Center-tapped winding current ratings?

[92Volts]

I'm confused about PT ratings. Imagine a PT advertised as "600 VAC center-tapped 200ma".

This can be bridge-rectified to 848 VDC at 141ma if we assume 200ma refers to 200ma@600VAC.

Now the INTENDED use is 300-0-300 rectified to 424 VDC... this is "obvious" to tube amp builders (though not explicitly described in the PT specs).

Can I assume it's capable of watts suggested by 600VACx200mA which gives me 281ma DC (minus a safety margin)?

There are 2 reasons it may be smaller-- 140mA DC. First is the current rating is a hard peak limit for any OR both sides, even though each side is in use 1/2 of the cycle under this configuration.
Or I guess the original wattage rating could be calculated with the winding "meant to be used as 300VAC" and a 2x conversion factor or peak vs. average current restrictions... all of it has already been taken into account... this wouldn't surprise me since it's so "obviously" the intended use of the transformer, but neither Edcor nor other companies explicitly describe what hookup is described by their voltage rating.

Edcor's response to this inquiry is "all ratings are AC-- 600V is AC and 200ma is AC" and that didn't 100% answer my question. That does suggest my first interpretation (works out to 280ma DC at the lower voltage configuration). But the few Edcor PTs with suggested or packaged tube/OT choices either seem counter to this, or unclear.

[pompeiisneaks]

I'm curious about this myself, my general understanding (but I want someone to confirm/deny) is that the 200mA is a pretty fixed rating, the AC to DC conversion is that the AC value coming out of the transformer is what you'll get, when you rectify it, it becomes higher or lower depending on the rectification means (tube vs solid state etc plus voltage doubler if used etc.).  The mA rating is what you should keep under, as the wire itself will overheat and could fail if the current is over it for any extended period of time.  I think generally stepping up voltage in a circuit reduces the current capacity.  Lowering it increases it.  Therefore, if you're going to use solid state rectification and your DC voltage goes up, you should expect the current ability of the system to go down a bit.  (200mA/1.4 the SS rectifier multiplier).  But the current capacity of this type may not change with rectification, only with transformers.  (I.e. if you can handle 200mA at 120V but up it to 240, then you can only handle 100mA usually on the secondary etc.)  That's the grey area I'm not sure of.  Does the 'as voltage goes up, current ratings go down, and as voltage goes down, current rating goes up' only apply to tranformers?

~Phil
 

[SILVERGUN]

Posting for reference in case you haven't seen this:

[SILVERGUN]

Does the 'as voltage goes up, current ratings go down, and as voltage goes down, current rating goes up' only apply to tranformers?

~Phil

No, it's meant to be taken into account for voltage/amperage relationships everywhere.
The more voltage you have, the less amperage you will draw to perform the same work.
Example:
-I have a welding machine that can be run off of 120VAC or 240VAC - I just have to change jumpers internally
-It can and will put out 200 amps of welding current in either configuration - so the output is constant
-On the 120VAC input setting it will draw 40 amps from the supply
-On the 240VAC input setting it will draw 20 amps

You can then use those numbers to calculate wattage
Volts X Amps = Watts
120 x 40 = 4,800 W
240 x 20 = 4,800 W
Same input power used to perform the same function

That relation stays constant...
Guess how may amps you would draw if you were able to set that same machine up for 480VAC input?

[92Volts]

The AC to DC conversion can be accounted for as Silvergun says... in short, watts in (from the transformer) equals watts used (by the load, or losses).

Effective DC voltage is higher than RMS AC voltage-- capacitors hold the DC value near the peak, not average, of the AC wave-- and the extra power needed to deliver x amount of current at a higher average voltage needs to come from somewhere.

The whole picture is complicated. The average current across the cycle might not appear higher, but the current is being drawn disproportionately at the high-voltage peaks and not being drawn at the lower-voltage portions of the cycle. Overall, higher power is being drawn than we'd expect for the same average current distributed "normally" across the power cycle.

