Planning my BA-6A

[letterbeacon]

Hi Brian,

Unfortunately I'm on a rather large project at work at the moment, with all my free-time and head-space devoted to it, so the BA-6 project is on hold for the time being.

I've bought almost all the tubes (NOS RCA), all the transformers (minus the Sowter interstage), almost all the capacitors (mostly PIO as per the original BOM) so I've invested a lot of time and money into the project already, so don't worry it will happen at some stage, but not for another month or so!

Thanks for the interest!

 

[letterbeacon]

I had a spare bit of time today to do some work on this project.  I have most of the main components and decided to do a rough layout to see if it'll all fit on the chassis.  I've attached a picture and wondered what you guys think.

I'm going to use a 2U rack chassis and place the transformers and tubes and the meter on top of the chassis, much like the original BA-6A.  This will make it a 5U beast!

As this is my first point to point layout, I've roughly laid it out according to the schematic, which should make things slightly easier.

In the picture, the card is the chassis and we're looking down on it.  The top of the picture is the front.  The Russian PIO capacitors and the choke will be mounted underneath the case.

Do you think having the output tx fairly far away from the output XLR is a problem?  I'm hoping that using shielded wire should solve any noise issues.

The power switch will sit next to the PT, as well as the fuse.  That way the mains doesn't have to travel to the front of the chassis and therefore there's less chance of noise.

Please let me know your thoughts...

 

[lassoharp]

Looks good.  My first reaction is to turn the bell of the PT towards the input (turn lams away from) but it may not be an issue with the shielded input, which can also be rotated if need be.


I doubt you will have problems running normal long leads from OT secondary to output jack considering the very high signal level.

 

[letterbeacon]

Great, thanks for your input. I think I'll rotate the PT, just to be on the safe side.

I just realised I haven't put the balance pots on there yet, but I suppose I could always put them on the front panel.  Or I could move the input tx and the interstage back a bit and put them in between the 6SK7s and the 6J7s.

 

[Biasrocks]

Here's what the real deal looks like.

http://www.flickr.com/photos/8596658@N05/524539237/
http://farm1.staticflickr.com/222/524539271_01d1622653_z.jpg?zz=1

Regards,
Mark

 

[letterbeacon]

I'm having a bit of a rethink about the power transformer, mostly because of lack of space.  Because I want to use 6.3Vdc to heat the tube filaments I need to use a separate toroidal transformer to supply 6-0-6Vac to a rectifying circuit.  Space is at a premium and I'm thinking that I might get a custom power transformer made that has all the windings I need, rather than have two separate ones.

My ideal power transformer needs to supply the following:
The B+
The tube rectifier heater (5R4GY which requires 2A)
The rectifier circuit to make the 6.3Vdc for the other tubes' heaters (2x 6SK7, 2x 6J7, 2x 6V6, 1x 6H6 = 2.5A).  I intend to use DaveP's circuit that he used in his BA-6A
The balance circuit in the BA-6A
A lamp to go in the meter (3.5V @ .3A)

So I am going to get a quote on the following:

Primary: 0-230Vac (UK)
Secondaries:
275-0-275V @ 150mA (for the B+)
2.5-0-2.5 @ 3A (for the 5R4GY heater)
6-0-6v @ 3A (for the DC heaters circuit)
0-3.15V @ 1A (for the balance circuit and the lamp)

How does that look?

My questions are:
I have 1A extra on the 5V winding for what the 5R4GY needs - is that enough?
I have .5A extra on the 12V winding.  Is this too little?  I imagine the rectification circuit will use up some amps.  What's a good safety margin for this?
How can I work out how many amps the balance circuit in the BA-6A will use?  I imagine it must be some application of Ohm's Law, but I can't work out what figures to use.

Thank you very much!

 

[DaveP]

Letterbeacon,

The first thing I notice is that you do not want 2.5-0-2.5v for the rectifier because one end of the winding must be at B+ so where would the 0 go?  The idea is that the heater/cathode insulation is not strained by excessive voltage differences, so you want 5-0, its elevated to the B+.

Also you don't need a separate winding for the balance test voltage, it uses microamps at 3.15V.  Just make a simple voltage divider from the heater winding for the balance voltage, it is just a signal after all.  In fact the BA-6A uses the reactance of the two caps to provide the correct level and you can do the same.  Just use smaller caps for use with a higher voltage like 6V.  Use ohms law on the reactance of the caps at 50/60Hz.

You should also specify the regulation required in the transformer.  Poor regulation will have much higher off load voltages which drop right down on load.  10~12% is good.

