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Topic: Question regarding transformers and size (amp's and windings) (Read 3100 times) previous topic - next topic

oddbeck

This is more of a theoretical question:

If I have two transformers, "BIG_T" with windings ratio (100'000 : 10'000) and "LITTLE_T" with ratio 1000:100) will the effect
be that "BIG_T" can have say 250 v 5 amps in and 25 volts 50 amps out and "LITTLE_T" can have 250 v 1amp in and 25 v 10 amps out?

I assume that the amount of wiring is what makes a transformer able to deliver/handle more current?

And then the none theoretical question:

I opened an unused "motion sensor light switch" the other day, as I was curious, and I think I found something to the extent of a small
transformer, it said something like 250 VAC 12 V - I don't have the parts number here but I'm gonna check it out when I get the parts number,
but could this be the case?

TomGeorge

Hi, yes you have the basics right, power in = power out + power losses.
The wire gauge dictates the amount of current that the transformer can be use for.
The amount of turns usually dictates the operating voltage.
Both of your examples have 10:1 turns ratio, the BIG_T will be design to operate at a higher voltage primary voltage than SMALL_T.
There are equations out there and other considerations to be made when designing a transformer.
Which is why its best left to the experts, parameters such as volts per turn, heat losses, eddy current losses, core types, operating frequency and insulation considerations.
Usually the smaller the transformer the higher the losses depending on operating frequency, but this is where core materials comes in.

These days with switch mode power supplies transformer technology is quite fascinating.

You may not find a transformer in the motion sensor switch, most use the impedance of a capacitor to drop the circuit voltage to about 30V, so the circuit is still "live" even though its low voltage. So caution if you apply power while its open.
Tom.
Everything runs on smoke, let the smoke out, it stops running....

Grumpy_Mike

Quote
I assume that the amount of wiring is what makes a transformer able to deliver/handle more current?

Yes the more wire it has the less current it can deliver.

oddbeck

#3
Sep 06, 2013, 10:23 am Last Edit: Sep 06, 2013, 10:27 am by oddbeck Reason: 1
I thought the only thing that could change voltage from 250v (~220v) to 12v was a transformer?

...And I'm not planning on implementing my own transformer, that'll be up to the experts, I was curious as to how the curcuit was set up while trying to identify items on the light switch motion sensor.

Grumpy_Mike

Quote
I thought the only thing that could change voltage from 250v (~220v) to 12v was a transformer?

No a switch mode power supply is another way of changing voltage up or down. A transformer is just the simplest.

Quote
And I'm not planning on implementing my own transformer

They are quite easy, just two coils wound on a nail will do. Mind you that will have quite a high loss compaired to one with a proper laminated core.

oddbeck

I've tried to read up on how to lower AC voltage from ~220VAC to ~12VAC (or even DC) but I found very few examples and explanations, maybe I'm not using the correct keywords.

If you connect 220 VAC to a capacitor, and resistors in series, wouldn't it then contain 220 V, just that the current would flow at a different speed (amps)?
Would you have to induce elements that consume voltage in series in order to get down to the desired voltage?

I'm trying to keep it simple and not saying that I require either 12 VAC or 12 VDC, just curious as how the different components would
make this happen... I do (think I) understand how transformers work, and up till now I thought I had figured out capacitors and resistors too...

nickgammon

I'm no expert, but read up on buck converters. They reduce voltage in a different way than a transformer.

http://en.wikipedia.org/wiki/Buck_converter
Please post technical questions on the forum, not by personal message. Thanks!

More info: http://www.gammon.com.au/electronics

michael_x

Quote
If you connect 220 VAC to a capacitor, and resistors in series, wouldn't it then contain 220 V, just that the current would flow at a different speed (amps)?

Just for your understanding, this is how I understand it -without the complex AC maths - :
If you connect a capacitor to 250V AC ( assuming it's rated for that voltage ), it will allow a small AC current flowing.
The current is determined by voltage, capacitance and frequency.
If you add a resistor in series, you can measure a voltage across the resistor according to V = I * R.

The voltage across the capacitor and the current are not in phase, so it is not producing heat according to DC Voltage*Current, which is the advantage over a resistor voltage divider.

