I'm stumped......

[MichaelGP3]

I ran into something today I can't explain. Transformer (120V to 24V) in its own integral housing installed on the top of a NEMA enclosure, wired to this rectifier,

http://www.partsexpress.com/images/050-060m.jpg

installed inside the NEMA enclosure. The 24VDC+ output of this rectifier is wired to the common (aux. contact) terminal in an analog addressable smoke detector base. The N. O. terminal feeds this DC+ voltage (during an alarm condition) to the coil of a 24VDC 2 pole form 'C' interposing relay. The 24V DC- wire goes directly from the rectifier to the coil of the interposing relay, which is used to light a 24VDC incandescent lamp. Wiring is THHN 14 gauge stranded, in 3/4" EMT.

When the wires are removed from the DC output side of the rectifier, they read as free of foreign or induced voltages, shorts, and grounds.

Using a Fluke model 83 VOM, with a clear fire alarm (no off normal conditions) I'm reading the following:

Output side of transformer/input side of rectifier: 25.5 VAC
Output side of rectifier, with field wires not connected: 23 VDC
Output side of rectifier, with field wires connected: 33 VDC

These readings are stable.......any ideas? With no load on the wires, I would have predicted no change in the rectifier's output voltage.

 

[HighWirey]

You have a lot of words in your post, however, what is the problem?

Best Wishes Everyone

 

[76nemo]

I ran into something today I can't explain. Transformer (120V to 24V) in its own integral housing installed on the top of a NEMA enclosure, wired to this rectifier,

http://www.partsexpress.com/images/050-060m.jpg

installed inside the NEMA enclosure. The 24VDC+ output of this rectifier is wired to the common (aux. contact) terminal in an analog addressable smoke detector base. The N. O. terminal feeds this DC+ voltage (during an alarm condition) to the coil of a 24VDC 2 pole form 'C' interposing relay. The 24V DC- wire goes directly from the rectifier to the coil of the interposing relay, which is used to light a 24VDC incandescent lamp. Wiring is THHN 14 gauge stranded, in 3/4" EMT.

When the wires are removed from the DC output side of the rectifier, they read as free of foreign or induced voltages, shorts, and grounds.

Using a Fluke model 83 VOM, with a clear fire alarm (no off normal conditions) I'm reading the following:

Output side of transformer/input side of rectifier: 25.5 VAC
Output side of rectifier, with field wires not connected: 23 VDC
Output side of rectifier, with field wires connected: 33 VDC

These readings are stable.......any ideas? With no load on the wires, I would have predicted no change in the rectifier's output voltage.

What, what, what????? 33 with wiring connected??? What, what, what????

 

[gar]

080614-0856 EST

MichaelGP3:

Assuming the meter is in the DC position for the DC measurements, then the rectifier alone reading implies a bridge or full wave rectifier. When the load is applied the reading implies a capacitor in the load.

The full wave rectified average voltage is 0.636/0.707 of the RMS AC voltage. In your case 25.5 *0.9 = 22.9. but under some resistive load it would be less by two diode drops.

With a capacitor input filter you get a DC voltage equal to the AC peak voltage. In your case 25.5/0.707 = 36 V. Diode drop will probably take about 2 V off of this, and the magnitude of ripple may further reduce the average value. Diode drop is a function of load current and other factors. Thus, the 2 V is a ballpark value for 2 diodes.

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