Measuring relay voltage

[The_Gaz]

I'm using the full wave bridge scheme in Doug's section on relays, and just want to make sure I'm getting the right voltage to my 5 volt relay. The relay is switching properly, so that may be a clue 

With the switch closed I measure just about 5vdc across the coil. When the switch is open, however, I read approx. 9vdc from positive to ground. I'm not sure if the "open" reading is critical, or only the closed reading across the coil is the one I should be concerned with.

Thanks.

[navdave]

Should be ok. Are you using the 180Ω resistor like Doug does?
I've ran two relays and an LED using Doug's scheme works fine but I did have to remove the 180Ω resistor.

[The_Gaz]

I'm only using one relay, and no LED.

[eleventeen]

Looks utterly right to me. How close to 6.3 is your heater supply, with all your tubes plugged in? That's IMO where, if I felt the overwhelming need to worry about something is where I'd start. With no load on your heater windings I could see 9 vdc across the open relay, but that's maybe a volt high with a loaded filament winding. Nothing to worry about.

[The_Gaz]

Thanks for the reply. Yeah, my wall voltage may be a little high tonight, but I just wanted to make sure that the open relay voltage should be higher than the closed relay voltage, and that I wasn't about to fry my new relay! 

[RicharD]

With an unregulated supply, voltage will rise when you pull the load off.  You don't hafta have the LED but you really really need the diode across the relay coil.

[The_Gaz]

That makes sense, and it looks like I'm good shape then. Thanks for the explanation, Butter.

And eleventeen, just realized why my voltages looked like an unloaded filament winding... because I'm using a separate 6.3v PT for the relay! - effectively an 'unloaded filament winding'  I was having strange noise issues using the filament lines, which could not be alleviated by floating the ground, etc, etc, etc, etc, and so on.

It's okay because I can get a 200ma PT from Mouser for $5.00, (more than enough for my 30ma relay, right?) - Well worth the headache I was getting, and cheaper than a bottle of Excedrin to boot!

[The_Gaz]

...and I've got the diode across the relay.

[EL34]

The 180 ohm is just the current limiter for the LED
If you don't use a LED indicator, you don't need the 180ohm

Voltages drop when there is a load and rise when there is no load.
Relay coil on = Load
Relay coil off = No load

[The_Gaz]

Thanks, Doug! The 180R I was referring to was the one in the last schem in the relay tutorial, which makes a Pi filter. You say there that that resistor was to drop the voltage a bit.

[The_Gaz]

Would this guy http://www.mouser.com/Search/ProductDetail.aspx?R=41FD200virtualkey21980000virtualkey41FD200 have enough juice to power this 'lil guy http://www.mouser.com/ProductDetail/Omron/G5V-2-H1-DC5/?qs=B3tblJ0Nlt9vPpFeBhhwGQ%3d%3d? I think it would, but just wanted to run it by someone else as I'm not too saavy with datasheets. Thanks.

[sluckey]

yes

[The_Gaz]

thanks for the confirm, sluckey.

[PRR]

9VDC is reasonable for an UN-loaded rectifier from a 6.3V winding. 6.3VAC should be 8.9V DC minus diode losses; but with NO load the diodes and stray line spikes tend to wander high, sometimes higher than the sinewave math implies.

The momentary 9V at turn-on is not a problem. The "5V" relay may be rated to stand 8V. Anyway the 9V will sag (thanks to 180r in the Pi) to 5V long-long-long before smoke comes out of the relay coil. It might be unwise to try a 200V supply sagging to 5V, but a 9 to 5 sag won't hurt.

A diode across the relay coil reduces abuse on the switch. There is a turn-off spike maybe 10 times the operating voltage. For 12V relays goosed by 40V transistors, it is nearly essential. If you use one of those 2mm PCB mount tactile switches, the diode is probably wise. For a 5V relay operated by what amounts to a lamp-switch, the diode may not be needed at all.