PT Current & Votage Calculation

[jojokeo]

I saw this written by someone but it has me thinking the current availibility is not correct? His thinking was the PT's 55mA supplying each half cycle therefore yielding 110mA. Sorry, it's late & I'm confused at the moment, please read below.

Calculating Transformer Output Voltage & Current:

Input 115-120 volts 50 or 60Hz,
For output 480V CT (240-0-240) @ 55ma.  Full wave rectified, that’s 480/2*1.41= 340v @ 110 ma available and with solid-state diodes, or about 320v w. tube rectifier or choke.

*Wouldn't the mA output rating be the same 55mA on a PT whether it's 240-0-240 or 480 w/no CT?

[jjasilli]

The center tapped secondary is 2 windings wraped with coated wiring, out of phase.  That's 2X 240 windings.  They're wrapped so that the 2X -0- windings come out the middle; and the 2X 240 winding come out the far sides.  In the old days sometimes the 2X -0- winding wires came out of the tranny togehter, side-by-side.  You had to strip away the coating from the 2 wire ends; then twist the ends together.  Today that junction is usually made internally, and only one CT wire exits the tranny; but it still serves both secondary windings.

Ea 240-0 pair handles 22.5mA.  Used together they handle 55mA.  If you meter the 2 outside wires (240VAC) you'll read 480VAC @ 55mA.  Either end to CT will read 240VAC @ 55mA.  After full wave rectification:  240VAC X 1.414 = 340VDC less any Voltage Drop caused by the Rectifier's Impedance.  

[kagliostro]

For my knowledge in PT

55mA each branch give 240v-0-240v at 55mA or 480v at 55mA

only if it will be possible to put the windings in parallel we obtain 240v 110mA (but for what I know to put in parallel the windings of a 240v-0-240v PT is not possible)

Hope someone can confirm or explain why this is wrong (if is wrong)

don't be able to understand well >>

Ea 240-0 pair handles 22.5mA.  Used together they handle 55mA.

because the windings are in series with a CT so is difficult to me to have two 22.5mA windings in series giving 50mA

may be I need a further explanation

Kagliostro

[jjasilli]

In the xxx -0- xxx tranny it seems that the 2 secondary windings are in parallel:  http://www.hammondmfg.com/5CHook.htm

However, I believe Kaghilostro is correct that the ea winding must be at the full current rating, here 55mA.

[PRR]

It depends who wrote the specification.

HV CT windings are often specified for AC volts but DC current with two diodes and a cap-input filter. Rectification conversion is included in the spec. If this is the case, then it really means 340V 55mA total DC power.

OTOH, generic low-volt PTs are specified for AC load. The designer is expected to allow for rectification conversion.

> Wouldn't the mA output rating be the same 55mA on a PT whether it's 240-0-240 or 480 w/no CT?

No. The 2-diode plan is transformer abuse. Each half-winding is used only half the time. The consequences in sag and heat are complicated; that's maybe why these HV CT PTs are listed by DC current, not AC current as in most other PTs.

_IF_ you needed 680V DC and used a FWB (ignoring the CT)... well, first you must wonder if the CT area is insulated for 340V above the PT shell. If breakdown is not a problem, the nominal rating is the same total Power at twice the voltage, therefore half the current. 28mA. However now both windings are used on both half-cycles. Sag and loss are reduced. How much? Not sure. I'd call it at least 30mA, but nowhere near 55mA.

(I have thought about using "Bassman" iron to get 800V 100mA DC for a transmitter tube. Double voltage and half current. I think it would work. I think this is the easiest part of a hi-power freak SE amp.)

[jojokeo]

I think this may be getting more complicated than necessary (atleast for me)? Lets stay w/ the original transformers & specs for the two examples being one with a CT and one without. Both could be common Hammond transformers intened for guitar amp use only.

Ex1) 240-0-240 @ 55mA <-this one uses a standard full wave /2 diode to typical resevoir cap (pi filter or totum pole) like we normally use.

Ex2) 240 (no CT) @ 55mA <-this one uses a standard full wave bridge to typical resevoir cap (pi filter or totum pole) like we normally would use.

With either transformer set-up, should you get the same result being 240 * 1.41 = 340 @ 55mA (and not 110mA as said earlier)?

[jjasilli]

Oh, then:  http://www.hammondmfg.com/pdf/5c007.pdf 

[jojokeo]

A few things seem off or unclear from that page compared to past experiences.

Ex1) For full wave capacitive input load: it says v(peak)dc= .7x sec. vac
This isn't correct for we normally get 1.4xsec vac just like the 240 no CT tranny? If what they are saying is true then we would get 240x0.7=168vdc but atleast get our full 55mA? Doesn't seem correct.

Ex2) For the full wave bridge example the voltage is correct (v(peak)dc=1.41x sec vac) BUT idc= 0.62 sec iac - what!? So they're implying that my 55mA is only a useable 34.1mA???

That page just got things even uglier and clouded. 

[jjasilli]

Don't worry, they tricked you with the arrows.  Note in the CT example they are reading outside to outside, NOT: outside -0- outside.

That outside measurement yields 2X the voltage as usually stated, so they halved the forward conversion factor from 1.414 to .71.  I agree that this method seems confusing. But it's free!   

[jojokeo]

Mucho gusto Jjasilli - I usually just use trannies that are known good for situations and/or w/ more than what could be expected ratings-wise anyway but if wanting to design down to the "gnat's ass" I need better understanding on a few things. Ironically, it's the micro-scaling that's macro-working my cranium.  :book1:

[DummyLoad]

(I have thought about using "Bassman" iron to get 800V 100mA DC for a transmitter tube. Double voltage and half current. I think it would work. I think this is the easiest part of a hi-power freak SE amp.)

sounds like fun... wanna borrow a breadboard?   

807/5933? 813? 826? 829? 3CX300A? 8150? 4CX250?

[PRR]

HK-257 has SE A1 specs for 25W-30W outputs (see below). It also red-plates at normal safe power (transmitter tubes ARE different). The sheet shows two extremes; you can work anywhere in-between as long as V*I is just under 75W and your load is roughly V/I or a bit less, and you set G1 and G2 voltages to split-the-diff values. So 750V (800 counting B+ and cathode R losses), 100mA, 400V on G2 (conveniently got from the PT CT), about 32V bias or 300r 5W. Load could be 7K, or 5K, no big difference.

[PRR]

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