Beginners ohm's law question

[ikcdab]

I have been making solenoids to ring a single strike bell. The coil I have made has a resistance of 2ohms measured with my multimeter.
I have attached 12v wall wart power supply and everything works fine. I am pleased with the outcome. I close the switch and the bell receives a single strike.
My question, ohm's law tells me that 12v divided by 2ohms resistance means a current of 6 amps. However, the wallwart label says it supplies 400 milliamps.
So why doesn't the wallwart blow up when I connect it to my solenoid?

[rstofer]

This would be a perfectly good time to make some measurements.  How much current is actually flowing?  What is the voltage at the solenoid?  What is the voltage at the wall wart?  What happens if you hold the button down for some unreasonable period of time?  Careful, the test might smoke the solenoid and will very likely destroy the wall wart.

The wall wart likely has an output capacitor that will kick up the short circuit current and help overcome the initial effect of the inductance (which doesn't want to increase current flow above 0 amps).

A little more than half way down this page, there is a yellow box with the exact equation for current flow:

https://www.electronics-tutorials.ws/inductor/lr-circuits.html

After some long period of time, the current should be as simple as I = V / R but it won't be that high initially because the inductance is working to prevent a change in current.

[bob91343]

The wall wart is being overloaded and the output voltage is a lot less than 12V.  If you let it run, it will overheat and probably fail.  Better to wind more solenoid turns and make a more efficient system.  If you can get 12 Ohms wound with smaller wire, the wall wart will stay cool and the system will work better.

[james_s]

Or use a lower voltage supply, or drive the solenoids with a constant current regulator. Definitely do take measurements, do not rely solely on calculations.

[bdunham7]

So why doesn't the wallwart blow up when I connect it to my solenoid?

Because things don't always 'blow up' when you overload them.  Can you measure the voltage across the coil when it is powered?

[Zero999]

The coil will also have some inductance, which will limit the initial current.Obviously if power is continuously applied, the current will reach a level determined by the resistance of the coil and the power supply. If it's a modern, switched mode power supply.rather than the old mains transformer type, it might limit the current, by reducing the duty cycle of the switching transistor, which will reduce the output voltage.

My advice is to design a monostable multivibrator circuit, to limit the on time, to just enough to strike the bell. A 555 timer, or even a couple of transistors can do this. The coil will need a diode connected in reverse polarity, to protect the driver circuit.

[Badwolf]

As Zero999 said, the solenoid behaves the same as an inductor. It's not like a simple resistance. Like capacitors, it hates changes and so opposes. The resistance seen by the power supply is then very high. But like a capacitor it will gradually charge and in the process decrease the resistance seen by the power supply to reach 2 ohms. I advise you to find tutorials on the use of relays driven by a BJT, often they explain the phenomenon very well, in particular the use of a flyback diode which allows the relay coil to be unloaded (here your solenoid), because once the coil is charged, it will oppose cutting the power and may cause sparks in your switch (or burn out transistors when they are used)

An other good tutorial : https://en.wikipedia.org/wiki/Flyback_diode

[DavidAlfa]

Because the power supply has overcurrent protection. 2ohms is very low, you need to use a thinner wire or make more turns.

[MarkKn]

For completeness I had to ask if its an ac supply, but I believe others have analyzed the situation accurately.

[CatalinaWOW]

The inductance issues mentioned do affect initial pull in, but after a fraction of a second the solenoid will be in steady state current.  The actual current depends on the exact configuration of your wall wart. 

The simplest and oldest have a simple rectifier and significant internal resistance.  The spec means that when 700 mA are drawn approximately 12 volts will be at the output of the supply.  At open circuit these will typically supply somewhere between fifteen and twenty two volts.  If you measure this open circuit voltage you can infer the internal resistance as the open circuit voltage minus twelve divided by 0.7 amp.  The typical voltages I mentioned indicate a range of this resistance between about 3 ohms to a bit more than 13 ohms.   

The steady state current in your solenoid would then the the open circuit voltage divided by the sum of the solenoid resistance and the internal resistance. 

These are all approximations.  There are other things going on, and the specs on wall warts are notorious for their inaccuracy.

Newer switching wall warts can have better internal regulation, over current protection and a number of other "features".  It would be instructive to measure the voltage with various resistances in the 1 to 20 ohm range.  It is conceptually easy to do this, but be careful.  These tests will dissipate  ots of power in the load resistors.  Use the power equation, power equals volts squared divided by resistance to find the predicted power and make sure the resistors are up to the task.  Also recognize that they can get hot enough to burn you.  There are better ways to generate the loads, but are out of scope for a true beginner, and do have some issues of their own.

The bottom line, your wall wart doesn't blow up because one way or the other they have been designed to not cause a fire when misused or accidentally shorted.  The fun part is figuring out how this was achieved.  There are many solutions, all with different costs and benefits.  Think about why one of the simplest answers, a fuse, is often not used in wall warts.

