Go Down

Topic: magnetic door stop (Read 2889 times) previous topic - next topic

dc42

The adapter will have a large smoothing capacitor across the output. If it's not connected to anything and it doesn't also have a discharge resistor, it could take quite a while for the charge in that capacitor to leak away.
Formal verification of safety-critical software, software development, and electronic design and prototyping. See http://www.eschertech.com. Please do not ask for unpaid help via PM, use the forum.

clankill3r

can i speed it up by holding it against a radiator or something?

dc42

You can speed it up by connecting a resistor across the output. 1K should be about right.
Formal verification of safety-critical software, software development, and electronic design and prototyping. See http://www.eschertech.com. Please do not ask for unpaid help via PM, use the forum.

clankill3r

Ok got it ready.
When i check the voltage of the adapter it gives 37.9 (although it should be 30V according to the information on it). When i put a 100ohms resistor in between then my multimeter still gives 37.9 instand of something around 24V, why's that?
Also if i do a check on the mA then at the 10A setting on my multimeter it gives nothing, it should give o,4A. If i check at the 200mA then it breaks the fuse (i tried cause the 10A gave me nothing...), so what's up with that?

dc42


Ok got it ready.
When i check the voltage of the adapter it gives 37.9 (although it should be 30V according to the information on it).


That's not unusual for the open-circuit voltage, the 30v will be quoted at full load (1.5A I think you said).


When i put a 100ohms resistor in between then my multimeter still gives 37.9 instand of something around 24V, why's that?


I presume you mean that you connected your door lock to the supply with a 100 ohm series resistor. You need to measure the voltage across the door lock in that condition, not the output from the supply.


Also if i do a check on the mA then at the 10A setting on my multimeter it gives nothing, it should give o,4A. If i check at the 200mA then it breaks the fuse (i tried cause the 10A gave me nothing...), so what's up with that?


How exactly were you connecting the meter when you had it on the 200mA range and on the 10A range?
Formal verification of safety-critical software, software development, and electronic design and prototyping. See http://www.eschertech.com. Please do not ask for unpaid help via PM, use the forum.

clankill3r



Ok got it ready.
When i check the voltage of the adapter it gives 37.9 (although it should be 30V according to the information on it).


That's not unusual for the open-circuit voltage, the 30v will be quoted at full load (1.5A I think you said).


The doorstop is 1.5watt, the adapter is 12watt (so is a 1/2 watt resistor fine as mentioned?), it gives 400mA (according to the information on the adapter). And why is it no unusual? I mean if 30v get's quoted at full load then 37 volt should be overload according to my logic (which probably sucks).



When i put a 100ohms resistor in between then my multimeter still gives 37.9 instand of something around 24V, why's that?


I presume you mean that you connected your door lock to the supply with a 100 ohm series resistor. You need to measure the voltage across the door lock in that condition, not the output from the supply.


I first wanted to make sure i have the correct voltage and mA before i break the door lock. But if i hear you that's not possible?



Also if i do a check on the mA then at the 10A setting on my multimeter it gives nothing, it should give o,4A. If i check at the 200mA then it breaks the fuse (i tried cause the 10A gave me nothing...), so what's up with that?


How exactly were you connecting the meter when you had it on the 200mA range and on the 10A range?


I had it for the 10A like on the attachment picture, for the 200mA the black probe was moved from the 10A input to the COM input and the rotate button was set to 200mA of course.

dc42

OK, so you had the meter connected in series with a 100 ohm resistor, and that combination connected across the supply.

With the meter set to read voltage, it has a very high input resistance, so the load on the power supply is tiny and the voltage does not drop significantly from the 37.9 volts. If you replace the meter with the door stop, it will make a huge difference.

With the meter set to read current, it has a very low resistance, so effectively you put 100 ohms across the supply. The voltage will be somewhere between 30 and 37.9 volts, so the current will be between 300mA and 379mA. That is why the fuse blew on the 200mA range. On the 10A range, you should be able to see some current, unless you have previously blown the fuse for the 10A range.

Connecting your door stop to a little more than 24v for just a few minutes isn't going to harm it (connecting it to more than 24v for an extended period might cause it to overheat). It's not like an Arduino that has sensitive electronic components, it's just a solenoid. I suggest you connect the door stop in series with the 100 ohm resistor, connect the series combination to the power supply, and measure the voltage across the door stop, then the voltage from the power supply. You'll find around 24v or a little more across the door stop, and the power supply voltage will have dropped from 37 towards 30.
Formal verification of safety-critical software, software development, and electronic design and prototyping. See http://www.eschertech.com. Please do not ask for unpaid help via PM, use the forum.

clankill3r

Is a diode test on the probe's itself a way to test if the 10A fuse is broken?

Also i believe that it doesn't matter if you place the resistor in before or after the door lock but correct me if i'm wrong.
Anyway if i try it like on the image then it should be ok right? (door locks are quite expensive...)

dc42


Is a diode test on the probe's itself a way to test if the 10A fuse is broken?


No. However, if you remove the 10A fuse, the resistance range of the multimeter may still work so you can use it to test the fuse.


