Hi Everyone,

I have a circuit board that is quite simple looking. It's function is to take a 240v AC input and convert it to 24v DC to switch a relay.

I've inspected the board and traced everything and compiled the attached schematic.

The problem I have is that the relay stopped working and without a second thought I replaced it to find that the new one doesn't work either. Both have now been tested with a 24v DC supply and they switch.

When the board is powered the measured voltage at the relay coil is 12v DC for some reason. The resistors check out correctly and I think the diodes are also ok as I'm getting DC voltage, which leads me to believe that the 0.68uf capacitor is at fault. Unfortunately I have no way of checking the capacitor with my meter. Is it possible that the capacitor can go faulty in some way and cause this problem?

Thanks.

- Neil



Mods Note:
The schematic was copied from another duplicated thread which had been deleted and converted to a .png file.

 

Sorry for the double post everyone, I don't know how this happened.

 

Schematic?

 

Copied from the duplicate thread:

https://forum.allaboutcircuits.com/..._ys-module_sheet-1_20180512183939-pdf.152393/

It is indeed a voltage divider that uses a capacitor as one of the impedances. The problem with this is that the output voltage is not fixed, it is more like a current source. If the new relay coil needs twice the current, the voltage will be 12V instead of 24V. So you need to get a relay that needs the same current as the original, or you need to double the value of the capacitor to bring the voltage up to 24V at the new current.

Bob

 

Hi Bob,

Thanks for the reply, the new relay is an exact replacement for the old one, this is why I'm puzzled.

- Neil

 

If one diode is OK and the other is open you will get a DC voltage, but it will not be high enough. You need to do a forward and reverse check on each diode.Probably you will discover that one of them is open. If you publish the circuit we can provide a more help. Or at least a description of the circuit, like I do many times.

 

You should really get a meter that can measure capacitance. Even my cheap (relatively cheap) ragu multimeter can measure capacitance, diode, and all that important stuff. https://www.amazon.com/17B-Multimet...id=1526172985&sr=8-1&keywords=ragu+multimeter

A schematic would really help here.

 

SCHEMATIC?

 

SCHEMATIC?

Bob shared it in post 4.

 

 

nothing to see here.
lock it up.

 

View attachment 152419

Funny, it was there earlier when I clicked through. Maybe the duplicate thread got locked and/or deleted and is no longer accessible. I've copied it into LTSpice so I could play with values a little, so here it is - the only change I made in the first one is making a guess as to the relay coil resistance value in order to get the output to ~24V:


If my simulation is doing a reasonable job, then degradation of cap C1 might explain the lower output voltage. In my simulation, dropping C1 from the original 0.68uF to 0.34uF reduces output voltage by roughly half, matching the drop reported by the thread starter:



It's also worth noting that this circuit involves dangerous voltages and that if the circuit were powered up without the relay coil connected, very high voltages could be developed very quickly:

 

P.S. Here it is in its original form:

 

1. Given that the fundamental problem is that the input voltage is way to high for the load, why use a voltage doubler circuit?

2. If you delete R3 and R4, and move the GND point to AC-L2, you can see the output characteristics from the point of view of the output.

3. What is the output voltage under no load?

4. Isn't this a forbidden topic?

ak

 

Sorry for the double post everyone, I don't know how this happened.

If you didn't do it then it could be the bounce that it came from your left key of mouse.

 

The contents of the thread violated with the User Agreement, so this thread will be locked.

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