# What happens if a motor wants to pull more that the source can supply?

I have a solenoid that's rates at 12vdc and 2.5amps.

I have a small lead acid battery to power it, I took it out of a computer ups. But eventually I would like to power it from a wall socket. I found a walwart that's 12vdc and 1A.

So I'm wondering what would happen to the power adapter if the solenoid demands it's rated 2.5A?

You could burn out the wallwart.
It could get hot and cause a fire, then burn your house down, then your wife divorces you, your children hate you and you end up in a home all alone.

.

I agree with @LarryD

Use a properly sized power supply.

…R

The load will over-load the supply resulting in a drop in output voltage. This means something within the supply will convert the lost volts into heat. If the supply has an output fuse or a thermal overload trip (unlikely in a cheap one) this may rupture and effectively disconnect the load. However, on cheap wallwarts the likely outcome on a 150% overload is failure of the step-down transformer. If you are lucky, it'll be a simple failure, if you're unlucky it may ignite.

If you simply require a short-duration pulse to drive the solenoid you could install a large capacitor on the wallwart output to act as a reservoir. This will charge during the idle phase and supply the pulse of current required when you trigger the solenoid. This method is used by model railway enthusiasts to drive their track switching points. To control the charge rate of the capacitor you should connect it to the wallwart via a load limiting resistor, say around 10 ohms

Switching wallwarts are "smarter", they will just turn off the output until the short circuit condition goes away.
http://www.dipmicro.com/store/DCA-1210
Get the right size - motor at startup looks like a stalled motor and needs stall current to get it going

Ok thanks...Ill go for the wall adapter thats 12VDC-2.5A.

Now out of curiosity, when using a cap, how would I calculate that? The walwart is from a link sys router I guess. I cracked it open and it has a 25V/2200uF cap in it. I tested it and it still works because after being warned that opening up a UPS a while back that I could have gotten shocked by the caps, after testing this walwart which is supposed to be 12V, but actually read 17.5V when I plugged it in, after unplugging it from the wall, since the cables are peeled, i was worried about letting them hang and possible touch each other, so I left the meter on and sure enough they still had a 17.5V potential. I touched them to the solenoid and quickly discharged the cap.

So how would I know to calculate the cap to have it charge up and be able to supply the 2.5A jolt the solenoid might pull from it?

With a solenoid you have a pick voltage and a holding voltage.
i.e. it might pull in at 12 volts, then still be pulled in until the voltage goes below 9 volts.

A solenoid has resistance and there are formulas that are used to calculate time constants with inductors.
However, it is probably much easier to experimentally add capacitors in parallel until you get close to your desired results, just forget about fancy calculations.

Too much capacitance on the output of your reclaimed wallwart may damage it as caps are a short circuit when the wallwart first turns on.

Just buy the proper sized power supply.

The equation for a capacitor is I dt = C dV, or put another way dV/dt = I/C (rate of change of voltage is current/capacitance).

So 2.5A for 100ms, for a voltage drop of 3V means C = 2.5*0.1/3 = 0.08F = 80mF = 80,000uF,
a huge value for only 0.1s of charge...

You can think of it another way, the power the load takes is 122.5 = 30W, a capacitor stores energy
equal to 0.5 C V^2, so the 0.08F cap at 12V holds 0.5
0.081212 = 6 joules, which would drain to 0V
in 0.2s at 30W.

Marciokoko:
Ok thanks...Ill go for the wall adapter thats 12VDC-2.5A.

I would be much more comfortable with a 4amp or 5amp power supply so it is not running at its max.

...R

jackrae:
The load will over-load the supply resulting in a drop in output voltage.

Yeah... if the voltage drop is due to the DC power (voltage) supply's own internal resistance, then the drop in the supply output voltage will be proportional to the supply current. Higher the current.... the lower the output voltage for cases where there's no automatic method to compensate.

The sometimes confusing thing is the word 'load'. If they say '10 Ohm resistive load'.... then it usually means a power absorbing component with a 10 Ohm resistance.

And then they say things like '5 Watt load'... which refers to a power absorbing component that absorbs 5 Watt of power.

Generalising with resistive loads and DC voltage supply ....the higher the "load" (if referring to power or current draw) .... the lower the load resistance. Confusing.

Yes, for a 2.5A load you need at least a 3A rated supply, a 2.5A supply is total gamble!