DC measurement

[stgeorge]

At my work place I have an electric door issue. it is activated when you swipe your ID badge through the card reader and the door unlatches. It was diagnosed as the olenoid that was faulty so a new solenoid was installed but is failing to operate correctly. The system is labelled as 24VDC. I tested the DC circuit and am reading -25.6 VDC. What does the minus sign mean? Does this explain why the door will not work.
Thank you all.

 

[K2500]

The system is labelled as 24VDC. I tested the DC circuit and am reading -25.6 VDC. What does the minus sign mean? Does this explain why the door will not work.
Thank you all.

It may mean that you applied your meter leads backwards. Or that the system polarity has been reversed.

 

[LarryFine]

The system is labelled as 24VDC. I tested the DC circuit and am reading -25.6 VDC. What does the minus sign mean? Does this explain why the door will not work.

The minus sign means that if you swapped the meter leads, it would read without the minus. It's not relevant for non-polarity-sensitive devices, such as a solenoid.

I suggest making sure the solenoid coil isn't open. You can disconnect one lead and measure for voltage across the open point, and maybe even check coil current.

You can also see if it is magnetized. Try checking for plunger pull. You might have to disconnect the linkage to the lock. Your issue may be mechanical, not electrical.

 

[physis]

I agree with everything Larry has said.

I don't think there's an issue with polarity here, not to say that the DC current specs for a solinoid will give you the same machanical strength if driven by AC, but I don't think that's your problem.

I'd be interested in three things.

What is the current available at the solinoid coil, and I mean at the solinoid's terminals?

Does the thing operate freely when the door is open?

And, mostly, is the door binding the thing when the door is closed?

 

[stgeorge]

Thank you all. Your responses have helped.
The solenoid is magnetic. When activated it pulls the piston back unlocking the latch. In place before the solenoid there is what the manufacturer calls "a black box" a small 1" X 1" electrical device which the current runs through before the solenoid. When this is in place nothing happens but when bypassed the piston is being magnetised but not enough to open the latch. You can hear the solenoid being energised but it has not the strength to work fully. I am replacing this device in the hope that the lock will work again.

 

[glene77is]

StGeorge,
Please let us know what happens! :smile:

 

[gar]

090320-1333 EST

stgeorge:

You need to define the "black box" a little more. How many leads? Are there any + or - minus symbols?

A solenoid designed for AC operation will have a shading coil somewhere in the magnetic circuit. This is a slug of copper forming a shorted 1 turn coil around part of the magnetic core. If this does not exist, then it is not an AC solenoid. An AC solenoid or relay will have considerably less DC resistance in its coil compared to a comparable DC solenoid.

You may have a DC solenoid and the black box may be a rectifier. If it is a bridge rectifier it has 4 leads. Two for the AC input, one for + output, and one for - output.

What is the source for power to operate the solenoid? Is it just a transfomer followed by the black box and the solenoid? Where in this circuit is the switch that activates the solenoid? Physically is the switch in the card reader, or where? Where did you measure the 26 V? At the solenoid terminals or at the input to the balck box?

Draw your total circuit on paper in a schematic form so you can see the logic of the circuit and what to expect at various measurement points.

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