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rockazella

Transformer question

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I have to disagree with Gazza.  Since the voltage ratio rockazella inquired about is the same for both input voltages and both are below the voltage rating for the transformer, it will be fine for stepping 120 volts down to 60 volts, as Ldanielrosa said.  You don't need and couldn't use a tap to do the job unless there was a 50% tap on both primary and secondary.

The primary disadvantage would be excess weight due to more than the necessary amount of steel and copper in the transformer.  There might also be slightly greater fixed losses due to the larger volume of steel to be magnetized, but it's been a long, long time since I studied transformers so I may be mistaken on that.  If that is true, it would show up mainly in slightly greater fixed losses at low current draw.

Even though you would be using the transformer at well below its power rating, you shouldn't exceed the rated secondary current because the wire gauge was selected by the designer for the design current.

awright

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I have to disagree with Gazza.  Since the voltage ratio rockazella inquired about is the same for both input voltages and both are below the voltage rating for the transformer, it will be fine for stepping 120 volts down to 60 volts, as Ldanielrosa said.  You don't need and couldn't use a tap to do the job unless there was a 50% tap on both primary and secondary.

The primary disadvantage would be excess weight due to more than the necessary amount of steel and copper in the transformer.  There might also be slightly greater fixed losses due to the larger volume of steel to be magnetized, but it's been a long, long time since I studied transformers so I may be mistaken on that.  If that is true, it would show up mainly in slightly greater fixed losses at low current draw.

Even though you would be using the transformer at well below its power rating, you shouldn't exceed the rated secondary current because the wire gauge was selected by the designer for the design current.

awright



My bad, I read the question too fast. You are 100% correct.

If the turns ratio is .5, the output voltage will be half of the input voltage. Which is the case here.

My appologies.

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I think the voltage might step down okay with a light load. A power transformer has an iron core to keep the inductance not so high, which makes the reactance lower, which makes the current high. 120Vac can't generate as much current. 1 Watt on the primary can't provide 2W on the secondary. Bring the voltage up 240Vac, and you get 2 Watts on the primary, 2 Watts on the secondary.

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I think the voltage might step down okay with a light load. A power transformer has an iron core to keep the inductance not so high, which makes the reactance lower, which makes the current high. 120Vac can't generate as much current. 1 Watt on the primary can't provide 2W on the secondary. Bring the voltage up 240Vac, and you get 2 Watts on the primary, 2 Watts on the secondary.


I don't think you know what you are talking about. Once you have the turns ratio which is equal to

a=V2/V1 or a=N2/N1

You can find the voltage on the opposite side of the transformer.

If you have a 240 to 120V transformer, the turns ratio is equal to 120/240 = .5

Therefore if you feed the primary side of the transformer with 120 volts, the secondary voltage will be

.5= v2 / 120
v2 = 120 x .5
v2 = 60

Since current is inversely proportional to the voltage and power in must equal power out, the opposite is true, if you have 1 amp of current at 120V you must have twice as much current at 60V.

The iron core is to help mitigate losses in the transformer, I suggest you look into the equivalent circuit of a transformer.

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Okay. All you need is turns ratio and resistance. They say use 60 Hz, but that's only a general rule of thumb. I use 120 Hz.



Resistance has absolutely nothing to do with the turns ratio. What do you mean you use 120Hz?

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Alright. I took a 3:1 transformer, 34Vac input 10.3Vac output with 100 ohm load. That's 2.1 Watts peak power. I reduced the load to 25 Ohms. 34Vac input 6Vac output. That's 2.9 Watts peak power. The transformer is rated at 120Vac, and put's out 40Vac when plugged into the outlet. That's 3:1. A 100 ohm load keeps it a 3:1 transformer. A 25 ohm load makes it a 6:1 transformer, and the the current peaks at only 340mA. Well within current rating.

A light load of 100ohms works. The energy is not available for the 25 ohm load. Else I would get 10Vac on 25 ohms for 9 Watts peak power.


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Kevin,
A transformer has a max VA rating or a secondary max current rating. You forgot to tell us the rating of your transformer.

It produces its rated output voltage if it has its rated input voltage and has its rated output current. It is made to produce a higher output voltage without a load because the resistance of its windings will cause its output voltage to drop with its rated load.
Maybe you overloaded it.

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There's not much information written on the transformer itself, just some numbers. But it came from a power supply that had a 120Vac 60Hz input. It weighs about as much as an AC adaptor 4.5V .8A output DC. The AC adaptor says 6W also.

The 100ohm load with a 34Vac input stepped down 3:1, just like with 120Vac. A 25 ohm load took it 6:1.

