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Author Topic: Bucking Xformer Power Limitations?  (Read 840 times)
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Edward Cain
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« on: April 26, 2012, 11:06:36 AM »

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
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KE6DF
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« Reply #1 on: April 26, 2012, 11:25:45 AM »

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.
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Edward Cain
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« Reply #2 on: April 26, 2012, 12:59:15 PM »

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
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k4kyv
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Don
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« Reply #3 on: April 26, 2012, 01:07:57 PM »


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.
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Don, K4KYV                                       AMI#5
Licensed since 1959 and not happy to be back on AM...    Never got off AM in the first place.

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Edward Cain
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« Reply #4 on: April 26, 2012, 03:28:20 PM »

Thanks Don. As I said, forgive my ignorance. After thinking about it a bit more with the circuit visualized, it was "obvious".
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WA1GFZ
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« Reply #5 on: April 26, 2012, 03:58:44 PM »

I would think you need a some watts to charge up the core so I would derate the current by 10% or so.
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k4kyv
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Don
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« Reply #6 on: April 26, 2012, 08:08:54 PM »

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.
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Don, K4KYV                                       AMI#5
Licensed since 1959 and not happy to be back on AM...    Never got off AM in the first place.

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k6hsg
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« Reply #7 on: April 26, 2012, 09:02:08 PM »

Hi Don,
A little off the subject but could you give us a long form explanation of Power Factor?
73, John in Tucson
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John, KC0YAI<br />Bona, Missouri
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« Reply #8 on: April 26, 2012, 10:02:01 PM »

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
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KM1H
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« Reply #9 on: April 27, 2012, 08:33:16 PM »

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

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