The rectification circuit should not use up much current.  Use oversize low forward voltage schottky diodes (a double in a TO220 package is good) then the voltage drop times the current will be minimal.  Two diodes plus centre tap will waste less than a bridge with twice the diode drop.

The lamp, if you must use one, should be across the heater supply,  You need to know that the heaters are on, because if they fail then the full HT will be across the components.

Personally, having made a BA-6A, I would use the circuit I designed for my Groupdiy 26C if I made another.  I would use readily available toroidal transformers rather than fork out for a custom made job which will cost hundreds of pounds.

good luck

DaveP

 

[letterbeacon]

Thanks very much for this Dave -I feel like I'm learning a great deal, and I haven't even started the actual construction of it yet!

Also you don't need a separate winding for the balance test voltage, it uses microamps at 3.15V.  Just make a simple voltage divider from the heater winding for the balance voltage, it is just a signal after all.  In fact the BA-6A uses the reactance of the two caps to provide the correct level and you can do the same.  Just use smaller caps for use with a higher voltage like 6V.  Use ohms law on the reactance of the caps at 50/60Hz.

The caps that provide this reactance are C12 and C13, right?  The .012uf and .027uf in on the balance switch.  If I used the voltage divider to bring the 5Vac down to 3.15Vac, I wouldn't need to change the value of the caps though, right?

The rectification circuit should not use up much current.  Use oversize low forward voltage schottky diodes (a double in a TO220 package is good) then the voltage drop times the current will be minimal.  Two diodes plus centre tap will waste less than a bridge with twice the diode drop.

I was going to use 1N5401 for the rectification circuit.  They're rated for 100V @ 3A.  I'll look into the schottky ones though.

Personally, having made a BA-6A, I would use the circuit I designed for my Groupdiy 26C if I made another.  I would use readily available toroidal transformers rather than fork out for a custom made job which will cost hundreds of pounds.

The trouble with using a couple of toroidal TXs is the space!  Because I'm insisting on using large PIO caps and a tube power supply (a folly perhaps -it's more an exercise in building a true replica than any tone thing), real estate on my enclosure is scarce!  The price for a custom job is about £100.  The Hammond is £60 + £20 for a toroid for the DC heaters and that's nearly that much anyway.[/quote]

Thanks again Dave!

 

[DaveP]

Letterbeacon,

Forget that 5V winding its all at B+ volts!  The 0V is not connected to earth, its just across the heater!  Use one of the 6V windings, almost any two similar resistors will do to halve the voltage, 1k or 10k, its only the voltage you need, virtually no current required.  As long as you have 3V like the original unit then you can use the original cap values.

Look up the data sheets on various diodes and check the charts of Vf versus current.  You don't want Vf's of 0.7 or 1V, a good schottky can be as low as 0.25V.  The trick is to use the lowest voltage you can safely get away with, for 6V windings even 20V diodes will be ok but stick to 40V to be on the safe size.  Its the highest current versions that are best, not the highest Voltage.

I understand about making a close copy, if you can get a PT made cheap enough then thats cool.

best
DaveP

 

[letterbeacon]

I had a browse through a few datasheets last night and came up with this: http://uk.rs-online.com/web/p/rectifier/5430737/

According to the picture attached, there'll be about a 0.4 voltage drop across it at @ 2.5A (if I'm reading it correctly.

So, if I am using your 6.3Vdc circuit with these diodes, these are the values I get:

The 6Vac is rectified to 5.6Vdc.
The smoothing caps increases the the DC to 1.4 x 5.6 = 7.84Vdc.  So I'll need to drop 1.54V with the dropping resistor.
R = E/ I
.6r = 1.54 / 2.5
So I'll need a 1.5r and a 1r 3W in parallel.

Is my maths correct?

Thank you!

 

[DaveP]

Letterbeacon,

Yes you are on the right track.

However, there are several other variables to be taken into account.  The regulation of the transformer will probably give you 10% extra voltage over that specified and the DC you get will be affected by the size of the caps you use.

In practice, it is better to have a small range of suitable resistors available so that you can fine tune it when its on.  In fact I usually try all the tubes on DC before I wire up the rest of the circuit.  You can then fit suitable 25W metal clad resistors in place before the rest of the components.

Basically, the calculation is this:-  12V/root2 = 8.48V 

8.48-6.3 = 2.18V.        2.18V/2.5A = 0.87 ohms  This is the ballpark figure.  Depending on the diode drop, the regulation and the cap used it could be more or less than this value.

They are about £1.64 each from RS so buy a few to play around with, they will always come in handy.

best
DaveP

 

[letterbeacon]

Thanks Dave (again!).

