Question about Low DCR power transformer for filament supply

[Tube747]

Hello RATS!
I am looking for a low DCR power transformer for filament supply of rectifier
tube because some one told me that the low DCR power transformer the better
regulation it is.

I want to use the 6D22S rectifier tube (2 for full wave rectification).
Therefore, the power transformer for the filament of this tube is going to be
6.3V 4A, will it be ok?

Each 6D22S with the heater spec. 6.3V 0.9A.

Also, if I order a power transformer which has a lowest possible DCR on the
secondary winding with centre tap and lowest possible ACR on the primary
winding, will it be damage the tube (6D22S)?

I saw a transformer like that from my friend's place for the 6D22S which look
like a 300VA transformer and quite heavy as a filament transformer for only
6.3V @ 4A. Any problem if using such as transformer with the rectifier tube
like 6D22S?

Thanks!

[Fred Nachbaur]

Tube747 wrote:
> Hello RATS!
> I am looking for a low DCR power transformer for filament supply of rectifier
> tube because some one told me that the low DCR power transformer the better
> regulation it is.
>
> I want to use the 6D22S rectifier tube (2 for full wave rectification).
> Therefore, the power transformer for the filament of this tube is going to be
> 6.3V 4A, will it be ok?
>
> Each 6D22S with the heater spec. 6.3V 0.9A.
>
> Also, if I order a power transformer which has a lowest possible DCR on the
> secondary winding with centre tap and lowest possible ACR on the primary
> winding, will it be damage the tube (6D22S)?
>
> I saw a transformer like that from my friend's place for the 6D22S which look
> like a 300VA transformer and quite heavy as a filament transformer for only
> 6.3V @ 4A. Any problem if using such as transformer with the rectifier tube
> like 6D22S?
>
> Thanks!

The DC resistance of a filament winding has very little to do with
anything, at least directly. Since the load is essentially constant,
there's no real concern about "regulation" of a filament transformer.

By choosing a transformer that has the correct voltage at the rated
current, you will automatically take into account DC resistance and
other transformer losses. If you choose a transformer with a far higher
rated current than what you actually need, your voltage will tend to be
higher. For 6.3 volts at 1.8 amperes a 4A transformer would be more than
adequate, the voltage might be a bit higher than ideal, but if you get
one from a reputable manufacturer it shouldn't exceed the tubes' rating.

On the other hand, the DC resistance of the plate winding (HT winding)
can play an important part in determining final output voltage,
regulation, and even the maximum safe input capacitance of the filter
network. The effects of this are easy to simulate using a program like
Duncan Munro's "PSU Designer II".

Cheers,
Fred
--
+--------------------------------------------+
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| Projects, Vacuum Tubes & other stuff: |
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[Ronald]

6.3V should have rang the bell .... It's indirect heated , so forget about
my argument !!
So it's a heater supply , not a fiament supply !!!

Ronald .

"Ronald" > schreef in bericht
...
> I would use a 12.6V 0.9A and take the CT between the tubes , not on the
> tranny .
> Gives a cleaner sound since the trash from the mains get less chance
.......
>
> "Mikkel C. Simonsen" > schreef in bericht
> ...
> > Tube747 wrote:
> > >
> > > Hello RATS!
> > > I am looking for a low DCR power transformer for filament supply of
> > > rectifier tube because some one told me that the low DCR power
> > > transformer the better regulation it is.
> >
> > I would say that regulation is a non-issue here. The load is constant so
> > regulation doesn't matter.
> >
> > > I want to use the 6D22S rectifier tube (2 for full wave
> > > rectification). Therefore, the power transformer for the filament of
> > > this tube is going to be 6.3V 4A, will it be ok?
> >
> > Yes, probably. What you need is a transformer that will supply the tubes
> > with 6.3V with a 1.8A load. It doesn't really matter if the transformer
> > has a 2A or 200A rating as long as the voltage is correct at 1.8A.
> >
> > I would get a 3 or 4A transformer.
> >
> > Best regards,
> >
> > Mikkel C. Simonsen
>
>

[Mikkel C. Simonsen]

Ronald wrote:
>
> 6.3V should have rang the bell .... It's indirect heated,
> so forget about my argument !!
> So it's a heater supply , not a fiament supply !!!

It seems a lot of people don't know the difference between a heater and
a filament... ;-)

Best regards,

Mikkel C. Simonsen

[FC5687]

>Each 6D22S with the heater spec. 6.3V 0.9A.

Sorry I make a mistake, the tube has a heater spec. 6.3V 1.9A, not 0.9A!

