4=1.04545

20LOG(1.04545)= 0.386dB reduction  Not a lot!

You won't get a 6dB reduction until you have reached halfway with a linear taper, which is why we use log audio tapers.

This is another way of saying that all the useful switch positions will be at one end.

When you have reverse calculated the resistors to get meaningful dB steps you will have mastered it!
best
DaveP

 

[Brian Roth]

Or, wire the 5.11K resistor in series into the 23 position switch.  That means the "wide open" switch position will actually be -2 dB attenuation, which I would guess would be acceptable in most any application.  The question is "does anyone actually run the BA-6 with the attenuator wide open?"

Bri

 

[letterbeacon]

Matt,

This is a simple if tedious application using the voltage divider calculation (R1+R2/R1) x 20LOG

You've got 23 positions right? maybe the last is grounded or a similar resistor to ground? I don't know the details.  Anyway, taking the first position from full on as an example.

Full on = 23x1.2k=27.6k, next position is minus 1.2k =26.4k

27.6/26.4=1.04545

20LOG(1.04545)= 0.386dB reduction  Not a lot!

You won't get a 6dB reduction until you have reached halfway with a linear taper, which is why we use log audio tapers.

This is another way of saying that all the useful switch positions will be at one end.

When you have reverse calculated the resistors to get meaningful dB steps you will have mastered it!
best
DaveP

Thanks Dave -that's good to know for future reference.  Why can't they explain it like that on any of the website that come up in Google??

Or, wire the 5.11K resistor in series into the 23 position switch.  That means the "wide open" switch position will actually be -2 dB attenuation, which I would guess would be acceptable in most any application.  The question is "does anyone actually run the BA-6 with the attenuator wide open?"

Bri

Thanks Brian.  This would mean that I wouldn't have to throw away 2 hours worth of work, or £25 worth of resistors!

 

[DaveP]

Matt,

Brian's solution is the best idea, there is gain to spare on a BA-6A.

best
DaveP