I have this idea in my head that if I use an SS rectifier (& Voltage Regulator?) & R/C filter network across the 6.3v C.T. transformer sec. for a 12.6v DC supply. Will the tubes care if the heaters are fed DC instead of AC? Will this cause any problems, sonic or otherwise, to the input signal? I'm just considering this because there are times when it could be nice to have readily available DC supply for incorporating an SS boost circuit or a microprocessor controlled relay to switch amp channels or would I be better off using an additional transformer wired (in parallel?) with the amp's main transformer for a DC supply?
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Is it ok to convert 6.3v C.T. sec. to DC?
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Running the filaments off DC is common practise in hi-fi because it improves the noise performance. However, you won't be able to get 12.6 VDC unless you use a voltage doubler. Then you face the problem of how to get 6.3V for the tubes that need it (eg, the output tubes). Furthermore, even 12.6 VDC is often not enough if you want to maintain adequate headroon in the audio signal chain.
All in all, it's usually easier to just use another transformer...
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Originally posted by TransLucid View PostThen you face the problem of how to get 6.3V for the tubes that need it (eg, the output tubes).
You can also leave the heaters AC, then run the DC supply right off the source. That's how I do my relay supplies with 5V relays.
Now for a microprocessor circuit you need regulated 5V so you'll need to use a 7805 regulator to get that once it's rectified and filtered once. Common practice is to use a 9VAC transformer for a +5V regulated application to minimize the chance of regulator dropout. But since most uC driven relays are transistor driven, the uC won't draw enough current from the regulator to cause dropout so you'll be good there. You'll just need a separate transformer and DC supply to power the relays.
Now the other thing is that since most heater windings use a grounded center tap, you MUST not let the negative side of the DC supply that's powered from the heater winding contact ground. It MUST be a floating DC supply. If you use a separate DC supply to power the relays and you want to drive them with a uC via transistors, the relay supply negative side must be referenced to the negative side of the 5V regulated supply that's powered via the heater supply.Last edited by Wilder Amplification; 01-23-2011, 05:47 AM.Jon Wilder
Wilder Amplification
Originally posted by m-fineI don't know about you, but I find it a LOT easier to change a capacitor than to actually learn how to play wellOriginally posted by JoeMI doubt if any of my favorite players even own a soldering iron.
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Originally posted by mooreamps View PostI don't normally post something like this. But, this is the circuit I use...
- the neon only comes on when the fuse is blown
- the output voltage is not 12.6Vdc as shown; it's more like 16.4 before loading and ripple and sags from there
Since the idea of using DC heaters is to not put AC on your heaters, you'd want to suppress the rectification ripple to a pretty substantial degree, keeping the average voltage from sagging a lot. The voltage will sag because of the high peak currents of the rectifiers. How much it will sag depends a lot on the transformer and rectifiers.
If the transformer were perfect, the 12.6Vac would be converted by the full wave rectification to 12.6*1.414 = 17.8Vdc, and you'd lose two diodes' drop in the rectifiers, about 1.4V, giving 16.4Vdc. That's at nominal line and before transformer sag. That still leaves almost four volts of DC that need to be dropped before it's healthy for 12V-rated tube filaments to eat it. Maybe the transformer sags 3.8/17.8 =
0.213... or 21.3%. It's pretty rough treatment to make a large transformer sag that way.
16.4 is a good DC voltage to run a 12V regulator from, so a 7812 would make a nice addition.
Running a heavily loaded DC rectifier filter to make it sag leaves you open to overvolting the tube heaters connected to the supply if you remove a tube or two; it's the loading that keeps the voltage down (if the circuit works as drawn).
So, unless the circuit is drawn incorrectly, or there is a lot of detail left out about how to get the DC down to 12.6Vdc, I would advise caution about using it to power expensive tubes. They may not last long.Amazing!! Who would ever have guessed that someone who villified the evil rich people would begin happily accepting their millions in speaking fees!
Oh, wait! That sounds familiar, somehow.
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Originally posted by R.G. View PostIt's worth noting that for the circuit as drawn:
- the neon only comes on when the fuse is blown
- the output voltage is not 12.6Vdc as shown; it's more like 16.4 before loading and ripple and sags from there
Since the idea of using DC heaters is to not put AC on your heaters, you'd want to suppress the rectification ripple to a pretty substantial degree, keeping the average voltage from sagging a lot. The voltage will sag because of the high peak currents of the rectifiers. How much it will sag depends a lot on the transformer and rectifiers.
If the transformer were perfect, the 12.6Vac would be converted by the full wave rectification to 12.6*1.414 = 17.8Vdc, and you'd lose two diodes' drop in the rectifiers, about 1.4V, giving 16.4Vdc. That's at nominal line and before transformer sag. That still leaves almost four volts of DC that need to be dropped before it's healthy for 12V-rated tube filaments to eat it. Maybe the transformer sags 3.8/17.8 =
0.213... or 21.3%. It's pretty rough treatment to make a large transformer sag that way.
16.4 is a good DC voltage to run a 12V regulator from, so a 7812 would make a nice addition.
Running a heavily loaded DC rectifier filter to make it sag leaves you open to overvolting the tube heaters connected to the supply if you remove a tube or two; it's the loading that keeps the voltage down (if the circuit works as drawn).
So, unless the circuit is drawn incorrectly, or there is a lot of detail left out about how to get the DC down to 12.6Vdc, I would advise caution about using it to power expensive tubes. They may not last long.Jon Wilder
Wilder Amplification
Originally posted by m-fineI don't know about you, but I find it a LOT easier to change a capacitor than to actually learn how to play wellOriginally posted by JoeMI doubt if any of my favorite players even own a soldering iron.
