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### Network # How much current can my small transformer provide?

E

#### ErikBaluba

Jan 1, 1970
0
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.

E

J

#### John Popelish

Jan 1, 1970
0
ErikBaluba said:
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.

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.

K

#### Kitchen Man

Jan 1, 1970
0
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.

Power factor is irrelevant to the transformer rating, as it is given
in VA, or "Volt-Amps." A secondary rated at 6V @ 0.175VA will give,
using standard mathematic grouping of terms [Volt-Amps divided by
Volts], 0.175 / 6 = 29mA. Your calculation below is thus correct, and
not really surprising because as you said, the transformer is tiny.
6VAC is typical for powering a vacuum tube cathode heater, is that
where this came from?
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.

If you want to find Watts, now you have to know the power factor of
your load, done by calculating X vs. R at line frequency. Watts is
the measure of actual work done, or volt-amps into the resistive part
of the load. Volt-amps delivered into the reactive component of the
load is wasted (no work is done), and is measured in Volt-Amps
Reactive, or VARs. If you are only getting 2/3 of your energy
converted to Watts, then you have a rather inefficient load.

B

#### Bob Eldred

Jan 1, 1970
0
ErikBaluba said:
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.

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

E

#### ErikBaluba

Jan 1, 1970
0
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 ?
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

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

J

#### John Popelish

Jan 1, 1970
0
ErikBaluba wrote:
(snip)
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?
Yes.

So the reason
we have the concept of apparent power and real power is the
inductive/capacitive characteristics of a load.

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

This applies only to AC circuits, not DC.
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?

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.
- 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?

That doesn't work, once you put a rectifier in there.
Also, can I combine the two taps in some way to get one 6V output with
double VA rating, thus 2x0.175VA ?

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.
Wouldn't that mean that I = Vrms/30, or more than 100mA? That should surely
make it hot yes...

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

Yes. The time it takes to reach an over temperature condition depends
on the thermal mass of the transformer and the amount of overload.
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 Make sure there are no grounded surfaces within reach, is a good idea,
also.

E

#### ErikBaluba

Jan 1, 1970
0
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.

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

J

#### John Popelish

Jan 1, 1970
0
ErikBaluba said:
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?

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.
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.

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.
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...

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.

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