Dumb Q About Center Tapped Power XFMR

[Paul Marossy]

Not a huge loss. Remember your cap-input voltage went UP, 12VAC to 17VDC. TANSTAAFL. Current MUST be less or a law of energy would be broken. This fact alone means current must be reduced by 1/1.414 or to 0.707. The ugly wave-form in cap-input shaves that to 0.62 (1/1.6). In practical small-work, "half" is the safer bet.

Makes sense now that I had the RMS current revelation.

Can we touch on the effects of adding a voltage regulator?  Say, for example, we stick an LM7812 on the end of this....

Yes that would be interesting to know.

[EBK]

My confidence is still only slowly recovering, but I'm willing to venture a guess that the regulator output current is roughly the same as its input current (minus a very small quiescent current of 5-8mA).  Because our one-half factor rule was fairly conservative, we can probably safely say that you can draw 250mA out of the regulator.

[Phoenix]

We've also got to remember that with a capacitor input rectifier, the load is not purely resistive, but reactive, this accounts for the difference between the VA (volt-amp) rating of the transformer, and the W (watt/power) we can extract from it.

A properly specified transformer (and they are NOT all properly specified) will state its VA (voltage in times current in at max), its regulation (percentage the voltage out will increase when unloaded), the voltage of each winding under full load, and what that load is (current) for each winding. The sum of the ratings of all the secondaries (voltage times current) should equal the VA rating of the transformer.

Now, if we load a transformer with a purely resistive load, we can get the full VA out of the transformer in W, power. If we bridge-rectify the AC and load that with a resistor, we can also extract the full VA rating in power (although the load will be slightly non-linear).
However, we can also load the transformer with a capacitor (no rectifier) - and this capacitor still presents an impedance to the transformer, even though no power is being dissipated.
Remember the calculation for capacitive reactance:
Xc=1/(2PifC)
So if we put a 32uF capacitor on a 100V transformer run on 50Hz, Xc=1/(2Pi*50*32*10^^-6) = ~100ohm
Plugging that into ohms law we get I=V/R=100/100=1A!
Calculating for power we get P=VI^²=100*1=100VA!
We need a 100VA transformer to do no actual work! This is called "apparent power" and is measured in VA. It is what you would get if you measured the power going into the transformer. In this example, we're getting no actual work out of this transformer, but if there were, we would make the same power measurements for the load, and this is called "real power" and is measured in watts.
If we take these two measurements, real power and apparent power, we can determine their ratio, the "power factor", by dividing real power by apparent power, which will always be between 0 and 1. In the above example, the power factor is 0.
Typically, a capacitor input rectifier will have a power factor of between 0.5 and 0.7, with bigger capacitors making power factor lower.

Now calculating the power factor that you'll get with a cap input rectifier requires a bit more effort, you need to calculate the RMS ripple current in the capacitor, and the load current, but instead you can approximate using those figures I noted above 0.5-0.7, because we can just guestimate that we'll need a transformer the inverse bigger VA rating than the power factor we expect, so twice as large if the power factor is 0.5, and 1.43 times larger if the power factor is 0.7.

Anyway, I hope that has cleared up some confusion and hasn't caused more. It should at least put people on the right track if they want to learn more.

[R.G.]

My confidence is still only slowly recovering, but I'm willing to venture a guess that the regulator output current is roughly the same as its input current (minus a very small quiescent current of 5-8mA).  Because our one-half factor rule was fairly conservative, we can probably safely say that you can draw 250mA out of the regulator.

Yes, regulators are designed to pass currents out as equal to current in as they can. In the case of the 78xx and 79xx family, the loss to the regulator itself is quite small. In the case of the LM317, the loss is deliberately designed to be down in the microamps.

One thing that may have been hit above but I didn't notice it with my first cup of coffee is that the smaller the transformer the worse the "regulation". A transformer that gives 6-0-6 is specified at some max current. What they don't tell you is that it only gives that voltage --at-- that specified max current. The no-load voltage is deliberately higher so that when it's fully loaded, it sags down to the spec. Paul touched on this, but I wanted to hit another aspect of it.

