Insight into "U47-inspired" with EF12 tube?

[soapfoot]

After showing someone my first draft schematic (pictured above), they expressed concern with all the resistance on the B+ rail.

Do you smart folks think that's a problem?  Is there anything else about my idea that looks wrong?

Thanks again,

--brad

 

[boogietube]

Nothing looks wrong to me, but I'm not really qualified. The only problem might be sourcing 370 and 420 ohm 1 watt resistors....

 

[moamps]

Hi,
that 370 ohms in front of the 6V/200mA heater doesn't look correct to me.
Regards,
Milan

 

[MagnetoSound]

Doesn't look right to me.

B+, what DC voltage are you expecting to see after rectification/smoothing?

Once you have this you can work out the total resistance required for the drop to 105 @ 0.55mA, and then work out the required drop at each filter stage.

The trimmer if you want one, can be simply a rheostat towards the end of the string, you don't need to have a shunt there, except maybe a high value to bleed off in case of no load.

 

[soapfoot]

I was looking for 105VDC.

I asked the advice of a guy who taught me some stuff.  I haven't talked to him in years and he's busy.  But he said he'll draw me up something when he has time.  I hope he does. 

 

[gary o]

Thankyou Odbfs I have been breadboarding without the board bit of a scary mess but helps me to understand whats going on I no Im a bit slow but am getting there

quick drawn schematic.....requirements little bit different to Soap foot looks like he is doing ok there good luck soap foot.....I have 240VAC 250ma transformer I want to use to power a MK47 mic 105VDC 40Ma Im getting close with this circuit dont know if Im doing anything wrong with the resistances here

http://good-times.webshots.com/photo/2160397500103387484DfTAMC

from learning from this thread I now suspect a mic PSU I made some time back is under powering one of my mics....

Question what will happen if a tube mic cannot draw enough current from the PSU for its B+ ?

Anyway I have extra voltage in my PSU so I want to add a 120VDC output for a C12 type tube mic Im still a little confused about current draw....Im not sure what the C12 B+ requires but Im sure it will be a lot less than 40ma....so do I need to limit the current or does the mic only take what it needs?

thankyou for your patience

 

[0dbfs]

If you close that 2k3 switch (or just connect the 2k3 across the output of the supply) that approximates the load the mic presents and you should be able to tweak until you get your 105V.

The mic will only draw the current it needs. If your supply can provide more that is fine. If you try to draw more than is available, something will most likely burn up (resistor, transformer, etc).

Cheers,
j

 

[MagnetoSound]

I was looking for 105VDC.

Yes, you want 105V at the output but you need to know what the voltage will be after the rectifier, at the first smoothing cap. Then you can calculate the drop you'll need to get your 105V (use Ohms Law again, drop voltage/0.55mA, to find the total R).

Rectifier relationships are good to know: http://www.davidbridgen.com/rects.htm

With a bridge rectifier you are going to have a DC of approx 1.4 times the RMS voltage at the secondary, so ... ?

 

[gary o]

Thanks J yep i did the 2K3 load resistor thing I can adjust my last resistor before the pattern section to provide the 105VDC....Im now getting greedy and can see the opportunity to get 120VDC for the plate of my C12 type mic....so as you say the C12 or any load draws only what it needs in power as long as enough is available ........

Soapfoot can you bread board your circuit....

cheers

 

[soapfoot]

MagnetoSound: That's useful info.  Thanks again, sincerely.

I'm going to digest this and try again.  I admit that I'm getting a little bit overwhelmed/frustrated so I might take a step back and come back tomorrow with a clear head and try again.  I wish I had more knowledge so I knew what I was doing.

Thanks again for all your help so far.

 

[soapfoot]

Soapfoot can you bread board your circuit....

I intend to eventually, but not yet.  I want to make sure I at least have some clear idea and some better understanding so that I make sure to at least purchase the correct "expensive" parts (i.e. choke, transformer). 

 

[soapfoot]

I was looking for 105VDC.

