AC Ice cube relay and chatter

PLCProf posted:

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Practically, this means that unfiltered 1/2 wave rectified 18V gives about 12.7 rms volts, plenty close enough to use with automotive 12V lamps and all that. It will also not overheat a 12VDC relay. and will pull it in just fine. It will probably buzz, however, in many cases the presence of the flyback diode is more than enough to solve that.

Tell me more.  By flyback diode do you mean D2 in diagram?  By "in many cases" what are the cases in which you can eliminate buzzing in an unfiltered (no capacitor) half-wave driven coil using a diode?

Hi Stan-

Yes, D2 is the flyback diode. When D1 stops conducting, the magnetic field in the relay starts to collapse, generating an emf counter to the applied voltage, so the bottom of the coil becomes positive with respect to the top in your drawing. This turns on D2, so current continues to flow in the same direction until the magnetic field is discharged. Or, in the vernacular, the "flywheel effect" of the inductance tries to maintain the current flow and D2 gives the current a path to follow. The effect is helped due to the fact that the drop-out voltage of a DC relay is generally much less than the pick-up due to the fact that the air gap is much smaller or non-existent when the armature is against the core. Sounds counter-intuitive, but try it! You might be surprised.

I have tried this with a variety of small relays, ice-cube size and a bit larger with 12 and 24 VDC coils. If you  are seriously interested contact me off line and I can get exact part numbers and particulars for you. Try it with a 12VAC supply, and a 12 VDC ice-cube relay. I will bet that it works.

While the diode may prevent the chatter due to the unfiltered waveform, it doesn't give you any protection for longer dropouts as with dirty track/wheels, etc. 

You are 100% correct. 

PLCProf posted:

... The effect is helped due to the fact that the drop-out voltage of a DC relay is generally much less than the pick-up ...

I have tried this with a variety of small relays, ice-cube size and a bit larger with 12 and 24 VDC coils.

Understood.  It seems that some relays might not have sufficient coil inductance (stored energy) to release during the re-circulation half-cycle.  What is counter-intuitive is that D2 "clamps" the terminal voltage to less than 1V (diode drop) which would seem below any 12/24 V relay's drop-out voltage... but of course it's the current that's doing the work!

I wonder if anyone ever made low-voltage AC-coil relays that used the thermal bi-metallic approach (automotive turn- signal relays) to effectively store energy to perform this hysteresis/delay function?

stan2004 posted:

PLCProf posted:

... The effect is helped due to the fact that the drop-out voltage of a DC relay is generally much less than the pick-up ...

I have tried this with a variety of small relays, ice-cube size and a bit larger with 12 and 24 VDC coils.

Understood.  It seems that some relays might not have sufficient coil inductance (stored energy) to release during the re-circulation half-cycle.  What is counter-intuitive is that D2 "clamps" the terminal voltage to less than 1V (diode drop) which would seem below any 12/24 V relay's drop-out voltage... but of course it's the current that's doing the work!

 

Well, neglecting the drop of D2, you have the current source of the inductance in series with the winding resistance, both of these being internal to the coil!. So, at T+, the coil "sees" full voltage, even though the external voltage is zero. The current and hence the voltage then decay due to the losses in the winding and diode, 

GRJ,

Now that you have pointed me to the 'Klingon' (cloaked) resistor in your schematic, does 200 ohms (1/2 watt) sound about right for 18 vac track power and 0.09 amps on the relay coil at 12vdc?

Not even close.   Let's assume you're dealing with roughly 18 volts after the half-wave rectification and filter cap.  You want to drop around 6 volts at 90 ma.  That would be 66 ohms and the power is over half a watt, so I'd probably size the resistor at 1 watt at least.

I typically use relays with 20-25ma coil current, so I use a 220 ohm resistor at 1/4W for mine and that does the trick.  That's why the note in the schematic mentions coil current.  For your values...

6 volts / .09 amps = 66.67 ohms

6 volts * .09 amps = .54 watts.

Thanks, I goofed that up. I used the full 18 volts instead of just the 6 that's being dropped. I did get the .54 right, but I didn't upsize the resistor wattage. I will have to get some 1 watt resistors, I only have 1/2 and 1/4 watt on hand. Or maybe use different relays which I think I might have.

rtr12 posted:

. I will have to get some 1 watt resistors, I only have 1/2 and 1/4 watt on hand.

Use 2 1/2 Watt resistors to make a 1 Watt.  So if you need 66 ohms, put 2 33 ohm 1/2 W resistors in series and you now have 66 ohm 1 W.

I like using lower coil current relays.  Signals don't need a relay that has a 90ma coil current!

Thanks, Stan, I think I can come up with the 1/2 watt ones.

GRJ, I agree and would like less current too. I didn't know these drew that much, I had never checked them before.