A small 6-0-6 transformer will give 12*1.414V peak into a filter cap (minus diode losses) at nominal-ish full current. If the load is lower than full current, it will be higher than that, and may cause issues for low-voltage regulators. Small trannies can have "regulations" of 25% or worse, so the no-load voltage may be 25% higher than the nominal voltage you'd think.
So the 17V of a fully loaded 12V transformer may be 10% higher for AC line variation, and then 25% higher on top of that for "regulation". That gets you to 23.3V. Your regulator has to be designed to stand that amount of voltage.

Your regulator also has to stand the heat from passing out as DC the required current while dropping the larger voltage that it may have coming it.

A quick review of "Power Supplies Basics at geofex would be a good start here.
http://www.geofex.com/Article_Folders/Power-supplies/powersup.htm

[Paul Marossy]

A small 6-0-6 transformer will give 12*1.414V peak into a filter cap (minus diode losses) at nominal-ish full current.

In this circuit I am playing with I have a 6V-0-6V 450mA RadioShack transformer that is apparently powering 300mA worth of 12AX7 filament heaters at 6V, a 20mA panel LED and a 12V 130mA brushless fan running at half speed... so I'm guessing that I'm getting somewhere around 350-370mA of current after the full wave rectification and filter caps. Unloaded that transformer was putting out something like 17.7 volts IIRC. AC in is usually about 120-121 volts.

Oh, and hello R.G., long time "no see". 

[EBK]

Yes, regulators are designed to pass currents out as equal to current in as they can. In the case of the 78xx and 79xx family, the loss to the regulator itself is quite small. In the case of the LM317, the loss is deliberately designed to be down in the microamps.

I'll take that as a minor confidence boost.  As to the rest, I swear I must have went beyond ideal transformers in school (I have a very vague recollection of studying various types of losses and interactions) and I know I studied power factor before, but I have to confess that if, at this very moment, I was asked to design a simple power supply, I'd struggle at choosing an appropriate transformer from a catalog.

A quick review of "Power Supplies Basics at geofex would be a good start here.
http://www.geofex.com/Article_Folders/Power-supplies/powersup.htm

Thanks, R.G., I'll take a look.

EDIT: R.G., that article helped immensely!
And thanks, Greg! Now that I've given what you wrote another read, I think I see the whole picture much clearer. 

[PRR]

> max DC current to be roughly half the spec'd AC current value.

The MAX current is 5 or 10 times higher. That's when it sags to zero voltage and starts to heat. Which would be fairly pointless for most electronics. (But doorbell transformers can be worked to large sag and would heat if rung continuously.)

> cheap, small transformers

Unless you paid a lot for specific specs, the "small" is the important feature. By basic geometry, small transformers perform less-well than big ones, unless woefully over-size (essentially a large tranny for a small job).

Yeah, expect 1 Amp AC to give OK performance at 0.5A DC.

> stick an LM7812 on the end of this....

Transformer does not know what it is powering. It does not have eyes to see.

You do have a choice. A low-sag raw supply of acceptable regulation (does not compensate wall-voltage wanders), or a saggy supply plus a regulator for impeccable regulation(*). With linear regs this means wasting some or a lot of the transformer's power to blow-off the excess voltage. It basically leads to a larger transformer. $/Watt you are worse off. The added $ must be justified only by load-circuit simplification when offered steady voltage.

Note a very common mistake. We say Vdc is 1.414*Vac, minus diode and ripple losses. But a regulator has a minimum drop voltage. What we actually need is the DIP of the ripple voltage. 12VAC may be showing 15V average DC, but maybe only 13V in the dips. A "12V" regulator will not go over 12V but will dip to 11V 120 times a second, a raucous buzz.

[PRR]

Here is a PSUD sim of a 12V 1A 20% transformer delivering 0.5A DC.

https://s29.postimg.org/3xzl474tj/AC_DC_factor.gif

This indicates the AC RMS is "only" 0.827A, or 1.65 times the DC current. This is in-line with derived theory.

The RMS seems high because the *peak* AC current is over 1.8 Amps, and peaks multiply in RMS heating losses. Moreso than the extended periods when AC current is zero.

Note that in real-life the quibble over 1.6, 1.8, or 2 is rather moot. Stock transformers are not offered in just-right sizes. But you do have to remember to multiply-up your DC current by a substantial amount. If I needed 1A DC, I would be looking near 2A AC. If a 1.5 Amp AC part jumped at me, I might try that. But if I needed to be "sure" of hitting my output goal, I would aim higher, over 2A.