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Wall outlet current draw Answered

I have built a transformer, that will take 120v, and step it down to 2.5v. I used all the right parts, and the correct rated wire for this. My only issue, is that when i connect the primary coil, to a call outlet, at 120v, it draws more current then my 5A fuse can handle, and the power goes out in my home. I need to get 1.5 amps, @120v going through my primary coil, in order to get the desired output on the secondary. I was wondering, how things "draw" current from the wall, as well as if i where to use a resistor, wouldn't it still draw too much amperage, and just dissipate it in heat, to give the desired output (i cannot use, because it draws more than 5 amps). i could not find a resistor out their, that could handle over 100 watts, so i was also wondering how things like vacuum cleaners only draw a small amount of current. Would i need some sort of current regulator (if that's even a thing)? if you had a light-bulb, and a battery, would the circuit the light-bulb is attached to drain less current, if their was a resistor, and thus having the battery last longer, or would it draw the same amount, and just dissipate energy through the resistor in the form of heat, to give the light-bulb its self less current?



2 years ago

It's complicated, but this stuff can be understood...

Any device has characteristics, specs, etc. This includes a rated operational voltage, and a specific amount of current that's drawn at that voltage (it may not be fixed, as some devices such as motors draw more current when under physical load).

We could say that the "nature" of the device dictates the amount of current required. A 6V motor generally draws more current than a 6V lamp. If the device does more work, it will draw more current. Current draw is a characteristic of the device itself.

We can learn a lot with Ohms Law. Let's look at the specifics of your case: a 5A fuse is pretty small. When I had a fusebox, we'd normally use 15, 20 and 30W fuses. Power (wattage) is V * A, so the maximum wattage your house circuit can handle is 120 * 5, or 600 watts. An unusually small amount of power capacity for a household circuit.

Now, look at your transformer--a good capable step-down transformer will convert voltage with more than 90% efficiency (somewhat higher, generally).

So, what is your 2.5V device, and how much current should it draw? Does it sound right to you that the 2.5V device and the transformer draw more than 5A @ 120V (600W)? Or is there a problem somewhere, such as with your transformer...maybe a short?

(It's somewhat possible to limit current, but Ohms Law WILL be satisfied, no matter what. The electrical energy is all converted to heat (eventually), and this too cannot be cheated.)

The 5A fuse is not the one in my fuse box, my apologies, it is the one in the power strip I am using. I am using the power strip because it has a 5A fuse. It is in place both because I do not want my transformer drawing more current then my house circuit breakers can handle, as well as that my transformer can not handle anything above 5A. so for the circuit is basically the wall outlet (power source), and and inductor (primary coil). How I would have set up my circuit, is to have a 80 ohm resistor (resistance to give me the 1.5 amps) and then the inductor (primary coil) the only issue with this is that I would need a resistor that could handle at least 180 watts, at 80 ohms, which does not exist. Even if it did, wouldn't my circuit still draw over 5 amps of power before it reached the resistor, and the resistor dissipates the energy as hear bring the amperage at my inductor (primary coil) to 1.5 amps? how can things such as phone charges, that are very small have a resistance so great, at a tolerance so great, for their size (resistors that can handle high wattage are huge). How then if I don't use a resistor could I limit the current?

Unless I misunderstand, what does adding a resistor to the primary (in series?) have to do with the load on the secondary? And the total current draw?

I.E., why do that?

You may drop the voltage across the primary, but if you're trying to adjust the output voltage like that, you're using the wrong transformer for your application. Trying to change the secondary (voltage or current) with a series resistor/inductor pair primary probably isn't going to work how you think. If you measured the resistance of the primary, the inductance (AC resistance) of the transformer doesn't equal it's DC resistance. In all likelihood, your transformer can't work with any efficiency like that -- it's being prevented from doing it's function.

To muddy things more, devices in series (like an inductor/resistor pair) always draw the same amount of current.

Again -- can't tell from your description if the resistor is in series with the primary...

Remember, whatever current your rig is drawing depends on the load, not the supply. You don't (can't) control the current draw with the power supply (well, under some circumstances, but Ohms Law can't be violated and that generally means the voltage is dropped to compensate) -- a power supply amperage rating is it's capacity only. Voltage you can set "supply side" but not current.

Seems some assumptions are being made here about how to reach your goal (which itself is still foggy). Information still incomplete here...