When we account for this by stating 1.4x greater AC current comes from the transformer than goes to the DC load, we are assuming the transformer doesn't really "care" about instantaneous current draw, only current draw in context of voltage... in other words, it cares about power. And we don't need complicated simulations to understand that power in equals power out, and if we understand the DC voltage is 1.4x greater than the RMS voltage we can assume the current must be 1.4x smaller.

Some things to note:
1. Bridge rectifier or center-tapped, peak current pulse is alreadyastronomically higher than average current, depending on capacitors.
2. When using half the winding alternately in a grounded-center-tap configuration, each half's average current (and power) are limited since it's unused for half the cycle. Logically this lets us double the current draw for the half of the cycle when it is used, then do the same with the other winding when it takes over. Overall this lets us pull 2x more power at 1/2 the voltage compared to a full-bridge rectifier, which seems to check out.

However, if we assume #2 is OK we depend on the assumption that average power (or current) matters and peak doesn't.

And for all of this we need to make some assumptions about what the manufacturer was trying to convey with their ratings, which I guess is what I was asking originally

[SILVERGUN]

Sorry for straying somewhat off topic there 92V, but I wanted to give Phil that explanation.
 
And this too...
Phil, (and anyone else) here's how I was trying to give 92V his answer by posting the Hammond rectifier selection guide.
 
The bottom spec on this pic is I D.C. = 1.00 X sec. I A.C.
I = current
 
So, in the case of a FW capacitor input load the current capacity is a 1:1 ratio, AC to DC

[DummyLoad]

http://www.hammondmfg.com/pdf/5c007.pdf


see FW with CT with capacitor load - AC : DC rating are same. FWB changes things. choke loading changes things.



--pete

[92Volts]

Thanks, that may give an answer for Hammond transformers at least. I am a bit confused by some info on that sheet though-- With capacitor input, full wave, it shows:
V(Peak) D.C. = 0.71 X Sec.V A.C.
V(Avg) D.C.= 0.45 X Sec.V A.C

That sounds a lot like an unfiltered or poorly filtered DC output. I'd expect average to be nearly equal to peak.

In fact, the same sheet shows the same full-wave recto into a resistive load (no filtering) and gives DC voltage of 0.45 X Sec.V A.C... equal to the V(Avg)D.C. in the FW/capacitor example. That suggests the capacitor's average figure is wrong, or assumes a capacitor so small it isn't really "filtering" to the extent we expect in most tube power supplies.

Weirder still is the DC current into a resistor from the same configuration is higher (1.27x vs 1.00x the AC current) than the one with capacitor filtering... I can't think of a reason why this could be true. The cap should be holding the average voltage higher but never lower than the secondary voltage, thanks to the diodes. It may little or nothing depending on size, but should never decrease the current.


EDIT: I reread that document and it sounds like Hammond is already taking everything into account. Current into a resistive load is not "higher" because this setup applies a higher voltage, therefore pushing more current through the same resistor. Instead, the same transformer can safely source more current into a resistive load than it can into a capacitive filter, because the resistive load is "easier" to drive-- less peak current for the same average. This is actually a very direct guide for selecting a transformer... for Hammond products at least.

[pompeiisneaks]

ahh cool all great info, I kinda suspected the current rating wouldn't change much, because the current drawn at the AC outlet is based on the current the wire can handle in either/or the primary/secondary windings, so that's the limiting factor, the current drawn isn't going to change, the transformer itself changes the capability of the secondary based on the winding ratio.   Although DummyLoad's note that chokes change this, how?  Does it limit/reduce it more?  I would guess it can cause spikes in current when it resists the change due to the inductance?

~Phil

[92Volts]

Actually, the choke reduces spikes in current as it resists sudden changes. It can handle substantially more current than direct to a capacitor or resistor, though DC voltage is lower than a capacitor (near average AC signal instead of peak). So more current doesn't equal more power, it's just another way to get full power from the transformer.

