Bias supply filtering

[Bruce / Mission Amps]

Scott - Both. This amp has a hum problem,AND I'm curious how to pull this off in general without messing up the supplies that are drawn off the heaters, because there are many applications for this.


Bruce........I appreciate the response but you're not answering the question I asked. I'm not asking how to elevate the heaters, thats covered elsewhere on the forum. I know how to do that, at least when a heater center tap is present, which is the case on this amp. I am curious how to do it when no heater center tap is present, but thats a different issue and not the question I'm asking here.

Look again at the 2005 schem you attached. See the switching supply? See where it says "From fillament supply"? If you elevate the heaters and run it through that, you are going to end up not with +/-3V, but with +43v/-3v are you not? That will fry that switching chip in nothing flat. The only way I see around that is maybe adding blocking caps between the elevated heater supply and the switching rectification, but again, STILL, no one seems to have an answer as to whether that will work or not.

I'm *NOT* asking how to elevate heaters. I'm asking how to avoid kiling the switching system that is run off the heater supply while elevating the heaters. Seems no one is really clear on that point.

Use a .22uF@400v coupling cap from each of the filament supply wires to the diodes to block that section from any DC bias.

[wizard333]

There we go!

Thats what I was thinking, but wanted some confirmation from someone who'd looked at this issue before.

Thanks!

[Enzo]

And keep in mind that hum is not generic, and any hum remedy only works on its own kind of hum. SO adding filters to the bias supply will do nothing to reduce heater hum, for example. And elevating heaters will have no effect on hum from ripple currents through ground.

It pays to know where all the sources of hum in your amp are located. Elevating heaters will ONLY fight hum that comes from heater to cathode coupling in the tubes.

[Steve Conner]

Bruce's idea seems generally the right approach, but it won't work, because capacitors block DC, and they're in series with the diodes, whose sole purpose is to generate DC. So the diodes won't be able to supply any current to the load.

You'd need another two diodes to ground ala charge pump/voltage doubler to make it all add up.

Also I think they ought to be a good deal bigger than .22uF. Maybe 22uF or even 220.

[wizard333]

The diodes are after the caps though so they'd still see the AC and rectify it wouldnt they? THe only thing the caps should be blocking is the DC elevation; the normal AC signal from the heaters should pass right on by and go to the diodes shouldn't it?

Sequence being: DC Elevated Heater Supply (6vAC & +40v DC) - BLocking Caps (Should block the +40VDC and leave the 6vAC) - Rectifiers (Shoud leave ~+/-3v DC + ripple) Filter caps (filter the ripple and leave fairly clean +/-3vDC) - Switching Chip and/or Relay.

What am I missing there?

Also, do you have an example schematic of what you're talking about with the extra 2 diodes?

[Bruce / Mission Amps]

Bruce's idea seems generally the right approach, but it won't work, because capacitors block DC, and they're in series with the diodes, whose sole purpose is to generate DC. So the diodes won't be able to supply any current to the load.

You'd need another two diodes to ground ala charge pump/voltage doubler to make it all add up.

Also I think they ought to be a good deal bigger than .22uF. Maybe 22uF or even 220.

OK ya got me! ha ha...
I haven't tried this of course, but note the bias supply in the AC100 diagram included here... .it is a feed from the hi-v tap through a .25uF cap... seems like something like this should work with a diode from each side and the diodes reversed too.
Scaled down to 6.3vac, wouldn't it also work for the other circuit?

[wizard333]

Still looking for a clarification of what Steve said.

[BackwardsBoB]

The current through the diode is in one direction (better be), and it charges the cap until the voltage on the cap keeps the anode of the diode negative compared to the cathode for the full AC cycle, and the diode stops conducting. With any load, you'll only get a nidge of voltage or current on the diode output to compensate for leakage, though the blocking cap will stay nicely charged.

[Steve Conner]

Basically what Backwards Bob said, but I've added a picture to clarify the solution.

Once you've added the two diodes, the new configuration of diodes is just a bridge rectifier, with the two heater lines connected via capacitors to the AC terminals, and the DC output taken from the DC terminals. I've indicated the polarity, in case you want to use electrolytics.

[wizard333]

Thanks Steve.

The stock schematic shows what you drew out but with the 2 extra diodes; in the case of the power supply for my lead boost I'm already using a self contained bridge rectifier, with the AC leads connected to pins 2&7 of the power tube, then filtering the rectified + and - sides through 1kuf filters and powering a relay to switch a separate midrange pot in and a transistor to switch in a CK.

In that case, if I'm putting the bridge rectifier after the caps, do I need extra diodes?

[Steve Conner]

No, all you should need to do is just put the two DC block caps in between pins 2&7 and the bridge rectifier's AC input terminals.

The stock schematic shows what you drew out but with the 2 extra diodes

I'm confused. If you mean the "stock schematic" of a 2203 that you posted earlier, the two extra diodes aren't in it. If they had been, we wouldn't be having this conversation.

[wizard333]

No I meant having added the 2 extra diodes to what was already there. Sorry for the confusion.

If you were using a large cap like a 22uf as a blocker, and using polarized electrolytics, which side would get the negative? The tube socket pin or the bridge rectifier input? And why?

Thanks!

[Steve Conner]

The bridge rectifier input would get the negative, and the heater wires the positive, as I showed in my picture.

This is because the heater elevation voltage appears across the capacitors, and it's positive.

As you might guess the voltage rating of electrolytics needs to be greater than the elevation voltage, so 63V etc.

[wizard333]

That makes sense. I should have puzzled that out but its late and my brain isn't firing on all cylinders.

Thanks Steve!