Power Trans Question

[19rick52]

I'd like to try one of these 5b6 amps for grins.
I got ahold of a donor amp. Schematic indicated 36-0-360 volts, a touch high but thought I'd try it.
The tranny actually measures 317-0-317 volts.
This is quite abit lower than indicated.,
Do I dare try this build with this transformer?

[Blackburn]

It all depends on the current ratings. The voltage rating sounds good. Just verify that it can handle that circuit and you're good.

[Phil_S]

If it is 317-0-317 without a load, it is going to be less under load, maybe 300-0-300. You will need to assess whether it can hold up for the power and filament needs of the 5B6. I think you only need about 100mA for the HT secondary. Because it is on the low side voltage-wise, I think I'd go with a solid state rectifier to maximize B+. For the HT winding, someone gave me this chart years ago. I don't know if it is valid, but it is a decent place to start your assessment.

For a 120 volt AC supply the VA rating and primary resistance is as
follows.
30 VA = 30 to 40 ohms
50 VA = 13 to 16 ohms
80 VA = 7 to 9 ohms
120 VA = 5 to 6 ohms
160 VA = 2.5 to 3.5 ohms
225 VA = 1.8 to 2.2 ohms
300 VA = 1.0 to 1.3 ohms
500 VA = 0.45 to 0.55 ohms
Simply multiply all ohmage values by two for a 230 / 240 volt supply. Derate to 65%, which is probably reasonable and to allow 15VA for the filament windings.

The VA rating is for the primary, but after you subtract the allowance for the filament supply, you should have a good idea about the HT secondary. I hope this helps.

Note: Edited to correct the note on conversion for use with 240V mains power to use a multiplier of 2x, not 4x as originally posted.

[Reeltarded]

Really? Those values snoop out unknown current capacity?

[19rick52]

Thanks for the replies. The 317 is no load, at the rectifier. I am somewhat dissapointed as the schem indicated 360. Might have to build something else.

[VacuumVoodoo]

Really? Those values snoop out unknown current capacity?

Basically, yes. You are given power rating.
Tables like this are usually meant to help you see if you're on the right track to Chattanooga, nothing more. They are based on "centuries of empirical data".
Also values for 230/240V mains are 2x not 4x those in the table.
One of the assumptions made is that a "universally accepted" standard current density in low power (less than a couple kW) transformer windings is in the range between 2 and 3A per mm² of wire cross section area.
One must be aware that suspiciously cheap transformer might either be wound for 5A/mm² current density or with aluminium wire. In both cases expect them to run hot and sag up to 25% from idle to full load. And you won't like your B+ dropping to 325V from 450V. Playing that amp would feel like riding a bicycle in the desert on flat tyres.

[Reeltarded]

That is most useful to know. Suddenly I am not totally in the dark.. maybe.



Thank you.

[M Fowler]

Playing that amp would feel like riding a bicycle in the desert on flat tyres.

Ah you guys saw me on Wide World of Sports didn't you

Mark

[Reeltarded]

Are you the guy in the brown slacks? You got some air.

[Blackburn]

I was pretty far off. I took a look and preferable voltages from the PT are around 350-0-350 for the 5B6. I was thinking 300-0-300, like in the octal Deluxe, for some strange reason. I'm with Phil S for using SS rectification to get where you'd like to be, but for the sake of the original amp, a lot of mojo is in using a 5U4. I guess it depends on whether you want to stay true to original design or not. Really sweet amps these are!

[Phil_S]

Really? Those values snoop out unknown current capacity?

It beats letting the smoke out. As Aleksander says, there is some math involved with the cross section of the wire, etc that I don't understand very well. The table allows you to make a reasonable assumption about current. From there, if you want to load test it, you've got a decent suggestion about just how much [1] of a load to put on it before you are getting near the point of letting the smoke out.

[1] Badly said...a load of fewer ohms is going to draw more current. When I do this, I use a primitive breadboard rig with a FW rectifier and a couple of caps. I hang a resistive load between the two B+ wires and use Ohms Law to calculate current as a function of voltage drop. I have several 25W wirewound resistors that I use in series or parallel to get to the target load, starting high and working down. I've learned to load test for the intended use and not to find the maximum mA, as the only reliable way to do that is to either unwind the thing or let the smoke out.

Other hints as mA capacity are the physical size of the lams and the size of the core. As you'd expect, bigger is better, but this can be deceiving, as well. Hints are just that.

Thank you, Aleksander, for correcting the record about using a 2x multiple for 240V and not 4x. I often wondered about that and am not surprised to learn the information as presented to me was not correct. I didn't pursue this because I live in the US. I am going to edit the earlier post just to be sure no one picks it up and is accidentally mislead.

[Phil_S]

Thanks for the replies. The 317 is no load, at the rectifier. I am somewhat dissapointed as the schem indicated 360. Might have to build something else.

If you are saying it is 317VDC at the rectifier without a load, then yes, it is not going to produce the intended voltages for a 5B6. That doesn't mean you won't get something good if you build it. In that range, you still have a viable B+ supply and you might end up with something you like very much. I don't think you've got much to lose if you build it. Maybe you build it with a pair of 6V6 instead of 6L6. That's actually not so bad, as the 30W or so you get from a pair of 6L6 is really quite loud. You are likely to see 20W+ from a pair of 6V6.

[19rick52]

Yes I think I'll try it. Like you say I may end up using 6V6s'. Really not much to loose using a no cost donor amp and a few parts.

[martin manning]

There was an article "Rating Unknown Power Transformers" by H. Q. Duguid posted here a while ago that described a method of estimating the current rating. Duguid used a factor of four on primary resistance, stating that a 220VAC winding would have twice the wire length and half the wire cross section.

In Duguid's method the HT current rating is found directly from the HT secondary resistance (DCR) and unloaded voltage (VAC) by reading a plot of full-load current IDC vs. DCR/VAC. Converting that plot to an equation, I get IDC = 0.046*(DCR/VAC)^-0.6. This is for a full-wave, capacitor input power supply. A FWB would be 0.62 times that. The fully-loaded VAC is estimated as Unloaded VAC - 2*DCR*IDC.

[Reeltarded]

Arrest this man. He talks in maths.