[FC5687]

>On the other hand, the DC resistance of the plate winding (HT winding)
>can play an important part in determining final output voltage,
>regulation, and even the maximum safe input capacitance of the filter
>network. The effects of this are easy to simulate using a program like
>Duncan Munro's "PSU Designer II".

Hi Fred!
If I order a power transformer for the HT supply with a low DCR on the
secondary winding, is it necessary to have a low "ACR" on the primary winding?

Cheers,
Tube747

[Fred Nachbaur]

FC5687 wrote:
>>On the other hand, the DC resistance of the plate winding (HT winding)
>>can play an important part in determining final output voltage,
>>regulation, and even the maximum safe input capacitance of the filter
>>network. The effects of this are easy to simulate using a program like
>>Duncan Munro's "PSU Designer II".
>
>
> Hi Fred!
> If I order a power transformer for the HT supply with a low DCR on the
> secondary winding, is it necessary to have a low "ACR" on the primary winding?
>
> Cheers,
> Tube747

No, you don't need to worry about it. As far as your circuit is
concerned, you can lump the primary and secondary DCR's, as well as all
other losses, into a single equivalent resistance on the secondary side.
In other words: Measure open-circuit voltage, then put a reasonable load
onto it and measure the loaded voltage. Subtract loaded voltage from
open-circuit voltage, and divide by load current. This gives the
effective series resistance on the secondary side, which is the value
you'd put into your simulator.

Transformers are usually rated in terms of output voltage at a given
maximum current. It's very rare that a manufacturer would bother
publishing the DC resistances, and other loss factors, because it would
not be meaningful to very many end users.

The thing you have to watch out for, that trips up a lot of people
building power supplies, is that the DC current you can get out of a
transformer (using a capacitor-input filter) is somewhat less than the
maximum rated AC current. This is because of the high current spikes
required to recharge the capacitor; since power and therefore heating is
proportional to the square of the current, there will be a lot more loss
during these high-current spikes.

I stay with reasonable input capacitor values (usually around 100 uF)
and choose a transformer with 50% higher rating than the DC current I'll
be drawing, so far that rule of thumb has always worked out. However, if
I'm using Hammond transformers, I'll push them closer to the limit
because they are more conservatively rated.

Cheers,
Fred
--
+--------------------------------------------+
| Music: http://www3.telus.net/dogstarmusic/ |
| Projects, Vacuum Tubes & other stuff: |
| http://www.dogstar.dantimax.dk |
+--------------------------------------------+

[Mikkel C. Simonsen]

Fred Nachbaur wrote:
>
> I stay with reasonable input capacitor values (usually around 100 uF)
> and choose a transformer with 50% higher rating than the DC current I'll
> be drawing, so far that rule of thumb has always worked out. However, if
> I'm using Hammond transformers, I'll push them closer to the limit
> because they are more conservatively rated.

As far as I remember the current ratings on Hammond transformers (at
least the plate winding) are DC ratings based on a cap input filter.

Best regards,

Mikkel C. Simonsen

[Tube747]

>No, you don't need to worry about it. As far as your circuit is
>concerned, you can lump the primary and secondary DCR's, as well as all
>other losses, into a single equivalent resistance on the secondary side.
>In other words: Measure open-circuit voltage, then put a reasonable load
>onto it and measure the loaded voltage. Subtract loaded voltage from
>open-circuit voltage, and divide by load current. This gives the
>effective series resistance on the secondary side, which is the value
>you'd put into your simulator.
>

Are you meaning that the ACR is not what I concern as long as power transformer
goes and DCR on the secondary winding is what I need to pay attention on?

>Transformers are usually rated in terms of output voltage at a given
>maximum current. It's very rare that a manufacturer would bother
>publishing the DC resistances, and other loss factors, because it would
>not be meaningful to very many end users.
>
>The thing you have to watch out for, that trips up a lot of people
>building power supplies, is that the DC current you can get out of a
>transformer (using a capacitor-input filter) is somewhat less than the
>maximum rated AC current. This is because of the high current spikes
>required to recharge the capacitor; since power and therefore heating is
>proportional to the square of the current, there will be a lot more loss
>during these high-current spikes.
>
>I stay with reasonable input capacitor values (usually around 100 uF)
>and choose a transformer with 50% higher rating than the DC current I'll
>be drawing, so far that rule of thumb has always worked out. However, if
>I'm using Hammond transformers, I'll push them closer to the limit
>because they are more conservatively rated.
>

I am currently using tube rectification and choke input right after the power
transformer in seqeunce. Therefore, do I have to worry about the rectifier tube
damage by inrush current due to the power transformer with low DCR on the
secondary winding?


Thanks again!


>Cheers,
>Fred