Yes, you want 105V at the output but you need to know what the voltage will be after the rectifier, at the first smoothing cap. Then you can calculate the drop you'll need to get your 105V (use Ohms Law again, drop voltage/0.55mA, to find the total R).

Rectifier relationships are good to know: http://www.davidbridgen.com/rects.htm

With a bridge rectifier you are going to have a DC of approx 1.4 times the RMS voltage at the secondary, so ... ?

OK, so here's what I have so far...

Using a choke input filter for the filament/bias supply, given a PT secondary of 12VAC and a filament current draw of 200mA, after the choke I should have 10.8VDC and the current draw should now be 188mA.

Using a capacitor input filter for the HT supply, given a PT secondary of 160VAC and a current draw of .55mA, after the first capacitor I should have 225.6VDC, and a current draw of 0.34mA.

Is this correct?

If so, what's the best way to drop 120 volts to get to the final 105V supply?  High resistor values in the pi filters?  A simple voltage divider or series resistor at the end?  It seems that 360K of total series resistance would drop that much, but I don't know if that's the right way to go.  Something tells me "probably not."

Thanks again.  I'm ready to attack it again.  My "step back" was short-lived.  I want to learn and get this thing.

 

[gary o]

Im struggling with the ohms law maths myself so I might not be much help but Im not sure the current you want to draw when you say .55ma & .34ma so thats in the region of half of 1 milliamp?...... are you doing your ohms law sums in Amps.....Im my case I want 40ma which is I think    .045A

so 1 mA = 0.001 A  ....so your .55 = 0.00055 Amps

Sorry if I got this completly wrong

 

[gary o]

Just been fiddling with the numbers & im adding up the same as you Soapfoot In your B+ circuit.....takes so long to sink into my brain

 

[soapfoot]

OK, I tried again.

Does this look any better?

 

[soapfoot]

By the way, my friend who offered to help me was really talking up regulation using 3-pin regulators.

However, over on PSW in a discussion about power supplies, David Bock says that he doesn't care for that type of design.

So who knows.  I'm kinda lost, but at least I'm learning.  I don't know what I'm going to do in the end.

 

[soapfoot]

So I was looking at the old Neumann NG schematic.  This PSU has a really good rep for being very good and quiet.  It appears to have some sort of twin-coil choke.

So I looked and found a Triad choke that's just a few bucks.  15H at 20mA.  I drew up a schematic with two of these on the B+ side.  I figured the 20mA rating would be plenty, since unlike the original U47, the B+ side of this circuit would only draw about half a mA of current.  

Aside from being a slight extra expense (about 25 bucks extra) and maybe a bit of overkill, anyone see anything wrong with this?

 

[soapfoot]

Updated the schematic in the post above--

I downloaded an emulator for my computer so that I could download a simulator for the circuit. 

Downloaded PSU Designer II from Duncan Amps. http://www.duncanamps.com/psud2/index.html

Had to download a program to allow me to emulate Windows 98 on my Mac, though, but I got it working.  Handy little program, that is.

This allowed me to correct a few voltages on the schematic, and to see that I wouldn't have been able to get the voltage up where I needed it running just an LC filter for the first stage of the filaments like I had originally planned.  So I'll sacrifice a little bit of regulation to gain a little more voltage, and run a Pi filter instead. 

 

[zebra50]

Something funny going one... that looks like a lot of current through the 820R! Imagine a scenario where you have 105V and that pot in the middle. R = 820 + 50 R

V=IR, I = 105V / 870R = 0.12 A

Power = I-square-R = 0.12 * 0.12 * 870 = 12.7 watts!

Did you mean 100K and 820K for those positions?


Why not use the MK7 / MK47 design and tweak the resistors to get your 105V? It's very close in function to what you've drawn, and it is well tested.

 

[soapfoot]

Good eye.  Yeah, I might do 100k and 820k instead.  I might also use/adapt the MK7 design.

Just curious about that one though... what is the purpose of the .1 cap and 3W 4.7R resistor before the first Pi filter?  I've never seen that configuration before and it looked strange to me.