[PRR]

The file on Hammond's site was not devised by Hammond, is wrong in several key details.

*Hammond's* published specs on high voltage CT windings give DC for the 2-diode connection in cap-input. They can be worked some harder in choke-input (we never do that). It would be half the current if you actually wanted twice the voltage (4 diode affair).

[Merlin]

Current into a resistive load is not "higher" because this setup applies a higher voltage, therefore pushing more current through the same resistor. Instead, the same transformer can safely source more current into a resistive load

Correct.

Just to clear things up, for a typical cap input rectifier:

VOLTAGE:
For a non centre-tapped transformer you get an ideal DC voltage of 1.4 times the AC voltage. However, in practice at full load it is usually closer to 1.35 times.
300Vac --> 420Vdc in theory, or about 400Vdc in practice (diode drop further subtracts from this figure).

For a centre-tapped transformer it is exactly the same if you're talking about one half of the winding.
300 - 0 - 300Vac --> 420Vdc in theory, or about 400Vdc in practice (diode drop further subtracts from this figure).

CURRENT:
At full load, for a non-centre tapped transformer the RMS current coming out of the rectifier (which is the same current flowing in the transformer) is typically about 1.5 times larger than the DC current you draw from the capacitor (i.e. power factor is around 1/1.5 = 0.66).
150mA RMS --> 100mA DC

At full load, for a centre tapped transformer the RMS current coming out of the rectifier is exactly the same, i.e. about 1.5 times larger than the DC current you draw from the capacitor. But the RMS current in one-half of the transformer winding is not half but 1/(sqrt 2) times the total RMS current.
105mA RMS in one half of the winding --> 150mA RMS total --> 100mA DC

That's why the Hammond guide shows Idc = 1.0 x Sec Iac for the centre tapped transformer, but 1.54 x Sec Iac for the ordinary transformer. (They split hairs by using 1.54 instead of 1.5)


"600 VAC center-tapped 200ma. The INTENDED use is 300-0-300 rectified to 424 VDC"
This transformer winding is therefore rated for 300-0-300V at 200mA RMS per half, or 120VA total. If you used a bridge rectifier to get close to 840V DC you could safely pull about 130mA DC. Notice that you get 30% more load power when using it in non-centre-tapped mode. This is NOT the same as when using it to generate two rails (e.g. bipolar supply), which would still be limited to 200mA total, e.g. 100mA per rail, or 150mA for one rail and 50mA for the other etc.

[jjasilli]

Lets get back to the actual question posed: "Can I assume it's capable of watts suggested by 600VACx200mA which gives me 281ma DC (minus a safety margin)?"

 No. If a tranny is rated for 200mA, then that's its Current rating. Period, end of story. 


The problem lies in conflating Current handling with Watts.  I agree with the 1st sentence of Reply 1, but then I submit that pompeii goes on to inadvertently contradict himself. 

A Transformer has wire windings of a certain Gauge.  The wire Gauge determines Current handling NOT Watts handling.  If you increase Current, while lowering Voltage and keeping Watts the same, that's NOT OK for the wire Gauge. 

Whatever type of circuitry follows the transformer is Irrelevant to the transformer.  If that circuitry pulls more than 200mA through the tranny, then its current rating is exceeded. 

(BTW: exceeding the current rating is OK with me, 'cause I like sag.)

[pompeiisneaks]

Lets get back to the actual question posed: "Can I assume it's capable of watts suggested by 600VACx200mA which gives me 281ma DC (minus a safety margin)?"

 No. If a tranny is rated for 200mA, then that's its Current rating. Period, end of story. 


The problem lies in conflating Current handling with Watts.  I agree with the 1st sentence of Reply 1, but then I submit that pompeii goes on to inadvertently contradict himself. 

A Transformer has wire windings of a certain Gauge.  The wire Gauge determines Current handling NOT Watts handling.  If you increase Current, while lowering Voltage and keeping Watts the same, that's NOT OK for the wire Gauge. 