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Alright. I took a 3:1 transformer, 34Vac input 10.3Vac output with 100 ohm load. That's 2.1 Watts peak power. I reduced the load to 25 Ohms. 34Vac input 6Vac output. That's 2.9 Watts peak power. The transformer is rated at 120Vac, and put's out 40Vac when plugged into the outlet. That's 3:1. A 100 ohm load keeps it a 3:1 transformer. A 25 ohm load makes it a 6:1 transformer, and the the current peaks at only 340mA. Well within current rating.

A light load of 100ohms works. The energy is not available for the 25 ohm load. Else I would get 10Vac on 25 ohms for 9 Watts peak power.






The resistance has NOTHING to do with the turns ratio of a transformer. What you are seeing is something totaly different.

PIn=POut therefore if you have a transformer that is 5:1 with a primary of 120V and a 60ohm load.

The secondary voltage is equal to 120/5= 24V
The current is = 24/60 = 400 mA

Now, if you reduce the load to 30 ohms

The secondary voltage is equal to the primary voltage divided by the turns ratio, it does not change 120/5 = 24V
Current = 24/30 = 800mA

The secondary current is a function of the load, and is not used to determine the turns ratio.

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I haven't gone back over the entire thread to remind myself of how we got to this point, but I think that Gazza"s analysis is a little simplistic.  I also do not claim to be a transformer engineer.

Pin = Pout ignores the losses in the transformer which can become significant for a small transformer near or beyond its rated load.  Notice that a small transformer loaded to around its rated load gets warm.  Some get downright hot.  That power going into heat comes from the input power.  While properly designed transformers are remarkably efficient compared to most electronic devices, they are not perfect and they become very imperfect when marginally designed and heavily loaded.

Part of the losses are in the hysteresis of the steel core and part of the losses are the I^2® resistive heating in the wire of the primary and secondary windings.  My vague recollection from school decades ago are that the hysteresis losses in the steel are constant, independent of current, while the I^2® losses are obviously totally dependent upon loading.

The result of these losses is REGULATION, the deviation of true output power from theoretical output power, which is seen as the output voltage being lower than the turns ratio would predict.  Prediction of the actual deviation from perfection involves careful evaluation of transformer parameters.

By the way, what is "peak power?"  Are you selling shop vacs or home air compressors?

awright

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While I've never designed an iron core transformer, I have designed more ferrite core transformers than I care to remember. Ferrite core transformers are typically 98% efficient. If core losses are any higher, it's considered a bad design. Core loss is a function of frequency and volume (and this is stupid... in KW/M^3... I've NEVER seen a one cubic meter ferrite core). Then theres skin affect (not a big problem at 60Hz, but at 500KHz that's a different story), eddy current losses, I^2*R losses, coupling losses if it's not wound well. Voltage, current , inductance and power are all proportional to the turns ratio.

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Gazza, that is the problem, the turns ratio should have stayed the same. A power transformer needs to transfer energy from the primary to the secondary. How are you going to get power in the primary if the reactance is so high that you don't get any current in the primary? Conversely, how are you going to get current in the primary if you don't have enough voltage?


I use peak power to avoid confusion. V*V/R can give you different results depending on what numbers you use.

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Kevin,
You want a transformer primary winding to have a high reactance, when the secondary doesn't have a load. Then the transformer doesn't draw much current when it shouldn't.
When you add a load to the secondary then you want the magnetic coupling to transfer power from the primary to the secondary.

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What is the primary current in ANY transformer?

The primary current will be the secondary current reflected to the primary by the turns ratio plus the magnetizing current.

Don't confuse reactance with resistance!!

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Hi Indulis,
Sure a transformer without a load has a high reactance in its primary winding.
If you remove its core then it is just a very low resistance piece of wire.

A resistive load on the secondary is reflected and stepped up or down to the primary in parallel with the very high reactance. Since the reactance is much higher than the reflected resistance then the reactance is neglegible and the primary appears resistive.

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Guru

My comments were intended for Kevin...

You mean that the inductive reactance dominates (XL) the impedance (Z).

Without a core it's still an inductor (air core) with mutual coupling to another inductor... sure the coupling coefficient is crappy, but it's still "more" than just a piece of wire.

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I attempted to construct a step down autotransformer one time with magnet wire. I figured that without a load, if I got the turns right, I would have a simple voltage divider. Being totally reactive for the most part, it would consume very little power.

The primary reactance was about 1000 ohms, the secondary reactance was about 1000 ohms. I took a 20 ohm resistor and attached it to the secondary. Is that a problem? A 20 ohm resistor in parallel with a 1000 ohm reactance. Will the autotransformer still be 2:1.

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The power rating for a mains transformer depends on the size of the core and wire, the number of turns and a few other things. If the input is 120V and the output is 60V then a 20 ohm load would be 180W. It must be a pretty big transformer.

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What is the primary current in ANY transformer?

The primary current will be the secondary current reflected to the primary by the turns ratio plus the magnetizing current.

Don't confuse reactance with resistance!!



Im glad some one gets it.

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