I just want to say thanks for all the help so far, it has been very helpful to me. My issue is not getting the correct current in the secondary coil of the transformer, my issue is that the primary coil (the one directly attached to the wall), draws more current then my 5A fuse can handle. This I believe is because the resistance between the two leads on the primary coil have a resistance of less than 1 ohm, and thus should draw more than 120 amps. I don't understand how the load on the secondary coil, can effect the amount of current going through the primary coil, because they are two separate coils of wire, that are not attached to each other in any way. It is because the resistance is so low, that I said I would put a resistor in series with the primary coil, to increase the resistance, and thus decrease the current. The main concept I don't understand is if my load has a resistance of less than 1 ohm, and my voltage is 120v, how is it possible for it to draw any current less than 120v? doesn't that break ohms law, because isn't the current determined by the voltage divided by the resistance no matter what?

There's your first flawed assumption: that the DC resistance of the primary is the working impedance. It's not.

Transformers are essentially "electromagnetic machines" with nearly ideal characteristics. Their primary impedance changes with the load (on the secondary), drawing more current from the source only when a larger load is connected.

Impedance is "AC resistance" and for inductors is dependent on the rising and falling magnetic flux in the core (and other stuff like frequency). The impedance (working resistance) with AC is much higher than you think, in fact that's not present at all with a DC test. Brings home that fact that transformers only work with AC...

So... The impedance of the primary drops when a load is applied (to the secondary); the greater the load, the more the impedance drops. Devices in series always draw equal current; as your transformer tries to work, your series resistor steals an equal amount of current -- the harder the transformer works, the more the resistor takes. AND it's preventing the transformer from performing it's task efficiently. This may be limiting the current on the secondary, but by multiplying current on the primary.

Assumption two: that a way exists to limit current to a motor that's performing a large amount of work (on a household circuit, no less).

Yes, you can limit the current. But it cannot do the work without the current. Work is work. It's not free.

Volts and amps, amps and volts. Put them together and you get power. It's unusual for a large motor to be low voltage/high amperage, even though it can perform work equal to a high voltage/low amperage motor. Why? Because high-power transformers are as expensive as the motor itself. Easier to work with existing voltage supplies.

Was there a custom PS for this motor, like a dedicated engine/generator? If so, replacing that with a transformer might be more expensive than replacing the motor....which is why you made your own transformer....

I (we'd) like to see your DIY transformer! Sounds cool, even it it's not equal to the task... Good luck, BTW.

Thanks, I did not realize the difference between impedance and resistance. Is their a way to calculate the impedance of a circuit? My apologize for not saying what I was connecting it to I am connecting it to a Bridge rectifier, and the output of the transformer is actually like 5v, so that the bridge rectifier brings it down to around 2.5 volts dc. Even if I connect the transformer with nothing on the secondary coil (aside from my voltmeter), it blows the fuse. I think this is similar to putting a paperclip in the wall outlet, accept the wire can handle more than 5A. Shouldn't it draw very little current on the primary coil if the secondary coil has nothing connected to it? and isn't connecting the primary coil to the wall outlet, the same as just connecting the two wall pins with a long wire (as in it draws the same amount of current)? So the load on the secondary coil determines the current through the primary coil? and if so why if their is no load on the secondary coil does the fuse blow? Thank you allot everyone for your help its really appreciated.

Yeah, the impedance can be calculated theoretically or measured directly. But as discussed above, the load on the secondary creates a "reflected impedance" on the primary. So the inherent impedance of a transformer doesn't matter much compared to it's impedance when loaded.

You are correct, a properly designed transformer shouldn't draw much current without a load. Think of the magnetic flux field that builds (and collapses 'cause it's AC) as a force that resists current flow. Convert some of that field (induce current) back into electricity on the secondary, and the impedance on the primary drops (begins to drift toward the DC resistance value).

I think you have a problem somewhere in the transformer. Maybe the secondary has a short. A well-designed power has closely coupled mutual inductance between the primary and secondary. Maybe there's a design issue.

In any case, your 80 ohm resistor shouldn't be necessary. Without a load drawing current through the primary, not much should flow. If it does, you'll have to track down the problem...

Thanks, but should connecting the primary coil with nothing on it be the same as basically putting a paperclip in the wall, and directly connecting the hot wire with the other lead? and doesn't that draw more than the circuit breakers in your house handle?

You weren't wrong about an 80 ohm resistor drawing about 1.5A with 120V... But the 5A breaker tripped anyway, which shouldn't have happened, if our assumptions are correct. Something isn't quite right (already). And I don't think that resistor is helping.