Also i believe that it doesn't matter if you place the resistor in before or after the door lock but correct me if i'm wrong.


Correct.


Anyway if i try it like on the image then it should be ok right? (door locks are quite expensive...)


That's fine for measuring the current, with the meter set to the 200mA range (assuming you have replaced the fuse). But to measure the voltage across the door lock, you need to connect the meter in parallel with the door lock, not in series with it.
Formal verification of safety-critical software, software development, and electronic design and prototyping. See http://www.eschertech.com. Please do not ask for unpaid help via PM, use the forum.

clankill3r

thanks for all your answers, hope you don't get sick of me.

Quote
That's fine for measuring the current, with the meter set to the 200mA range (assuming you have replaced the fuse).

Are you sure i don't need the 10A range?

Quote
But to measure the voltage across the door lock, you need to connect the meter in parallel with the door lock, not in series with it.

Why is that?

dc42


Are you sure i don't need the 10A range?


You measured the resistance of the door stop at 377 ohms, plus 100 ohms in series makes 477. Even if the supply still produces 37.9 with that load, the current will be 37.9/477 = 79mA, well within the 200mA range of the multimeter.


Quote
But to measure the voltage across the door lock, you need to connect the meter in parallel with the door lock, not in series with it.

Why is that?


That's how you use a meter:

- to measure current you connect the meter in series with whatever is drawing the current (the meter has a low resistance in current mode so it doesn't affect the current much)
- to measure voltage you connect the meter in parallel with whatever you want to measure the voltage across. In voltage mode the meter has high resistance so that it doesn't draw appreciable current and therefore doesn't significantly affect the voltage you are trying to measure,

Think of current like the rate at which water is flowing through a narrow pipe, and voltage as the pressure difference between the two ends of the pipe that is causing the water to flow. To measure the flow, you put a flow meter in series with the pipe. To measure the pressure, you connect a differential pressure gauge between the two ends, in parallel with the pipe.
Formal verification of safety-critical software, software development, and electronic design and prototyping. See http://www.eschertech.com. Please do not ask for unpaid help via PM, use the forum.

kf2qd

Those adapters often have a capacitor across the output pins to filter some of the ripple in the DC that is left after rectifying the AC. A capacitor with hold a charge for a bit of time after the unit is unplugged from the wall if it has nothing attached to the load side.

Now you have some more things to study - capacitors and rectifiers. Keep this up and you will have the stuff I paid good money to learn in my first semester when I was in college.

Don't feel too bad about being snowed under by all the terminology, you can look on wikipedia and other places on the web and you should be able to learn what all this means. Not knowing your background, but it shouldn't take too long if you want to learn it.

And don't feel too bad, some of the old timers manage to mess it up and have to go back to class to remember some of this fundamental stuff that makes up the basics for all these electrical toys we play with.


Rokkit

In the USA the "door stop" is referred to as a "Magnetic Door Holder".   Like you say, they're used to hold a fire door open until the fire alarm activates, which removes power to the door holder.  Most common are 24VAC, though there are some that work on 120VAC, 24VAC/DC, 12VDC, and I know of at least one of them that work on 24AC or 120VAC.

Generally you'd use a 24VAC transformer to power these things.  A 24VAC transformer will sometimes read more than 35VAC, and these door holders can take that kind of voltage without any problem.

Lastly, most meters that have a 10 amp scale have the ability to test the fuse by simply putting the meter on the ohms scale.  Leaving the meter leads where they would normally be when checking ohms, stick the positive lead (the probe) into the Amps connector (the hole where you'd plug in the lead when measuring Amps).  The meter should read pretty close to a dead short.
"Every day is a good day, some are just gooder than others."   D. Fowler

clankill3r

#28
Jan 27, 2012, 10:44 am Last Edit: Jan 27, 2012, 10:54 am by clankill3r Reason: 1
Ok, the door lock is working!

++++++(A)++++[ resistor]++++(B)++++[door lock] --------(C)--------------------

But, i tested the voltage (in paralel), it draws around 35.4V if i measure between A and C.
Between B and C i get 27.8V with a 100ohms resistor and 26.8V with a 120ohms resistor so i need a bit more to get the wanted 24V.

Also i tested the current (after the door lock, in series (C C)). It broke my 200mA fuse...why? !@^#($    :(

dc42


Ok, the door lock is working!

++++++(A)++++[ resistor]++++(B)++++[door lock] --------(C)--------------------

But, i tested the voltage (in paralel), it draws around 35.4V if i measure between A and C.
Between B and C i get 27.8V with a 100ohms resistor and 26.8V with a 120ohms resistor so i need a bit more to get the wanted 24V.


By my calculation, 180 ohms will give you very close to 24v across the door lock.


Also i tested the current (after the door lock, in series (C C)). It broke my 200mA fuse...why? !@^#($    :(


That is very strange. With the 100 ohm resistor, the current was (35.4 - 27.8)/100 = 0.076A.
Formal verification of safety-critical software, software development, and electronic design and prototyping. See http://www.eschertech.com. Please do not ask for unpaid help via PM, use the forum.

Go Up