[perieanuo]

I have been making solenoids to ring a single strike bell. The coil I have made has a resistance of 2ohms measured with my multimeter.
I have attached 12v wall wart power supply and everything works fine. I am pleased with the outcome. I close the switch and the bell receives a single strike.
My question, ohm's law tells me that 12v divided by 2ohms resistance means a current of 6 amps. However, the wallwart label says it supplies 400 milliamps.
So why doesn't the wallwart blow up when I connect it to my solenoid?

like other pointed, that is not a purely rezistive circuit, ohm's law is not adequate for complex impedance (loads with capacitance and/or inductance), and that's your case, impedance.
second, don't ever asume something like your supply is 12V before measuring or make the mental exercise considering he can fall when the load increase over the supply capacity. all ps's are reacting to overload by shutting down completely or on periodical basis, or decreasing in value and eventually die.
one good exercise is to connect your multimeter and see what's happening when you force the button pressed for some seconds.

[ikcdab]

Ok thank you for your replies.
The device I have built is for a single strike bell. I wound copper wire around an iron core and connected a momentary switch. When energised it attracts an iron rod that strikes the bell. It gives a satisfying "ding". The coil is only energised momentarily, I suggest less than quarter of a second.
I have measured the wall wart and although it says 12v, it is sending out 16v. (As a side comment, I did connect my bell to a true 12v supply and had the same loud ding, so it is not too voltage sensitive)
I then connected the original ps directly to an ammeter and it read a steady 2amps. Although the ps is labelled 400ma.
I also connected the ammeter in circuit with the bell and original ps.
When I close the switch for 1 second, the ammeter climbs rapidly to 1.72amps and stays there. When I open the switch, the ammeter gradually falls back to zero over 30secs or so.
Repeating the exercise gave various other values depending how long I waited between "dings". After giving 5 dings in rapid succession, the ammeter momentarily showed 11amps. But then fell quickly to zero when I stopped.
I have tested the bell a lot over the last few days (to my wife's annoyance) and it is working well and does what I want. The power supply is warm, but not hot.
The bells are for a model railway layout and when installed will be wired to a 2amp 12v supply.
So I am no overly worried, but I am very curious about what is actually happening.
Ian C


Edited to say that I have just tried it on 5v and the ding is still just as good

[iMo]

fyi - long time back there were so called "door bell transformers" sold (in a small black or brown bakelite box). The trafos were made with a special "soft core" such a permanent short at its secondary winding does not cause a smoke or fire in your household. My very first power supply I built as a kid was with that trafo.. Current limit built into the metal core..

[james_s]

With a basic multimeter you can take some measurements and quickly see exactly what is happening. Otherwise we can only guess, but I am fairly confident the voltage out of your power supply is sagging under the load.

[ikcdab]

With a basic multimeter you can take some measurements and quickly see exactly what is happening. Otherwise we can only guess, but I am fairly confident the voltage out of your power supply is sagging under the load.

Thanks, did you see my comments above about the measurements I have taken? The ps is outputting 16v rather than the advertised 12v.

[bdunham7]

Thanks, did you see my comments above about the measurements I have taken? The ps is outputting 16v rather than the advertised 12v.

That's very common to see a higher voltage with no load, closer to the rated voltage at the full rated load, then lower if you are over the rated load.  What you want is to measure the output voltage under load so you can see how much the voltage sags with the coil connected.

[ikcdab]

Thanks, did you see my comments above about the measurements I have taken? The ps is outputting 16v rather than the advertised 12v.

That's very common to see a higher voltage with no load, closer to the rated voltage at the full rated load, then lower if you are over the rated load.  What you want is to measure the output voltage under load so you can see how much the voltage sags with the coil connected.

Ok, so i have tested this. 16.8v with no load. Drops to 4.1v when i close the switch.
Ian C

[bdunham7]

OK, so (very approximately) you have a 2 ohm coil, the power supply has an open circuit voltage of 16 volts and an apparent internal resistance of 6 ohms.  When you close the switch, about two amperes of current flows, which combined with your 2 ohm coil resistance and Ohms law, gives you the 4 volts you measure.  The other 12 volts is dropped across the 6 ohm internal resistance of the power supply itself.  Note that I haven't measured the internal resistance of the power supply, I've deduced it from the voltage change under load and the load current.  You can try and measure as closely as you can and see if you can put a few more decimals of accuracy in, but that's the general idea.

[james_s]

It may be more accurate to describe it as source impedance rather than internal resistance. The old iron transformer wall warts were usually impedance protected, I suspect a significant aspect of that was rather loose magnetic coupling between the primary and secondary in addition to the DC resistance of the windings.

[TimFox]

With a basic multimeter you can take some measurements and quickly see exactly what is happening. Otherwise we can only guess, but I am fairly confident the voltage out of your power supply is sagging under the load.

Thanks, did you see my comments above about the measurements I have taken? The ps is outputting 16v rather than the advertised 12v.

It is very common for a transformer or unregulated power supply to be rated for the output voltage at a specific current.  If so, it is likely to have a higher output voltage when unloaded (this is especially true with low-power transformers), the advertised voltage (approximately) at the specified current, and a lower voltage at higher current.