I'd say remove the resistor, and anything connected to the secondary and try it. The breaker should prevent any serious issues, and it's backed up with the breaker in the circuit. The circuit breakers in your house are there to prevent the house from burning down (if someone sticks a fork in the outlet).

If it trips, there's a problem somewhere within your transformer. Additionally, maybe test the breaker in the strip with the 80 ohm resistor only. Do that first, if you don't trust it.

And seriously be careful around this stuff. We haven't stressed safety here. 120V mains are dangerous!

Im sorry, did not test it with a 80 ohms resistor, because I do not have one that can handle 180 watts, they are large and expensive. However, without any sort of resistor, or load on the secondary coil, the fuse blows anyway, so does that mean that their is something wrong with the transformer? (if so what), and isnt connecting just a primary coil (basically just a long piece of wire) similar to putting a fork into the wall (in how much current it draws)?

Well, if your strip breaker tripped the resistor was drawing major current anyways...

No, it should not be like a fork. If transformers acted like a dead short, why would we use them? They were once present in 80% of home appliances (now are being replaced with switching supplies). The primary of the transformer has a larger impedance with AC current than the measured DC resistance, as we said previously. Especially unloaded.

People leave old-school wallwarts plugged in all the time, and they don't draw much current when unconnected. Ten years ago almost all of those were simple transformer-driven power supplies. They feel warm when in use, and cold when not in use (but still plugged in outlet). That heat change is evidence of the difference in current flow, loaded vs unloaded.

Note that correct transformer operation depends on correct design decisions, and includes factors such as the number of turns, the wire gauge, the core material, the reflected secondary load, etc.

If there are too few turns in the primary, or the wire is too heavy for the AC supply, that's an issue. An other possibility is nicked or missing insulation on the coils (internal short).

so the reason it is not basically a short is because of the impedance? If so is their a way for me to calculate the size and number if wires in order to get the desired impedance

The input impedance of an open circuit transformer, if it was perfect would be infinite. It isn't, quite, because of inefficiencies in some of the internals. The PRIMARY turns are not set randomly, they are calculated from considering the saturation flux density in the core. What you are doing is over stressing the core, so it saturates, and the resulting transformer has a very LOW input impedance, so it blows the fuse.

So I need to have a larger core in order to increase the impedance? In that case, does changing the thickness of the coils, as well as quantity of the wrappings affect the impediance?

First thing: specify the input volts, output volts and output current.

If you have a convenient transformer you want to use, measure the total cross sectional area of the core, if you know what power the original transformer handled, that's a clue as to what the iron will actually handle. It won't handle MORE than its original rating. If it came from a microwave, it'll be a similar rating to the original rating of the microwave. A 700W microwave is likely to have a 900 VA rated transformer.

I am trying to build my own transformer, and make it as small as possible. I just tried making the iron core much larger but it did not work. My set up is as follows now. My primary coil is 40 wrappings of 18 gauge wire around a large piece of steel. Then, on the other side I have about 5 coils of 18 gauge wire as my secondary coil (I know the ratios are not right, im just trying to get something to work). The secondary coil is connected to my voltmeter and my primary coil is connected to a power-strip with a 15 amp fuse (not 5A as I previously thought). Each time I turn the power strip on, it blows the fuse, which means that my transformer is drawing more than 15A. That is an issue. What, if anything, is worong with my set up. I believe that the piece of iron core i am using is large enough, because it is bigger than the one for a microwave transformer. However, it is a piece of steel, not iron, im not sure if that makes a difference.

The magnetic circuit you have is useless for a transformer. Look up what a real transformer looks like - there's a reason for it, the flux paths have to form easy closed circles.

Your fuse is blowing because your magnetic circuit is virtually non-existent, so it looks to the supply like a resistive short circuit.

And steel is, for complex reasons, no use for a transformer.

Why is steel no use for a transformer, cause it would be allot easier for me to build that shape out of steel bars, then it would to find iron bars, and build it (since home depot doesn't have iron bars I don't think) ?

Steel has the property of permanent magnetisation. Alternating magnetic fields create large losses in the metal, and lump metal anyway wastes large losses in eddy currents - which is why transformers are always made from metal strips.

In answer to your other question: Yes, it IS possible to calculate the iron you need, but only when you answer my very first question: What power is the transformer to handle ????