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The output voltages were measured when running a full load. Running a regulator would not do anything... You really have to spec the current rating of the transformer for the amount of load you put on it ; as with any other un-regulated PS. Besides, I like having the fault lamps illuminate when power is energized...
Besides, why would you "remove" a tube or 2 ?
-gLast edited by mooreamps; 01-23-2011, 09:24 AM.
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Originally posted by mooreamps View PostThe output voltages were measured when running a full load.
Originally posted by I said:If the transformer were perfect, the 12.6Vac would be converted by the full wave rectification to 12.6*1.414 = 17.8Vdc, and you'd lose two diodes' drop in the rectifiers, about 1.4V, giving 16.4Vdc. That's at nominal line and before transformer sag. That still leaves almost four volts of DC that need to be dropped before it's healthy for 12V-rated tube filaments to eat it. Maybe the transformer sags 3.8/17.8 = 0.213... or 21.3%. It's pretty rough treatment to make a large transformer sag that way.
It's also rough treatment for the caps to make the ripple across them so large if the sag is obtained that way. But if it works for you and you're happy with the reliability you get, as one famous man said, it's no skin off my nose.
Running a regulator would not do anything...
However, if the DC didn't sag that far, a regulator *would* give you predictability of operation over line and load changes, which is engineer-speak for what happens when you either pull a tube or two out (more on that one later) or when the AC line voltage changes from an 85V brownout condition to 130Vac like it sometimes does. I suspect the filament voltages you derive from this circuit will do some gynmastics when the AC line changes. Has to, in fact.
You really have to spec the current rating of the transformer for the amount of load you put on it ; as with any other un-regulated PS.
Besides, I like having the fault lamps illuminate when power is energized...
Originally posted by as I said:- the neon only comes on when the fuse is blownBesides, why would you "remove" a tube or 2 ?
When either of those happens, the heater voltages on the other tubes rise because the only thing holding the voltage down is the current pulled by the total load of the tube heaters. Remove one or have one burn out, and the other heaters get more voltage, and hence more current, and this stabilizes with the other heaters running hotter and burning out even sooner than they would otherwise.
This is in fact the problem with non-regulated power supplies - their load voltage varies with the load unless the load is insignificantly small (i.e. large resistance) compared to the voltage source resistance. In that case, the voltage variations are large. That becomes a problem when the load is degraded or damaged by voltage variations, like heater filaments are.
Hope that helps.Amazing!! Who would ever have guessed that someone who villified the evil rich people would begin happily accepting their millions in speaking fees!
Oh, wait! That sounds familiar, somehow.
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Well, he asked a question, and I gave him an answer.... More specific answers require more specific questions...
-g
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I agree that regulation (and running the filaments at the proper voltage) is preferable.
As for providing 6.3V for the output tubes, wiring the filaments in series seems reasonable (as long as no one pulls a tube as above). How about, as an alternative, running one output tube filament off of the 6.3V output (to ground) and the other off the 12.6V and 6.3V outputs? This would raise the filament of one tube by 6.3VDC, but that shouldn't make a difference. You could also run the output tube filaments off of 6.3VAC directly from the PT secondary.
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Originally posted by TransLucid View PostHow about, as an alternative, running one output tube filament off of the 6.3V output (to ground) and the other off the 12.6V and 6.3V outputs?
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In the case of running the heaters for power tubes directly off AC and the preamp tubes off of a regulated DC supply, how is the Filament CT from the transformer handled relative to the power tubes? Is it just grounded as usual even though the DC supply is grounded seperately at the rectifier?
Thanks for the advice.Mandopicker
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Originally posted by mooreamps View PostWell, he asked a question, and I gave him an answer.... More specific answers require more specific questions...
It's kind of a be-good-to-the-beginners program. That's why they ask. Help them out where you can.
In the case of running the heaters for power tubes directly off AC and the preamp tubes off of a regulated DC supply, how is the Filament CT from the transformer handled relative to the power tubes? Is it just grounded as usual even though the DC supply is grounded seperately at the rectifier?
There is another way. You can tie the heater AC winding to ground directly, make DC from this balanced AC supply, and ground the middle of the DC through resistors or use a dual capacitor setup to "ground" the middle of the derived DC. That works too.
You can also perch this CT/ground on a positive DC voltage to help suppress electrons back into the heater.
Mostly you have to be sure SOME point on the heater supply, AC or DC side, is tied to ground through a wire or low-value resistor.Amazing!! Who would ever have guessed that someone who villified the evil rich people would begin happily accepting their millions in speaking fees!
Oh, wait! That sounds familiar, somehow.
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Originally posted by R.G. View PostThat's true. I always try to err on the side of providing beginners more information than they have asked for, because they don't yet have the experience to wonder why THIS does THAT. They're prone to simply building something directly from the schematic, not knowing that all schematics have unstated assumptions behind them.
It's kind of a be-good-to-the-beginners program. That's why they ask. Help them out where you can.
I sometimes use analogies that relate complex electronics concepts to the simplest of things that normal people encounter in their everyday lives. After a few times of explaining it to them in this fashion the light bulb usually comes on.Jon Wilder
Wilder Amplification
Originally posted by m-fineI don't know about you, but I find it a LOT easier to change a capacitor than to actually learn how to play wellOriginally posted by JoeMI doubt if any of my favorite players even own a soldering iron.
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Well, then ahhh,, hopefully we can rely on you and the others to help fill in the gaps... :}
-g
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