PT question

[rutledj]

I'm using a Hammond 372jx xformer that is rated at 300-0-300v 250ma to run a 2x 6l6 amp. Looking at the Hammond specs, this is about the highest rated (current) available until you get to a 400-0-400 output.

It appears, however, that this doesn't provide enough current to run 2 6l6's. With the bias set around 33ma the PT gets quite hot when playing with vol around noon. The remaining ckt contains 5 12ax7's, silicon bridge ckt, choke.

I guess my question is should I be concerned about the heat on the xformer?

Thanks,
Rut

[sliberty]

Don't EL34's pull more current than 6L6's? The reason I ask is that 250mA is enough for an Express with EL34's, so I am thinking it should be enough for a pair of 6L6's as well. In fact, if you think about some of the SF Fenders, like the Bandmaster, that ran 6L6's, their power trannies were probably considerably smaller. I am thinking your problem is not that the trannie has too low a current rating, but rather, something else is going on in that amp.

[Andy Le Blanc]

12ax7 require .3 A each..... so thats 1.5 A ...... two 6l6 at .9 A is 1.8A total
so the tubes are useing 3.3A.....the 372jx has a 8 A 6.3v winding .....thats ok
you said a bridge rectifier......... this tranny has a 4 A 5v winding ...are those leads secure?
this tranny has a universal primary for 100v ac to 240v ac..... is it wired correctly for your area?
the rest of your description gets fuzzy.... do you mean a full wave bridge with
a choke input load ?....... the limits for tube bias are published as static plate
dissipation in WATTS .... what does 33ma mean? in relation to what?
put a 1 ohm resistor in the circuit, measure the ma. across this .....
measure the plate VOLTS....... convert the ma. to AMPS "A" and multiply by "V" to get "W"
then youll know ....... then present your data so that the rest of can double
check it and go ..... hmmmmmm...

sorry about the rant.... its not ment as a snubb....

you should worry about excessive heat..... the 372jx has a VA of 243
and a rated DC ma. of 250..... this should supply a pair of 6l6 ...
what type of 6l6?.... a full wave bridge rectifier with a capacitor input load
will give a DC ma. of .62 X the rated ma. of the secondary....
that means your supply may be only in the nieborhood of 150 ma.
that'll be fine for most 6l6 types but the 6l6-GC at higher plate voltages
a pair can draw over 200 ma. under maximum-signal plate current conditions
this might be whats going on with your tranny..... you can get more available
ma. by going to a full wave v.s. a full wave bridge but your dc volts will drop to half
give us some better readings with plate volts and a static dissipation figure
or a zero signal plate current

[bluefireamps]

The Hammonds are designed to use a full wave rectifier, not a bridge rectifier. That transformer shouldn't get hot with that tube compliment.
Dave

[Ears]

from O'Connor's "extended " Hammond specs
http://www.londonpower.com/hammond/classic1.htm

372JX

High voltage plate supply:
end-to-end voltage and total power rating: 600VCT--165.6VA ;

rated mADC assuming capacitive input filter and half-bridge full-wave rectification: (250mADC-RATED);

DC output possibilities using both half-bridge full-wave rectification (low output voltage) and full-bridge full-wave rectification (high voltage):

300V--552mA or 600V--276mA
424VDC-390mADC 848VDC-195mADC

[Ears]

What confuses me about the figures from OConnor above is the figures

250mADC-RATED (assuming capacitive input filter and half-bridge full-wave rectification)

and 300V--552mA (for the DC possibility with half-bridge full-wave rectification (low output voltage))

Why are they different?

[Andy Le Blanc]

that does sound confusing..... half wave is exactly that .... rectification of half the ac wave form......
full wave means what it says..... and the bridge with four diodes..... etc
the added description is confusing....... a full wave configuration does look like half a bridge
what important is the load that the rectifier "see's".....resister..... capacitor.... or choke....
so......
a full wave rectifier..... thats two rectifiers and a center tapped secondary
like your used to seeing in old fenders..... with a tube or silicon diodes
with a resistive load gives you....

V D.C. = 0.45 x sec. V A.C.
I D.C. = 1.27 x sec. I A.C.

with a capacitor input......