My apologize, I forgot to include that, the power I need it to handle is an input of 120 volts @ 2 amps. I have decided not to use a transformer and rectifier as my power source, and instead will be using a buck converter. I have a tone of questions about it, so please check out my topic on it https://www.instructables.com/community/Switch-mode-power-supplys/

Also, is their a way to calculate the exact volume of the iron core I use, since i will be building one custom? If it is too large, is it going to draw more power? If it is too small not enough? Also, if you have two iron cylinders and wrap one primary coil on one, and a secondary coil on the other, and put them in close proximity to each other, would it work, or do the cores have to be connected?

Steveastrouk, you are an EE, yes?

merlinj, listen up...

Yes, and I specialised in electric machine theory at school, many, if you'll forgive the expression, moons ago.

Cool. Good to get E info from someone in the know. I am familiar with moons ago...

Yes, and that impedance stems from it's inductance. The magnetic field generated in a transformer (or any inductor) wants to oppose the current trying to flow through the coil, or really it's change. That's why inductors are an AC phenomenon, and impedance can't be measured simply with DC current like resistance.

Search for stuff like "transformer primary impedance" and "transformer turns ratio" and similar. Transformer design is EE (Electronic Engineer) level stuff, but you'll probably find some practical examples...

Also, are we sure the resistor hasn't failed...closed? (I.E., fused into something with near zero resistance?)

Maybe it would be easier if you try to source something like this locally.
Otherwise provide all the details we might need on your project, although I highly doubt going the transformer way for such currents is the right way.
If you still want to continue this way try a microwave tranformer.
Remove the secondary winding and put a few turns of really thick wire in there as your new secondary.
Start with just 4 turns and measure the output, if not enough voltage increase by one winding and check again.
A MOT should give you in excess of 100A if you use the right wire to handle this current.

I read Downunder35m's comment and just assumed your end application is a high-capacity motor -- sorry! I guess we still don't know what you're powering...

Don't get me wrong but using a transformer for this is like using a 100ton excavator to do the yearly cleanup in your garden beds.
Unless you need really insane power levels go with a switchmode inverter style power supply.
Even a modern computer PSU can deliver over 50A on the 5V rail and the circuits are quite simple.
If you study the circuits for the standard computer PSU you might even be able to use the info and guts of one to make your own for 2.5V.
For a simple test you can even use some high current diodes in series on the 5V rail to get to around 2.5V as each diode drops the voltage by around 0.7V.

I am only using a transformer because of the emmence amount of power I need on the other end. I need at least 70 amps, which I did not believe a switch power supply could provide. I also am asking because I think their is a flaw in my understanding of load and ohms law.

Have any pictures of your setup? May I ask what you're trying to power with the secondary?

If you used the right size of wire on the primary witch I doubt the transformer would have blown before the fuse. off the top of my head something like 500 turns of 28AWG wire.

What is the formula you were using?

is their any specific formula I should be using, because I made as many turns of 18 gauge wire as I could ( I used 18 because it is rated for the amperage i'm using). Then i set up my coil ratio accordingly, so I made the secondary coil a ratio relative to the primary. Should i use something like 28 gauge wire? if so doesn't that exceed its rated amperage, and thus the wire would blow?

I worked in rewiring motors, alternators, and transformers.

18 gauge way to big for 120 volts 1.5 amps, the largest it should be is 24 gauge.

It is all ratio of change

To make a long story short I use this shortcut for laminate steel core just because I can remember 10, 12, 16, or 10 amps, 120 volts & turns, 16 AWG wire 1200 watts.

You want same voltage and power down or 120 volts 1.5 amps 180 watts

voltage up, turns up, wire diameter down, current down, power same.

If you change the voltage to 220 you change the turns to 220 and the wire to 20 AWG and the power remains the same.

You want voltage same, turns up, wire diameter down, current down, power down.

Your ratio is 1200 watts divided by 180 watts is 6.66.

120 turns times 6.66 is 800 turns.

Current 1.5 amps 28 to 24 gauge wire.

That makes your primary 120 volts 1.5 amps 180 watts with 800 turns of 28 to 24 gauge wire.

For anything in the home remember this your current determines wire diameter.

12 AWG 20 amps

14 AWG 15 amps

16 AWG 10 amps

18 AWG 8 amps

20 AWG 5 amps

22 AWC 4 amps

24 AWG 2 amps

26 AWG 1 to 2.2 amps depending on application.

28 AWG .75 to 1.5 amps depending on application.

The transformer you built is a 8 amp dead short under 140 turns.