current draw equation

[jeff]

I have a device that I want to reuse as a tube preamp. The device is marked
 "115V 50-60 CYCLES .48AMPS"
and uses seven 12AU7s three 12AZ7s and one 6SC6.
 Checking the tube charts that's(7X.3 + 3X.45 + .3 =) 3.75 A@6.3V{EDIT: I orignally put mA thanks for catching my mistake}. (Although it is a 6.3-0-6.3 heater tap with half the tubes connected to one side and half the tubes connected to the other side with the CT grounded. for the heater.)
The B+ measures 172V unloaded(no CT).

Is there any way to calculate approx. current I can draw(=the orignial device drew) using this info?

 Thanks
  Jeff

[jjasilli]

That's 3.57 Amps (not mA), for the filament winding. 

The HT supply is a separate winding (or maybe a separate tranny??).  The GE tube chart for the 6CS6 (note correct spelling) is rated to typically draw 2.3 mA (plate + screen).  The 12AZ7 is a toss-up depending on circuit design: either, say, 2X 4mA or 2X 10mA.  Depending upon use as amplifiers or oscillators, the 12au7's can draw from about 10 - 20mA ea with as much as 300mA peak draw.

Figuring most conservatively for draw of the B+ supply, that's:  2.3mA + (3X 4mA) + (7X 10mA) = 84.3mA (Please check my math!!).  However, the voltage is rather low.  Might be interesting to find a  power tube to match, or just use as a preamp.

[eleventeen]

3.75 mA@6.3V.

Those would be AMPS, not ma = milliamps. Obviously the tranny can power that many tubes, filament-wise.

Are you sure the "6SC6" is not a 6CS6? This is a mixing heptode, used in some transceivers. Bunch of grids, still a low current tube, 2 ma or so.
The 6CS6 is another .3 amp heater. They all run off 6 volts, so, across a transformer running from 120 volts, we would divide their req'd current by 20. So filament wise they would need 3.75 / 20 = .1875 amps powered from 120 volts and use 3.75 * 6.3 = 23.63 watts.

[is there a 6.3 volts pilot lamp typically 150 ma?]

Let's call it 25 watts for the fils.

That leaves 23 watts for your plates, subtracting 25 watts from the nameplate 48 watts.

One figures the 12A_7s use 1-2-3 mils in terms of plate current and same for your 6CS6. So you have 11 tubes using 2 ma each and there you go, 22 mils of plate current required by the whole gang. Very light.

How do we know a garden variety 12AX7 in a preamp uses such low current? Because we go look at a Deluxe Reverb and see an early-stage cathode sitting on top of a 1500 ohm resistor reading 1.3 volts. 1.3 / 1500 = .0008666 amps or .86 ma. 1 mil, basically.

But your question is about "how much can I draw". In my experience, we generally do not encounter power trannies with such large (comparatively) filament capacity with such paltry B+ capacity. This is assuming that the mfr of your device would have much preferred to buy an off-the-shelf transformer instead of having a custom one built. So I would guess that you have some spare B+ capacity. I also assume you are not going to have (as many as) 11 little tubes in your preamp build. Every one you delete eliminates .3 * 6.3 = 1.89 watts of power pulled from the tranny.

Now you have to kind of add all this together. Suppose you wanted to build a SE 6V6 amp in this. You decide to eliminate 4 preamp tubes. So you get back 4 * 1.89 watts of fil power, but your 6V6 will want .6 amp (equivalent to 2 preamp tubes) to light up. Net gain, 2 * 1.89 watts = 3.78 watts. At 250 volts, that's about 15 mils, not really enough for a 6V6. Do we have 15 ma of spare capacity in the tranny? Probably.

If your plan is to build a preamp (only) I can't see there being any problem.

[jeff]

3.75 mA@6.3V.

Those would be AMPS, not ma = milliamps. Obviously the tranny can power that many tubes, filament-wise.

Are you sure the "6SC6" is not a 6CS6? This is a mixing heptode, used in some transceivers. Bunch of grids, still a low current tube, 2 ma or so.

You are correct on both accounts

Pilot light runs off the 120VAC.

[sluckey]

Is there any way to calculate approx. current I can draw(=the orignial device drew) using this info?

Why calculate? You already have the answer.

The device is marked
 "115V 50-60 CYCLES .48AMPS"

[eleventeen]

But that 48 watt nameplate number indicates what the current device requires/draws, not the ampacity of the power tranny. Unless you assume that the tranny was being operated at 100% of its ratings....not a general practice. It's also unusual, as I said, to have enough juice to light up 11 tubes (essentially, the filament requirements of a Super Reverb @ 6 * .3 amps for tubelets plus 2 * .9 amps for dual 6L6 without *some* power capability (beyond 25 measly ma) in the B+. I've taken apart reasonable numbers of crunched surplus radios (talking about receivers, not transmitters) and generally the PTs look promising based on size but very often are only rated 70-75 ma. Still, that 70-75 ma is a lot more than the cheesy 25 or so mils we would calculate based on adding up the filament power we KNOW has to be there with "whatever's left" in terms of nameplate watts. 

If you took a Super Reverb and threw a Twin Reverb tranny into it, changing nothing else, the amp would not draw any more power other than a tiny amount we might speculate would represent the increased losses of the larger windings on the Twin tranny (which could go the other way just as easily) 

[sluckey]

But that 48 watt nameplate number

It's not 48 watts. It's .48 amps.

[jeff]

Is there any way to calculate approx. current I can draw(=the orignial device drew) using this info?

Why calculate? You already have the answer.

The device is marked
 "115V 50-60 CYCLES .48AMPS"

OOps I meant what "(=the original 175 secondary supplied)" not what the 120V primary drew. Sorry for the confusion.

[tubeswell]

Is there any way to calculate approx. current I can draw(=the orignial device drew) using this info?

Why calculate? You already have the answer.

The device is marked
 "115V 50-60 CYCLES .48AMPS"

OOps I meant what "(=the original 175 secondary supplied)" not what the 120V primary drew. Sorry for the confusion.

For an approximate guesstimate, work from Pr for the total VA of PT, and then subtract the various bits of VA from the secondary that you know about

i.e.: 115V @ .48A = 55.2VA

Heater winding = 6.3V@3.75A = 23.625VA

That leaves 31.575VA for the HT winding. Divide 31.575 by whatever the loaded VAC of the HT winding is to get the current draw for the HT.

(My 2CW of what I think about it anyhow.)