Bucking Xformer Power Limitations?

[Edward Cain]

Forgive my ignorance. If I use a xformer with a 12.6 v secondary rated at 3.0 A as a bucking xformer, what will be the max wattage I can supply my equipment? And why?

Thanks,
Ed

[KE6DF]

It's a question of current.

If the bucking transformer can handle 3 amps, since it's in series with the equipment being powered, you can run equipment drawing a maximum of 3 amps.

Assuming a power factor of 1.0, you could supply:

(original line voltage - 12.6) x 3 watts.

Often the power factor is less than 1, then the watts delivered to the equipment would be reduced by that factor.

[Edward Cain]

Thanks for the reply, 'DF.

So, if used as a simple step down xformer, it would be limited to 12.6 V x 3 A or  about 38 watts but when used in a bucking arrangement it can supply on the order of 300 watts?

Ed

[k4kyv]

So, if used as a simple step down xformer, it would be limited to 12.6 V x 3 A or  about 38 watts but when used in a bucking arrangement it can supply on the order of 300 watts?

That's right; you should be able to pull close to 300 watts.  The bucking voltage is in series with the line voltage, so assuming a reasonably good power factor, power (watts) ≈ (line voltage − 12.6) × 3A. The bucking transformer is still handling only about 38 watts.

[Edward Cain]

Thanks Don. As I said, forgive my ignorance. After thinking about it a bit more with the circuit visualized, it was "obvious".

[WA1GFZ]

I would think you need a some watts to charge up the core so I would derate the current by 10% or so.

[k4kyv]

That's an interesting thing about a variac.  You are pulling the most power through the core winding somewhere an an intermediate voltage level, well below maximum.  Near maximum voltage, close to line voltage, it is mostly acting as a choke across the power line.  The buck or add voltage a few volts above or below line voltage is pulling very little power through the autotransformer.

[k6hsg]

Hi Don,
A little off the subject but could you give us a long form explanation of Power Factor?
73, John in Tucson

[KE6DF]

Hi Don,
A little off the subject but could you give us a long form explanation of Power Factor?
73, John in Tucson

Don is so much better than I at explaining things, I'll be interested in reading his explaination too.

But meanwhile, I'll give it a shot.

Powerfactor applies to the phase angle between current and voltage in AC circuits.

When you attach a pure resistive load to an AC circuit, like an incandescant bulb for example, the current and voltage in the circuit are in phase. The RMS volt * amps (VA) flowing in the circuit equals the power in Watts delivered to the load and the power factor is 1.0.

Now if instead, you have a purely reative load, like attaching an inductor or capacitor across the AC circuit, you also have a current flowing.

But in this case, the current and the voltage are out of phase. What is happening, is during part of the cycle as the voltage rises, the current decreases until part of the way through the cycle, the current reverses and then flows the other way during the other part of the cycle.

So during 1/2 the cycle power is being delivered by the wall socket to the load and during the other half, the load is delivering power back into the wall socket.

In a purely reactive load, the VA may be the same as with a resistive load -- and a current meter would read the same -- but the power factor is 0 so no power is consumed from your utility company and your electric meter doesn't charge you anything.

In effect, you are charging the capacitor (in the case of a capacitive load) across the line for half the cycle, and then the capacitor is discarging and putting power back into the line during other half the cycle -- with the net result that no power is consumed.

Most real world devices have a power factor between 0 and 1 and are either partially capacitive or partially inductive.

In these cases, you need to use the formula below to determine the power consumed by the device:

Power consumed in Watts = RMS voltage of the AC circuit x Amps x F

Where F is the power factor.

Dave

[KM1H]

http://en.wikipedia.org/wiki/Power_factor

http://www.allaboutcircuits.com/vol_2/chpt_11/3.html

http://masterslic.tripod.com/FAQ-2/22.html

Those are only the first 3 in a Google. Why ask questions that have established and well published answers?

Carl

[K6JEK]

Excuse me for firing up an old thread but I realize I just don't understand something.  When using filament transformers as bucking transformers we talk about the current rating of the secondary, 6.3V at 3A, that kind of the thing. But what about the primary? It's in series with the secondary. Shouldn't we also be concerned with the current rating of primary which we often don't know?

[WZ1M]

Carl, good question.........

[W2VW]

The primary goes in parallel with the existing transformer's primary.

[WD5JKO]

The primary goes in parallel with the existing transformer's primary.

  Not quite. The primary of the boost/buck transformer goes across the AC supply. The secondary of the boost/buck transformer is in series with the existing transformer primary. This series combination is also across the AC supply. The phasing of the boost/buck transformer determines whether the power supply output is increased or decreased.

  A boost / bucking transformer only needs to be rated for the Volts X Amps boosted or bucked. A 10V @ 5A transformer is 50VA, yet can add or subtract 10V at up to 5A at the existing transformers primary.  

  Power factor is real power in Watts / VA. VA is just Volts X Amps. For a sine wave source, Real power in watts is VA * Cosine of the angle between Volts and Amps. If the volts are in phase with the amps, then the angle is zero, and the Cosine of 0 = 1...So the power factor is VA/VA or 1. If the volts and amps are 90 degrees out of phase, the Cosine of 90 = 0, so the real power is also zero, and therefore the power factor is zero.