V ( avg) D.C. = 0.45 x sec. V A.C. V (peak) D.C. = 0.71 x sec. V A.C.

I D.C. = 1.00 x sec. I A.C.

and with a choke input...

V ( avg) D.C. = 0.45 x sec. V A.C. V (peak) D.C. = 0.45 x sec. V A.C.

I D.C. = 1.54 x sec. I A.C.

this is most likely what is ment by possibilities

[mlp-mx6]

If you have a bridge rectifier connected to ground AND the center tap connected to ground, that is NOT GOOD. Convert to a full-wave non-bridge rectifier and ONLY connect the center tap to ground.

[Ears]

that does sound confusing..... half wave is exactly that .... rectification of half the ac wave form......
full wave means what it says..... and the bridge with four diodes..... etc
the added description is confusing....... a full wave configuration does look like half a bridge
what important is the load that the rectifier "see's".....resister..... capacitor.... or choke....
so......
a full wave rectifier..... thats two rectifiers and a center tapped secondary
like your used to seeing in old fenders..... with a tube or silicon diodes
with a resistive load gives you....

V D.C. = 0.45 x sec. V A.C.
I D.C. = 1.27 x sec. I A.C.

with a capacitor input......

V ( avg) D.C. = 0.45 x sec. V A.C. V (peak) D.C. = 0.71 x sec. V A.C.

I D.C. = 1.00 x sec. I A.C.

and with a choke input...

V ( avg) D.C. = 0.45 x sec. V A.C. V (peak) D.C. = 0.45 x sec. V A.C.

I D.C. = 1.54 x sec. I A.C.

this is most likely what is ment by possibilities

Thanks Andy but I'm still confused. I'm familiar with those equations except the def'n of avg (see following)

Of all topics electrical and electronic that I've come across I have consistently found power transformer specs the most confusing. They take the biscuit every time.

More confusing than Maxwell's equations.
More confusing than Z transforms.
More confusing than boolean algebra etc etc etc.

No one even seems to agree on basic terminology.
For example, "DC (avg) - excuse me? what's the average of a DC value?

Why do Hammond and others second guess the end-users use of the device with their "specs"?
Why not just supply the turns ratio/s, taps; the winding DC resistances; secondary VA and internal power dissipation limitations; and the no-load voltages/currents.

O'Connor gives five figures for most of these transformers.
I assume four of them are for the following: two figures for a full wave rectifiers (my assumtion is that one figure being after a voltage doubler) and two for a full wave bridge (again one direct and one for a doubler arrangement).

Then he confuses it all by also providing a figure for "rated" current draw assuming a full wave capacitive input rectifier. It is different from any of the other four figures.

End of frustrated rant.

[Ears]

Server failure => D-Post

[Andy Le Blanc]

you are the end user..... so the maker may give you what is the most
important or usable rateing of its product.... the hammond product comes with several published caveats.... the one most applicable here may be
"H.V. DC current, measured with cap. input filter (full wave - 2 diode, C.T. rectifier circuit)."
this is from a hamond manufacturing publication 5c-00... you as the designer
must dig to find the best product for the application your designing for....
as "bluefireamps" posted..... "rutledj" may not have made a good design choice...
its alway nice to think that you can plug-and-play...... but you really have to research your design choices....
ive encountered many problems that were based on my poor assumptions....
just because you can assemble an electrical appliance does not mean you can design one based on that experience.....
just because you hear that this forum is awesome and you can chat with an expert
does not mean that it is a valid form of peer review which is what is needed for your design choices
just because you found you could make a living selling books...
does not me you fully understand all aspects of the field or wrote a good book....
research the bibliography in O'Connor's books.... then research the bibs in those
then you will have valid refferences to draw from

[Ears]

research the bibliography in O'Connor's books.... then research the bibs in those
then you will have valid refferences to draw from

Actually, taking O'Connor at his written word can COST you big time.
For example, in POP he recommends the 278X as PT for a 100W, 2 x 6550.
Do that and you'll fall short at least 20W and be out of pocket (at this end of the world) over $300.

I'm amazed he hasn't published a page/s of errata, it would be helpful.