Power Transformer current draw if not using additional taps

[leroy]

Hi All

I have a question about transformer current draw.

The Transformer in question is the hammond 290EUX
https://au.mouser.com/datasheet/2/177/EDB290UEX-951043.pdf

The secondary has a HT tap with 308v @ 250ma Max (77VA). It also has another tap 39 volts @ 250ma (10VA)

My question is if i don't use the 10va tap can i then draw the extra 10va from the HT tap?? Giving the HT 87VA of available power.

There's so much conflicting stuff on the net about how much current a PT actually needs for a tube type.

That PT is for a fender running 2 6L6's and i'm looking at using GLGGC's or EL34's
 
Thanks
Leroy
 

[PRR]

> looking at using GLGGC's or EL34's

Insufficient information.

A pair of 6L6GC/EL34 can make 25 Watts out or 60W-100W out (yes, there is a 100W/pair condition for EL34).

The DC power will be more, probably 2X.

The AC VA rating usually must be more, nominally 1.7X. (However Hammond's ratings sometimes allow for this.)

[Merlin]

My question is if i don't use the 10va tap can i then draw the extra 10va from the HT tap??

Almost certainly yes. You'll get a little bit more sag, but meh, you're presumably only using full power under peak-music conditions.

[pdf64]

Just to be clear, the 39V winding is a separate winding to the 308V secondary, rather than being a tap from it (eg as per a regular Fender bias tap).
The info says 120mA rated, 250mA max, so probably best to take them at their word and not exceed 250mA.
Bear in mind that with the FWB rectifier necessary here, the ac current will be much higher than the dc current.

[mschock]

personally, I would assume a hard 250 mA. I am curious what other tubes you would be in the system. correct me if I am wrong, when you said GLGGC's I took a huge leap and assumed you meant 6L6GC's and if so, a pair of these are a current hogs with 210 mA combined( if I can recall correctly)

[jjasilli]

I'm going out on a limb and disagreeing with Merlin.  The tranny has 3 separate secondary windings which are independent of each other.  The specs for ea secondary winding are clearly stated on the spec sheet.  The specs for one winding do not vary in relation to the use or non-use of the other winding(s).

However Merlin is right that you can probably exceed spec and get away with it, but at your own risk.  Max voltage & current draw, from which VA can be calculated, is stated on the spec sheet.


EDIT:  It appears that I'm basically repeating pdf64's Reply in my own words.  However I think the AC current point he makes is not relevant.  I.e., the max spec DC current (after rectification) is 250mA.  Whatever the AC current was before that has already been taken into account by the Manufacturer, and need not concern us.


Also the VA is of no concern and is a misleading red herring.  This is the source of the confusion! The criteria here is Current NOT Power.  The current handling is determined by wire gauge. (This  is not strictly true in the more complex matter of the coil of a winding of wire.  But it's true enough for our purposes.)  IOW, a 400 VA tranny might be wound to handle 1V @ 400A; or be wound to handle 400V @ 1A.  Both tranny's would be 400VA!  The wire gauge to handle 1A would be far slimmer than that needed to handle 400A.  Putting 400A through a 1A tranny would instantly fry it, even thought the VA works-out "right".

[PRR]

> The current handling is determined by wire gauge.

Actually by Heat.

The copper won't melt, even if seriously under-gauge.

The *insulation* gets toasted, eventually fails.

The heat conductivity of copper-varnish-paper is good (not great). What tends to matter is the total heat. An un-used winding adds no heat. Conceptually you could run another winding a little over-spec, have the same total heat, and maybe the hot-spot on the overworked winding is not too bad.

However there are many known-good PTs. And OP has not revealed if the two 6L6 are loafing or groaning. Any advice is a guess. All I would say is: it will work for many minutes. If another OT is awkward, try this one and see how hot it gets.

[leroy]

Thanks for all of the Responses.

I'm not sure what PRR is referring to when he asking is it "loafing or groaning"???

The amount of power out isn't a concern to me. It doesn't bother me if its 30 or 50 watts.

The output transformer is 4K primary impedance.

I'd like to use this transformer as it has free shipping from Mouser. More powerful transformers have $60 dollars shipping tacked on.

The linked power transformer is a replacement for a Fender hot rod deluxe running 2 x 6L6's

I've always been confused as to how amps with the same output valves can run at 30 watts or 80 watts.

Plus how do you calculate what the minimum current required is? I've seen so much conflicting evidence and the "correct" specs vary wildly depending on who's explaining.

The Hammond spec sheets also dont specify if the secondary current is AC or DC?? 

Any advice would be greatly appreciated.

Thanks
Leroy

[DummyLoad]

I'm not sure what PRR is referring to when he asking is it "loafing or groaning"???

groaning - under extreme electrical loading; application of an underrated part.

loafing - under light electrical loading; application of an overrated part.

hot rod deluxe is a 40W amp with SSR and 4.25K OT reflected Z. --- use the transformer it will work. copy the fender HRD power supply.

EL34's will load the heater winding more by over 1A more than 6L6GC. use 6L6GC if the heater voltage falls below 6.0V when loaded with EL34 powered with rated line voltage.

--pete

[leroy]

Ok

I think I've got it.

Current required = (No of tubes * (PT secondary AC Voltage - VSAT) * 1.41)/ Output transformer impedance.

So with a pair of EL34's and the 290EUX we get

2 * (306 - 50) * 1.41 = 722.

722/4000 = 180ma

Add 10ma for the preamp = 190ma

and maybe add 10% for safety = 210ma DC.

So if the 290EUX's current ratting of 308v @ 250ma max is DC current then it should work??

How did I do?

Leroy

[jjasilli]

I've always been confused as to how amps with the same output valves can run at 30 watts or 80 watts.

For Push-Pull power amps, Power out in Watts is determined by the Power Formula:  W = V² / R; R = the impedance of the primary of the OT.  If The PT puts out 308VAC; for SS rectification 308VAC  X 1.414 = 436VDC.  (For tube rectification use the voltage drop multiplier for that tube rectifier). Your OT has 4000 Ohm primary impedance. W = 435² / 4000 =
190,096 / 4000 = 48W.  (EDIT:  that's unloaded.  Tube current draw would drop that voltage by about 10 - 15% to about 400VDC which, when re-calculated, nails 40W.)

So the primary factor is plate voltage, assuming you use spec primary impedance.  But you could double or halve output W by "cheating" and halving or doubling primary impedance.

Plus how do you calculate what the minimum current required is?  Don't calculate; use the tube charts.  Add plate current + screen current  @ max signal.  Ignore preamp tube current draw; it's not even a rounding error. 

Checkout Hammond Power Transformer guide - attached - can't find it online