Edit: An old fashion AC induction motor such as a washing machine motor (before modern variable speed RFI generators of today) when unloaded is almost pure inductive, so the current lags the voltage nearly 90 degrees. Overcoming frictional losses, and the copper I^2 * R losses adds a real part to the load (some R) so perhaps the current lags the voltage by 80 degrees instead of the ideal 90 (if the motor were perfect). So that motor will have close to the same VA unloaded as compared to the rated Horsepower load. What changes as we load it is the phase lag drops steadily towards zero as we approach rated load. As you can imagine the power factor unloaded is very low (maybe .3), and perhaps around .9 at rated load.

  Then comes ELI the ICE man. If the voltage E leads the current I, the current is lagging the voltage..The circuit is inductive. With a capacitive load, the current leads the voltage. So a circuit with a low power factor can be either an inductive, or a capacitive circuit. A circuit with a resistive load has unity power factor (1).

Edit: It was asked about what is the boost/buck transformer primary current? So lets say we are using a 12.5V 5A transformer to buck the AC mains voltage down from 125 - 12.5 to 112.5 VAC.  Although the transformer might be rated at 5 amps, perhaps the load is only 3 amps (whatever the main transformer draws with 112.5 vac on the primary). The boost/buck transformer has a 10:1 turns ratio which is the same as the voltage ratio 125/12.5. The current on the buck / boost transformer primary is also reduced exactly by the turns ratio, so with 3 amps secondary, the primary current will be 1/10th that, or 300ma. Knowing this primary current is useful should you want to fuse the primary. Of course if the fuse blows, the main transformer has still got voltage; the fuse only protects the buck / boost transformer.

Jim
WD5JKO

[W2VW]

Stated it incorrectly. Thanks.

[w8khk]

Actually, I agree with both VW and JKO.

My buck/boost box has a switch to select either 3.15 or 6.3 12.6 buck or boost, and it also has a switch to choose whether the input to the buck transformer primary is on the input side or the load side.  This provides a slight increase or decrease in the output voltage, between the buck transformer secondary steps.  Maybe overkill, but when doing measurements on receivers, it gets me closer to the rated input voltage of the device under test.

[K6JEK]

I must have been asleep when I asked that question. Nice discussion anyway, guys. Thanks.

It's for a '40's Zenith console (10S668) I got working for friend. It's been living a sheltered existence on a Variac at my place but is about to enter the harsh outside world of 125 VAC. Maybe working with all that crumbling insulation in the radio crumbled some insulation in my brain.

[K4RT]

http://en.wikipedia.org/wiki/Power_factor
http://www.allaboutcircuits.com/vol_2/chpt_11/3.html
http://masterslic.tripod.com/FAQ-2/22.html

Those are only the first 3 in a Google. Why ask questions that have established and well published answers?

Carl

The way I see it, posting questions here gives the rest of us an opportunity to consider the issue, the actual or proposed application, and benefit from the answer(s), which often include personal experiences we might learn from.

Brad

[K5MIL]

My line voltage always runs on the high side, anywhere from 122-124 volts. I had a doctor friend who retired some months ago and gave me some electrical medical equipment from his office. One piece of equipment was a cardiac difibrallitor unit, probably of 1960's vintage. The unit consisted of two chassis, one the cardiac syncronizer panel and the other contained a huge transformer and the bank of capacitors that were charged to administer the shock.

Upon examining the transformer it looked like it had a low voltage secondary that could carry a lot of current. The secondary winding is square buss wire. The output voltage measured 8 vac. I thought this would make a good bucking transformer that would easily handle everything in my ham station.

I wanted to knock the voltage down to around 117 volts or so, so I tapped the secondary to produce 5 vac to buck the 122 vac line voltage.

I also constructed an ac control panel that would allow switching ten ac sockets so I could plug the rigs into the control unit and switch all rigs on and off, thus eleminating wear and tear on the on/off switches in the rigs from having to switch any load. The on/off switches in the Collins rigs sometimes go bad and are not easy to find and can be expensive.

I used TRIACS to do the ac switching instead of relays as TRIACS are much less expensive. The TRIACS are driven with optical isolator ICs.

A relay, controlled from the Master switch on the control panel controls the AC in the bucking unit with a relay. There is also a low voltage supply that provides 12 vdc for the relay and 6 vdc for the TRIAC driver board.

Below is a link that shows pictures of the two chassis and the construction.

http://radioremembered.org/controlpanel.html

[WD5JKO]

Bill,

   That story you told was very inspiring. Having the triac zero crossing switches is neat as the rigs should never "thump" when turned on. I wonder if some equipment loads (reactive loads) result in the triac turning off somewhere mid half cycle?  In the 70's I used to design triac pulse train gate drive circuits when the load was highly inductive, or capacitive. The MCO3010's are pretty good though, and that arrangement you got is so simple, yet I bet effective.

   Good Job!

Jim
WD5JKO