How much current can my small transformer provide?

I suppose each secondary can provide .175/6=.029A AC RMS.

If you rectify the outputs and filter with a capacitor, you can expect the transformer to get that hot when you use about half that current as DC. This is because the capacitor filter gets all its current in short pulses near the peak voltage, each half cycle. This makes the windings hotter than when the load current is a sinusoid of the same average (of absolute) current.

They must be really tiny transformers. I don't think I have ever seen a 0.35 VA mains transformer.

Hi,

I have a tiny isolated mains transformer with one primary winding described as 230V, and two secondary taps, both rated 6V and 0.175VA It doesn't say anything about power factor.

What current can this transfomer provide? If I use it to power a DC circuit I guess it can provide 0.175 watt, or 0.175/6 = 30mA only? That's barely enough to drive a couple of leds? I also read somewhere that often the VA rating may be reduced to aproximately 2/3 to get the real power in Watts.

described

circuit

Are you sure? I can't imagine a mains transformer that is only 0.175VA. That is vanishingly small. The smallest one I can find in my junk box is 5 Watts, 6V @ 300 mA. I suspect the 0.175 VA is really current, 0.175A, 175 mA and it is mislabled or you are reading it wrong.

Put a 30 ohm load on it and see what it does. What is the output voltage loaded and unloaded and how warm does it get after an hour or so? That should answer your question.

ErikBaluba wrote: (snip)

Yes.

That is one application of power factor. Nonlinear loads like rectifiers that generate harmonic currents are another.

This applies only to AC circuits, not DC.

No. The power factor of the rectifier with a capacitor filter and resister load will be a low power factor load due to the harmonics the rectifiers generate when they charge the cap up with narrow pulses at the voltage peaks.

That doesn't work, once you put a rectifier in there.

You can just parallel the two windings, as long as both windings put out the same polarity. Test this by putting them in series, first. if they add up to about zero volts this way, just connect the free ends together and you are done. If they put out about 12 volts, break the connection between them, and remake the connection using the other end of one winding and test again to see if you have achieved the first case.

When fully loaded, the output of the windings should be 6 volts, if the line is at the rated voltage.

If the parallel combination is rated for only about 58 ma, yes, it will get hot, and the voltage will be less than 6.

Yes. The time it takes to reach an over temperature condition depends on the thermal mass of the transformer and the amount of overload.

Make sure there are no grounded surfaces within reach, is a good idea, also.

Thanks for the answers guys, I was out for a few days, so late follow-up. By the way, the transformer is really that small, its a mains transformer in a blue plastic shield, about the size of a grape. It was very cheap also, about $1. I can mail you a picture of it if you want :-) I bought these small transformers in a local store. I plan to build my own home automation system and I want to place small RF controlled relays around my house, between the electrical-outlets and the appliances. I need to provide the RF receivers and the logic with appropriate voltage, so I thought a really tiny 6V transf. would be the thing. I guess 29mA is enough to power some simple RF receiver from Laipac etc.. I just saw an article in an Electronics Magazine that described a extremely useful small circuit to step-down the mains voltage to TTL level by just using a X2 safety capictor, a FET and a few compos, no transformator. Perhaps i should just use that instead.

Anyway, the concept of powerfactor confused me at first because I read that powersupply equipment is often given a powerfactor rating. So is such a rating typically given for a particular load or ampere usage? Let me follow up on that a bit so I get this right. When a load is purely resistive the powerfactor is 1, right? So the reason we have the concept of apparent power and real power is the inductive/capacitive characteristics of a load. I read that if my load has a capacitive reactance I should be able to add some coil/inductance to compensate/cancel out that, in theory at least.

So, if I build a simple powersupply with my transformator, using a full rectifier and a filter cap (no regulator), can you comment on the following

- Assuming the Cap is a big electrolyte, the powerfactor I get when connecting a resistive load would be a good indication of the quality of my transformer?

- If the transformer and the cap where "ideal", I could select the cap to "match" my transformer so that my setup would have powerfactor 1?

Also, can I combine the two taps in some way to get one 6V output with double VA rating, thus 2x0.175VA ?

Wouldn't that mean that I = Vrms/30, or more than 100mA? That should surely make it hot yes... What is the likely senario when overloading a transformator like this? I guess the heat can melt the isolation on the copper wires and create a short circuit in the windings? I think I need to set up a few things in my work environment before I start fooling around with mains stuff. Rubber gloves and mat comes to mind :-)

cheers, erik

Not necessarily. It just means there are currents at frequencies that add into the line RMS current that do not add into the average DC output current. Here is my attempt to show the line current of a capacitor filtered full wave rectifier load (vies with fixed width font, like Courier):

^ | | | |

----- ----- ----- | | | | V

That kind of waveform is made up of the line fundamental and many odd harmonics. The RMS current is the square root of the sum of the squares of the RMS current of each of those harmonic frequencies. But the average DC is just the average of the absolute magnitude of the waveform. For a given load resistance, the larger the capacitor, the shorter the pulses and the higher the harmonic content in the current waveform, even though the average DC current will be about the same. The higher RMS value of the large, narrow pulses also have an increased heating effect on the transformer winding resistances.

All those harmonic currents have to be supplied from the AC source (through the distribution system), even though they do not relate directly to making the DC.

Neither. It is because the capacitor receives current in narrow pulses twice per cycle. Switching supplies with power factor correction circuits are attempting to spread the line current out over the entire cycle, making the load look more like a resistor to the distribution system.

Everything just scales up in frequency, as long as the diode conduction time remains the same fraction of the cycle time. Changing to an inductor input filter makes the current almost constant, that produces a different mix of line harmonics, but a mix that contains more fundamental and less harmonics, so the power factor is better.

Ok, I partly understand that. Bear with me if the following are lame questions and assumptions.

So the cap is only being charged when the input voltage is near peak, because the cap only have time to discharge a little bit between peeks. But can you give a more "layman's" explanation of what is meant when harmonic currents are generated? Does it mean that the current delivered by the cap has a sinus-derived form? When something generates harmonics, is that generally understood as noise in the circuit? In this example, is it because diodes are not perfect that you get harmonics, or is it because of the inductance in the transformer trying to resist the increasing current to the cap? I am just trying to understand what it actually means.

Also, if the frequency of the input AC was higher, what would that mean for the harmonics? I mean the the cap would not have time to discharge as much in each recharging cycle...But then again, the higher frequency would generate a higher impedance in the transformer...

thanks for your help...

erik

to

surely

short

start