Whatever type of circuitry follows the transformer is Irrelevant to the transformer.  If that circuitry pulls more than 200mA through the tranny, then its current rating is exceeded. 

(BTW: exceeding the current rating is OK with me, 'cause I like sag.)

I thought so, and the contradiction was due to my lack of full understanding, and I think you've just helped me reaffirm what I was thinking I understood, (Or so I hope lol) thanks!

~Phil

[pompeiisneaks]

Current into a resistive load is not "higher" because this setup applies a higher voltage, therefore pushing more current through the same resistor. Instead, the same transformer can safely source more current into a resistive load

Correct.

Just to clear things up, for a typical cap input rectifier:

VOLTAGE:
For a non centre-tapped transformer you get an ideal DC voltage of 1.4 times the AC voltage. However, in practice at full load it is usually closer to 1.35 times.
300Vac --> 420Vdc in theory, or about 400Vdc in practice (diode drop further subtracts from this figure).

For a centre-tapped transformer it is exactly the same if you're talking about one half of the winding.
300 - 0 - 300Vac --> 420Vdc in theory, or about 400Vdc in practice (diode drop further subtracts from this figure).

CURRENT:
At full load, for a non-centre tapped transformer the RMS current coming out of the rectifier (which is the same current flowing in the transformer) is typically about 1.5 times larger than the DC current you draw from the capacitor (i.e. power factor is around 1/1.5 = 0.66).
150mA RMS --> 100mA DC

At full load, for a centre tapped transformer the RMS current coming out of the rectifier is exactly the same, i.e. about 1.5 times larger than the DC current you draw from the capacitor. But the RMS current in one-half of the transformer winding is not half but 1/(sqrt 2) times the total RMS current.
105mA RMS in one half of the winding --> 150mA RMS total --> 100mA DC

That's why the Hammond guide shows Idc = 1.0 x Sec Iac for the centre tapped transformer, but 1.54 x Sec Iac for the ordinary transformer. (They split hairs by using 1.54 instead of 1.5)


"600 VAC center-tapped 200ma. The INTENDED use is 300-0-300 rectified to 424 VDC"
This transformer winding is therefore rated for 300-0-300V at 200mA RMS per half, or 120VA total. If you used a bridge rectifier to get close to 840V DC you could safely pull about 130mA DC. Notice that you get 30% more load power when using it in non-centre-tapped mode. This is NOT the same as when using it to generate two rails (e.g. bipolar supply), which would still be limited to 200mA total, e.g. 100mA per rail, or 150mA for one rail and 50mA for the other etc.

I just got done reading this in your book on HIFI design, hehe.  BTW, when is the power supply design book coming out again!!! :D

[jjasilli]

I thought so, and the contradiction was due to my lack of full understanding, and I think you've just helped me reaffirm what I was thinking I understood, (Or so I hope lol) thanks!
~Phil
 

[PRR]

> 200mA, then that's its Current rating. Period, end of story. 

No, it depends on context.

The wire alone will stand far more than a winding of the same gauge. The limit is heat in the winding.

The 2-diode CT affair works each half winding half the time.

The 4-diode form works the whole winding both cycles.

As Merlin says, the 4-diode form is more efficient of Copper.

The 2-diode form is more economic in Cathode Material (when we only had vacuum cathodes), since it can be contrived to use one large cathode (one spray-job) and two plates (cheap stampings).

[jjasilli]

Yes. Unlike typical wiring, for a winding, gauge alone is not the sole factor which determines current-handling capacity.  However, it's an important factor; and a more in-depth look into inner tranny function does not change the answer to the question posed.  Transformer theory is endlessly complex, and frankly above my pay grade.  But the answer to the question posed is simple and direct: the stated mA rating should not be conflated with watts.  Current handling does not equal watts.  That's why, for example, fuses are rated in amps & not watts.  No one wants to turn their tranny into a "blown fuse".