I notice that you use a slightly different DC heater circuit on your Bridge Compressor project - less smoothing caps.  Just out of interest, why is that?


I have one more (probably slightly silly) question.  I would like to use a 600r bridged T-attenuator on the output and I was thinking of using NYD's schematic: http://electronicdave.myhosting.net/miscimages/600bridgedt.gif

I read that the BA-6A puts out 10W on the output.  The last few resistors (-16dB, -18dB, -20dB) should be rated for 10W, but the others can be rated at less, yes?  Because when resistors are in parallel they sort of 'share' the power dissipation.  Is this correct?

Thank you.

 

[emrr]

You can use 1/4 of the size for power rating on the output, it's a generally accepted standard for audio power dissipation.  Think about it; the only time you'd see 10W in a meaningful way is if you cranked a sine wave full blast.    Might be worth looking at the Collins 26U output method.  It uses a fixed T pad with higher rated resistors, followed by a variable control.  If you took that approach, you could use finer steps with the variable, or use the 600 T as seen in everyone's 1176 builds. 

 

[letterbeacon]

Thanks Doug, using a fixed pad and a variable attenuator sounds like a plan.

So looking at the 26U, they used 2W resistors on the balanced T-pad.

I think I'm going to use a fixed T-pad to attenuate 20dB and then NYD's stepped attenuator to attenuate a further 20dB, bringing the total attenuation to 40dB, if I wanted it.

Using this site: http://www.nu9n.com/tpad-calculator.html I've calculated what I need for a balanced T-pad with a total resistance of 600r. I've drawn it out and attached.

I will use 2W resistors as the 26U did.

Am I able to connect NYD's stepped attenuator directly to the output of the fixed T-pad?

Thank you.

 

[DaveP]

Letterbeacon,

Regarding the slight differences on DC supply.

Because this is a project for others too, I have tried to make it as affordable as possible, you can use a smaller cap for the first position if you wish.

best
DaveP

 

[letterbeacon]

So I've finally found a suitable meter for my project. An old Simpson one that looks just like the original! However, it's only 0-50uA DC rather than the 0-500uA DC that the BA-6A would like, so I'll need to use a shunt resistor to fix it.

I've done some reading up on the equation that's used to calculate the shunt resistor and it seems fairly straight forward, but I'm not sure how to find out the meter's internal resistance. Some website say you need to set up a test right with batteries and potentiometers, while others say putting a multi-meter across it will be fine.

What do you guys recommend?

Thank you!

 

[DaveP]

Letterbeacon,

50uA is just about the most sensitive meter you can get and putting low res meter ranges across it might bend the needle.

Best to connect a 1.5V battery with a big pot in series, say 50k, that means it can't pass any more than 30uA.  Turn the pot until you get 50uA FSD. then measure the voltage across the meter.  Meter V/50uA  gives you the meter resistance.

Play safe with that meter.

best
DaveP

 

[letterbeacon]

Thanks Dave!  It's winging its way over to me from the States at the moment, but I'll be careful when it gets here.

I was having another browse on the internet before I read this post and found this pdf which shows a method using two pots:  http://www.jaycar.com.au/images_uploaded/shuntmul.pdf

What's the difference?  Is the first pot used to find out the range of the meter?  But as I already know it (0-50uA), I don't need to use it?

 

[DaveP]

Letterbeacon,

I've got that pdf too, its very good.

The first pot is to control the current, the second is to find the resistance, you need both.  My method just used Ohms law, their way might be easier.  You will need to fully understand the guide to sort out a suitable shunt, just follow the instructions.  The shunt will be approximately a tenth of the meter resistance.

best
DaveP

 

[letterbeacon]

I just want to double check I have the equation for calculating the shunt resistor correct in my head.

Here are the equations, according to the pdf:

Is = Itotal -Im

Where Is is the current taken by the shunt, Itotal is the total current equal to the new effective FSD we want to give the meter, and IM is the meter’s own basic FSD.  My meter is 50uA and I want it to read 500mA

450mA = 500uA - 50uA

The next equation is for the shunt resistor.  I don't know the meter resistance of my meter yet, but let's put 3500r for now as that's the resistance they give their 0-50mA meter in the pdf.  So the equation is:

Rs = Rm / (Is/Im)

Where Rs is the shunt resistor value and Rm is the meter resistance.  So I work that out as

70r = 3500r / (500uA / 50uA)

So, in this instance, I would need a shunt resistor of 70 ohms.

Is that correct?  Do I need to write the uA as A?  ie. .000500A?