But for currents above a very few mA and a voltage of 250V @ 50Hz you'd need big expensive capacitors, so this approach is not useful if you're looking for a real power supply.
And the initial current when switching on independent of the AC voltage phase, might be way beyond the steady state current.

Besides, there's no galvanic separation between the 250V and your low voltage, and every hint should start with a "Children Don't Do This At Home" disclaimer.
Rather use LTSpice simulation to see that theoretically, if you're interested.

In reality, you can't and should not compete with wall warts, which nowadays have a small transformer/inductance running much more efficiently at high frequency switching.


polymorph

A capacitor in series with a load will cause the voltage drop across the load resistor to change, less voltage with more load (less resistance), so not a very good way to reduce voltage.

There is also the problem of plugging it in - if you happen to connect the plug or flip the ON switch just as the line voltage is at a peak, you have just connected an uncharged capacitor across 170V (USA) or 340V (UK), causing a huge spike of current through the switch, and that 170V or 340V will appear in its entirety across the load resistor.
Steve Greenfield AE7HD
Drawing Schematics: tinyurl.com/23mo9pf - tinyurl.com/o97ysyx - https://tinyurl.com/Technote8
Multitasking: forum.arduino.cc/index.php?topic=223286.0
gammon.com.au/blink - gammon.com.au/serial - gammon.com.au/interrupts

MarkT

You can use a capacitive divider to prevent the voltage spike across the resistor, but it won't prevent the
current spike at switch on, for which snubber components might be a solution (the simplest being a resistor
in series with the top capacitor - thus inventing the 10x oscilloscope probe!)
[ I will NOT respond to personal messages, I WILL delete them, use the forum please ]

Grumpy_Mike

I used a capacitor as the top leg of a potential divider to light the LED in an opto isolator to detect any mains failure in a set top box I designed. They made 250,000 of those boxes. I even have a patent on the method of detecting mains failure to allow essential information to be stored in EEPROM in a set top box context. Well I say me but all the rights to it are owned by the company I worked for at the time.

polymorph

Did you have just a capacitor to the LED side of the optoisolator? Or was there a resistor there, too? How did you protect the LED inside the OI from reverse breakdown?

Remember, we have a lot of people here new to electronics.
Steve Greenfield AE7HD
Drawing Schematics: tinyurl.com/23mo9pf - tinyurl.com/o97ysyx - https://tinyurl.com/Technote8
Multitasking: forum.arduino.cc/index.php?topic=223286.0
gammon.com.au/blink - gammon.com.au/serial - gammon.com.au/interrupts

nickgammon


I even have a patent on the method of detecting mains failure to allow essential information to be stored in EEPROM in a set top box context.


If you could just give us the patent idea, Mike, I'm sure everyone would be happy. :)
Please post technical questions on the forum, not by personal message. Thanks!

More info: http://www.gammon.com.au/electronics

Grumpy_Mike

The idea was simple. If the mains disappeared through mains failure or the customer unplugging the set top box, things like the current channel and settings like mute, volume, channel you were on, and a load of other stuff had to be saved. It was in the spec for the box. Other boxes monitored the supply voltage to the processor and when that started to drop it fired an interrupt and you did the saving to flash. This gave you about 6 to 10 mS to do the saving which was not always enough. By monitoring the mains input to the power supply you could get 250mS before the set went down, this was an age by comparison.

The arrangement was mains live through a capacitor -> resistor -> mains neutral. From the capacitor resistor junction was a diode that fed into the LED of an opto isolator LED, other end of LED to mains neutral.
The output transistor of the opto isolator was across a capacitor that was being charged to 5V through a resistor. Each times there was a mains cycle the transistor would turn on and discharge the capacitor. The time constant was such that you could miss three mains cycles without triggering the interrupt, because the box had to be able to survive  three missing mains cycles and still work (again part of the spec of the box). If more than three cycles were missing the capacitor would charged up sufficiently to fire the interrupt. The charge in the smoothing capacitors kept the box going while the saving took place.

nickgammon

Please post technical questions on the forum, not by personal message. Thanks!

More info: http://www.gammon.com.au/electronics

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